I need creative ideas.....
If you just need an on / off optocoupler, use a neon bulb (with suitable
resistor) and a phototransistor -- heatshrink the two together.
Original Message
> If I have a resistor, say 25k in series with a pair of diodes in inverse
> parallel (inside an optocoupler pair chip), and this whole little circuit
> across 240VAC, fully 180 degrees out of phase, it should be easy to size
the
> resistor for power, right?
> Evidently not so. Here's what I did and why I think my result is off....
>
> I=E/R, so I = 240/25k, right? well, almost. the two led's each drop about
> 1.5V, so let's make it I = 237/25k = .00948, or about 9 1/2 mA. perfect
for
> the LED side of the optocoupler.
> Whether I use P = IE, .00948 * 237 = 2.25 Watts, or P=I squared *R,
> .00008987 * 25,000 = 2.25 Watts, you can see I get the same result, 2.25W.
>
> The problem is this: I started with a 5W resistor, and noticed immediately
> that it got too hot. I am now up to a 10W, which is running at 105 degrees
> C - also way too hot. These should be running about 60C.
>
> If the capacitive reactance of the optocoupler input is a factor, it
should
> be raising the effective resistance, or impedance, and I would have LESS
> real power, no?
>
> So can anyone offer an idea that may help find what I have done wrong???
resistor) and a phototransistor -- heatshrink the two together.
Original Message
> If I have a resistor, say 25k in series with a pair of diodes in inverse
> parallel (inside an optocoupler pair chip), and this whole little circuit
> across 240VAC, fully 180 degrees out of phase, it should be easy to size
the
> resistor for power, right?
> Evidently not so. Here's what I did and why I think my result is off....
>
> I=E/R, so I = 240/25k, right? well, almost. the two led's each drop about
> 1.5V, so let's make it I = 237/25k = .00948, or about 9 1/2 mA. perfect
for
> the LED side of the optocoupler.
> Whether I use P = IE, .00948 * 237 = 2.25 Watts, or P=I squared *R,
> .00008987 * 25,000 = 2.25 Watts, you can see I get the same result, 2.25W.
>
> The problem is this: I started with a 5W resistor, and noticed immediately
> that it got too hot. I am now up to a 10W, which is running at 105 degrees
> C - also way too hot. These should be running about 60C.
>
> If the capacitive reactance of the optocoupler input is a factor, it
should
> be raising the effective resistance, or impedance, and I would have LESS
> real power, no?
>
> So can anyone offer an idea that may help find what I have done wrong???
Comments
And, yes, this is slightly OT, but there has been a Stamp connected to this
project on & off several times now. OK, qualifying aside, it goes back to
2nd semester EE - AC Theory, if I recall:
If I have a resistor, say 25k in series with a pair of diodes in inverse
parallel (inside an optocoupler pair chip), and this whole little circuit
across 240VAC, fully 180 degrees out of phase, it should be easy to size the
resistor for power, right?
Evidently not so. Here's what I did and why I think my result is off....
I=E/R, so I = 240/25k, right? well, almost. the two led's each drop about
1.5V, so let's make it I = 237/25k = .00948, or about 9 1/2 mA. perfect for
the LED side of the optocoupler.
Whether I use P = IE, .00948 * 237 = 2.25 Watts, or P=I squared *R,
.00008987 * 25,000 = 2.25 Watts, you can see I get the same result, 2.25W.
The problem is this: I started with a 5W resistor, and noticed immediately
that it got too hot. I am now up to a 10W, which is running at 105 degrees
C - also way too hot. These should be running about 60C.
If the capacitive reactance of the optocoupler input is a factor, it should
be raising the effective resistance, or impedance, and I would have LESS
real power, no?
So can anyone offer an idea that may help find what I have done wrong???
Thanks,
Chris
Based on your math, which is quite correct, the heating problem suggests you
are using a 2500 resistor. Is that possible?
Sid
No, really, thanks, but I already have boards made for the opto chips
intended.
