Hysteresis needed in 741 Op-Amp circuit

[CF]

Yes sorry, I had spelled this out on an earlier draft of the web page, then
deleted it along with other superfluous stuff...

This thermostat circuit is hooked up to a cooling fan, so I want the relay
contacts to close (turn the fan on) when the temperature is high and the
thermistor resistance is low.

CF


rpaisley4@cogeco.ca (Rob Paisley) wrote in
news:bb39a577.0406131205.6fd50e9@posting.google.com:

A basic design question if I may; Do you want the relay to turn ON or
OFF when the temperature is HIGH (the thermistor resistance is LOW)?

Rob.

------------

CF <form@web.site> wrote in message
news:<Xns9507BF920FE079dkjdj7u3wjedlkr84jh@211.29.133.50>...
Excellent Rob! Thanks for simplifying the layout of my circuit. I
knew it was getting messy. I'd just finished drawing my own new
version, and then saw yours and had to go back and improve it a bit
more (still messy, and amateurish, but oh well). I think your layout
is fundamentally sound. The only thing that worries me is a note that
comes with the National Semiconductor spec sheet -- to the effect
that when driving a ground- referred load, the inputs are reversed.

Also, looking at the spec sheet today, I had, as I said, just
finished revising my own circuit by doing away with the transistor
entirely. Apparently the LM311's output can sink up to 50mA at 50V,
so why bother with an extra transistor for a 30mA relay? See this
page for what I've come up with, and a few snippets of the spec sheet
including that note I mentioned:

www.copperleife.com/craig/tech/thermo/rp.htm

Regards,

CF


rpaisley4@cogeco.ca (Rob Paisley) wrote in
news:bb39a577.0406121903.2f51014d@posting.google.com:

Here is my take on the circuit.

http://home.cogeco.ca/~rpaisley4/zFile.html

I have reduced the setpoint control to is minimum and maximum
values.

Rob.

 

[CF]

Thanks for the circuit Bill, demonstrating the positive feedback for
hysteresis that we've been talking about. Interested to see the 339's
output feeding into the transistor, I looked it up. It's output is
classified as "Open Drain/Collector" (which I presume is similar to the
331). Thus, your output should be like Rob Paisley's at:
http://home.cogeco.ca/%7Erpaisley4/zFile.html

However, you seem to have a pullup resistor attached to the same pin that
is apparently sourcing current for the transistor, so I'm confused. What is
to stop current bypassing the 339 altogether by running directly through
the 2.2k pullup resistor into the transistor's base?

Also, you don't have the first IC marked. Is it also an LM339?

Regards,

CF


wrongaddress@att.net (Bill Bowden) wrote in
news:ad025737.0406130337.105fc04c@posting.google.com:

There are a couple thermostat circuits using the LM339 quad comparator
on my website here:

http://ourworld.compuserve.com/homepages/Bill_Bowden/page2.htm#therm.gif

-Bill

 

[CF]

I have no idea what I'm talking about, but I'm not happy with the way the
311-based circuits that we've been discussing take their positive feedback
from an output that is constantly pulled up to a positive voltage when the
output is supposedly "off". Surely this undermines the whole concept of
positive feedback?

The earlier 741-based circuits, those modelled by Terry, and Bill Bowden's
399 circuit for that matter, are quite different in this respect. They seem
to provide feedback only when it is appropriate.

With this in mind, I've tried to take advantage of the option to use a
"ground-referred" load on the 311's output, and redesigned the circuit at
the following page. However, I'm uncertain about the polarity of the pins,
as they are supposedly reversed under these conditions, yet are drawn in
the same configuration in the spec brochure (see same page). If anyone has
any insight into this, please let us know.

See:
www.copperleife.com/craig/tech/thermo/groundref.htm

 

[Terry Pinnell]

CF <form@web.site> wrote:

Terry, no, I haven't got around to this yet, being seduced by the theory of
a more elegant revision based on an LM311 comparator IC. However, I would
like to get the 741 circuit working before butchering it.

Your second mod looks good, but I'm worried: By hooking up the feedback pot
as you have it -- where you've got the non-inverting input going through
its lower portion -- won't that drastically change the calibration of my
temperature pot? You'd be changing the total resistance leading into pin 3
on the IC by large amounts whenever you adjust the hysteresis feedback. At
present I have temperature gradations marked around the circumference of a
knob mounted on a front panel.

By the way, I get a "file not found" error on your third link, the one to
the 741HysteresisWave.gif image.

Me too! In fact, all 3 files seemed to have vanished. Very odd, as I
checked them all after posting. Anyway, I've just uploaded them again
and at present they are all accessible.
http://www.terrypin.dial.pipex.com/Images/741Hysteresis.gif
http://www.terrypin.dial.pipex.com/Images/741Hysteresis2.gif
http://www.terrypin.dial.pipex.com/Images/741HysteresisWave.gif

Without your thermistor to hand I can't be sure, but if it's that
sensitive then I expect you're right about the modification making
adjustment more difficult. I was primarily aiming to demonstrate a
practical way of adjusting hysteresis over a wide range, as you hadn't
specified how much you wanted.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[CF]

Terry Pinnell <terrypinDELETE@THESEdial.pipex.com> wrote in
news:qblqc0p2iu45fhnl8tgb1g3e4oro16ah50@4ax.com:

CF <form@web.site> wrote:

Terry, no, I haven't got around to this yet, being seduced by the
theory of a more elegant revision based on an LM311 comparator IC.
However, I would like to get the 741 circuit working before butchering
it.

