Pulse Conditioning for BS2
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Posts: 46,084
I am in the process of building a data logger for my car. It has an
onboard computer already, so many of the signals are there just
waiting to be logged. One of the more challenging channels has
turned out to be the rpm's. After toying with other ideas, I found
the signal going to the tach as the likely (relatively transient
free) signal source.
Here is what I know about the signal:
It is a square wave of some kind (per the cryptic manual), although I
do not know if it drives the tach via pulse width modulation or
frequency of pulses.
I probed the signal and found this:
at 850 rpms: 9.5vdc per multimeter, 85% duty cycle, 30ms positive
pulse, 5.2ms negative pulse
at 2000 rpms: 8.4vdc on multimeter, 75% duty cycle, 11.2ms pos pulse,
3.8ms neg pulse
at 3000 rpms: 7.2vdc, 63.5% duty cycle, 6.4/3.7 pos/neg pulses.
4 cylinder car, two sparks per revolution. Can anyone tell me how
the signal relates to rpms? Once I understand the signal, I need to
scale it down to fit into a BS2 input pin. ANy suggestions there?
Two resistors in series and a zener diode?
p.s. I am not sure I understand my multimeter pulse units. It
says "ms" but also seems to be auto-ranging (range 4? --Fluke), so it
may be some multiple of 10 times the readings quoted above.
onboard computer already, so many of the signals are there just
waiting to be logged. One of the more challenging channels has
turned out to be the rpm's. After toying with other ideas, I found
the signal going to the tach as the likely (relatively transient
free) signal source.
Here is what I know about the signal:
It is a square wave of some kind (per the cryptic manual), although I
do not know if it drives the tach via pulse width modulation or
frequency of pulses.
I probed the signal and found this:
at 850 rpms: 9.5vdc per multimeter, 85% duty cycle, 30ms positive
pulse, 5.2ms negative pulse
at 2000 rpms: 8.4vdc on multimeter, 75% duty cycle, 11.2ms pos pulse,
3.8ms neg pulse
at 3000 rpms: 7.2vdc, 63.5% duty cycle, 6.4/3.7 pos/neg pulses.
4 cylinder car, two sparks per revolution. Can anyone tell me how
the signal relates to rpms? Once I understand the signal, I need to
scale it down to fit into a BS2 input pin. ANy suggestions there?
Two resistors in series and a zener diode?
p.s. I am not sure I understand my multimeter pulse units. It
says "ms" but also seems to be auto-ranging (range 4? --Fluke), so it
may be some multiple of 10 times the readings quoted above.
Comments
cycle for each half revlotion of the motor. The math works perfectly.
(At 2k rpms for example, the cycle is 15ms. Since there are 60k ms in
a minute, there are 4k cycles in the signal per minute, or 2k rpms
per minute. Same math works for other data given below as well.)
BS2 questions:
1) I would like to use pulsin to measure the pulse width on this
signal, although I will need to read the entire cycle and not just
the "on" time. How do I read the complete cycle time on a BS2? (I
want to measure the time it takes go from one "low to high" to the
next "low to high." ) Do I need to set up a flip flop? Any easier all-
software way?
2) The peak voltage of the signal is probably 12vdc, although the
pulsing makes it read lower on the multimeter. Is it safe to read
this signal directly to a BS2 input pin for the pulsin command?
Assuming not, can anyone suggest a means of reducing the peak voltage
to 5 volts without affecting the pulse cycle time?
Thanks!
--- In basicstamps@y..., "tomatlarge" <tomatlarge@y...> wrote:
> I am in the process of building a data logger for my car. It has an
> onboard computer already, so many of the signals are there just
> waiting to be logged. One of the more challenging channels has
> turned out to be the rpm's. After toying with other ideas, I found
> the signal going to the tach as the likely (relatively transient
> free) signal source.
>
> Here is what I know about the signal:
>
> It is a square wave of some kind (per the cryptic manual), although
I
> do not know if it drives the tach via pulse width modulation or
> frequency of pulses.
>
> I probed the signal and found this:
>
> at 850 rpms: 9.5vdc per multimeter, 85% duty cycle, 30ms positive
> pulse, 5.2ms negative pulse
>
> at 2000 rpms: 8.4vdc on multimeter, 75% duty cycle, 11.2ms pos
pulse,
> 3.8ms neg pulse
>
> at 3000 rpms: 7.2vdc, 63.5% duty cycle, 6.4/3.7 pos/neg pulses.
>
> 4 cylinder car, two sparks per revolution. Can anyone tell me how
> the signal relates to rpms? Once I understand the signal, I need to
> scale it down to fit into a BS2 input pin. ANy suggestions there?
> Two resistors in series and a zener diode?
>
> p.s. I am not sure I understand my multimeter pulse units. It
> says "ms" but also seems to be auto-ranging (range 4? --Fluke), so
it
> may be some multiple of 10 times the readings quoted above.
Take two Pulsin measurements, one with the hi/low setting set to 0, and the
other set to 1. Add the values, and you'll have a cycle length. Those two
measurements won't be from the same cycle, but they'll be close enough for
what you are doing.
Use a voltage divider, a series of diodes, or an attenuator to drop the
voltage. You'll want to make sure your circuit doesn't affect your tach.
By the way, for solving problems you mentioned in your first email, I've
found that Excel helps a lot to find out if signals are linear. You can
punch in the pulse times for your three tach readings and look at a graph.
Cliff
Original Message
From: "tomatlarge" <tomatlarge@y...>
To: <basicstamps@yahoogroups.com>
Sent: Friday, December 28, 2001 8:04 PM
Subject: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
Ok, sorry I rambled below. The signal I found generates one full
cycle for each half revlotion of the motor. The math works perfectly.
(At 2k rpms for example, the cycle is 15ms. Since there are 60k ms in
a minute, there are 4k cycles in the signal per minute, or 2k rpms
per minute. Same math works for other data given below as well.)
BS2 questions:
1) I would like to use pulsin to measure the pulse width on this
signal, although I will need to read the entire cycle and not just
the "on" time. How do I read the complete cycle time on a BS2? (I
want to measure the time it takes go from one "low to high" to the
next "low to high." ) Do I need to set up a flip flop? Any easier all-
software way?
2) The peak voltage of the signal is probably 12vdc, although the
pulsing makes it read lower on the multimeter. Is it safe to read
this signal directly to a BS2 input pin for the pulsin command?
Assuming not, can anyone suggest a means of reducing the peak voltage
to 5 volts without affecting the pulse cycle time?
Thanks!
--- In basicstamps@y..., "tomatlarge" <tomatlarge@y...> wrote:
> I am in the process of building a data logger for my car. It has an
> onboard computer already, so many of the signals are there just
> waiting to be logged. One of the more challenging channels has
> turned out to be the rpm's. After toying with other ideas, I found
> the signal going to the tach as the likely (relatively transient
> free) signal source.
>
> Here is what I know about the signal:
>
> It is a square wave of some kind (per the cryptic manual), although
I
> do not know if it drives the tach via pulse width modulation or
> frequency of pulses.
>
> I probed the signal and found this:
>
> at 850 rpms: 9.5vdc per multimeter, 85% duty cycle, 30ms positive
> pulse, 5.2ms negative pulse
>
> at 2000 rpms: 8.4vdc on multimeter, 75% duty cycle, 11.2ms pos
pulse,
> 3.8ms neg pulse
>
> at 3000 rpms: 7.2vdc, 63.5% duty cycle, 6.4/3.7 pos/neg pulses.
>
> 4 cylinder car, two sparks per revolution. Can anyone tell me how
> the signal relates to rpms? Once I understand the signal, I need to
> scale it down to fit into a BS2 input pin. ANy suggestions there?
> Two resistors in series and a zener diode?
>
> p.s. I am not sure I understand my multimeter pulse units. It
> says "ms" but also seems to be auto-ranging (range 4? --Fluke), so
it
> may be some multiple of 10 times the readings quoted above.
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>cycle for each half revlotion of the motor. The math works perfectly.
>(At 2k rpms for example, the cycle is 15ms. Since there are 60k ms in
>a minute, there are 4k cycles in the signal per minute, or 2k rpms
>per minute. Same math works for other data given below as well.)
>
>BS2 questions:
>
>1) I would like to use pulsin to measure the pulse width on this
>signal, although I will need to read the entire cycle and not just
>the "on" time. How do I read the complete cycle time on a BS2? (I
>want to measure the time it takes go from one "low to high" to the
>next "low to high." ) Do I need to set up a flip flop? Any easier all-
>software way?
It is really a question of two full cycles. That could be handled by
a divide-by-four circuit, or by one of the available pulse
measurement modules. To do it all-software, you have to do four
pulse measurements in succession and then combine them,assuming that
the rpm does not change much in 2 revolutions.
pulsin 0,0,[noparse][[/noparse]H1]
pulsin 0,0,[noparse][[/noparse]H2]
pulsin 0,1,[noparse][[/noparse]L1]
pulsin 0,1,[noparse][[/noparse]L2]
totalpulse=H1+H2+L1+L2
>2) The peak voltage of the signal is probably 12vdc, although the
>pulsing makes it read lower on the multimeter. Is it safe to read
>this signal directly to a BS2 input pin for the pulsin command?
>Assuming not, can anyone suggest a means of reducing the peak voltage
>to 5 volts without affecting the pulse cycle time?
You can connect it to the stamp through a voltage divider
12v---/\/\-o-/\/\---Vss
20k | 10k
`
p0
In a pinch you can connect it through a single resistor
12v
/\/\
P0
47k to 100k
But do _*not*_ connect it direct without a resistor or some kind of
protection. If there is noise on the signal, you can put a small
capacitor (~100pf or 1000pf) from the Stamp input to Vss to help.
-- regards,
Tracy Allen
electronically monitored ecosystems
mailto:tracy@e...
http://www.emesystems.com
system failed I did a search for the model number and came up with a whole
pile of info about the diagnostics, including pinout details.