I need to figure out why the series resistor power defies gravity. I can
re-size the resistor somewhat, but I can't redesign the board - too much $
invested in it...
CL
>
Original Message
> From: Rodent [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=6ILZv79M8JorzTsjCxxM7IC6qly1PyApehd-9buEr8RKHk-rAvMyTKOF4ZOhDcKwsNeZPEujt8_Eifw]daweasel@s...[/url
> Sent: Friday, June 29, 2001 5:22 PM
> To: basicstamps@yahoogroups.com
> Subject: Re: [noparse][[/noparse]basicstamps] I need creative ideas.....
>
>
> If you just need an on / off optocoupler, use a neon bulb
> (with suitable
> resistor) and a phototransistor -- heatshrink the two together.
>
>
Original Message
>
> > If I have a resistor, say 25k in series with a pair of
> diodes in inverse
> > parallel (inside an optocoupler pair chip), and this whole
> little circuit
> > across 240VAC, fully 180 degrees out of phase, it should be
> easy to size
> the
> > resistor for power, right?
> > Evidently not so. Here's what I did and why I think my
> result is off....
> >
> > I=E/R, so I = 240/25k, right? well, almost. the two led's
> each drop about
> > 1.5V, so let's make it I = 237/25k = .00948, or about 9 1/2
> mA. perfect
> for
> > the LED side of the optocoupler.
> > Whether I use P = IE, .00948 * 237 = 2.25 Watts, or P=I squared *R,
> > .00008987 * 25,000 = 2.25 Watts, you can see I get the same
> result, 2.25W.
> >
> > The problem is this: I started with a 5W resistor, and
> noticed immediately
> > that it got too hot. I am now up to a 10W, which is running
> at 105 degrees
> > C - also way too hot. These should be running about 60C.
> >
> > If the capacitive reactance of the optocoupler input is a factor, it
> should
> > be raising the effective resistance, or impedance, and I
> would have LESS
> > real power, no?
> >
> > So can anyone offer an idea that may help find what I have
> done wrong???
>
>
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed with. Text
> in the Subject and Body of the message will be ignored.
>
>
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http://docs.yahoo.com/info/terms/
I just re-checked - they are 25kJ's cement power type. I even checked with a
meter - 24.88k. 2.5k, or 2k5 would have popped the opto by exceeding If.
So, I'm as lost as ever. I even re-washed the test board, in case some flux
was conducting some leakage, but to no avail...
Thanks, Sid.
>
Original Message
> From: Newzed@a... [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=dVxqjFlnMRYWHVPUamzZGHJ9eElV-VGvGc8cFSj_i-jjky27AOdnXp9fDWVw9QNEyxGpQZuC]Newzed@a...[/url
> Sent: Friday, June 29, 2001 5:36 PM
> To: basicstamps@yahoogroups.com
> Subject: Re: [noparse][[/noparse]basicstamps] I need creative ideas.....
>
>
> For Chris@tampabay
>
> Based on your math, which is quite correct, the heating
> problem suggests you
> are using a 2500 resistor. Is that possible?
>
> Sid
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed with. 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/
>
>
Your calculations are correct .. 9.6ma and 2.25 watts.
Wirewound resistors are designed to run *HOT*. You can use either the 5 or
10 watt resistor. Power resistors are often mounted off-board, or, at least
on extended leads to keep from burning the circuit-board. 50 to 60 deg C is
about the max temp that you can keep your finger on. Since almost of the
240 volts is dropped across the resistor, you have a simple resistive
circuit, where V and I are in phase. Capacitive reactance of the
optocoupler little effect on the current because the coupler voltage drop is
negligible .
You could use a capacitor to drop line voltage, which would produce
essentially no heat. Use a cap instead of the wirewound resistor. Assume
you want 10 ma through the coupler. Ignoring the coupler drop, you want 240
volts/.01 amps=24K of capacitive reactance. Xc=1/(2*pi*f*C), or solving for
C:
C=1/(2*pi*f*Xc)
for f=60 hz, C=0.11 uf. Try a 0.1 uf paper or mylar capacitor.
