# Need help understanding how to use opto-isolator

M

#### mjohnson

Jan 1, 1970
0
I want to build an automatic garage door closer with an alarm clock and
my garage door remote. I realize that I could just buy something but I
want to build it so I can learn something and have some fun (and
frustration).

Here is a block diagram of what I am imagining:
http://img98.echo.cx/img98/3411/phase12nt.jpg

My question is what do I need to do to take the output voltage at the
clock's buzzer to activate the interface circuit. The voltage I read
on the buzzer when it's going off is 395mV (.395V). If for example, I
just want to turn on an LED (baby steps right) what would I need to do
to couple the alarm clock to the LED circuit?

I'm assuming that the actual coupling of the LED circuit to the buzzer
will represent a new load to the alarm clock which it wasn't designed
to take. So my guess is that I would need an opto-isolator and run the
LED circuit on it's own power supply? But is 395mV is enough to drive
the opto-isolator?

thanks for your time and help!

L

#### Larry Brasfield

Jan 1, 1970
0
mjohnson said:
I want to build an automatic garage door closer with an alarm clock and
my garage door remote. I realize that I could just buy something but I
want to build it so I can learn something and have some fun (and
frustration).

Here is a block diagram of what I am imagining:
http://img98.echo.cx/img98/3411/phase12nt.jpg

My question is what do I need to do to take the output voltage at the
clock's buzzer to activate the interface circuit. The voltage I read
on the buzzer when it's going off is 395mV (.395V). If for example, I
just want to turn on an LED (baby steps right) what would I need to do
to couple the alarm clock to the LED circuit?

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.
I'm assuming that the actual coupling of the LED circuit to the buzzer
will represent a new load to the alarm clock which it wasn't designed
to take.

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.
So my guess is that I would need an opto-isolator and run the
LED circuit on it's own power supply? But is 395mV is enough to drive
the opto-isolator?

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.
thanks for your time and help!
You're welcome.

J

#### John Fields

Jan 1, 1970
0
I want to build an automatic garage door closer with an alarm clock and
my garage door remote. I realize that I could just buy something but I
want to build it so I can learn something and have some fun (and
frustration).

Here is a block diagram of what I am imagining:
http://img98.echo.cx/img98/3411/phase12nt.jpg

My question is what do I need to do to take the output voltage at the
clock's buzzer to activate the interface circuit. The voltage I read
on the buzzer when it's going off is 395mV (.395V). If for example, I
just want to turn on an LED (baby steps right) what would I need to do
to couple the alarm clock to the LED circuit?

I'm assuming that the actual coupling of the LED circuit to the buzzer
will represent a new load to the alarm clock which it wasn't designed
to take. So my guess is that I would need an opto-isolator and run the
LED circuit on it's own power supply? But is 395mV is enough to drive
the opto-isolator?

---
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
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?

M

#### mjohnson

Jan 1, 1970
0
Specifically, on your remote control are the OPEN and CLOSE functions
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?
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?

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.

http://img195.echo.cx/img195/4791/diagram6yf.jpg

Front side of the remote:
http://img195.echo.cx/img195/1594/remotefront5nd.jpg

Back side of the remote (red lines are show the traces)
http://img195.echo.cx/img195/6156/remoteback8tt.jpg

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
(www.vishay.com/document/83519/83519.pdf) between the clock and between
the remote.

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

J

#### John Fields

Jan 1, 1970
0
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?

---
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.
---
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.

---
Before anything can happen we need to find out definitively what's
happening at the buzzer when the alarm goes off, or we need to find a
signal somewhere in the clock which changes state when the alarm goes
off. Either that or couple to the clock acoustically in order to
detect the sound of the buzzer and use that to trigger the chain of
events leading to the activation of the remote.
---
Here's a new diagram, as well as front/back images of the remote.

http://img195.echo.cx/img195/4791/diagram6yf.jpg

Front side of the remote:
http://img195.echo.cx/img195/1594/remotefront5nd.jpg

Back side of the remote (red lines are show the traces)
http://img195.echo.cx/img195/6156/remoteback8tt.jpg

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?

---
No, it looks like you only have to get across the terminals with trace
connected to them. But, since we don't know how the switch is being
used in the circuit, it would be best to use relay contacts (instead
of the out of an optocoupler) across the switch terminals in order to
activate the remote. Test it by shorting the terminals momentarily
and see if it works the door. If it does, then relay contacts will be
fine.
---
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 wouldn't use an opto because of the current required for its LED and
the uncertainty of being able to use its transistor output to trigger
the remote.