Regards,
Tony Wells
Original Message
From: "tomatlarge" <tomatlarge@y...>
To: <basicstamps@yahoogroups.com>
Sent: Saturday, December 29, 2001 3:20 AM
Subject: [noparse][[/noparse]basicstamps] Pulse Conditioning for BS2
> I am in the process of building a data logger for my car. It has an
> onboard computer already, so many of the signals are there just
> waiting to be logged. One of the more challenging channels has
> turned out to be the rpm's. After toying with other ideas, I found
> the signal going to the tach as the likely (relatively transient
> free) signal source.
>
> Here is what I know about the signal:
>
> It is a square wave of some kind (per the cryptic manual), although I
> do not know if it drives the tach via pulse width modulation or
> frequency of pulses.
>
> I probed the signal and found this:
>
> at 850 rpms: 9.5vdc per multimeter, 85% duty cycle, 30ms positive
> pulse, 5.2ms negative pulse
>
> at 2000 rpms: 8.4vdc on multimeter, 75% duty cycle, 11.2ms pos pulse,
> 3.8ms neg pulse
>
> at 3000 rpms: 7.2vdc, 63.5% duty cycle, 6.4/3.7 pos/neg pulses.
>
> 4 cylinder car, two sparks per revolution. Can anyone tell me how
> the signal relates to rpms? Once I understand the signal, I need to
> scale it down to fit into a BS2 input pin. ANy suggestions there?
> Two resistors in series and a zener diode?
>
> p.s. I am not sure I understand my multimeter pulse units. It
> says "ms" but also seems to be auto-ranging (range 4? --Fluke), so it
> may be some multiple of 10 times the readings quoted above.
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed. Text in the Subject and
Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
the time from one "low to high" to the next "low to high"? Isn't
that just two pulsin commands? (Maybe I confused things by using the
word "cycle" -- I meant one high pulse and one low pulse.) As for a
voltage divider, my concern is that I not draw current from the car's
signal. I'd like the stamp to behave as much like a high-impedance
multi-meter as possible, to avoid any issues with the car's computer.
How much current does the stamp need to register a high state?
Could I, for example, use resistors near 1meg for the voltage divider?
--- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> >Ok, sorry I rambled below. The signal I found generates one full
> >cycle for each half revlotion of the motor. The math works
perfectly.
> >(At 2k rpms for example, the cycle is 15ms. Since there are 60k ms
in
> >a minute, there are 4k cycles in the signal per minute, or 2k rpms
> >per minute. Same math works for other data given below as well.)
> >
> >BS2 questions:
> >
> >1) I would like to use pulsin to measure the pulse width on this
> >signal, although I will need to read the entire cycle and not just
> >the "on" time. How do I read the complete cycle time on a BS2? (I
> >want to measure the time it takes go from one "low to high" to the
> >next "low to high." ) Do I need to set up a flip flop? Any easier
all-
> >software way?
>
> It is really a question of two full cycles. That could be handled
by
> a divide-by-four circuit, or by one of the available pulse
> measurement modules. To do it all-software, you have to do four
> pulse measurements in succession and then combine them,assuming
that
> the rpm does not change much in 2 revolutions.
> pulsin 0,0,[noparse][[/noparse]H1]
> pulsin 0,0,[noparse][[/noparse]H2]
> pulsin 0,1,[noparse][[/noparse]L1]
> pulsin 0,1,[noparse][[/noparse]L2]
> totalpulse=H1+H2+L1+L2
>
>
> >2) The peak voltage of the signal is probably 12vdc, although the
> >pulsing makes it read lower on the multimeter. Is it safe to read
> >this signal directly to a BS2 input pin for the pulsin command?
> >Assuming not, can anyone suggest a means of reducing the peak
voltage
> >to 5 volts without affecting the pulse cycle time?
>
> You can connect it to the stamp through a voltage divider
> 12v---/\/\-o-/\/\---Vss
> 20k | 10k
> `
p0
>
> In a pinch you can connect it through a single resistor
> 12v
/\/\
P0
> 47k to 100k
>
> But do _*not*_ connect it direct without a resistor or some kind of
> protection. If there is noise on the signal, you can put a small
> capacitor (~100pf or 1000pf) from the Stamp input to Vss to help.
>
> -- regards,
> Tracy Allen
> electronically monitored ecosystems
> mailto:tracy@e...
> http://www.emesystems.com
>the time from one "low to high" to the next "low to high"? Isn't
>that just two pulsin commands? (Maybe I confused things by using the
>word "cycle" -- I meant one high pulse and one low pulse.)
I thought what you want is the time for one revolution. According to
your data, the electrical signal has two full cycles during each
revolution of the shaft. So to get the time for one full revolution,
you have to measure two LOW times and two HIGH times.
Here was your data:
>at 2000 rpms: 8.4vdc on multimeter, 75% duty cycle, 11.2ms pos pulse,
>3.8ms neg pulse
The 11.2ms plus 3.8 ms add up to 15 ms. But the time for one
revolution at 2000rpm is 30 milliseconds. That is where the factor
of two comes from. I don't know how this is tied together
mechanically. You might get away with timing one cycle and
multiplying by two.
The program works by sampling low and high pulses over two revolutions:
(view with monospace font)
|.....2 revs....|
---_---_---_---_---_---_---_---_---_
a b c d e f g h
pulsin 0,1,[noparse][[/noparse]H1] ' measures HIGH pulse a
pulsin 0,1,[noparse][[/noparse]H2] ' measures HIGH pulse c
pulsin 0,0,[noparse][[/noparse]L1] ' measures LOW pulse f
pulsin 0,0,[noparse][[/noparse]L2] ' measures LOW pulse h
totalpulse=H1+H2+L1+L2
that should be the same as a+b+c+d for one revolution.
>As for a
>voltage divider, my concern is that I not draw current from the car's
>signal. I'd like the stamp to behave as much like a high-impedance
>multi-meter as possible, to avoid any issues with the car's computer.
>How much current does the stamp need to register a high state?
>Could I, for example, use resistors near 1meg for the voltage divider?
You can try an experiment by putting a 100kohm resistor across the
tach signal, and see on your voltmeter if there is any change. (This
is another instance of measuring output resistance of the "black
box", like we talked about earlier in this thread.) My guess is that
the tach signal is at a low impedance level and 100kohm would not
load it significantly. Its sourcing and sinking capabilities may be
different. A bigger issue is ground loop noise in the nasty
automotive environment. It might be wise to use an opto isolator;
the tach would have to be able to source or sink a couple of
milliamps to make that feasible.
The stamp input is very high impedance, which means it needs
practically zero current to register a high state. You can try it
with one megaohm, but the signal might be too sluggish, due to stray
capacitance in the circuit.
-- regards,
Tracy Allen
electronically monitored ecosystems
mailto:tracy@e...
http://www.emesystems.com
high" to the next "low to high") and you were trying to get the time
for the complete revolution. I think I'll be fine just multiplying
the half rev time by 2, but I think I understand it now, so thank you
very much.
When we last spoke of impedance, we were talking about an input on a
black box. I put a resistor between my voltage signal and the black
box and confirmed that there was no difference in voltage on either
side of the resistor, and concluded it was a high impendance input,
which gave me comfort that I would not blow it out by feeding it too
much current. This time around, we are talking about an output on the
same black box. Where are you suggesting I put the 100k resistor for
this purpose? Just hang it off the end of the tach signal and see if
I get less voltage on either side of the resistor? If the voltage is
the same, then are you saying that the output from the black box
is "low impendance?" And, if so, does that necessarily mean it can
handle a voltage divider using resistors more in the 10k range? To
me, logically, it just means that the voltage divider will not throw
off the voltage signal running to the rest of the original circuit,
but does not necessarily mean the black box's output can provide the
current needed to run the voltage divider.
And, to check my understanding, when I ran the first test to conclude
the black box input was high impedance, then I think I also obtained
evidence (unknowingly at the time) that the output of the voltage
follower ciruit was low impedance. Yes?
--- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
>
> >Thanks for the suggestions. Why would I need 4 pulsin commands to
get
> >the time from one "low to high" to the next "low to high"? Isn't
> >that just two pulsin commands? (Maybe I confused things by using
the
> >word "cycle" -- I meant one high pulse and one low pulse.)
>
> I thought what you want is the time for one revolution. According
to
> your data, the electrical signal has two full cycles during each
> revolution of the shaft. So to get the time for one full
revolution,
> you have to measure two LOW times and two HIGH times.
>
> Here was your data:
> >at 2000 rpms: 8.4vdc on multimeter, 75% duty cycle, 11.2ms pos
pulse,
> >3.8ms neg pulse
>
> The 11.2ms plus 3.8 ms add up to 15 ms. But the time for one
> revolution at 2000rpm is 30 milliseconds. That is where the factor
> of two comes from. I don't know how this is tied together
> mechanically. You might get away with timing one cycle and
> multiplying by two.
>
> The program works by sampling low and high pulses over two
revolutions:
> (view with monospace font)
>
> |.....2 revs....|
> ---_---_---_---_---_---_---_---_---_
> a b c d e f g h
>
> pulsin 0,1,[noparse][[/noparse]H1] ' measures HIGH pulse a
> pulsin 0,1,[noparse][[/noparse]H2] ' measures HIGH pulse c
> pulsin 0,0,[noparse][[/noparse]L1] ' measures LOW pulse f
> pulsin 0,0,[noparse][[/noparse]L2] ' measures LOW pulse h
> totalpulse=H1+H2+L1+L2
>
> that should be the same as a+b+c+d for one revolution.
>
> >As for a
> >voltage divider, my concern is that I not draw current from the
car's
> >signal. I'd like the stamp to behave as much like a high-impedance
> >multi-meter as possible, to avoid any issues with the car's
computer.
> >How much current does the stamp need to register a high state?
> >Could I, for example, use resistors near 1meg for the voltage
divider?
>
> You can try an experiment by putting a 100kohm resistor across the
> tach signal, and see on your voltmeter if there is any change.