Your second mod looks good, but I'm worried: By hooking up the
feedback pot as you have it -- where you've got the non-inverting
input going through its lower portion -- won't that drastically change
the calibration of my temperature pot? You'd be changing the total
resistance leading into pin 3 on the IC by large amounts whenever you
adjust the hysteresis feedback. At present I have temperature
gradations marked around the circumference of a knob mounted on a
front panel.

By the way, I get a "file not found" error on your third link, the one
to the 741HysteresisWave.gif image.

Me too! In fact, all 3 files seemed to have vanished. Very odd, as I
checked them all after posting. Anyway, I've just uploaded them again
and at present they are all accessible.
http://www.terrypin.dial.pipex.com/Images/741Hysteresis.gif
http://www.terrypin.dial.pipex.com/Images/741Hysteresis2.gif
http://www.terrypin.dial.pipex.com/Images/741HysteresisWave.gif

Without your thermistor to hand I can't be sure, but if it's that
sensitive then I expect you're right about the modification making
adjustment more difficult. I was primarily aiming to demonstrate a
practical way of adjusting hysteresis over a wide range, as you hadn't
specified how much you wanted.

I can see the images now too. They're very nice outputs that seem to say it
should work.

Yes, the thermistor is extremely sensitive, even if not completely linear
over a wide range. I think I'll try hooking up the feeback pot to pin 3
without running the main inputs through it. I presume this will still throw
my dial off very slightly, but if I set the hysteresis pot and then leave
it, I can recalibrate once and be done.

Of course, then I'll be tempted to pull it all apart and try the LM 311.

Thanks heaps for your help.

CF

 

[Terry Pinnell]

CF <form@web.site> wrote:

Terry Pinnell <terrypinDELETE@THESEdial.pipex.com> wrote in
news:qblqc0p2iu45fhnl8tgb1g3e4oro16ah50@4ax.com:

Me too! In fact, all 3 files seemed to have vanished. Very odd, as I
checked them all after posting. Anyway, I've just uploaded them again
and at present they are all accessible.
http://www.terrypin.dial.pipex.com/Images/741Hysteresis.gif
http://www.terrypin.dial.pipex.com/Images/741Hysteresis2.gif
http://www.terrypin.dial.pipex.com/Images/741HysteresisWave.gif

Without your thermistor to hand I can't be sure, but if it's that
sensitive then I expect you're right about the modification making
adjustment more difficult. I was primarily aiming to demonstrate a
practical way of adjusting hysteresis over a wide range, as you hadn't
specified how much you wanted.


I can see the images now too. They're very nice outputs that seem to say it
should work.

'Should'? "And here is a typical actual result from that breadboarded
circuit (using a triangle wave as input)" Note 'actual'.

Yes, the thermistor is extremely sensitive, even if not completely linear
over a wide range. I think I'll try hooking up the feeback pot to pin 3
without running the main inputs through it. I presume this will still throw
my dial off very slightly, but if I set the hysteresis pot and then leave
it, I can recalibrate once and be done.

To repeat my earlier recommendation, experiment on breadboard is IMO
by far the most effective way of developing this sort of circuit. And
to avoid the trouble (and time) of heating and cooling the thermistor
for various settings of the hysteresis pot, use another pot. You could
choose its value, or rather, its value and that of a couple of
typically equal fixed resistors in series with it, top and bottom.
Maybe use John Field's formulae, or just start with a 10k pot and
maybe 6k8 or so series resistors and see how it works. Takes only
seconds to swap the components and try again.

Of course, then I'll be tempted to pull it all apart and try the LM 311.

All good experience!

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[CF]

Terry Pinnell <terrypinDELETE@THESEdial.pipex.com> wrote in
news:up5tc0lkim5sso8m3a84n5eq25r8601cfc@4ax.com:

CF <form@web.site> wrote:

Terry Pinnell <terrypinDELETE@THESEdial.pipex.com> wrote in
news:qblqc0p2iu45fhnl8tgb1g3e4oro16ah50@4ax.com:

Me too! In fact, all 3 files seemed to have vanished. Very odd, as I
checked them all after posting. Anyway, I've just uploaded them
again and at present they are all accessible.
http://www.terrypin.dial.pipex.com/Images/741Hysteresis.gif
http://www.terrypin.dial.pipex.com/Images/741Hysteresis2.gif
http://www.terrypin.dial.pipex.com/Images/741HysteresisWave.gif

Without your thermistor to hand I can't be sure, but if it's that
sensitive then I expect you're right about the modification making
adjustment more difficult. I was primarily aiming to demonstrate a
practical way of adjusting hysteresis over a wide range, as you
hadn't specified how much you wanted.


I can see the images now too. They're very nice outputs that seem to
say it should work.

'Should'? "And here is a typical actual result from that breadboarded
circuit (using a triangle wave as input)" Note 'actual'.

Yes, the thermistor is extremely sensitive, even if not completely
linear over a wide range. I think I'll try hooking up the feeback pot
to pin 3 without running the main inputs through it. I presume this
will still throw my dial off very slightly, but if I set the
hysteresis pot and then leave it, I can recalibrate once and be done.

To repeat my earlier recommendation, experiment on breadboard is IMO
by far the most effective way of developing this sort of circuit. And
to avoid the trouble (and time) of heating and cooling the thermistor
for various settings of the hysteresis pot, use another pot. You could
choose its value, or rather, its value and that of a couple of
typically equal fixed resistors in series with it, top and bottom.
Maybe use John Field's formulae, or just start with a 10k pot and
maybe 6k8 or so series resistors and see how it works. Takes only
seconds to swap the components and try again.

Of course, then I'll be tempted to pull it all apart and try the LM
311.

All good experience!