Here's what I see as a much simpler solution, with only the clock
output needing to be defined in order to make it work:

INTERFACE
BATTERY
CLOCK 4.5V REMOTE
BATTERY | BATTERY
3V +---------+---------+ 3V
| | | | |
+--+--+ +--+--+ +--+--+ +--+--+ +--+---+
|CLOCK|---|ALARM|---|1SEC |---|REED |---|REMOTE|
+-----+ | DET | |MONO | |RELAY|---|SWITCH|
+-----+ +-----+ +-----+ +------+

L

#### Larry Brasfield

Jan 1, 1970
0
John Fields said:
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.

(To the OP I concur with that good advice.

....
Before anything can happen we need to find out definitively what's
happening at the buzzer when the alarm goes off, or we need to find a
signal somewhere in the clock which changes state when the alarm goes
off.

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).

....
---
No, it looks like you only have to get across the terminals with trace
connected to them. But, since we don't know how the switch is being
used in the circuit, it would be best to use relay contacts (instead
of the out of an optocoupler) across the switch terminals in order to
activate the remote. Test it by shorting the terminals momentarily
and see if it works the door. If it does, then relay contacts will be
fine.

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.
Here's what I see as a much simpler solution, with only the clock
output needing to be defined in order to make it work:

INTERFACE
BATTERY
CLOCK 4.5V REMOTE
BATTERY | BATTERY
3V +---------+---------+ 3V
| | | | |
+--+--+ +--+--+ +--+--+ +--+--+ +--+---+
|CLOCK|---|ALARM|---|1SEC |---|REED |---|REMOTE|
+-----+ | DET | |MONO | |RELAY|---|SWITCH|
+-----+ +-----+ +-----+ +------+

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).

J

#### John Fields

Jan 1, 1970
0
(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).

---
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.
---
...

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.

---
Not necessarily, CTR falls off quickly as LED forward current
diminishes and there are temperature effects which need to be taken
into consideration which can largely be ignored with a comparator-reed
switch solution. Also, with the reed switch solution there is no
saturation voltage VS LED If problem since it's either on or off.
---
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.

---
Either solution will require the generation of a timed pulse to the
remote, so that's probably a wash.
---
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).

---
Could be. I'll defer judgement and wait until the OP comes back with
something definitive on the buzzer signal to post my design. If he
doesn't, I can always fall back on the acoustic thing I've already
posted. You may want to ask him about the current being conducted by
the remote's switch switch to see whether you can use an opto in
there. An easy way to determine the current would be to jump the
switch contacts with a milli/microammeter...

L

#### Larry Brasfield

Jan 1, 1970
0
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.
---
Not necessarily, CTR falls off quickly as LED forward current
diminishes and there are temperature effects which need to be taken
into consideration which can largely be ignored with a comparator-reed
switch solution. Also, with the reed switch solution there is no
saturation voltage VS LED If problem since it's either on or off.

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.
---
Either solution will require the generation of a timed pulse to the
remote, so that's probably a wash.
Yes.

---
Could be. I'll defer judgement and wait until the OP comes back with
something definitive on the buzzer signal to post my design. If he
doesn't, I can always fall back on the acoustic thing I've already
posted.

Seems reasonable.
You may want to ask him about the current being conducted by
the remote's switch switch to see whether you can use an opto in
there. An easy way to determine the current would be to jump the
switch contacts with a milli/microammeter...

Yes.

J

#### John Fields

Jan 1, 1970
0
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.

---
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 |
|<-
O |
|
GND

In this case, if the load on Eout is insignificant, (CMOS, say) Eout
will be either +V or, assumong GND is at 0V, 0V.

However, in this case:

+V
|
[100k]
|
+----->Eout
|
C A
B <--[LED]
E K
|
GND

Eout will depend on the collector-to-emitter resistance of the
transistor, and if it isn't driven low enough (if the current through
the LED isn't high enough) Eout may not cross the switching threshold
of the driven device.

The problem can also be exacerbated by the use of cheap conductive
rubber switches which require a substantially lower value pullup.

Additionally, if the remote used pull-down resistors, the opto's
transistor would be operating as a follower which, with only a 3V
supply available in the remote, would complicate matters even more.

All of these problems go away with the mechanical contacts of a reed
relay.
---
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.