(This
> is another instance of measuring output resistance of the "black
> box", like we talked about earlier in this thread.) My guess is
that
> the tach signal is at a low impedance level and 100kohm would not
> load it significantly. Its sourcing and sinking capabilities may
be
> different. A bigger issue is ground loop noise in the nasty
> automotive environment. It might be wise to use an opto isolator;
> the tach would have to be able to source or sink a couple of
> milliamps to make that feasible.
>
> The stamp input is very high impedance, which means it needs
> practically zero current to register a high state. You can try it
> with one megaohm, but the signal might be too sluggish, due to
stray
> capacitance in the circuit.
>
> -- regards,
> Tracy Allen
> electronically monitored ecosystems
> mailto:tracy@e...
> http://www.emesystems.com
a .1volt difference on the voltmeter on either side of the resistor.
(The volt reading reflects an 85% "on" pulsed 12vdc signal, so the
actual voltage drop may be slightly greater.) For reference, the tach
signal is still fully functioning in its circuit. It goes from the
black box directly to the tach, and also directly to another black
box that uses it to create the ignition timing signal. I just tapped
into it to check the voltage/pulse. The voltage was .1vdc lower when
I checked it through a 100k resistor.
--- In basicstamps@y..., "tomatlarge" <tomatlarge@y...> wrote:
> I see. I was content timing half a revolution (i.e. from one "low
to
> high" to the next "low to high") and you were trying to get the
time
> for the complete revolution. I think I'll be fine just multiplying
> the half rev time by 2, but I think I understand it now, so thank
you
> very much.
>
> When we last spoke of impedance, we were talking about an input on
a
> black box. I put a resistor between my voltage signal and the black
> box and confirmed that there was no difference in voltage on either
> side of the resistor, and concluded it was a high impendance input,
> which gave me comfort that I would not blow it out by feeding it
too
> much current. This time around, we are talking about an output on
the
> same black box. Where are you suggesting I put the 100k resistor
for
> this purpose? Just hang it off the end of the tach signal and see
if
> I get less voltage on either side of the resistor? If the voltage
is
> the same, then are you saying that the output from the black box
> is "low impendance?" And, if so, does that necessarily mean it can
> handle a voltage divider using resistors more in the 10k range? To
> me, logically, it just means that the voltage divider will not
throw
> off the voltage signal running to the rest of the original circuit,
> but does not necessarily mean the black box's output can provide
the
> current needed to run the voltage divider.
>
> And, to check my understanding, when I ran the first test to
conclude
> the black box input was high impedance, then I think I also
obtained
> evidence (unknowingly at the time) that the output of the voltage
> follower ciruit was low impedance. Yes?
>
> --- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> >
> > >Thanks for the suggestions. Why would I need 4 pulsin commands
to
> get
> > >the time from one "low to high" to the next "low to high"? Isn't
> > >that just two pulsin commands? (Maybe I confused things by
using
> the
> > >word "cycle" -- I meant one high pulse and one low pulse.)
> >
> > I thought what you want is the time for one revolution. According
> to
> > your data, the electrical signal has two full cycles during each
> > revolution of the shaft. So to get the time for one full
> revolution,
> > you have to measure two LOW times and two HIGH times.
> >
> > Here was your data:
> > >at 2000 rpms: 8.4vdc on multimeter, 75% duty cycle, 11.2ms pos
> pulse,
> > >3.8ms neg pulse
> >
> > The 11.2ms plus 3.8 ms add up to 15 ms. But the time for one
> > revolution at 2000rpm is 30 milliseconds. That is where the
factor
> > of two comes from. I don't know how this is tied together
> > mechanically. You might get away with timing one cycle and
> > multiplying by two.
> >
> > The program works by sampling low and high pulses over two
> revolutions:
> > (view with monospace font)
> >
> > |.....2 revs....|
> > ---_---_---_---_---_---_---_---_---_
> > a b c d e f g h
> >
> > pulsin 0,1,[noparse][[/noparse]H1] ' measures HIGH pulse a
> > pulsin 0,1,[noparse][[/noparse]H2] ' measures HIGH pulse c
> > pulsin 0,0,[noparse][[/noparse]L1] ' measures LOW pulse f
> > pulsin 0,0,[noparse][[/noparse]L2] ' measures LOW pulse h
> > totalpulse=H1+H2+L1+L2
> >
> > that should be the same as a+b+c+d for one revolution.
> >
> > >As for a
> > >voltage divider, my concern is that I not draw current from the
> car's
> > >signal. I'd like the stamp to behave as much like a high-
impedance
> > >multi-meter as possible, to avoid any issues with the car's
> computer.
> > >How much current does the stamp need to register a high state?
> > >Could I, for example, use resistors near 1meg for the voltage
> divider?
> >
> > You can try an experiment by putting a 100kohm resistor across
the
> > tach signal, and see on your voltmeter if there is any change.
> (This
> > is another instance of measuring output resistance of the "black
> > box", like we talked about earlier in this thread.) My guess is
> that
> > the tach signal is at a low impedance level and 100kohm would not
> > load it significantly. Its sourcing and sinking capabilities may
> be
> > different. A bigger issue is ground loop noise in the nasty
> > automotive environment. It might be wise to use an opto
isolator;
> > the tach would have to be able to source or sink a couple of
> > milliamps to make that feasible.
> >
> > The stamp input is very high impedance, which means it needs
> > practically zero current to register a high state. You can try
it
> > with one megaohm, but the signal might be too sluggish, due to
> stray
> > capacitance in the circuit.
> >
> > -- regards,
> > Tracy Allen
> > electronically monitored ecosystems
> > mailto:tracy@e...
> > http://www.emesystems.com
>a .1volt difference on the voltmeter on either side of the resistor.
>(The volt reading reflects an 85% "on" pulsed 12vdc signal, so the
>actual voltage drop may be slightly greater.) For reference, the tach
>signal is still fully functioning in its circuit. It goes from the
>black box directly to the tach, and also directly to another black
>box that uses it to create the ignition timing signal. I just tapped
>into it to check the voltage/pulse. The voltage was .1vdc lower when
>I checked it through a 100k resistor.
That experiment allows you to estimate the input resistance. I think
you added the 100k resistor as follows:
"black box 1 output" ----/\/\---- "black box 2+3 input"
100k
The voltage on the left side of the resistor was about 0.1 volt
greater than the voltage on the right, say 10 volts and 9.9 volts.
Yes? That lets you estimate the input resistance of black box 2+3.
The current through the 100kohm resistor is V/R = 0.1/100000 =~ 1
microamp. That 1 microamp also flows through the input resistance of
black box #2+3, which has 9.9 volts across it. So the input
resistance is, R=V/I=9.9 megaohms. Roughly 10 megaohms, give or take
a few. Pretty high.
To measure the output resistance of black box 1, connect the 100kohms
as follows:
"black box 1 output" --o
"black box 2+3 input"
|
|
Vss---/\/\
'
100kohm
That is, direct from the output to ground. What is the voltage
before and after you connect in the resistor? E.g., 10.0 before, 9.9
after. If there is no change, then try 10k or even 1k. If there is
no change at all, then the output resistance is much much less than
the value of resistance you are trying. If the output drops to 1/2
of its original value, then the output resistance equals the added
resistance. The point is, if you add your own "black box" to the
circuit, you do not want to load it down too much. The test give you
an estimate of how much is too much.
-- regards,
Tracy Allen
electronically monitored ecosystems
mailto:tracy@e...
http://www.emesystems.com
to the tach signal going from one black box to the other. Using a
multimeter on the tapped line (red lead to tapped line, black lead to
ground), I found the pulse and voltages I listed yesterday. Today, I
put a 100k resistor on the end of the tapped line (in series between
the tapped line and the red lead of the multimeter). With the
resistor in place, I have about.1vdc less than without the resistor.
I thought about doing it like your second suggestion (100k to ground)
but got paramoid. At any rate, I ended up making a voltage follower
out of an LM358, and sent my tach signal to it through a 100k
resistor. I then used a voltage divider on the 12v output of the
LM358, and everything seems to be working fine. I assumed (hopefully
correctly) that the LM358 with 100k resistor would have virtually no
effect on the rest of the circuit, and that the votage divider on the
output side of the LM358 (using 10k and 20k resistors) would not over-
tax the LM358 output. At least it all seems to work...sound ok? I am
reading up on basic electronics and am admittedly a bit stumped on
the impedance concept. Your help has been invaluable...I'll get
there. Now, I am starting to struggle with BS2 math calculations to
turn my pulse readings into actual rpms. I found your write-up in the
FAQ, but may be looking for help soon. I need to divide 30000 by the
pulse width (in milliseconds) I measure, and I'm trying to do it
without taking up too much cpu time. I'll play with that tomorrow.
Thanks again.
--- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> >Well, I put a 100k resistor on the tach signal and there was about
> >a .1volt difference on the voltmeter on either side of the
resistor.
> >(The volt reading reflects an 85% "on" pulsed 12vdc signal, so the
> >actual voltage drop may be slightly greater.) For reference, the
tach
> >signal is still fully functioning in its circuit. It goes from the
> >black box directly to the tach, and also directly to another black
> >box that uses it to create the ignition timing signal. I just
tapped
> >into it to check the voltage/pulse. The voltage was .1vdc lower
when
> >I checked it through a 100k resistor.
>
> That experiment allows you to estimate the input resistance. I
think
> you added the 100k resistor as follows:
>
> "black box 1 output" ----/\/\---- "black box 2+3 input"
> 100k
>
> The voltage on the left side of the resistor was about 0.1 volt
> greater than the voltage on the right, say 10 volts and 9.9 volts.
> Yes? That lets you estimate the input resistance of black box 2+3.
> The current through the 100kohm resistor is V/R = 0.1/100000 =~ 1
> microamp. That 1 microamp also flows through the input resistance
of
> black box #2+3, which has 9.9 volts across it. So the input
> resistance is, R=V/I=9.9 megaohms. Roughly 10 megaohms, give or
take
> a few. Pretty high.