I hear you Terry. A breadboard has got to be the way to go. I was just
looking at them in a catalogue today, admiring the concept.

Sorry about misreading your post. My old newsreader program decided to die
in the middle of this discussion and it threw things off a bit. I'm amazed
that you actually wired it up. Glad to know that it's a real result though,
as I've got the soldering iron heating up right now. I like the idea of
using a pot in place of the thermistor temporarily too.

We'll see what happens.

Cheers,

CF

 

[Bill Bowden]

CF <form@web.site> wrote in message news:<Xns950878A36932B9dkjdj7u3wjedlkr84jh@211.29.133.50>...

Thanks for the circuit Bill, demonstrating the positive feedback for
hysteresis that we've been talking about. Interested to see the 339's
output feeding into the transistor, I looked it up. It's output is
classified as "Open Drain/Collector" (which I presume is similar to the
331). Thus, your output should be like Rob Paisley's at:
http://home.cogeco.ca/%7Erpaisley4/zFile.html

However, you seem to have a pullup resistor attached to the same pin that
is apparently sourcing current for the transistor, so I'm confused. What is
to stop current bypassing the 339 altogether by running directly through
the 2.2k pullup resistor into the transistor's base?

When the 339 output is high, the output pin is completely disconnected
from the transistor base and resistor because it is a "open collector"
output. Therefore, the transistor will turn on since it sees a resistor
connected from base to +V. When the 339 output goes low, the transistor
turns off because the base is grounded.

Also, you don't have the first IC marked. Is it also an LM339?

Yes, same chip.

Regards,

CF


wrongaddress@att.net (Bill Bowden) wrote in
news:ad025737.0406130337.105fc04c@posting.google.com:

There are a couple thermostat circuits using the LM339 quad comparator
on my website here:

http://ourworld.compuserve.com/homepages/Bill_Bowden/page2.htm#therm.gif

-Bill

 

[CF]

Hooray! The 741-based version is working with real hysteresis. Please see
circuit at:
www.copperleife.com/craig/tech/thermo/final.htm

Terry, I found that the way you had it drawn, with the thermistor side
going into the inverting input, it seems to act as a heater control unit.
This might be because of the negative temperature coefficient of the
thermistor.

Also, I've got a 500K trimpot on the feedback, and it's barely enough to
keep hysteresis to 1 degree C. A larger pot would suit me to put this
figure closer to the middle of its range. By rigging it the way I have, it
only shifts my scale by less than half a degree compared to having no
feedback circuit.

I've found that a 20K resistor is enough to saturate the PNP transistor
that I'm using too, so the original 1K itemwas overkill.

Thanks to all who've contributed to this. Next step is to try the 311,
though I might save this for a few weeks hence. Eventually I'll draw up a
web page with the two versions detailed for the benefit of others.

CF

 

[Robert C Monsen]

"CF" <form@web.site> wrote in message
news:Xns950ACB916BA439dkjdj7u3wjedlkr84jh@211.29.133.50...

Hooray! The 741-based version is working with real hysteresis.
Please see
circuit at:
www.copperleife.com/craig/tech/thermo/final.htm

Terry, I found that the way you had it drawn, with the thermistor
side
going into the inverting input, it seems to act as a heater control
unit.
This might be because of the negative temperature coefficient of the
thermistor.

Also, I've got a 500K trimpot on the feedback, and it's barely
enough to
keep hysteresis to 1 degree C. A larger pot would suit me to put
this
figure closer to the middle of its range. By rigging it the way I
have, it
only shifts my scale by less than half a degree compared to having
no
feedback circuit.

I've found that a 20K resistor is enough to saturate the PNP
transistor
that I'm using too, so the original 1K itemwas overkill.

Thanks to all who've contributed to this. Next step is to try the
311,
though I might save this for a few weeks hence. Eventually I'll draw
up a
web page with the two versions detailed for the benefit of others.

CF

Just another thought, the 741 isn't going to be able to turn off the
current into your relay completely. If you measure the current through
the pass transistor, you will see some current leakage due to this.
This wastes power. If you put a 10k resistor from the output of the
741 to the 12V rail, it'll give the 741 a little lift, and so it'll be
able to turn off the PNP transistor more effectively.

Thus, I'd make the circuit look similar to this:

12V -------o--------o----------------------o----.
| | | |
| | | |
.-. .-. | |
10k | | | |10k .-. |
| | | | 10k| | |
'-' '-' | | |
| ___ | ___ '-' |
o--|___|----o--|___|--. | |
| 3960T | | 330k | | |
| R1 | | R2 | | |
| | | |\ | 3.3k | |
| o--|---|-\ | ___ | |<
| | | | >--o-|___|---o--| PNP
| | .---|+/ |\
| | |/ |
| | .------|
| o---. | .--+---.
NTC .-. .-. | | | Relay|
10k \ | | | |<-. - | Coil |
\ | | | ^ | |
'-' '-' 22k | '--+---'
| | Adjust for temp | |
| | | |
0V -----o--------o--------------------o------.
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

According to JF's post, the TC of the 10k NTC is about -4%. By
coincidence, for the circuit above, that works out to about a 1%/C
swing at the voltage divider, assuming temps are close to the center
of the range at 25C. That works out to 120mV of hysteresis per degree
C. So,

0.120V*T = 10V * R1/330k, so R1(T) := 3960*T.

where T is the temperature range you need.

These are ball park figures, based on the 4% NTC of the thermistor.
Different thermistors will have different TCs.

Regards
Bob Monsen

 

[CF]

Bob, you were absolutely spot-on. I measured the current leakage as you
said, and it was 2.3 mA. With a 10k resistor in place as you suggested, it
was 0. Fantastic!