---
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...
---

L

#### Larry Brasfield

Jan 1, 1970
0
John Fields said:
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.

---
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 |
|<-
O |
|
GND

In this case, if the load on Eout is insignificant, (CMOS, say) Eout
will be either +V or, assumong GND is at 0V, 0V.

However, in this case:

+V
|
[100k]
|
+----->Eout
|
C A
B <--[LED]
E K
|
GND

Eout will depend on the collector-to-emitter resistance of the
transistor, and if it isn't driven low enough (if the current through
the LED isn't high enough) Eout may not cross the switching threshold
of the driven device.

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.
The problem can also be exacerbated by the use of cheap conductive
rubber switches which require a substantially lower value pullup.

I thought pullup values would go up in that case,
at least when expressed in Ohms. What makes
you say they would go lower?
Additionally, if the remote used pull-down resistors, the opto's
transistor would be operating as a follower which, with only a 3V
supply available in the remote, would complicate matters even more.

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.
All of these problems go away with the mechanical contacts of a reed
relay.

Most or all of them go away on their own.
---
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...

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.

L

#### Larry Brasfield

Jan 1, 1970
0
[Brasfield had suggested an optoisolator in parallel with
a button on a remote.]
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:
[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.

J

#### John Fields

Jan 1, 1970
0
John Fields said:
On Fri, 22 Apr 2005 09:42:24 -0700, "Larry Brasfield"

...
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).
---
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.

---
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 |
|<-
O |
|
GND

In this case, if the load on Eout is insignificant, (CMOS, say) Eout
will be either +V or, assumong GND is at 0V, 0V.

However, in this case:

+V
|
[100k]
|
+----->Eout
|
C A
B <--[LED]
E K
|
GND

Eout will depend on the collector-to-emitter resistance of the
transistor, and if it isn't driven low enough (if the current through
the LED isn't high enough) Eout may not cross the switching threshold
of the driven device.

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.

---
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.
---
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 know you haven't suggested that, but it may well be the position the
opto finds itself in if there's not enough photocurrent to drive its
output into saturation. That is, operating linearly, its
collector-to-emitter resistance may be too high to cause the driven
device's switching threshold to be exceeded.
---
I thought pullup values would go up in that case,
at least when expressed in Ohms. What makes
you say they would go lower?

---
Yes, you're right. I got it backwards.
---
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.

---
That's not the point. If the driven device uses pull-downs and
expects its input to be driven high, then the opto's output (the
emitter) becomes a follower, of necessity.

You might be right though, if the LED drive current is high enough,
but with the microamp drive levels you're proposing, I don't think so.

Run this:

Version 4
SHEET 1 880 680
WIRE -192 208 -192 176
WIRE -192 400 -192 288
WIRE -112 176 -192 176
WIRE -16 400 -16 272
WIRE 32 176 -32 176
WIRE 32 272 -16 272
WIRE 288 240 224 240
WIRE 288 272 288 240
WIRE 288 400 288 352
WIRE 384 176 224 176
WIRE 384 192 384 176
WIRE 384 400 384 272
FLAG 384 400 0
FLAG 288 400 0
FLAG -192 400 0
FLAG -16 400 0
SYMBOL Optos\\4N25A 128 240 R0
SYMATTR InstName U1
SYMBOL voltage 384 176 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 3
SYMBOL res 272 256 R0
SYMATTR InstName R1
SYMATTR Value 100k
SYMBOL voltage -192 192 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 4.5
SYMBOL res -16 160 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R2
SYMATTR Value 100k
TEXT -226 506 Left 0 !.tran 0 .1 0
Most or all of them go away on their own.

---
I disagree. Doing it your way requires knowing the values of the
pullups and/or pull-down resistors, the contact resistance of the
switch(es), the switching thresholds of the driven device, the drive
level and voltage available from the buzzer to drive the OPTO LED (if
that's how you were planning to do it) and...

My way only requires me to know what the drive voltage to the buzzer
is, or if there's an ALARM ON signal available, what its voltage is.
---

M

#### mike

Jan 1, 1970
0
mjohnson said:
I want to build an automatic garage door closer with an alarm clock and
my garage door remote. I realize that I could just buy something but I
want to build it so I can learn something and have some fun (and
frustration).