>
> To measure the output resistance of black box 1, connect the
100kohms
> as follows:
>
> "black box 1 output" --o
"black box 2+3 input"
> |
> |
> Vss---/\/\
'
> 100kohm
>
> That is, direct from the output to ground. What is the voltage
> before and after you connect in the resistor? E.g., 10.0 before,
9.9
> after. If there is no change, then try 10k or even 1k. If there
is
> no change at all, then the output resistance is much much less than
> the value of resistance you are trying. If the output drops to 1/2
> of its original value, then the output resistance equals the added
> resistance. The point is, if you add your own "black box" to the
> circuit, you do not want to load it down too much. The test give
you
> an estimate of how much is too much.
>
> -- regards,
> Tracy Allen
> electronically monitored ecosystems
> mailto:tracy@e...
> http://www.emesystems.com
Perhaps I've missed something here. Forgive me if I have.
Wouldn't it be easier to count the number of cpu cycles between each pulse?
If the pulses are too short an interval, use a divide by 10 or divide by 100
counter.
Yes, you would have to calibrate it, but the code would be simpler, and you
can choose when to sample - say once per second.
Regards,
Tony
Original Message
From: "tomatlarge" <tomatlarge@y...>
To: <basicstamps@yahoogroups.com>
Sent: Sunday, December 30, 2001 10:43 AM
Subject: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
> Oops, I think I mistook your original suggestion. I connected a wire
> to the tach signal going from one black box to the other. Using a
> multimeter on the tapped line (red lead to tapped line, black lead to
> ground), I found the pulse and voltages I listed yesterday. Today, I
> put a 100k resistor on the end of the tapped line (in series between
> the tapped line and the red lead of the multimeter). With the
> resistor in place, I have about.1vdc less than without the resistor.
> I thought about doing it like your second suggestion (100k to ground)
> but got paramoid. At any rate, I ended up making a voltage follower
> out of an LM358, and sent my tach signal to it through a 100k
> resistor. I then used a voltage divider on the 12v output of the
> LM358, and everything seems to be working fine. I assumed (hopefully
> correctly) that the LM358 with 100k resistor would have virtually no
> effect on the rest of the circuit, and that the votage divider on the
> output side of the LM358 (using 10k and 20k resistors) would not over-
> tax the LM358 output. At least it all seems to work...sound ok? I am
> reading up on basic electronics and am admittedly a bit stumped on
> the impedance concept. Your help has been invaluable...I'll get
> there. Now, I am starting to struggle with BS2 math calculations to
> turn my pulse readings into actual rpms. I found your write-up in the
> FAQ, but may be looking for help soon. I need to divide 30000 by the
> pulse width (in milliseconds) I measure, and I'm trying to do it
> without taking up too much cpu time. I'll play with that tomorrow.
> Thanks again.
>
> --- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> > >Well, I put a 100k resistor on the tach signal and there was about
> > >a .1volt difference on the voltmeter on either side of the
> resistor.
> > >(The volt reading reflects an 85% "on" pulsed 12vdc signal, so the
> > >actual voltage drop may be slightly greater.) For reference, the
> tach
> > >signal is still fully functioning in its circuit. It goes from the
> > >black box directly to the tach, and also directly to another black
> > >box that uses it to create the ignition timing signal. I just
> tapped
> > >into it to check the voltage/pulse. The voltage was .1vdc lower
> when
> > >I checked it through a 100k resistor.
> >
> > That experiment allows you to estimate the input resistance. I
> think
> > you added the 100k resistor as follows:
> >
> > "black box 1 output" ----/\/\---- "black box 2+3 input"
> > 100k
> >
> > The voltage on the left side of the resistor was about 0.1 volt
> > greater than the voltage on the right, say 10 volts and 9.9 volts.
> > Yes? That lets you estimate the input resistance of black box 2+3.
> > The current through the 100kohm resistor is V/R = 0.1/100000 =~ 1
> > microamp. That 1 microamp also flows through the input resistance
> of
> > black box #2+3, which has 9.9 volts across it. So the input
> > resistance is, R=V/I=9.9 megaohms. Roughly 10 megaohms, give or
> take
> > a few. Pretty high.
> >
> > To measure the output resistance of black box 1, connect the
> 100kohms
> > as follows:
> >
> > "black box 1 output" --o
"black box 2+3 input"
> > |
> > |
> > Vss---/\/\
'
> > 100kohm
> >
> > That is, direct from the output to ground. What is the voltage
> > before and after you connect in the resistor? E.g., 10.0 before,
> 9.9
> > after. If there is no change, then try 10k or even 1k. If there
> is
> > no change at all, then the output resistance is much much less than
> > the value of resistance you are trying. If the output drops to 1/2
> > of its original value, then the output resistance equals the added
> > resistance. The point is, if you add your own "black box" to the
> > circuit, you do not want to load it down too much. The test give
> you
> > an estimate of how much is too much.
> >
> > -- regards,
> > Tracy Allen
> > electronically monitored ecosystems
> > mailto:tracy@e...
> > http://www.emesystems.com
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed. Text in the Subject and
Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
>put a 100k resistor on the end of the tapped line (in series between
>the tapped line and the red lead of the multimeter). With the
>resistor in place, I have about.1vdc less than without the resistor.
You have measured the input resistance of your multimeter!
>I thought about doing it like your second suggestion (100k to ground)
>but got paramoid.
All advice here is taken strictly at your own risk!
>At any rate, I ended up making a voltage follower
>out of an LM358, and sent my tach signal to it through a 100k
>resistor. I then used a voltage divider on the 12v output of the
>LM358, and everything seems to be working fine. I assumed (hopefully
>correctly) that the LM358 with 100k resistor would have virtually no
>effect on the rest of the circuit, and that the votage divider on the
>output side of the LM358 (using 10k and 20k resistors) would not over-
>tax the LM358 output. At least it all seems to work...sound ok?
Yes.
>Now, I am starting to struggle with BS2 math calculations to
>turn my pulse readings into actual rpms. I found your write-up in the
>FAQ, but may be looking for help soon. I need to divide 30000 by the
>pulse width (in milliseconds) I measure, and I'm trying to do it
>without taking up too much cpu time.
E.g., 30000/15=2000 rpm, but 30000/14=2143rpm, quite a big unit step
of 143, no? What resolution do you expect?
The pulse width from the pulsin commnad will be in units of 2
microseconds, e.g. 15 milliseconds will come out as PW=7500 counts.
The measurement of pulse width has lots of resolution, 4 digits. On
the face of it, 15000000/7500 = 2000 but the numerator is too big for
the Stamp to handle in one step. The Stamp can handle it as
(shortcut #2 at <http://www.emesystems.com/BS2math2.htm>)
' 150000/(PW/100) = 600000/(PW/25)
PW=PW/25 ' scale down
RPM = 60000/PW*10+(60000//PW*10/PW)
Thus when PW=7500, PW/25=300, RPM=2000
and when PW=7475, PW/25=299, RPM=2006
A unit step of 6.
There are more accurate methods, but they would take more CPU time.
-- regards,
Tracy Allen
electronically monitored ecosystems
mailto:tracy@e...
http://www.emesystems.com
enough. I seem to be having some real word "issues" however,
probably relating to my voltage follower. First, the program does not
want to work in the car unless I touch either or both of the power
supply leads...makes me think I need a capacitor somewhere. Second,
when the program is running, the pulsin command generates slightly
smaller pulse times than is actually on the tach signal (according to
my Fluke), resulting in slightly higher than real rpms. As I
mentioned, I used an LM358 as a voltage follower, and split the
output of the LM358 with a 10k and 20k resistor. Could that be
throwing off the timing? I am measuring one low and one high pulse
and adding them together, so I would think the result should be the
same even if the output is shifted a bit. For what it is worth, I may
try smaller resistors on the voltage dividers, since I cannot get my
Fluke to pick up the pulse at all when measuring in the middle of the
divider (which is where the Stamp is getting the pulse). I'm getting
there! Thanks.
--- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> >Today, I
> >put a 100k resistor on the end of the tapped line (in series
between
> >the tapped line and the red lead of the multimeter). With the
> >resistor in place, I have about.1vdc less than without the
resistor.
>
> You have measured the input resistance of your multimeter!
>
> >I thought about doing it like your second suggestion (100k to
ground)
> >but got paramoid.
>
> All advice here is taken strictly at your own risk!
>
> >At any rate, I ended up making a voltage follower
> >out of an LM358, and sent my tach signal to it through a 100k
> >resistor. I then used a voltage divider on the 12v output of the
> >LM358, and everything seems to be working fine. I assumed
(hopefully
> >correctly) that the LM358 with 100k resistor would have virtually
no
> >effect on the rest of the circuit, and that the votage divider on
the
> >output side of the LM358 (using 10k and 20k resistors) would not
over-
> >tax the LM358 output. At least it all seems to work...sound ok?
>
> Yes.
>
> >Now, I am starting to struggle with BS2 math calculations to
> >turn my pulse readings into actual rpms. I found your write-up in
the
> >FAQ, but may be looking for help soon. I need to divide 30000 by
the
> >pulse width (in milliseconds) I measure, and I'm trying to do it
> >without taking up too much cpu time.
>
> E.g., 30000/15=2000 rpm, but 30000/14=2143rpm, quite a big unit
step
> of 143, no? What resolution do you expect?
>
> The pulse width from the pulsin commnad will be in units of 2
> microseconds, e.g. 15 milliseconds will come out as PW=7500 counts.
> The measurement of pulse width has lots of resolution, 4 digits.
On
> the face of it, 15000000/7500 = 2000 but the numerator is too big
for
> the Stamp to handle in one step. The Stamp can handle it as
> (shortcut #2 at <http://www.emesystems.com/BS2math2.htm>)
> ' 150000/(PW/100) = 600000/(PW/25)
> PW=PW/25 ' scale down
> RPM = 60000/PW*10+(60000//PW*10/PW)
>
> Thus when PW=7500, PW/25=300, RPM=2000
> and when PW=7475, PW/25=299, RPM=2006
> A unit step of 6.
> There are more accurate methods, but they would take more CPU time.
>
> -- regards,
> Tracy Allen
> electronically monitored ecosystems
> mailto:tracy@e...