Along the same lines of avoiding wastage, I notice that my supposedly 30mA
(at 12v) relay coil is actually drawing 50mA. I suspect this is because my
supply voltage is a tad high. I measure 19.3 volts with an open circuit,
and 16.3 under load. I'm using a transformer I had lying around, which fits
nicely into my enclosure. I originally measured it at about 14v AC, but it
seems that the smoothed average DC is quite a bit higher. This might also
account for high-ish values of the resistors I've added lately too.

To avoid waste and possibly safely convert to an LM311 one day (sinks 50mA
max through output), what is the best way to regulate the voltage down to
say, 13-14 under load? I don't know much about the way that voltage
regulators and zeners work, but I hate the thought of a component
constantly throwing away power.

Thanks for the formula too.


CF



"Robert C Monsen" <rcsurname@comcast.net> wrote in
news:Jt1Ac.115699$Ly.50292@attbi_s01:

Just another thought, the 741 isn't going to be able to turn off the
current into your relay completely. If you measure the current through
the pass transistor, you will see some current leakage due to this.
This wastes power. If you put a 10k resistor from the output of the
741 to the 12V rail, it'll give the 741 a little lift, and so it'll be
able to turn off the PNP transistor more effectively.

Thus, I'd make the circuit look similar to this:

12V -------o--------o----------------------o----.
| | | |
| | | |
.-. .-. | |
10k | | | |10k .-. |
| | | | 10k| | |
'-' '-' | | |
| ___ | ___ '-' |
o--|___|----o--|___|--. | |
| 3960T | | 330k | | |
| R1 | | R2 | | |
| | | |\ | 3.3k | |
| o--|---|-\ | ___ | |
| | | | >--o-|___|---o--| PNP
| | .---|+/ |\
| | |/ |
| | .------|
| o---. | .--+---.
NTC .-. .-. | | | Relay|
10k \ | | | |<-. - | Coil |
\ | | | ^ | |
'-' '-' 22k | '--+---'
| | Adjust for temp | |
| | | |
0V -----o--------o--------------------o------.
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

According to JF's post, the TC of the 10k NTC is about -4%. By
coincidence, for the circuit above, that works out to about a 1%/C
swing at the voltage divider, assuming temps are close to the center
of the range at 25C. That works out to 120mV of hysteresis per degree
C. So,

0.120V*T = 10V * R1/330k, so R1(T) := 3960*T.

where T is the temperature range you need.

These are ball park figures, based on the 4% NTC of the thermistor.
Different thermistors will have different TCs.

Regards
Bob Monsen

 

[CF]

wrongaddress@att.net (Bill Bowden) wrote in
news:ad025737.0406151725.46e1ef39@posting.google.com:

CF <form@web.site> wrote in message
news:<Xns950878A36932B9dkjdj7u3wjedlkr84jh@211.29.133.50>...
Thanks for the circuit Bill, demonstrating the positive feedback for
hysteresis that we've been talking about. Interested to see the 339's
output feeding into the transistor, I looked it up. It's output is
classified as "Open Drain/Collector" (which I presume is similar to
the 331). Thus, your output should be like Rob Paisley's at:
http://home.cogeco.ca/%7Erpaisley4/zFile.html

However, you seem to have a pullup resistor attached to the same pin
that is apparently sourcing current for the transistor, so I'm
confused. What is to stop current bypassing the 339 altogether by
running directly through the 2.2k pullup resistor into the
transistor's base?

When the 339 output is high, the output pin is completely disconnected
from the transistor base and resistor because it is a "open collector"
output. Therefore, the transistor will turn on since it sees a
resistor connected from base to +V. When the 339 output goes low, the
transistor turns off because the base is grounded.

Aha! I see now. Current through the 2.2k pullup resistor alternately flows
to ground either via the transistor or the 339 depending on whether the
latter is high or low.

Also, you don't have the first IC marked. Is it also an LM339?


Yes, same chip.

Oh yes, of course!

Just curious, did you choose the LM339 quad-comparator over, say, the LM319
high speed dual comparator
(http://www1.jaycar.com.au/products_uploaded/ZL3319.pdf) for any particular
reason, or just familiarity?

Thanks for the explanation.

CF

Regards,

CF


wrongaddress@att.net (Bill Bowden) wrote in
news:ad025737.0406130337.105fc04c@posting.google.com:

There are a couple thermostat circuits using the LM339 quad
comparator on my website here:

http://ourworld.compuserve.com/homepages/Bill_Bowden/page2.htm#therm
.gif

-Bill

 

[Rich Grise]

"CF" <form@web.site> wrote in message
news:Xns950ACB916BA439dkjdj7u3wjedlkr84jh@211.29.133.50...

Hooray! The 741-based version is working with real hysteresis. Please see
circuit at:
www.copperleife.com/craig/tech/thermo/final.htm

Terry, I found that the way you had it drawn, with the thermistor side
going into the inverting input, it seems to act as a heater control unit.
This might be because of the negative temperature coefficient of the
thermistor.

Also, I've got a 500K trimpot on the feedback, and it's barely enough to
keep hysteresis to 1 degree C. A larger pot would suit me to put this
figure closer to the middle of its range. By rigging it the way I have, it
only shifts my scale by less than half a degree compared to having no
feedback circuit.

I've found that a 20K resistor is enough to saturate the PNP transistor
that I'm using too, so the original 1K itemwas overkill.

Thanks to all who've contributed to this. Next step is to try the 311,
though I might save this for a few weeks hence. Eventually I'll draw up a
web page with the two versions detailed for the benefit of others.