The peak voltage of 240 volts is 1.414*240=339 volts, so you should use at
least a 600 volt cap, allowing for line spikes.
Regards,
Ray McArthur
Original Message
From: Chris Loiacono (E-mail) <chris01@t...>
To: Basicstamps Egroup (E-mail) <basicstamps@yahoogroups.com>
Sent: Friday, June 29, 2001 5:25 PM
Subject: [noparse][[/noparse]basicstamps] I need creative ideas.....
> I=E/R, so I = 240/25k, right? well, almost. the two led's each drop about
> 1.5V, so let's make it I = 237/25k = .00948, or about 9 1/2 mA. perfect
for
> the LED side of the optocoupler.
> Whether I use P = IE, .00948 * 237 = 2.25 Watts, or P=I squared *R,
> .00008987 * 25,000 = 2.25 Watts, you can see I get the same result, 2.25W.
>
> The problem is this: I started with a 5W resistor, and noticed immediately
> that it got too hot. I am now up to a 10W, which is running at 105 degrees
> C - also way too hot. These should be running about 60C.
you better use a capacitor IN SERIES with a resistor and
use a zener diode and back diode in parallel across the
optocoupler input.
A 60Hz or 50Hz line can contain considerable high frequency
energy at times for a variety of reasons - mostly transient
spiking and the capacitor will pass this directly into the
optocoupler LED unless you give it another dissipative path;
e.g. a resistor into diodes. A properly selected resistor
can also act to limit destructive effects IF the capacitor
fails shorted - meantime the capacitor lowers power dissipated.
You may not like the neon tube and photo resistor type isolator
but that works very well! [noparse][[/noparse]commercial units from Vactec "Vactrol"].
They require very little power at high voltage - only drawbacks
are neon tube aging and slow (millisecond) response.
I saw a Beckman unit one time which used these for logic
to light the high voltage electroluminescent 7 segment display.
Getting Basic Stamps to control hundreds of volts is always
going to be a little touchy and needs to be handled with care.
On the matter of splipsticks.... assuming the site still works:
>Java RULEs....Here's the ultimate computer math tool:
>http://www.comcen.com.au/~adavie/javaslide/javaslide.html
Personally I like circular slide rules as the ends then meet
instead of having to swap.... :-)
Have fun, premena
>>>>>>
Date: Fri, 29 Jun 2001 22:31:42 -0400
From: "Ray McArthur" <rjmca@u...>
Chris:
Your calculations are correct .. 9.6ma and 2.25 watts.
Wirewound resistors are designed to run *HOT*. You can use either the 5
or
10 watt resistor. Power resistors are often mounted off-board, or, at
least
on extended leads to keep from burning the circuit-board. 50 to 60 deg C
is
about the max temp that you can keep your finger on. Since almost of the
240 volts is dropped across the resistor, you have a simple resistive
circuit, where V and I are in phase. Capacitive reactance of the
optocoupler little effect on the current because the coupler voltage drop
is
negligible .
You could use a capacitor to drop line voltage, which would produce
essentially no heat. Use a cap instead of the wirewound resistor.
Assume
you want 10 ma through the coupler. Ignoring the coupler drop, you want
240
volts/.01 amps=24K of capacitive reactance. Xc=1/(2*pi*f*C), or solving
for
C:
C=1/(2*pi*f*Xc)
for f=60 hz, C=0.11 uf. Try a 0.1 uf paper or mylar capacitor.
The peak voltage of 240 volts is 1.414*240=339 volts, so you should use
at
least a 600 volt cap, allowing for line spikes.
________________________________________________________________
GET INTERNET ACCESS FROM JUNO!
Juno offers FREE or PREMIUM Internet access for less!
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Won't the current lead the voltage by 90 degrees?
Then my opto will out of time, will it not?
I didn't mention that my goal was to have the opto output during each full
1/2 wave, in phase with it as closely as possible, hence the resistive-only
circuit.
