Need help understanding how to use opto-isolator

The voltage applied to the buzzer is undoubtedly higher than 395 mV. That might be its DC value, but if you measure AC, you will find quite a bit more RMS.

The LED can be driven with a small fraction of the power consumed by the buzzer, so your concern is not one that should take much more your time.

Put a 1k limiting resistor in series with the LED, put the combination across the buzzer, then measure DC voltage across the resistor when the buzzer goes off. You'll see more than you might have expected.

You're welcome.

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--Larry Brasfield
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I posted a circuit to another thread, (a ircuit to swith a relay by
alarm clock) which might work for you with a few modifications, so
could you post some more information about your system?  Specifically,
on your remote control are the OPEN and CLOSE functions separate or
does a single button toggle them?

What supply voltages do you have to work with?

Do you need the output which is going to the remote to stay on for as
long as the alarm clock provides an output or do you just need a
pulse?  If a pulse, how long a pulse?

Could you post some information as to what you used to measure the
395mV and how you went about it?  That is, was it an AC or DC meter?
Did you measure the voltage across the buzzer, or from one of its
terminals to ground?  

I\'m assuming that you\'re planning on paralleling the contacts on the
remote\'s keypad with whatever will be actuating it.  Am I right?

separate or does a single button toggle them?

It's a toggle switch, I need it "pressed" for a little less than a second for it to activate the garage door then I have to let it go or else the motor won't respond. So I would imagine I need a pluse that last about 800 to 900 milliseconds.

The alarm clock runs off 3V DC. I measured the DC voltage with a multimeter between the two solder points on the buzzer itself -- so it was across the buzzer. Should I be measuring it someplace else?

remote's keypad with whatever will be actuating it. Am I right?

I wanted to extend two leads from the solder points on the buzzer to an opto-isolator which would activate the "pulse" to the garage door remote through an intermediate circuit.

Here's a new diagram, as well as front/back images of the remote.

Front side of the remote:

Back side of the remote (red lines are show the traces)

I'm having a hard time getting a for the voltage accross the toggle switch on the remote. I read 0 no matter if the switch is toggled or not. I am probably not laying my probes across the switch correctly. Do I need to touch both pairs of connectors at the same time?

I'm assuming that I would use an opto-isolator to interface the pluse circuit to the remote and that the output of the opto-isolator will be enough to activiate the remote? I guess I'll just have to try it an see.

I was going to use an 4N27 opto-isolator

between the clock and between the remote.

thanks again, let me know if y'all have any thoughts or questions...

....

(To the OP:) I concur with that good advice.

....

The power needed to drive a buzzer will be many times larger than what needs to be picked off to activate another circuit, (many mW versus uW).

....

Some such uncertainty is warranted, but I suggest that there is reason to believe an opto-isolator will be fine. The cheap (typically membrane) switches used in many remotes are not asked to carry much current and, to conserve battery power, large value pull-{up,down} resistors are used. If a replacement for the contanct had to carry more than 100 uA, I would be surprised. The CTR (current transfer ratio) for opto-isolaters is often guaranteed to be 100% or better, so a similar current is all that the LED would need. Finally, the signal sent thru the opto-isolator can be time limited to just over what is needed for the remote in order to conserve the battery.

The opto-isolator would plug into that with little change except reduction of the 4.5V battery drain (unless my power surmises are completely wrong).

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--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
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Across the buzzer should be OK, but 395mV sounds awfully low, so I
suspect it\'s being driven by AC.  Measure it with your meter set to AC
VOLTS and see what you get.

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Depends.  The OP\'s advocating using the signal driving the buzzer to
also drive the LED in an opto, which will be milliwatts VS milliwatts.

I was not addressing use of the optoisolator in that position. However, if the current taken through the opto LED is limited to a few 100 uA, such usage would still be a small fraction of the buzzer power. As you point out, CTR would be reduced, but no more than a few uA of output would be needed.

All the alarms with buzzers I have heard are very loud (and annoying). It is hard to imagine getting that without using many 10s of mW.

I agree that the reed relay is simpler to apply. For that reason alone, it may well be most suitable for the OP's project. My suggestion about an optoisolator in its place is more like a feasable alternative than any kind of compelling improvement. Reed relays are fragile and, if their leads are not carefully heat-sunk during soldering, they can fail quickly or slowly as a result. That, together with a dislike of moving parts, made me think it might be an attractive alternative.

Yes.

Seems reasonable.

Yes.

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--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

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I've seen commercial uses of photo-transistors used as switches with 1M pullups. The transistors themselves do fine at such levels, suffering only slight beta reduction. Saturation resistance is very nearly inversely proportional to excess base current (or the equivalent photocurrent for a phototransistor), so I see no reason to expect the problem you allude to here.

As for using the phototransistor in a mode other than saturated (or nearly off), I have not suggested that. It might be useful, when the receiver can deal with a non-switching input, but that is not the case for the position I suggested the opto for.

I thought pullup values would go up in that case, at least when expressed in Ohms. What makes you say they would go lower?

The output transistor can be, and often is, used as a two terminal switch. Even if a base-emitter resistor is added to control switching speed or leakage, (not needed here), it can be used that way. Whether it pulls high or low is not a complication and no follower need be created.

Most or all of them go away on their own.

Ok, that's funny. I was thinking of the bare reed switches which, due to their glass envelope, are fragile. Obviously, they are not so fragile when packaged.