If you plan to change to a real comparator, do it right away. There's
no use in flailing away at a 741 circuit, trying to get a mediocre,
albeit robust, opamp - which they've tried to optimize for linear
operation - to act like a comparator is giving yourself an unnecessary
obstacle before you even start.

Have Fun!
Rich

 

[CF]

"Rich Grise" <null@example.net> wrote in
newshcAc.28924$H65.19579@nwrddc02.gnilink.net:

If you plan to change to a real comparator, do it right away. There's
no use in flailing away at a 741 circuit, trying to get a mediocre,
albeit robust, opamp - which they've tried to optimize for linear
operation - to act like a comparator is giving yourself an unnecessary
obstacle before you even start.

I wish I'd started with a real comparator in the first place with
hysteresis figured in. I now realise this is du rigueur. Instead, I have
already built the inellegantly-designed 741 circuit gleaned from the web,
and the source of much frustration. However, thanks to the great help here,
it's now at least functional (actually seems to work quite well), and best
of all, has taught me a lot about electronics that I would have
conveniently ignored with a good initial circuit.

Have Fun!

I'm having a ball. Part of me says to leave well enough alone, and the
other part is itching to rip the 741 out of there and do it properly. I
know which part will win in the long run.

CF

 

[Robert C Monsen]

"CF" <form@web.site> wrote in message
news:Xns950BA487143F39dkjdj7u3wjedlkr84jh@211.29.133.50...

Bob, you were absolutely spot-on. I measured the current leakage as
you
said, and it was 2.3 mA. With a 10k resistor in place as you
suggested, it
was 0. Fantastic!

Along the same lines of avoiding wastage, I notice that my
supposedly 30mA
(at 12v) relay coil is actually drawing 50mA. I suspect this is
because my
supply voltage is a tad high. I measure 19.3 volts with an open
circuit,
and 16.3 under load. I'm using a transformer I had lying around,
which fits
nicely into my enclosure. I originally measured it at about 14v AC,
but it
seems that the smoothed average DC is quite a bit higher. This might
also
account for high-ish values of the resistors I've added lately too.

Put an emitter resistor of about 150 Ohms between the supply voltage
and the pass transistor. That should drop it down to the appropriate
12V at 30mA.

To avoid waste and possibly safely convert to an LM311 one day
(sinks 50mA
max through output), what is the best way to regulate the voltage
down to
say, 13-14 under load? I don't know much about the way that voltage
regulators and zeners work, but I hate the thought of a component
constantly throwing away power.

You can buy regulated 12V supplies (wall warts) for a song. Or, you
can use a 7812 voltage regulator, which will drop the voltage to a
regulated 12V. You are going to lose the power anyway, though, no
matter what you do, unless you get a new transformer, or you build an
SMPS.

Getting 30mA through your pass transistor means you'll be wasting 4.3V
x 0.030A = 129mW. Not too bad.

Regards,
Bob Monsen

PS: Its customary to put replys inline, or at the bottom of the reply
posting, so that responses make sense to those who use lossy news
readers. Since almost nobody uses these lossy news readers anymore,
its not as critical, but folks still appear to insist on it as a
matter of taste.

 

[John Fields]

On 17 Jun 2004 06:10:29 GMT, CF <form@web.site> wrote:

....

Along the same lines of avoiding wastage, I notice that my supposedly 30mA
(at 12v) relay coil is actually drawing 50mA. I suspect this is because my
supply voltage is a tad high. I measure 19.3 volts with an open circuit,
and 16.3 under load. I'm using a transformer I had lying around, which fits
nicely into my enclosure. I originally measured it at about 14v AC, but it
seems that the smoothed average DC is quite a bit higher. This might also
account for high-ish values of the resistors I've added lately too.

To avoid waste and possibly safely convert to an LM311 one day (sinks 50mA
max through output), what is the best way to regulate the voltage down to
say, 13-14 under load? I don't know much about the way that voltage
regulators and zeners work, but I hate the thought of a component
constantly throwing away power.

....

Since the LM311 can safely handle supply voltages up to about 36VDC,
and its output can handle about 50V, there's really no need for a
regulated supply there. Neither is there a need for a regulator if
you choose the proper relay to use with the supply you have at hand.

If you measured the unloaded output voltage from the transformer and
it was 14VAC, then there's a good chance it's a 12V transformer with a
regulation of about 17% from no-load to full load.

With a 14VRMS output, rectifying that in a full-wave bridge and
smoothing it should result in about 18.4VDC across the filter cap,
with no other load across it. A 19VDC output smacks of half-wave
rectification, which is kind of OK for the light loads you're using,
but it makes the filter cap need to have twice the capacitance as if
you used a full-wave bridge, plus its ripple-current requirements also
double (or something like that, anyway).

Assuming that a 50mA load isn't really enough to change the output
voltage from the transformer, knowing that you're using a half-wave
rectifier, a reservoir capacitor and that your load varies from 0mA to
50 mA, with a 3 volt change in the output voltage we can figure out
what your filter cap looks like from:

I t
C = -----
Vr

Where C = capacitance in farads
I = the steady-state current into the load in amperes
t = the period of the ripple waveform in seconds
Vr = the amplitude of the ripple in volts

I t 0.05A*0.017s
so, C = ----- = --------------- = 0.0028F ~ 3000ľF
Vr 3V

Pretty close?