Chris
>
Original Message
> From: Ray McArthur [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=QdvD3beX6FG69htQaLnOQhuWPoB8ytWvOKjIx5OQcYYVeecxkdlbKRxKVMeNZuHl_38ae7JLd_N8jMF7]rjmca@u...[/url
> Sent: Friday, June 29, 2001 10:32 PM
> To: basicstamps@yahoogroups.com
> Subject: Re: [noparse][[/noparse]basicstamps] I need creative ideas.....
>
>
> Chris:
> Your calculations are correct .. 9.6ma and 2.25 watts.
>
> Wirewound resistors are designed to run *HOT*. You can use
> either the 5 or
> 10 watt resistor. Power resistors are often mounted
> off-board, or, at least
> on extended leads to keep from burning the circuit-board. 50
> to 60 deg C is
> about the max temp that you can keep your finger on. Since
> almost of the
> 240 volts is dropped across the resistor, you have a simple resistive
> circuit, where V and I are in phase. Capacitive reactance of the
> optocoupler little effect on the current because the coupler
> voltage drop is
> negligible .
>
> You could use a capacitor to drop line voltage, which would produce
> essentially no heat. Use a cap instead of the wirewound
> resistor. Assume
> you want 10 ma through the coupler. Ignoring the coupler
> drop, you want 240
> volts/.01 amps=24K of capacitive reactance. Xc=1/(2*pi*f*C),
> or solving for
> C:
>
> C=1/(2*pi*f*Xc)
>
> for f=60 hz, C=0.11 uf. Try a 0.1 uf paper or mylar capacitor.
> The peak voltage of 240 volts is 1.414*240=339 volts, so you
> should use at
> least a 600 volt cap, allowing for line spikes.
>
> Regards,
> Ray McArthur
>
>
Original Message
> From: Chris Loiacono (E-mail) <chris01@t...>
> To: Basicstamps Egroup (E-mail) <basicstamps@yahoogroups.com>
> Sent: Friday, June 29, 2001 5:25 PM
> Subject: [noparse][[/noparse]basicstamps] I need creative ideas.....
>
>
> > I=E/R, so I = 240/25k, right? well, almost. the two led's
> each drop about
> > 1.5V, so let's make it I = 237/25k = .00948, or about 9 1/2
> mA. perfect
> for
> > the LED side of the optocoupler.
> > Whether I use P = IE, .00948 * 237 = 2.25 Watts, or P=I squared *R,
> > .00008987 * 25,000 = 2.25 Watts, you can see I get the same
> result, 2.25W.
> >
> > The problem is this: I started with a 5W resistor, and
> noticed immediately
> > that it got too hot. I am now up to a 10W, which is running
> at 105 degrees
> > C - also way too hot. These should be running about 60C.
>
>
>
> To UNSUBSCRIBE, just send mail to:
> basicstamps-unsubscribe@yahoogroups.com
> from the same email address that you subscribed with. 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/
>
>
Ray McArthur
Original Message
From: Chris Loiacono (E-mail) <chris01@t...>
To: <basicstamps@yahoogroups.com>
Sent: Saturday, June 30, 2001 9:40 PM
Subject: RE: [noparse][[/noparse]basicstamps] I need creative ideas.....
> Won't the current lead the voltage by 90 degrees?
> Then my opto will out of time, will it not?
> I didn't mention that my goal was to have the opto output during each full
> 1/2 wave, in phase with it as closely as possible, hence the
resistive-only
> circuit.
Original Message
> From: Chris Loiacono [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=fFNRF-KaZOaktt8U6QNwV51LItqDdPpyQMOAlZ13IKmfGqgYZ6NJTCy1FE8uklFfclwvNpV5dwPl]chris@m...[/url
> Sent: Saturday, June 30, 2001 9:50 PM
> To: 'basicstamps@yahoogroups.com'
> Subject: RE: [noparse][[/noparse]basicstamps] Re: I need creative ideas.....
