Voltage Dividers
I want to measure DC voltages, but they will vary from 0 to 8 volts. Can I just
use a simple voltage divider to cut that voltage in half and then feed that into
ADC or is there a better way?
If a voltage divider will work is there an optimum value to use for the
resistors? Since there is no real current draw I would guess that 1Meg resistors
would work just fine.
Hank
use a simple voltage divider to cut that voltage in half and then feed that into
ADC or is there a better way?
If a voltage divider will work is there an optimum value to use for the
resistors? Since there is no real current draw I would guess that 1Meg resistors
would work just fine.
Hank
Comments
Vout=Vin*Rb/(Ra+Rb)
You must have an idea of the input resistance (Rin) of the D/A.
unless the Input resistance Rin >> Rb (at least a order of magnitude)
Rin will be in parallel with Rb, giving a different resistance ratio
than expected.
--- In basicstamps@y..., "rcvehicles" <rcvehicles@y...> wrote:
> I want to measure DC voltages, but they will vary from 0 to 8
volts. Can I just use a simple voltage divider to cut that voltage in
half and then feed that into ADC or is there a better way?
>
> If a voltage divider will work is there an optimum value to use for
the resistors? Since there is no real current draw I would guess that
1Meg resistors would work just fine.
>
> Hank
be very precise, only voltage dividers (no ADC) and the IN command
will also work.
I believe 1M may be a bit high. I use around 10K .
ACJacques
rcvehicles wrote:
>
> I want to measure DC voltages, but they will vary from 0 to 8 volts. Can I
just use a simple voltage divider to cut that voltage in half and then feed that
into ADC or is there a better way?
>
> If a voltage divider will work is there an optimum value to use for the
resistors? Since there is no real current draw I would guess that 1Meg resistors
would work just fine.
>
> Hank
>
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>Can I just use a simple voltage divider to cut that voltage in half
>and then feed that into ADC or is there a better way?
>
>If a voltage divider will work is there an optimum value to use for
>the resistors? Since there is no real current draw I would guess
>that 1Meg resistors would work just fine.
>
>Hank
On the face of it, yes, the 1 meg resistors should work fine. There
are a couple of other issues you need to consider.
* noise pickup. Keep the leads short.
* ADC glitches. Many ADCs shove out a little packet of charge
every time they look at the input. That pulse of current has to flow
through your voltage divider and generates an error voltage. The
error voltage will be high if you use 1meg resistors, lower if the
resistors in are, say, 10kohms. You can get around it if you
connect a capacitor, say 0.1uF from the ADC input to analog ground.
The capacitor charges up to the voltage of your input signal with the
voltage divider, but the capacitor also acts as a low pass filter,
limiting the speed of response. Also, the capacitor absorbs the
pulses of current from the ADC, and that limits the ADC sample rate,
because rapid sampling pulses will average out to become a
significant error voltage. If you need both high input impedance and
speed, there has to be a buffer amplifier (op-amp) in the circuit
between the signal source and the ADC.
-- regards,
Tracy Allen
electronically monitored ecosystems
mailto:tracy@e...
http://www.emesystems.com
circuit the +5 to - 5 volt supply
needed to power an Op amp? Can this be done with an extra battery back to back
with the stamp battery for example?
Michael
Tracy Allen wrote:
> >I want to measure DC voltages, but they will vary from 0 to 8 volts.
> >Can I just use a simple voltage divider to cut that voltage in half
> >and then feed that into ADC or is there a better way?
> >
> >If a voltage divider will work is there an optimum value to use for
> >the resistors? Since there is no real current draw I would guess
> >that 1Meg resistors would work just fine.
> >
> >Hank
>
> On the face of it, yes, the 1 meg resistors should work fine. There
> are a couple of other issues you need to consider.
> * noise pickup. Keep the leads short.
>
> * ADC glitches. Many ADCs shove out a little packet of charge
> every time they look at the input. That pulse of current has to flow
> through your voltage divider and generates an error voltage. The
> error voltage will be high if you use 1meg resistors, lower if the
> resistors in are, say, 10kohms. You can get around it if you
> connect a capacitor, say 0.1uF from the ADC input to analog ground.