> http://www.emesystems.com
for 30ms and off for 5ms, repeating, or faster. SO, I think I measure
the on or off pulses, or some collection of on and off pulses, and
calculate an rpm from that. I'm not sure what you mean when you
say "between each pulse"? The signal is plus 12 volts for a few ms
and 0 volts for a few milliseconds, over and over. Am I missing your
point?
--- In basicstamps@y..., tony.wells@a... wrote:
> Hi ,
>
> Perhaps I've missed something here. Forgive me if I have.
>
> Wouldn't it be easier to count the number of cpu cycles between
each pulse?
> If the pulses are too short an interval, use a divide by 10 or
divide by 100
> counter.
>
> Yes, you would have to calibrate it, but the code would be simpler,
and you
> can choose when to sample - say once per second.
>
> Regards,
>
> Tony
>
>
Original Message
> From: "tomatlarge" <tomatlarge@y...>
> To: <basicstamps@y...>
> Sent: Sunday, December 30, 2001 10:43 AM
> Subject: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
>
>
> > Oops, I think I mistook your original suggestion. I connected a
wire
> > to the tach signal going from one black box to the other. Using a
> > multimeter on the tapped line (red lead to tapped line, black
lead to
> > ground), I found the pulse and voltages I listed yesterday.
Today, I
> > put a 100k resistor on the end of the tapped line (in series
between
> > the tapped line and the red lead of the multimeter). With the
> > resistor in place, I have about.1vdc less than without the
resistor.
> > I thought about doing it like your second suggestion (100k to
ground)
> > but got paramoid. At any rate, I ended up making a voltage
follower
> > out of an LM358, and sent my tach signal to it through a 100k
> > resistor. I then used a voltage divider on the 12v output of the
> > LM358, and everything seems to be working fine. I assumed
(hopefully
> > correctly) that the LM358 with 100k resistor would have virtually
no
> > effect on the rest of the circuit, and that the votage divider on
the
> > output side of the LM358 (using 10k and 20k resistors) would not
over-
> > tax the LM358 output. At least it all seems to work...sound ok?
I am
> > reading up on basic electronics and am admittedly a bit stumped on
> > the impedance concept. Your help has been invaluable...I'll get
> > there. Now, I am starting to struggle with BS2 math calculations
to
> > turn my pulse readings into actual rpms. I found your write-up in
the
> > FAQ, but may be looking for help soon. I need to divide 30000 by
the
> > pulse width (in milliseconds) I measure, and I'm trying to do it
> > without taking up too much cpu time. I'll play with that tomorrow.
> > Thanks again.
> >
> > --- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> > > >Well, I put a 100k resistor on the tach signal and there was
about
> > > >a .1volt difference on the voltmeter on either side of the
> > resistor.
> > > >(The volt reading reflects an 85% "on" pulsed 12vdc signal, so
the
> > > >actual voltage drop may be slightly greater.) For reference,
the
> > tach
> > > >signal is still fully functioning in its circuit. It goes from
the
> > > >black box directly to the tach, and also directly to another
black
> > > >box that uses it to create the ignition timing signal. I just
> > tapped
> > > >into it to check the voltage/pulse. The voltage was .1vdc lower
> > when
> > > >I checked it through a 100k resistor.
> > >
> > > That experiment allows you to estimate the input resistance. I
> > think
> > > you added the 100k resistor as follows:
> > >
> > > "black box 1 output" ----/\/\---- "black box 2+3 input"
> > > 100k
> > >
> > > The voltage on the left side of the resistor was about 0.1 volt
> > > greater than the voltage on the right, say 10 volts and 9.9
volts.
> > > Yes? That lets you estimate the input resistance of black box
2+3.
> > > The current through the 100kohm resistor is V/R = 0.1/100000 =~
1
> > > microamp. That 1 microamp also flows through the input
resistance
> > of
> > > black box #2+3, which has 9.9 volts across it. So the input
> > > resistance is, R=V/I=9.9 megaohms. Roughly 10 megaohms, give or
> > take
> > > a few. Pretty high.
> > >
> > > To measure the output resistance of black box 1, connect the
> > 100kohms
> > > as follows:
> > >
> > > "black box 1 output" --o
"black box 2+3 input"
> > > |
> > > |
> > > Vss---/\/\
'
> > > 100kohm
> > >
> > > That is, direct from the output to ground. What is the voltage
> > > before and after you connect in the resistor? E.g., 10.0
before,
> > 9.9
> > > after. If there is no change, then try 10k or even 1k. If
there
> > is
> > > no change at all, then the output resistance is much much less
than
> > > the value of resistance you are trying. If the output drops to
1/2
> > > of its original value, then the output resistance equals the
added
> > > resistance. The point is, if you add your own "black box" to
the
> > > circuit, you do not want to load it down too much. The test
give
> > you
> > > an estimate of how much is too much.
> > >
> > > -- regards,
> > > Tracy Allen
> > > electronically monitored ecosystems
> > > mailto:tracy@e...
> > > http://www.emesystems.com
> >
> >
> > To UNSUBSCRIBE, just send mail to:
> > basicstamps-unsubscribe@y...
> > from the same email address that you subscribed. Text in the
Subject and
> Body of the message will be ignored.
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
> >
> >
> >
I may have missed some earlier postings, and I am a bit sleepy today (5
day-old baby). So if I'm just talking rubbish or repeating other posts then
just ignore me ;-)
This is a long posting......
What I'm trying to say is that for a given engine speed, the pulses are
occuring at a rate which is related to the engine speed - which you know
already. Using your example, you get an ON for 30ms then an OFF for 5 ms at
say 1000 rpm. If the speed is 2000 rpm then the pulses occur as an ON at
15ms and OFF for 5ms.
If this is the case, what you *can* do, is to cycle the CPU (eg count up
some variable) during the ON period and count the CPU cycles when the pin is
ON, and stop when the pin is OFF.
The CPU cycles, or count variable value then has a relationship to the
rpm.
Now the Basic stamp is slower than Mr. Tortoise in Tortoiseville. It is
entirely likely that it won't be able to cycle a counter many times at 30ms
or less, so the accuracy will be poo. The classic way of solving this is to
put the On/Off pulses into a divider chain - say a divide by 100.
This has three effects. Firstly it makes the on-off ratio of the pulses
50:50 - which can be quite handy when counting numbers. Secondly, and more
importantly, the pulses are now streeeeeeched out, and the slow CPU can
increment a counter more times during the ON period.
The third effect is extremely interesting. Because it will take 100 on/offs
as input before you get an output, and because the on/offs are never going
to be the same value (as the engine revs are never constant) then the
divider chain will sort of average up the values. This will produce a sort
of hysteresis effect, similar to a real needle on a analogue tacho.
Now, once you have installed the divide by 100 counter and identified the
count values for given rpms, you can then work out the relationship between
counts and RPM. You might have to fiddle with a timing constant when
counting to make the values neat.
Once the counting routine is ok, all you have to do is (say) once a second
run the counting routine - make the routine wait until a new pulse comes in,
then do your counting.
I hope all this makes sense.....
Regards,
Toyn Wells
Original Message
From: "tomatlarge" <tomatlarge@y...>
To: <basicstamps@yahoogroups.com>
Sent: Sunday, December 30, 2001 11:12 PM
Subject: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
> I'm not sure I follow this suggestion. The pulse is, for example, on
> for 30ms and off for 5ms, repeating, or faster. SO, I think I measure
> the on or off pulses, or some collection of on and off pulses, and
> calculate an rpm from that. I'm not sure what you mean when you
> say "between each pulse"? The signal is plus 12 volts for a few ms
> and 0 volts for a few milliseconds, over and over. Am I missing your
> point?
>
> --- In basicstamps@y..., tony.wells@a... wrote:
> > Hi ,
> >
> > Perhaps I've missed something here. Forgive me if I have.
> >
> > Wouldn't it be easier to count the number of cpu cycles between
> each pulse?
> > If the pulses are too short an interval, use a divide by 10 or
> divide by 100
> > counter.
> >
> > Yes, you would have to calibrate it, but the code would be simpler,
> and you
> > can choose when to sample - say once per second.
> >
> > Regards,
> >
> > Tony
> >
> >
Original Message
> > From: "tomatlarge" <tomatlarge@y...>
> > To: <basicstamps@y...>
> > Sent: Sunday, December 30, 2001 10:43 AM
> > Subject: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
> >
> >
> > > Oops, I think I mistook your original suggestion. I connected a
> wire
> > > to the tach signal going from one black box to the other. Using a
> > > multimeter on the tapped line (red lead to tapped line, black
> lead to
> > > ground), I found the pulse and voltages I listed yesterday.
> Today, I
> > > put a 100k resistor on the end of the tapped line (in series
> between
> > > the tapped line and the red lead of the multimeter). With the
> > > resistor in place, I have about.1vdc less than without the
> resistor.
> > > I thought about doing it like your second suggestion (100k to
> ground)
> > > but got paramoid. At any rate, I ended up making a voltage
> follower
> > > out of an LM358, and sent my tach signal to it through a 100k
> > > resistor. I then used a voltage divider on the 12v output of the
> > > LM358, and everything seems to be working fine. I assumed
> (hopefully
> > > correctly) that the LM358 with 100k resistor would have virtually
> no
> > > effect on the rest of the circuit, and that the votage divider on
> the
> > > output side of the LM358 (using 10k and 20k resistors) would not
> over-
> > > tax the LM358 output. At least it all seems to work...sound ok?
> I am
> > > reading up on basic electronics and am admittedly a bit stumped on
> > > the impedance concept. Your help has been invaluable...I'll get
> > > there. Now, I am starting to struggle with BS2 math calculations
> to
> > > turn my pulse readings into actual rpms. I found your write-up in
> the
> > > FAQ, but may be looking for help soon. I need to divide 30000 by
> the
> > > pulse width (in milliseconds) I measure, and I'm trying to do it
> > > without taking up too much cpu time. I'll play with that tomorrow.
> > > Thanks again.