Here is a block diagram of what I am imagining:
http://img98.echo.cx/img98/3411/phase12nt.jpg

My question is what do I need to do to take the output voltage at the
clock's buzzer to activate the interface circuit. The voltage I read
on the buzzer when it's going off is 395mV (.395V). If for example, I
just want to turn on an LED (baby steps right) what would I need to do
to couple the alarm clock to the LED circuit?

I'm assuming that the actual coupling of the LED circuit to the buzzer
will represent a new load to the alarm clock which it wasn't designed
to take. So my guess is that I would need an opto-isolator and run the
LED circuit on it's own power supply? But is 395mV is enough to drive
the opto-isolator?

thanks for your time and help!

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
--
Return address is VALID but some sites block emails
..
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Bunch of stuff For Sale and Wanted at the link below.
MAKE THE OBVIOUS CHANGES TO THE LINK
ht<removethis>tp://www.geocities.com/SiliconValley/Monitor/4710/

L

#### Larry Brasfield

Jan 1, 1970
0
John Fields said:
John Fields said:
On Sat, 23 Apr 2005 00:31:26 -0700, "Larry Brasfield"

On Fri, 22 Apr 2005 09:42:24 -0700, "Larry Brasfield"

...
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).
---
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.

---
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 |
|<-
O |
|
GND

In this case, if the load on Eout is insignificant, (CMOS, say) Eout
will be either +V or, assumong GND is at 0V, 0V.

However, in this case:

+V
|
[100k]
|
+----->Eout
|
C A
B <--[LED]
E K
|
GND

Eout will depend on the collector-to-emitter resistance of the
transistor, and if it isn't driven low enough (if the current through
the LED isn't high enough) Eout may not cross the switching threshold
of the driven device.

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.
---
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.
Agreed.
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 know you haven't suggested that, but it may well be the position the
opto finds itself in if there's not enough photocurrent to drive its
output into saturation. That is, operating linearly, its
collector-to-emitter resistance may be too high to cause the driven
device's switching threshold to be exceeded.

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.
You might be right though, if the LED drive current is high enough,
but with the microamp drive levels you're proposing, I don't think so.

LED current, suggesting it might be "limited to
a few 100 uA". More on this in a moment.
Run this:

Version 4
SHEET 1 880 680
WIRE -192 208 -192 176
WIRE -192 400 -192 288
WIRE -112 176 -192 176
WIRE -16 400 -16 272
WIRE 32 176 -32 176
WIRE 32 272 -16 272
WIRE 288 240 224 240
WIRE 288 272 288 240
WIRE 288 400 288 352
WIRE 384 176 224 176
WIRE 384 192 384 176
WIRE 384 400 384 272
FLAG 384 400 0
FLAG 288 400 0
FLAG -192 400 0
FLAG -16 400 0
SYMBOL Optos\\4N25A 128 240 R0
SYMATTR InstName U1
SYMBOL voltage 384 176 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 3
SYMBOL res 272 256 R0
SYMATTR InstName R1
SYMATTR Value 100k
SYMBOL voltage -192 192 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 4.5
SYMBOL res -16 160 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R2
SYMATTR Value 100k
TEXT -226 506 Left 0 !.tran 0 .1 0

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 disagree. Doing it your way requires knowing the values of the
pullups and/or pull-down resistors, the contact resistance of the
switch(es), the switching thresholds of the driven device, the drive
level and voltage available from the buzzer to drive the OPTO LED (if
that's how you were planning to do it) and...

My way only requires me to know what the drive voltage to the buzzer
is, or if there's an ALARM ON signal available, what its voltage is.

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.
---
Take a look at the contact life specs of any reed relay hot switching
microamps and you'll probably get a grin on your chops when you
consider how many centuries worth of garage openings and closings that
comes out to!^)

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!

M

#### mjohnson

Jan 1, 1970
0
The alarm clock is powered by two AAA batteries and has a piezo buzzer
so I don't think there's any AC involved.

L

#### Larry Brasfield

Jan 1, 1970
0
mjohnson said:
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.

T

#### Terry Pinnell

Jan 1, 1970
0
mjohnson said:
The alarm clock is powered by two AAA batteries and has a piezo buzzer
so I don't think there's any AC involved.

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.

L

#### Larry Brasfield

Jan 1, 1970
0
Terry Pinnell said:
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.

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

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.

M

#### mjohnson

Jan 1, 1970
0
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...

L

#### Larry Brasfield

Jan 1, 1970
0
mjohnson said:
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?

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).
Sorry for my ignorance...

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.