Now, here's a shocker for ya; if you switch from a half-wave rectifier
to a full-wave bridge, you won't need to use filtered DC to run the
relay, you can do it like this:


12VAC>-----------------------+
|
12VAC>--------------------+ |
| |
+-[1N4002>|]--+--|--[1N4002>|]-+
| | |
+-[1N4002>|]-----+--[1N4002>|]-+
| |
GND>--+-----+---[C1+]--------+--[|<1N4002]-+
| | |
| IN |
+-------------------[78L12] |
| OUT |
| | |
| | |
| | |
| +--------+--------+ +-------+
| | | | |K | O--> |
| [R1] [10.7K] | [1N4001] [COIL]- | - -|
| | | | | | | O--->COM
| | +-[634K]-------+-+-----+ |
| | | | | +-------->NO
| | | +------+ |
| | +----|+ C|----+
| | | |LM311 |
| [1000]<-----|----|- E|--+
| | | +------+ |
| | | | |
| [R2] [RT1] | |
| | | | |
+----+--------+--------+-----+

The reason being that the full-wave rectified AC will be pulsing on
and off 120 times a second, and the spring trying to return the
relay's armature to the non-energized position won't have enough time
to do it in between pulses. You may not need the 78L12 either, but
it's nice to have a steady voltage to drive the reference and the
thermistor string, and if you're going to do that you may as well
regulate the supply voltage to the comparator since it won't cost any
more to do that as well.

Looking at the relay supply, since you've got 14VRMS coming out of the
transformer, that would rectify and filter to about 19.8VDC with
perfect diodes and a perfect cap, but with the reality of two diode
drops that'll go to about 18.4V, which is about 13VRMS, which is what
will be driving the magnetic field being generated in the relay coil.
That may be fine, but you'll need to look at the relay spec's to make
sure that 1V of overdrive won't hurt the coil. Otherwise, what you'll
need to do is put a resistor in series with the relay coil to drop the
extra voltage. Since you'll want to drop 1 volt at 50 mA, you can
figure out the resistance you'll need from:

E 1V
R = --- = ------- = 20 ohms
I 0.05A


and the power the resistor must dissipate from:


P = I E = 0.05A*1V = 0.05W

so a standard 20 ohm, +/-5%, 1/4 watt resistor would be fine.

All that's left is to figure out the capacitance of the reservoir cap,
C1, and to do that we can use


I t
C = -----
Vr

if we can determine the load current of the circuitry attached to it
and specify the ripple coltage we can stand.

For the LM311 we'll have 7.5ma worst case, for the thermistor string
something less than 1mA, for the reference string (I'm not going to
figure it out, but assuming there'll be no more than 5V dropped across
the pot puts that current at 5mA), and 5.5mA for the 78L12 brings the
total to 19, say 20, mA. Just for grins let's say that we can live
with 50 millivolts of ripple. Then, plugging that in gives us

0.02A*0.0083s
C = --------------- ~ 0.0033F = 3300ľF
0.05V

Interestingly, just about what you had before!

Now, however, you've got a nice little supply with almost no ripple
and a pretty reliable relay driver.

We still need to look at the power the 78L12 is going to dissipate,
and if we look at the worst case unloaded DC across the filter we'll
have 18.4V into the 78L12 (so we'll have no headroom problems,
anyway!) and 12V out of it, so the voltage across it will be 5.6V, and
with 20mA flowing through it that'll be 112mW it needs to get rid of.

The 78L12 has a worst case junction-to-ambient thermal resistance of
160°C/watt, so with a 25°C ambient and 112mW being dissipated by the
device its junction temp will rise by

160°C
Tj = -------- * 0.112W ~ 18°C
1 watt

With an ambient temp of 25°C, then, the junction temp will go to 43°C.

Piece of cake! The thing won't even _think_ about going into thermal
overload protection there and, BTW, that's the _only_ power that'll be
wasted in the circuit, (except for the 50mW being dissipated in the
relay's series resistor, if you have to use one) which isn't too bad.


--
John Fields

 

[John Fields]

On Thu, 17 Jun 2004 14:11:20 -0500, John Fields
<jfields@austininstruments.com> wrote:

12VAC>-----------------------+
|
12VAC>--------------------+ |
| |
+-[1N4002>|]--+--|--[1N4002>|]-+
| | |
+-[1N4002>|]-----+--[1N4002>|]-+
| |
GND>--+-----+---[C1+]--------+--[|<1N4002]-+
| | |
| IN |
+-------------------[78L12] |
| OUT |
| | |
| | |
| | |
| +--------+--------+ +-------+
| | | | |K | O--> |
| [R1] [10.7K] | [1N4001] [COIL]- | - -|
| | | | | | | O--->COM
| | +-[634K]-------+-+-----+ |
| | | | | +-------->NO
| | | +------+ |
| | +----|+ C|----+
| | | |LM311 |
| [1000]<-----|----|- E|--+
| | | +------+ |
| | | | |
| [R2] [RT1] | |
| | | | |
+----+--------+--------+-----+

---
Oops... I forgot about the hysteresis. You'll have to refigure it
because of the increase in the relay driver's peak voltage, and you'll
need to add a cap to the comparator's output to smooth the 120Hz
pulses when the relay isn't turned on. Something like this this:

12VAC>-----------------------+
|
12VAC>--------------------+ |
| |
+-[1N4002>|]--+--|--[1N4002>|]-+
| | |
+-[1N4002>|]-----+--[1N4002>|]-+
| |
GND>--+-----+---[C1+]--------+--[|<1N4002]-+
| | |
| IN |
+-------------------[78L12] |
| OUT |
| | |
| | |
| | |
| +--------+--------+ +-------+
| | | | |K | O---> |
| [R1] [10.7K] | [1N4001] [COIL]- -|- - -|
| | | | | | | O--->COM
| | +-[634K]-------+-------+ |
| | | | | +--------->NO
| | | +------+ |
| | +----|+ C|----------+
| | | |LM311 | |
| [1000]<-----|----|- E|--+ |
| | | +------+ | |+
| | | | | [10ľF]
| [R2] [RT1] | | |
| | | | | |
+----+--------+--------+-----+-------+


--
John Fields

 

[CF]

"Robert C Monsen" <rcsurname@comcast.net> wrote in
news:7bkAc.129111$Ly.101820@attbi_s01:

"CF" <form@web.site> wrote in message
news:Xns950BA487143F39dkjdj7u3wjedlkr84jh@211.29.133.50...
Bob, you were absolutely spot-on. I measured the current leakage as
you
said, and it was 2.3 mA. With a 10k resistor in place as you
suggested, it
was 0. Fantastic!