> > >
> > > --- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> > > > >Well, I put a 100k resistor on the tach signal and there was
> about
> > > > >a .1volt difference on the voltmeter on either side of the
> > > resistor.
> > > > >(The volt reading reflects an 85% "on" pulsed 12vdc signal, so
> the
> > > > >actual voltage drop may be slightly greater.) For reference,
> the
> > > tach
> > > > >signal is still fully functioning in its circuit. It goes from
> the
> > > > >black box directly to the tach, and also directly to another
> black
> > > > >box that uses it to create the ignition timing signal. I just
> > > tapped
> > > > >into it to check the voltage/pulse. The voltage was .1vdc lower
> > > when
> > > > >I checked it through a 100k resistor.
> > > >
> > > > That experiment allows you to estimate the input resistance. I
> > > think
> > > > you added the 100k resistor as follows:
> > > >
> > > > "black box 1 output" ----/\/\---- "black box 2+3 input"
> > > > 100k
> > > >
> > > > The voltage on the left side of the resistor was about 0.1 volt
> > > > greater than the voltage on the right, say 10 volts and 9.9
> volts.
> > > > Yes? That lets you estimate the input resistance of black box
> 2+3.
> > > > The current through the 100kohm resistor is V/R = 0.1/100000 =~
> 1
> > > > microamp. That 1 microamp also flows through the input
> resistance
> > > of
> > > > black box #2+3, which has 9.9 volts across it. So the input
> > > > resistance is, R=V/I=9.9 megaohms. Roughly 10 megaohms, give or
> > > take
> > > > a few. Pretty high.
> > > >
> > > > To measure the output resistance of black box 1, connect the
> > > 100kohms
> > > > as follows:
> > > >
> > > > "black box 1 output" --o
"black box 2+3 input"
> > > > |
> > > > |
> > > > Vss---/\/\
'
> > > > 100kohm
> > > >
> > > > That is, direct from the output to ground. What is the voltage
> > > > before and after you connect in the resistor? E.g., 10.0
> before,
> > > 9.9
> > > > after. If there is no change, then try 10k or even 1k. If
> there
> > > is
> > > > no change at all, then the output resistance is much much less
> than
> > > > the value of resistance you are trying. If the output drops to
> 1/2
> > > > of its original value, then the output resistance equals the
> added
> > > > resistance. The point is, if you add your own "black box" to
> the
> > > > circuit, you do not want to load it down too much. The test
> give
> > > you
> > > > an estimate of how much is too much.
> > > >
> > > > -- regards,
> > > > Tracy Allen
> > > > electronically monitored ecosystems
> > > > mailto:tracy@e...
> > > > http://www.emesystems.com
> > >
> > >
> > > To UNSUBSCRIBE, just send mail to:
> > > basicstamps-unsubscribe@y...
> > > from the same email address that you subscribed. Text in the
> Subject and
> > Body of the message will be ignored.
> > >
> > >
> > > Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
> > >
> > >
> > >
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed. Text in the Subject and
Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
with a 5 day old baby. I assumed that the pulsin command was the most
accurate way of timing a pulse in the 30ms range, so it did not occur
to me to bypass that command in favor of manual counting. Stretching
the pulse is an appealing idea. I started toying with a flip flop
circuit yesterday, but managed to fry the only chip I had for that. I
do like the idea of taking many pulses at a time, to smooth out the
reading, but do worry about using too much cpu time. The overall plan
is to build a data logger for my car (DME controlled Porsche), so I
would not want to hamper the sampling rate just to calculate rpms.
Right now, I want to get a BS2 tach working in the car (have LCD
panel for read-out) so I know there are no issues with the sampling.
But, once I put the rest of the pieces in place, I may just have the
data logger capture the raw data (pulse time--raw or stretched), and
let a PC do the calculations after the fact at the time the results
are displayed. I'm still working on getting reliable data using 2
pulsin commands, but if that proves too unstable, I may very well
think along the lines you are suggesting.
Thanks!
--- In basicstamps@y..., <tony.wells@a...> wrote:
> Hi Tom,
>
> I may have missed some earlier postings, and I am a bit sleepy
today (5
> day-old baby). So if I'm just talking rubbish or repeating other
posts then
> just ignore me ;-)
>
> This is a long posting......
>
> What I'm trying to say is that for a given engine speed, the pulses
are
> occuring at a rate which is related to the engine speed - which you
know
> already. Using your example, you get an ON for 30ms then an OFF for
5 ms at
> say 1000 rpm. If the speed is 2000 rpm then the pulses occur as an
ON at
> 15ms and OFF for 5ms.
>
> If this is the case, what you *can* do, is to cycle the CPU (eg
count up
> some variable) during the ON period and count the CPU cycles when
the pin is
> ON, and stop when the pin is OFF.
>
> The CPU cycles, or count variable value then has a relationship to
the
> rpm.
>
> Now the Basic stamp is slower than Mr. Tortoise in Tortoiseville.
It is
> entirely likely that it won't be able to cycle a counter many times
at 30ms
> or less, so the accuracy will be poo. The classic way of solving
this is to
> put the On/Off pulses into a divider chain - say a divide by 100.
>
> This has three effects. Firstly it makes the on-off ratio of the
pulses
> 50:50 - which can be quite handy when counting numbers. Secondly,
and more
> importantly, the pulses are now streeeeeeched out, and the slow CPU
can
> increment a counter more times during the ON period.
>
> The third effect is extremely interesting. Because it will take 100
on/offs
> as input before you get an output, and because the on/offs are
never going
> to be the same value (as the engine revs are never constant) then
the
> divider chain will sort of average up the values. This will produce
a sort
> of hysteresis effect, similar to a real needle on a analogue tacho.
>
> Now, once you have installed the divide by 100 counter and
identified the
> count values for given rpms, you can then work out the relationship
between
> counts and RPM. You might have to fiddle with a timing constant when
> counting to make the values neat.
>
> Once the counting routine is ok, all you have to do is (say) once a
second
> run the counting routine - make the routine wait until a new pulse
comes in,
> then do your counting.
>
>
> I hope all this makes sense.....
>
> Regards,
>
> Toyn Wells
>
>
Original Message
> From: "tomatlarge" <tomatlarge@y...>
> To: <basicstamps@y...>
> Sent: Sunday, December 30, 2001 11:12 PM
> Subject: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
>
>
> > I'm not sure I follow this suggestion. The pulse is, for example,
on
> > for 30ms and off for 5ms, repeating, or faster. SO, I think I
measure
> > the on or off pulses, or some collection of on and off pulses, and
> > calculate an rpm from that. I'm not sure what you mean when you
> > say "between each pulse"? The signal is plus 12 volts for a few
ms
> > and 0 volts for a few milliseconds, over and over. Am I missing
your
> > point?
> >
> > --- In basicstamps@y..., tony.wells@a... wrote:
> > > Hi ,
> > >
> > > Perhaps I've missed something here. Forgive me if I have.
> > >
> > > Wouldn't it be easier to count the number of cpu cycles between
> > each pulse?
> > > If the pulses are too short an interval, use a divide by 10 or
> > divide by 100
> > > counter.
> > >
> > > Yes, you would have to calibrate it, but the code would be
simpler,
> > and you
> > > can choose when to sample - say once per second.
> > >
> > > Regards,
> > >
> > > Tony
> > >
> > >
Original Message
> > > From: "tomatlarge" <tomatlarge@y...>
> > > To: <basicstamps@y...>
> > > Sent: Sunday, December 30, 2001 10:43 AM
> > > Subject: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
> > >
> > >
> > > > Oops, I think I mistook your original suggestion. I connected
a
> > wire
> > > > to the tach signal going from one black box to the other.
Using a
> > > > multimeter on the tapped line (red lead to tapped line, black
> > lead to
> > > > ground), I found the pulse and voltages I listed yesterday.
> > Today, I
> > > > put a 100k resistor on the end of the tapped line (in series
> > between
> > > > the tapped line and the red lead of the multimeter). With the
> > > > resistor in place, I have about.1vdc less than without the
> > resistor.
> > > > I thought about doing it like your second suggestion (100k to
> > ground)
> > > > but got paramoid. At any rate, I ended up making a voltage
> > follower
> > > > out of an LM358, and sent my tach signal to it through a 100k
> > > > resistor. I then used a voltage divider on the 12v output of
the
> > > > LM358, and everything seems to be working fine. I assumed
> > (hopefully
> > > > correctly) that the LM358 with 100k resistor would have
virtually
> > no
> > > > effect on the rest of the circuit, and that the votage
divider on
> > the
> > > > output side of the LM358 (using 10k and 20k resistors) would
not
> > over-
> > > > tax the LM358 output. At least it all seems to work...sound
ok?
> > I am
> > > > reading up on basic electronics and am admittedly a bit
stumped on
> > > > the impedance concept. Your help has been invaluable...I'll
get
> > > > there. Now, I am starting to struggle with BS2 math
calculations
> > to
> > > > turn my pulse readings into actual rpms. I found your write-
up in
> > the
> > > > FAQ, but may be looking for help soon. I need to divide 30000
by
> > the
> > > > pulse width (in milliseconds) I measure, and I'm trying to do
it
> > > > without taking up too much cpu time. I'll play with that
tomorrow.
> > > > Thanks again.
> > > >
> > > > --- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> > > > > >Well, I put a 100k resistor on the tach signal and there
was
> > about
> > > > > >a .1volt difference on the voltmeter on either side of the
> > > > resistor.
> > > > > >(The volt reading reflects an 85% "on" pulsed 12vdc
signal, so
> > the
> > > > > >actual voltage drop may be slightly greater.) For
reference,
> > the
> > > > tach
> > > > > >signal is still fully functioning in its circuit. It goes
from
> > the
> > > > > >black box directly to the tach, and also directly to
another
> > black
> > > > > >box that uses it to create the ignition timing signal. I
just
> > > > tapped
> > > > > >into it to check the voltage/pulse. The voltage was .1vdc
lower
> > > > when
> > > > > >I checked it through a 100k resistor.