>
>
> I have seen the neon method work well back in school XX years
> ago....It's just too slow. I need 50 usec response on and off
> here. It's also common practice to use a pair of
> inverse-parallel diodes to protect against reverse current,
> and I have used a pair of opto's LED's to accomplish this.
> The cap will give me a phase shift, will it not? That would
> mess up my timing again.
> ON Semi & Fairchild Opto both recommend the single resistor
> method with an inverse diode pair only. I have seen many
> commercial boards done this way.
> I just can't figure out why the resistor is getting so hot!
> I was targeting 60C or less, and even when I over size the
> resistor by 100%, it still goes over 105C.
>
> Could there be a constant HF signal on the line causing this,
> rather than random spikes??
>
> chris
>
> >
Original Message
> > From: s premena [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=1nEhnm7Ezp9-U_DYX7y3LW3c8gYQhFMoF4jcO6LRN6bnBCFiCMTOTImv1hcrqemZEwJ2kIs2xg]premzee@j...[/url
> > Sent: Saturday, June 30, 2001 2:28 PM
> > To: basicstamps@yahoogroups.com
> > Subject: [noparse][[/noparse]basicstamps] Re: I need creative ideas.....
> >
> >
> > Whereas I agree substantially with Ray's analysis in practice
> > you better use a capacitor IN SERIES with a resistor and
> > use a zener diode and back diode in parallel across the
> > optocoupler input.
> >
> > A 60Hz or 50Hz line can contain considerable high frequency
> > energy at times for a variety of reasons - mostly transient
> > spiking and the capacitor will pass this directly into the
> > optocoupler LED unless you give it another dissipative path;
> > e.g. a resistor into diodes. A properly selected resistor
> > can also act to limit destructive effects IF the capacitor
> > fails shorted - meantime the capacitor lowers power dissipated.
> >
> > You may not like the neon tube and photo resistor type isolator
> > but that works very well! [noparse][[/noparse]commercial units from Vactec "Vactrol"].
> > They require very little power at high voltage - only drawbacks
> > are neon tube aging and slow (millisecond) response.
> >
> > I saw a Beckman unit one time which used these for logic
> > to light the high voltage electroluminescent 7 segment display.
> >
> > Getting Basic Stamps to control hundreds of volts is always
> > going to be a little touchy and needs to be handled with care.
> >
> > On the matter of splipsticks.... assuming the site still works:
> >
> > >Java RULEs....Here's the ultimate computer math tool:
> > >http://www.comcen.com.au/~adavie/javaslide/javaslide.html
> >
> > Personally I like circular slide rules as the ends then meet
> > instead of having to swap.... :-)
> >
> > Have fun, premena
> >
> > >>>>>>
> > Date: Fri, 29 Jun 2001 22:31:42 -0400
> > From: "Ray McArthur" <rjmca@u...>
> >
> > Chris:
> > Your calculations are correct .. 9.6ma and 2.25 watts.
> >
> > Wirewound resistors are designed to run *HOT*. You can use
> > either the 5
> > or
> > 10 watt resistor. Power resistors are often mounted
> off-board, or, at
> > least
> > on extended leads to keep from burning the circuit-board. 50
> > to 60 deg C
> > is
> > about the max temp that you can keep your finger on. Since
> > almost of the
> > 240 volts is dropped across the resistor, you have a simple
> resistive
> > circuit, where V and I are in phase. Capacitive reactance of the
> > optocoupler little effect on the current because the coupler
> > voltage drop
> > is
> > negligible .
> >
> > You could use a capacitor to drop line voltage, which would produce
> > essentially no heat. Use a cap instead of the wirewound resistor.
> > Assume
> > you want 10 ma through the coupler. Ignoring the coupler
> > drop, you want
> > 240
> > volts/.01 amps=24K of capacitive reactance. Xc=1/(2*pi*f*C),
> > or solving
> > for
> > C:
> >
> > C=1/(2*pi*f*Xc)
> >
> > for f=60 hz, C=0.11 uf. Try a 0.1 uf paper or mylar capacitor.
> > The peak voltage of 240 volts is 1.414*240=339 volts, so you
> > should use
> > at
> > least a 600 volt cap, allowing for line spikes.