I guess that's not so funny.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

--- That's not the point. As you've already stated, large value pullups or pull-down resistors may be used in the remote in order to conserve battery power during switching, and it's precisely that which makes using an opto in other than a saturated mode problematical. Consider:

+V | [100k] | +----->Eout | O | |Eout | C A B >>> the uncertainty of being able to use its transistor output to trigger

--- Yeah, right! I can just see millions of through-hole reed relays going through wave-solder machines with little heat sinks attached to their leads, LOL. Worse yet, millions of surface-mount units going through soldering ovens with no heat sinks attached...

---

-- John Fields Professional Circuit Designer

[Brasfield had suggested an optoisolator in parallel with a button on a remote.] [Points made and responded to elsewhere cut.] Brasfield once wrote: ... I suggest that there is reason to believe an opto-isolator will be fine. and later wrote: ... I agree that the reed relay is simpler to apply. For that reason alone, it may well be most suitable for the OP's project.

It occurs to me now that the most problematical attribute of the optoisolator in that position is its capacitance. If that had an effect only at button pushing speeds, it would hardly matter. But it is quite likely that the buttons on the remote at in a matrix and scanned via pulses applied to the matrix and responses sampled shortly after changes. The commonly large capacitance of the phototransistors used in optoisolators could interfere with such scanning.

The reed relay is more certain to work for that reason.

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--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

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Without adequate drive the phototransistor will never go into
saturation, so whether the output of the opto can pull the driven load
down (or up) far enough to cross the switching threshold becomes the
problem.

Why do you need the remote? If you're close enough to sense the door state, you're close enough to use the wired actuation.

Only you and your insurance company can determine what should be in the door sensor block.

I can tell you from personal experience that unattended operation of a garage door is a BAD idea. Once had the door hang up on a gasoline can. It hung up on the (sharp)edge of the door brace, so a sensor along the bottom of the door wouldn't have helped. Once had a broom handle hang up in the door guides. Kids, pets, newspaper, bicycle... It's no fun to have the door close as you're driving in from that late night on the town. Yes, if you put enough safety interlocks and timeouts, you can make it as safe as you think you need.

A much easier/safer way would be to use the door state sensor to inhibit the alarm. If the alarm goes off, get off your butt and go close the door. More technology/automation is not always a good thing.

If you're determined to do this, stick a scope on the buzzer and see what's actually there. Some clocks use DC on a buzzer module. Others drive the piezo directly with AC. YMMV.

Also do a LOT of testing on what happens to the clock with various power line glitches/outages, battery failure...Interesting things can happen to a battery powered clock as temperature variations, as might be encountered in a garage, change the battery voltage slightly near the cutoff point.

This is one of those situations with a very low probability of having a very BIG problem. Multiply the numbers and you'll feel very safe...until it fails. mike

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Agreed.

I think we can noisily agree that using sufficient current is necessary with the optoisolator and that it should be more or less saturated when used to replace a switch.

I think we have merely a terminolgy issue here. You are willing to call a two terminal circuit a follower. I use the term for common {emitter, source, cathode} amplifiers.

However, maybe you think the underlying facts impact the operation, and that being a follower limits the voltage in the pullup configuration more than in the pulldown configuration. In that case, I suggest you connect a 1T resistor from the base of the 4N25A in your simulation to ground and run it, then observe the base voltage. You may be surprised to see it more positive than the other two terminals by about a diode drop.

I've only made one explicit statement about the LED current, suggesting it might be "limited to a few 100 uA". More on this in a moment.

Interesting. It certainly demonstrates the reduction of CTR at lower currents. When I reduce the LED current setting resistor to 36K to get about 100 uA thru the LED, the output saturates nicely. And with that 1T resistor, the base goes to 3.53 V.

I've already agreed that the relay solution is easier to apply. My reason for stating so is exactly the sort of required knowing you mention. For that reason, (and the capacitance issue I've posted), I think your solution is entirely appropriate for the OP's purpose.

....

Ok, that's funny too. I could quibble about centuries being the proper unit of measurement. (What falls between millenia and eons?) I never claimed my dislike of moving parts was rational!

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--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

The alarm clock is powered by two AAA batteries and has a piezo buzzer so I don't think there's any AC involved.

The piezo device requires AC to do anything above DC. The sound you hear from it is at the same frequency applied to the buzzer.

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--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

Of course, AC is involved then, too.

Mr. Johnson has already mentioned observing 395 mV on the buzzer when it is sounding. That is not likely to be enough to run the latter kind of buzzer, so I conclude that his buzzer is being fed with narrow pulses at whatever voltage the clock's logic device is powered by.

It is also apparent that the repeated advice he has been given to measure the AC signal on the buzzer has not been taken. That ought to reduce interest in this thread, I would think.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

I guess I'm not sure how to measure the AC then. My multimeter has an ACV seeting for 200 and 750. I might not understand how to use the meter. Is that what what I use to measure the VC and do I measure it across at the two leads attached to the buzzer?

Sorry for my ignorance...

If that is 200 and 750 VAC, you'll have to observe carefully to see the few volts likely across that buzzer. It may not go that low if it is a moving needle type.

I suggest that you put a small diode in series with your meter, use the same DC setting you got the 395 mV with, and measure in both directions across the buzzer. You will probably see several volts when it sounds. That will represent approximately the peak voltage applied, minus a small diode drop (400 - 500 mV).

No need for that. Please understand that my earlier comment was a little frustration showing due to an excess of speculation when what is really needed is some real data.

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--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

That doesn't necessarily follow. Piezo buzzers come in two flavours:

1. Basic component, requiring connection to an external oscillator (which, of course, delivers AC).
2. With built-in oscillator, requiring DC.

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Terry Pinnell
Hobbyist, West Sussex, UK