> > > > >
> > > > > That experiment allows you to estimate the input
resistance. I
> > > > think
> > > > > you added the 100k resistor as follows:
> > > > >
> > > > > "black box 1 output" ----/\/\---- "black box 2+3 input"
> > > > > 100k
> > > > >
> > > > > The voltage on the left side of the resistor was about 0.1
volt
> > > > > greater than the voltage on the right, say 10 volts and 9.9
> > volts.
> > > > > Yes? That lets you estimate the input resistance of black
box
> > 2+3.
> > > > > The current through the 100kohm resistor is V/R =
0.1/100000 =~
> > 1
> > > > > microamp. That 1 microamp also flows through the input
> > resistance
> > > > of
> > > > > black box #2+3, which has 9.9 volts across it. So the input
> > > > > resistance is, R=V/I=9.9 megaohms. Roughly 10 megaohms,
give or
> > > > take
> > > > > a few. Pretty high.
> > > > >
> > > > > To measure the output resistance of black box 1, connect the
> > > > 100kohms
> > > > > as follows:
> > > > >
> > > > > "black box 1 output" --o
"black box 2+3 input"
> > > > > |
> > > > > |
> > > > > Vss---/\/\
'
> > > > > 100kohm
> > > > >
> > > > > That is, direct from the output to ground. What is the
voltage
> > > > > before and after you connect in the resistor? E.g., 10.0
> > before,
> > > > 9.9
> > > > > after. If there is no change, then try 10k or even 1k. If
> > there
> > > > is
> > > > > no change at all, then the output resistance is much much
less
> > than
> > > > > the value of resistance you are trying. If the output
drops to
> > 1/2
> > > > > of its original value, then the output resistance equals the
> > added
> > > > > resistance. The point is, if you add your own "black box"
to
> > the
> > > > > circuit, you do not want to load it down too much. The test
> > give
> > > > you
> > > > > an estimate of how much is too much.
> > > > >
> > > > > -- regards,
> > > > > Tracy Allen
> > > > > electronically monitored ecosystems
> > > > > mailto:tracy@e...
> > > > > http://www.emesystems.com
> > > >
> > > >
> > > > To UNSUBSCRIBE, just send mail to:
> > > > basicstamps-unsubscribe@y...
> > > > from the same email address that you subscribed. Text in the
> > Subject and
> > > Body of the message will be ignored.
> > > >
> > > >
> > > > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> > > >
> > > >
> > > >
> >
> >
> > To UNSUBSCRIBE, just send mail to:
> > basicstamps-unsubscribe@y...
> > from the same email address that you subscribed. Text in the
Subject and
> Body of the message will be ignored.
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
> >
> >
> >
resistors (both in the voltage divider and on the signal from the
black box), and added a cap across the BS2 input as Tracy suggested,
and the tach now reads extremely stable and accurate on my LCD. The
timing problem I mentioned earlier was a software glitch (forgot to
add the negative pulse.) Many thanks for all the help. Next up:
injector pulse timing, and a host of switch and voltage-based
sensors. Then I'll need to tackle writing data to BS2 memory for
later PC processing. [noparse]:)[/noparse]
--- In basicstamps@y..., "tomatlarge" <tomatlarge@y...> wrote:
> Ah, I see what you are saying. Very clear. Thanks. Not bad for a
guy
> with a 5 day old baby. I assumed that the pulsin command was the
most
> accurate way of timing a pulse in the 30ms range, so it did not
occur
> to me to bypass that command in favor of manual counting.
Stretching
> the pulse is an appealing idea. I started toying with a flip flop
> circuit yesterday, but managed to fry the only chip I had for that.
I
> do like the idea of taking many pulses at a time, to smooth out the
> reading, but do worry about using too much cpu time. The overall
plan
> is to build a data logger for my car (DME controlled Porsche), so I
> would not want to hamper the sampling rate just to calculate rpms.
> Right now, I want to get a BS2 tach working in the car (have LCD
> panel for read-out) so I know there are no issues with the
sampling.
> But, once I put the rest of the pieces in place, I may just have
the
> data logger capture the raw data (pulse time--raw or stretched),
and
> let a PC do the calculations after the fact at the time the results
> are displayed. I'm still working on getting reliable data using 2
> pulsin commands, but if that proves too unstable, I may very well
> think along the lines you are suggesting.
>
> Thanks!
> --- In basicstamps@y..., <tony.wells@a...> wrote:
> > Hi Tom,
> >
> > I may have missed some earlier postings, and I am a bit sleepy
> today (5
> > day-old baby). So if I'm just talking rubbish or repeating other
> posts then
> > just ignore me ;-)
> >
> > This is a long posting......
> >
> > What I'm trying to say is that for a given engine speed, the
pulses
> are
> > occuring at a rate which is related to the engine speed - which
you
> know
> > already. Using your example, you get an ON for 30ms then an OFF
for
> 5 ms at
> > say 1000 rpm. If the speed is 2000 rpm then the pulses occur as
an
> ON at
> > 15ms and OFF for 5ms.
> >
> > If this is the case, what you *can* do, is to cycle the CPU (eg
> count up
> > some variable) during the ON period and count the CPU cycles when
> the pin is
> > ON, and stop when the pin is OFF.
> >
> > The CPU cycles, or count variable value then has a relationship
to
> the
> > rpm.
> >
> > Now the Basic stamp is slower than Mr. Tortoise in Tortoiseville.
> It is
> > entirely likely that it won't be able to cycle a counter many
times
> at 30ms
> > or less, so the accuracy will be poo. The classic way of solving
> this is to
> > put the On/Off pulses into a divider chain - say a divide by 100.
> >
> > This has three effects. Firstly it makes the on-off ratio of the
> pulses
> > 50:50 - which can be quite handy when counting numbers. Secondly,
> and more
> > importantly, the pulses are now streeeeeeched out, and the slow
CPU
> can
> > increment a counter more times during the ON period.
> >
> > The third effect is extremely interesting. Because it will take
100
> on/offs
> > as input before you get an output, and because the on/offs are
> never going
> > to be the same value (as the engine revs are never constant) then
> the
> > divider chain will sort of average up the values. This will
produce
> a sort
> > of hysteresis effect, similar to a real needle on a analogue
tacho.
> >
> > Now, once you have installed the divide by 100 counter and
> identified the
> > count values for given rpms, you can then work out the
relationship
> between
> > counts and RPM. You might have to fiddle with a timing constant
when
> > counting to make the values neat.
> >
> > Once the counting routine is ok, all you have to do is (say) once
a
> second
> > run the counting routine - make the routine wait until a new
pulse
> comes in,
> > then do your counting.
> >
> >
> > I hope all this makes sense.....
> >
> > Regards,
> >
> > Toyn Wells
> >
> >
Original Message
> > From: "tomatlarge" <tomatlarge@y...>
> > To: <basicstamps@y...>
> > Sent: Sunday, December 30, 2001 11:12 PM
> > Subject: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
> >
> >
> > > I'm not sure I follow this suggestion. The pulse is, for
example,
> on
> > > for 30ms and off for 5ms, repeating, or faster. SO, I think I
> measure
> > > the on or off pulses, or some collection of on and off pulses,
and
> > > calculate an rpm from that. I'm not sure what you mean when you
> > > say "between each pulse"? The signal is plus 12 volts for a
few
> ms
> > > and 0 volts for a few milliseconds, over and over. Am I missing
> your
> > > point?
> > >
> > > --- In basicstamps@y..., tony.wells@a... wrote:
> > > > Hi ,
> > > >
> > > > Perhaps I've missed something here. Forgive me if I have.
> > > >
> > > > Wouldn't it be easier to count the number of cpu cycles
between
> > > each pulse?
> > > > If the pulses are too short an interval, use a divide by 10 or
> > > divide by 100
> > > > counter.
> > > >
> > > > Yes, you would have to calibrate it, but the code would be
> simpler,
> > > and you
> > > > can choose when to sample - say once per second.
> > > >
> > > > Regards,
> > > >
> > > > Tony
> > > >
> > > >
Original Message
> > > > From: "tomatlarge" <tomatlarge@y...>
> > > > To: <basicstamps@y...>
> > > > Sent: Sunday, December 30, 2001 10:43 AM
> > > > Subject: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
> > > >
> > > >
> > > > > Oops, I think I mistook your original suggestion. I
connected
> a
> > > wire
> > > > > to the tach signal going from one black box to the other.
> Using a
> > > > > multimeter on the tapped line (red lead to tapped line,
black
> > > lead to
> > > > > ground), I found the pulse and voltages I listed yesterday.
> > > Today, I
> > > > > put a 100k resistor on the end of the tapped line (in series
> > > between
> > > > > the tapped line and the red lead of the multimeter). With
the
> > > > > resistor in place, I have about.1vdc less than without the
> > > resistor.
> > > > > I thought about doing it like your second suggestion (100k
to
> > > ground)
> > > > > but got paramoid. At any rate, I ended up making a voltage
> > > follower
> > > > > out of an LM358, and sent my tach signal to it through a
100k
> > > > > resistor. I then used a voltage divider on the 12v output
of
> the
> > > > > LM358, and everything seems to be working fine. I assumed
> > > (hopefully
> > > > > correctly) that the LM358 with 100k resistor would have
> virtually
> > > no
> > > > > effect on the rest of the circuit, and that the votage
> divider on
> > > the
> > > > > output side of the LM358 (using 10k and 20k resistors)
would
> not
> > > over-
> > > > > tax the LM358 output. At least it all seems to work...sound
> ok?
> > > I am
> > > > > reading up on basic electronics and am admittedly a bit
> stumped on
> > > > > the impedance concept. Your help has been invaluable...I'll
> get
> > > > > there. Now, I am starting to struggle with BS2 math
> calculations
> > > to
> > > > > turn my pulse readings into actual rpms. I found your write-
> up in
> > > the
> > > > > FAQ, but may be looking for help soon. I need to divide
30000
> by
> > > the
> > > > > pulse width (in milliseconds) I measure, and I'm trying to
do
> it
> > > > > without taking up too much cpu time. I'll play with that
> tomorrow.