Along the same lines of avoiding wastage, I notice that my
supposedly 30mA
(at 12v) relay coil is actually drawing 50mA. I suspect this is
because my
supply voltage is a tad high. I measure 19.3 volts with an open
circuit,
and 16.3 under load. I'm using a transformer I had lying around,
which fits
nicely into my enclosure. I originally measured it at about 14v AC,
but it
seems that the smoothed average DC is quite a bit higher. This might
also
account for high-ish values of the resistors I've added lately too.

Put an emitter resistor of about 150 Ohms between the supply voltage
and the pass transistor. That should drop it down to the appropriate
12V at 30mA.


To avoid waste and possibly safely convert to an LM311 one day
(sinks 50mA
max through output), what is the best way to regulate the voltage
down to
say, 13-14 under load? I don't know much about the way that voltage
regulators and zeners work, but I hate the thought of a component
constantly throwing away power.

You can buy regulated 12V supplies (wall warts) for a song. Or, you
can use a 7812 voltage regulator, which will drop the voltage to a
regulated 12V. You are going to lose the power anyway, though, no
matter what you do, unless you get a new transformer, or you build an
SMPS.

Getting 30mA through your pass transistor means you'll be wasting 4.3V
x 0.030A = 129mW. Not too bad.

Regards,
Bob Monsen

PS: Its customary to put replys inline, or at the bottom of the reply
posting, so that responses make sense to those who use lossy news
readers. Since almost nobody uses these lossy news readers anymore,
its not as critical, but folks still appear to insist on it as a
matter of taste.

Thanks for this Bob, I see your point that it's wasted energy one way or
the other. I don't mind if it's unavoidable. An SMPS is not justified yet!

Cheers,

CF

PS: I'm all for standardization.

 

[CF]

John Fields <jfields@austininstruments.com> wrote in
news:bkd3d0dgfedbpkgkmg3oaaes2j4huf9teh@4ax.com:

On 17 Jun 2004 06:10:29 GMT, CF <form@web.site> wrote:

...

Along the same lines of avoiding wastage, I notice that my supposedly
30mA (at 12v) relay coil is actually drawing 50mA. I suspect this is
because my supply voltage is a tad high. I measure 19.3 volts with an
open circuit, and 16.3 under load. I'm using a transformer I had lying
around, which fits nicely into my enclosure. I originally measured it
at about 14v AC, but it seems that the smoothed average DC is quite a
bit higher. This might also account for high-ish values of the
resistors I've added lately too.

To avoid waste and possibly safely convert to an LM311 one day (sinks
50mA max through output), what is the best way to regulate the voltage
down to say, 13-14 under load? I don't know much about the way that
voltage regulators and zeners work, but I hate the thought of a
component constantly throwing away power.

...

Since the LM311 can safely handle supply voltages up to about 36VDC,
and its output can handle about 50V, there's really no need for a
regulated supply there. Neither is there a need for a regulator if
you choose the proper relay to use with the supply you have at hand.

If you measured the unloaded output voltage from the transformer and
it was 14VAC, then there's a good chance it's a 12V transformer with a
regulation of about 17% from no-load to full load.

With a 14VRMS output, rectifying that in a full-wave bridge and
smoothing it should result in about 18.4VDC across the filter cap,
with no other load across it. A 19VDC output smacks of half-wave
rectification, which is kind of OK for the light loads you're using,
but it makes the filter cap need to have twice the capacitance as if
you used a full-wave bridge, plus its ripple-current requirements also
double (or something like that, anyway).

Assuming that a 50mA load isn't really enough to change the output
voltage from the transformer, knowing that you're using a half-wave
rectifier, a reservoir capacitor and that your load varies from 0mA to
50 mA, with a 3 volt change in the output voltage we can figure out
what your filter cap looks like from:

I t
C = -----
Vr

Where C = capacitance in farads
I = the steady-state current into the load in amperes
t = the period of the ripple waveform in seconds
Vr = the amplitude of the ripple in volts

I t 0.05A*0.017s
so, C = ----- = --------------- = 0.0028F ~ 3000ľF
Vr 3V

Pretty close?


Now, here's a shocker for ya; if you switch from a half-wave rectifier
to a full-wave bridge, you won't need to use filtered DC to run the
relay, you can do it like this:


12VAC>-----------------------+
|
12VAC>--------------------+ |
| |
+-[1N4002>|]--+--|--[1N4002>|]-+
| | |
+-[1N4002>|]-----+--[1N4002>|]-+
| |
GND>--+-----+---[C1+]--------+--[|<1N4002]-+
| | |
| IN |
+-------------------[78L12] |
| OUT |
| | |
| | |
| | |
| +--------+--------+ +-------+
| | | | |K | O--> |
| [R1] [10.7K] | [1N4001] [COIL]- | - -|
| | | | | | | O--->COM
| | +-[634K]-------+-+-----+ |
| | | | | +-------->NO
| | | +------+ |
| | +----|+ C|----+
| | | |LM311 |
| [1000]<-----|----|- E|--+
| | | +------+ |
| | | | |
| [R2] [RT1] | |
| | | | |
+----+--------+--------+-----+

The reason being that the full-wave rectified AC will be pulsing on
and off 120 times a second, and the spring trying to return the
relay's armature to the non-energized position won't have enough time
to do it in between pulses. You may not need the 78L12 either, but
it's nice to have a steady voltage to drive the reference and the
thermistor string, and if you're going to do that you may as well
regulate the supply voltage to the comparator since it won't cost any
more to do that as well.