> >
> > ________________________________________________________________
> > GET INTERNET ACCESS FROM JUNO!
> > Juno offers FREE or PREMIUM Internet access for less!
> > Join Juno today! For your FREE software, visit:
> > http://dl.www.juno.com/get/tagj.
> >
> > To UNSUBSCRIBE, just send mail to:
> > basicstamps-unsubscribe@yahoogroups.com
> > from the same email address that you subscribed with. 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/
> >
> >
>
> re-size the resistor somewhat, but I can't redesign the board - too much $
> invested in it...
Well you did check my first idea ... i.e. measure the resistance. THe next
idea is to measure the current. If you have a multimeter capable of AC
current, put it in series and measure. It should be the 9 mA you see.
Also, try looking for a better opto (higher current transfer ratio) with the
same footprint. You can get opto's that take 0.1 mA on the primary side.
This would allow you to use an even higher value resistor on the promary
side and reduce the power consumption / current draw even more. Often they
cost 50-100% more per opto, but it might be a cheaper solution of your
heavily invested in the boards....
FWIW,
John <- who needs to go help the wife bag the recyling.... durn it [noparse];)[/noparse] ...
> resistance.
Yup. Verified as what it should be.
THe next
> idea is to measure the current. If you have a multimeter capable of AC
> current, put it in series and measure. It should be the 9 mA you see.
I;ll have to check my meters. I can also check the voltage on the scope and
take the actual measured resistance to get a semi-accurate current
measurement I suppose...But I'll check my meters first.
>
> Also, try looking for a better opto (higher current transfer
> ratio) with the
> same footprint. You can get opto's that take 0.1 mA on the
> primary side.
> This would allow you to use an even higher value resistor on
> the promary
> side and reduce the power consumption / current draw even
> more. Often they
> cost 50-100% more per opto, but it might be a cheaper solution of your
> heavily invested in the boards....
Good Idea. This is the kind of creativity I was hoping for.
I already have a transistor in a sort of common emitter scheme to increase
the CTR, and have biased it for fast switching. I guess I just hesitate at
the idea of going back and starting over..
I will look into this also...
>
> FWIW,
> John
It's worth quite a bit, especially when one's sanity is at risk.
Thanks very much!
Chris
<- who needs to go help the wife bag the recyling....
> durn it [noparse];)[/noparse] ...
>
Hey, it could have been worse - like mow the lawn, or something!!!
>
> I;ll have to check my meters. I can also check the voltage on the
> scope and
> take the actual measured resistance to get a semi-accurate current
> measurement I suppose...But I'll check my meters first.
??? Well, how's it going ??? Gotta give us feedback ya know *wink* .... you
got my curiousity up enough to write, so I want to know how it's going [noparse]:)[/noparse]
....
> > Also, try looking for a better opto (higher current transfer
> > ratio) with the
> Good Idea. This is the kind of creativity I was hoping for.
Good ... did you find it ???? I've done this way too often ... if you need
help, just tell me which opto your using (and package) and i'll try to help
....
> I already have a transistor in a sort of common emitter scheme to increase
> the CTR, and have biased it for fast switching. I guess I just hesitate at
????? Can you draw out what your talking about ???? I don't understand what
a "sort of a common emitter scheme" would look like in your application
.... are you talking about setting it up as a darlington ??? just missing
the picture here ....
> It's worth quite a bit, especially when one's sanity is at risk.
This I understand all too well ... and it almost always turns out to be
something ultra simple (I have had much experience with pulling my hair
out.... amazing I have any left [noparse]:)[/noparse]....
> Thanks very much!
De nada ...
> <- who needs to go help the wife bag the recyling....
> > durn it [noparse];)[/noparse] ...
> >
> Hey, it could have been worse - like mow the lawn, or something!!!
I did that the previous day [noparse]:)[/noparse] ... 90+ degrees, high humidity, in the sun,
etc.... yeay!!!!
[noparse]:)[/noparse]
John