> > > > > Thanks again.
> > > > >
> > > > > --- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> > > > > > >Well, I put a 100k resistor on the tach signal and there
> was
> > > about
> > > > > > >a .1volt difference on the voltmeter on either side of
the
> > > > > resistor.
> > > > > > >(The volt reading reflects an 85% "on" pulsed 12vdc
> signal, so
> > > the
> > > > > > >actual voltage drop may be slightly greater.) For
> reference,
> > > the
> > > > > tach
> > > > > > >signal is still fully functioning in its circuit. It
goes
> from
> > > the
> > > > > > >black box directly to the tach, and also directly to
> another
> > > black
> > > > > > >box that uses it to create the ignition timing signal. I
> just
> > > > > tapped
> > > > > > >into it to check the voltage/pulse. The voltage
was .1vdc
> lower
> > > > > when
> > > > > > >I checked it through a 100k resistor.
> > > > > >
> > > > > > That experiment allows you to estimate the input
> resistance. I
> > > > > think
> > > > > > you added the 100k resistor as follows:
> > > > > >
> > > > > > "black box 1 output" ----/\/\---- "black box 2+3 input"
> > > > > > 100k
> > > > > >
> > > > > > The voltage on the left side of the resistor was about
0.1
> volt
> > > > > > greater than the voltage on the right, say 10 volts and
9.9
> > > volts.
> > > > > > Yes? That lets you estimate the input resistance of black
> box
> > > 2+3.
> > > > > > The current through the 100kohm resistor is V/R =
> 0.1/100000 =~
> > > 1
> > > > > > microamp. That 1 microamp also flows through the input
> > > resistance
> > > > > of
> > > > > > black box #2+3, which has 9.9 volts across it. So the
input
> > > > > > resistance is, R=V/I=9.9 megaohms. Roughly 10 megaohms,
> give or
> > > > > take
> > > > > > a few. Pretty high.
> > > > > >
> > > > > > To measure the output resistance of black box 1, connect
the
> > > > > 100kohms
> > > > > > as follows:
> > > > > >
> > > > > > "black box 1 output" --o
"black box 2+3 input"
> > > > > > |
> > > > > > |
> > > > > > Vss---/\/\
'
> > > > > > 100kohm
> > > > > >
> > > > > > That is, direct from the output to ground. What is the
> voltage
> > > > > > before and after you connect in the resistor? E.g., 10.0
> > > before,
> > > > > 9.9
> > > > > > after. If there is no change, then try 10k or even 1k.
If
> > > there
> > > > > is
> > > > > > no change at all, then the output resistance is much much
> less
> > > than
> > > > > > the value of resistance you are trying. If the output
> drops to
> > > 1/2
> > > > > > of its original value, then the output resistance equals
the
> > > added
> > > > > > resistance. The point is, if you add your own "black
box"
> to
> > > the
> > > > > > circuit, you do not want to load it down too much. The
test
> > > give
> > > > > you
> > > > > > an estimate of how much is too much.
> > > > > >
> > > > > > -- regards,
> > > > > > Tracy Allen
> > > > > > electronically monitored ecosystems
> > > > > > mailto:tracy@e...
> > > > > > http://www.emesystems.com
> > > > >
> > > > >
> > > > > To UNSUBSCRIBE, just send mail to:
> > > > > basicstamps-unsubscribe@y...
> > > > > from the same email address that you subscribed. Text in
the
> > > Subject and
> > > > Body of the message will be ignored.
> > > > >
> > > > >
> > > > > Your use of Yahoo! Groups is subject to
> > > http://docs.yahoo.com/info/terms/
> > > > >
> > > > >
> > > > >
> > >
> > >
> > > To UNSUBSCRIBE, just send mail to:
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> > > from the same email address that you subscribed. Text in the
> Subject and
> > Body of the message will be ignored.
> > >
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> > >
> > >
> > >
>want to work in the car unless I touch either or both of the power
>supply leads...makes me think I need a capacitor somewhere.
Hi again Tom,
There should be a good bypass capacitor between pins 4 and 8 of the
LM358. Be sure all the grounds are well connected.
> Second,
>when the program is running, the pulsin command generates slightly
>smaller pulse times than is actually on the tach signal (according to
>my Fluke), resulting in slightly higher than real rpms.
How much slightly?
>As I
>mentioned, I used an LM358 as a voltage follower, and split the
>output of the LM358 with a 10k and 20k resistor. Could that be
>throwing off the timing? I am measuring one low and one high pulse
>and adding them together, so I would think the result should be the
>same even if the output is shifted a bit.
The LM358 is a relatively slow op-amp, when used in this way as a
comparator. Since this is a simple digital switch, a small signal
mosfet would be simpler and faster to translate the levels:
;
BS2 pin
|
||
o----/\/\----+5
0- ||-; 5k
+12--/\/\--||-o- Vss
1k VN10
Another option is the CD4050 hex cmos buffer. It can easily
translate high speed signals from 12 volts down to 5 volts.
-- regards,
Tracy Allen
electronically monitored ecosystems
mailto:tracy@e...
http://www.emesystems.com
Thanks again for all your help. I may indeed try the other circuits
you suggest below, but in the meantime, I was able to make things
work very well. The timing issue was my own fault (silly me, forgot
to add the neg pulse). And the capacitors seemed to fix the other
problems. It is amazing to me how much can be done with the Stamp and
the help of this list, with virtually no electronics training. I'm
working on the other input channels now, and should have a
functioning data logger shortly (with a little signal conditioning
tossed in for drivability)! --Tom
--- In basicstamps@y..., Tracy Allen <tracy@e...> wrote:
> >First, the program does not
> >want to work in the car unless I touch either or both of the power
> >supply leads...makes me think I need a capacitor somewhere.
>
> Hi again Tom,
>
> There should be a good bypass capacitor between pins 4 and 8 of the
> LM358. Be sure all the grounds are well connected.
>
> > Second,
> >when the program is running, the pulsin command generates slightly
> >smaller pulse times than is actually on the tach signal (according
to
> >my Fluke), resulting in slightly higher than real rpms.
>
> How much slightly?
>
> >As I
> >mentioned, I used an LM358 as a voltage follower, and split the
> >output of the LM358 with a 10k and 20k resistor. Could that be
> >throwing off the timing? I am measuring one low and one high pulse
> >and adding them together, so I would think the result should be the
> >same even if the output is shifted a bit.
>
> The LM358 is a relatively slow op-amp, when used in this way as a
> comparator. Since this is a simple digital switch, a small signal
> mosfet would be simpler and faster to translate the levels:
>
> ;
BS2 pin
> |
> ||
o----/\/\----+5
> 0- ||-; 5k
> +12--/\/\--||-o- Vss
> 1k VN10
>
> Another option is the CD4050 hex cmos buffer. It can easily
> translate high speed signals from 12 volts down to 5 volts.
>
> -- regards,
> Tracy Allen
> electronically monitored ecosystems
> mailto:tracy@e...
> http://www.emesystems.com
After reviewing all of the responces of this thread I believe you would be
very interested in Al Williams Pak VII. I have only reviewed these devices
but I believe they are exactly what you need. It is a co-processor that can
monitior the length of 8 diffeent pulses, HI/LO/ ,DURATION ,TOTALS,ETC. The
stamp can then retrieve this info as neede with a serin command, useing very
little of the stamps processing resources.
Shannon
[noparse][[/noparse]Non-text portions of this message have been removed]
--- In basicstamps@y..., moretraction@a... wrote:
> Tom,
> After reviewing all of the responces of this thread I believe you
would be
> very interested in Al Williams Pak VII. I have only reviewed these
devices
> but I believe they are exactly what you need. It is a co-processor
that can
> monitior the length of 8 diffeent pulses,
HI/LO/ ,DURATION ,TOTALS,ETC. The
> stamp can then retrieve this info as neede with a serin command,
useing very
> little of the stamps processing resources.
>
> Shannon
>
>
> [noparse][[/noparse]Non-text portions of this message have been removed]
>Sounds interesting. Do you have a link?
> http://www.al-williams.com/awce/pak7.htm
PAKs, use SHIFTIN/SHIFTOUT for communications. People often ask why the
PAKs don't use SERIN/SEROUT (the PAK-V and VI do because they have
different requirements). One of the benefits of a PAK is that it allows
a level of multitasking. If they used SERIN, the Stamp (and other non
interrupt processors) would have to listen for incoming data. With the
Shift scheme, you can issue a command, go do something, and then read
the results later. You can read the result at your own speed, stop in
the middle and result later, etc. You can't do that with SERIN.
The PAK-V only accepts input, so SEROUT is fine. The PAK-VI's whole
point is to convert PS/2 to RS232, so again, SERIN/SEROUT works there
too. Otherwise, all the PAKs use the same protocol.
You can find the manual and example code on our site.
Al Williams
AWC
* 8 channels of PWM
http://www.al-williams.com/awce/pak5.htm
>
Original Message
> From: moretraction@a... [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=iA5rf9Yi9Yum4khoxEUhd_jI7aCDKgvxL0-5_WB0zVcJZZSvdYdkuqx0RHkzQbNqYbu5Ri-SzPcqRw]moretraction@a...[/url
> Sent: Monday, December 31, 2001 11:10 PM
> To: basicstamps@yahoogroups.com
> Subject: Re: [noparse][[/noparse]basicstamps] Re: Pulse Conditioning for BS2
>
>
> Tom,
> After reviewing all of the responces of this thread I believe
> you would be
> very interested in Al Williams Pak VII. I have only reviewed
> these devices
> but I believe they are exactly what you need. It is a
> co-processor that can
> monitior the length of 8 diffeent pulses, HI/LO/ ,DURATION
> ,TOTALS,ETC. The
> stamp can then retrieve this info as neede with a serin
> command, useing very
> little of the stamps processing resources.
>
> Shannon
>
>
> [noparse][[/noparse]Non-text portions of this message have been removed]
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed. Text in the
> Subject and Body of the message will be ignored.
>
>
> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/
>