Looking at the relay supply, since you've got 14VRMS coming out of the
transformer, that would rectify and filter to about 19.8VDC with
perfect diodes and a perfect cap, but with the reality of two diode
drops that'll go to about 18.4V, which is about 13VRMS, which is what
will be driving the magnetic field being generated in the relay coil.
That may be fine, but you'll need to look at the relay spec's to make
sure that 1V of overdrive won't hurt the coil. Otherwise, what you'll
need to do is put a resistor in series with the relay coil to drop the
extra voltage. Since you'll want to drop 1 volt at 50 mA, you can
figure out the resistance you'll need from:

E 1V
R = --- = ------- = 20 ohms
I 0.05A


and the power the resistor must dissipate from:


P = I E = 0.05A*1V = 0.05W

so a standard 20 ohm, +/-5%, 1/4 watt resistor would be fine.

All that's left is to figure out the capacitance of the reservoir cap,
C1, and to do that we can use


I t
C = -----
Vr

if we can determine the load current of the circuitry attached to it
and specify the ripple coltage we can stand.

For the LM311 we'll have 7.5ma worst case, for the thermistor string
something less than 1mA, for the reference string (I'm not going to
figure it out, but assuming there'll be no more than 5V dropped across
the pot puts that current at 5mA), and 5.5mA for the 78L12 brings the
total to 19, say 20, mA. Just for grins let's say that we can live
with 50 millivolts of ripple. Then, plugging that in gives us

0.02A*0.0083s
C = --------------- ~ 0.0033F = 3300ľF
0.05V

Interestingly, just about what you had before!

Now, however, you've got a nice little supply with almost no ripple
and a pretty reliable relay driver.

We still need to look at the power the 78L12 is going to dissipate,
and if we look at the worst case unloaded DC across the filter we'll
have 18.4V into the 78L12 (so we'll have no headroom problems,
anyway!) and 12V out of it, so the voltage across it will be 5.6V, and
with 20mA flowing through it that'll be 112mW it needs to get rid of.

The 78L12 has a worst case junction-to-ambient thermal resistance of
160°C/watt, so with a 25°C ambient and 112mW being dissipated by the
device its junction temp will rise by

160°C
Tj = -------- * 0.112W ~ 18°C
1 watt

With an ambient temp of 25°C, then, the junction temp will go to 43°C.

Piece of cake! The thing won't even _think_ about going into thermal
overload protection there and, BTW, that's the _only_ power that'll be
wasted in the circuit, (except for the 50mW being dissipated in the
relay's series resistor, if you have to use one) which isn't too bad.

John, this is outstanding stuff. I think what I'll end up doing is letting
the bulk of the circuit run at the higher voltage, which is within range,
as you point out, for the 311 and even the current 741. I'll just drop the
voltage for the relay coil, which is what I'm most concerned about. This
way when the relay is off (most of the time), there's no extra component
making unwanted heat, etc.

My power supply circuit does use a full-wave bridge rectifier, but the
transformer is actually putting out 14.75 volts AC according to my latest
reading on my (non-true-RMS) multimeter. The reservoir cap is 2200uF. When
I spoke of the voltage under load, I was actually including a small fan
too, rated 12v, 0.11A. The transformer is not marked, but looks like a
typical 150-ish mW unit.

Talk about electronic detective work! I can't believe you could deduce so
much from so little info, though the quality of the info was bad.

Now I'm going to save and print out your formulae and play with it all.

Cheers,

CF

 

[CF]

John Fields <jfields@austininstruments.com> wrote in
news:1fs3d09jra7em3otgdudo6et4rqjj0cgou@4ax.com:

On Thu, 17 Jun 2004 14:11:20 -0500, John Fields
jfields@austininstruments.com> wrote:
---
Oops... I forgot about the hysteresis. You'll have to refigure it
because of the increase in the relay driver's peak voltage, and you'll
need to add a cap to the comparator's output to smooth the 120Hz
pulses when the relay isn't turned on. Something like this this:

12VAC>-----------------------+
|
12VAC>--------------------+ |
| |
+-[1N4002>|]--+--|--[1N4002>|]-+
| | |
+-[1N4002>|]-----+--[1N4002>|]-+
| |
GND>--+-----+---[C1+]--------+--[|<1N4002]-+
| | |
| IN |
+-------------------[78L12] |
| OUT |
| | |
| | |
| | |
| +--------+--------+ +-------+
| | | | |K | O---> |
| [R1] [10.7K] | [1N4001] [COIL]- -|- - -|
| | | | | | | O--->COM
| | +-[634K]-------+-------+ |
| | | | | +--------->NO
| | | +------+ |
| | +----|+ C|----------+
| | | |LM311 | |
| [1000]<-----|----|- E|--+ |
| | | +------+ | |+
| | | | | [10ľF]
| [R2] [RT1] | | |
| | | | | |
+----+--------+--------+-----+-------+

OK John, very nice. Any special type of capacitor? I presume I'd be getting
100Hz pulses where I live (50Hz mains).

CF