> The capacitor charges up to the voltage of your input signal with the
> voltage divider, but the capacitor also acts as a low pass filter,
> limiting the speed of response. Also, the capacitor absorbs the
> pulses of current from the ADC, and that limits the ADC sample rate,
> because rapid sampling pulses will average out to become a
> significant error voltage. If you need both high input impedance and
> speed, there has to be a buffer amplifier (op-amp) in the circuit
> between the signal source and the ADC.
>
> -- 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/
op amp with a gain <1. Then you don't need a voltage divider at all.
The thing that people forget with a voltage divider is that the input
resistance of the ADC is in parallel with "R2" -- so say you have a
10K/10K divider (1/2) but the input to the ADC is 5K. Really R2 is not
10K but 10K || 5K = 3.3K. So the divider is really 3.3/13.3 or 1/4!
That's an extreme case. If Rin>>R2 then no worries (in practice). Say
the ADC Rin was 1M. Now 10K || 1M = 9.9K so the actual ratio is
9.9K/19.9K = .4975 pretty close to .5 and closer than the component
tolerances, probably (less than .5%).
A few Web pages I have that might be of interest:
http://www.al-williams.com/wd5gnr/opampneg.htm - Getting that minus
supply.
http://www.al-williams.com/wd5gnr/basiccir.htm - Voltage dividers and
stuff.
Al Williams
AWC
* Control 8 servos at once
http://www.al-williams.com/awce/pak8.htm
>
Original Message
> From: Michael Gianturco [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=oduZ5feESrgesovN8zic-me_slDa1-4r8t3DSzrUoqhUZeHzEsJ-JUw4ZzH5Q0RDFB_q8tb7BkUTN2iE]michcg@m...[/url
> Sent: Sunday, December 30, 2001 7:24 AM
> To: basicstamps@yahoogroups.com
> Subject: Re: [noparse][[/noparse]basicstamps] Voltage Dividers
>
>
> Re the op-amp buffer. How should I introduce into the 0 to
> +5 volt basic stamp circuit the +5 to - 5 volt supply needed
> to power an Op amp? Can this be done with an extra battery
> back to back with the stamp battery for example?
>
> Michael
>
>
> Tracy Allen wrote:
>
> > >I want to measure DC voltages, but they will vary from 0
> to 8 volts.
> > >Can I just use a simple voltage divider to cut that
> voltage in half
> > >and then feed that into ADC or is there a better way?
> > >
> > >If a voltage divider will work is there an optimum value
> to use for
> > >the resistors? Since there is no real current draw I would
> guess that
> > >1Meg resistors would work just fine.
> > >
> > >Hank
> >
> > On the face of it, yes, the 1 meg resistors should work
> fine. There
> > are a couple of other issues you need to consider.
> > * noise pickup. Keep the leads short.
> >
> > * ADC glitches. Many ADCs shove out a little packet of charge
> > every time they look at the input. That pulse of current
> has to flow
> > through your voltage divider and generates an error voltage. The
> > error voltage will be high if you use 1meg resistors, lower if the
> > resistors in are, say, 10kohms. You can get around it if you
> > connect a capacitor, say 0.1uF from the ADC input to analog ground.
> > The capacitor charges up to the voltage of your input
> signal with the
> > voltage divider, but the capacitor also acts as a low pass filter,
> > limiting the speed of response. Also, the capacitor absorbs the
> > pulses of current from the ADC, and that limits the ADC
> sample rate,
> > because rapid sampling pulses will average out to become a
> significant
> > error voltage. If you need both high input impedance and
> speed, there
> > has to be a buffer amplifier (op-amp) in the circuit between the
> > signal source and the ADC.
> >
> > -- 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/
>
>
> 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/
>basic stamp circuit the +5 to - 5 volt supply
>needed to power an Op amp? Can this be done with an extra battery
>back to back with the stamp battery for example?
>
>Michael
Usually people use a single supply op-amp, like the LM358 or the
LT1078 or the LM10 etc etc.