# Help needed designing simple circuit

P

#### Paul G.

Jan 1, 1970
0

Unfortunately, your solution, while interesting, misses one of the
requirements of the whole deal: it needs to operate the
bell/chime/annunciator *momentarily*, not continuously.

Since you need 18-24vdc, instead of the hazard of components
directly connected to the powerline, use an approved AC adapter or
"wallwart" that would provide the necessary DC voltage. From the
output of the AC adapter, place a PTC (positive temperature
coefficient) thermistor in series with your load. It will be a bit
tricky to determine the right physical size and nominal resistance of
the thermistor, since you haven't specified the load.
When the adapter powers up, the thermistor will be cold, and will
run the "indicator" what ever that is. After the thermistor heats up,
it will reduce the current going to your device, hopefully to the
point where it won't be noticed. The "resetable fuses" work on this
idea, you could use one of them the same way, but you need to again
select based on current load, and power.
If you get it to work it's: simple
doesn't violate electrical safety
reliable
The disadvantage, is that once on, it needs some time to "reset"
(cool down). The thermistor does get hot, a poorly designed circuit
could get hot enough to be a problem.

There are relay circuits that do much the same, but I'd run them
off the ac adapter as well. (relay in series with large cap, operates
when cap charges up. cap was bleeder resistor in parallel so it
discharges prperly when power is removed). The relay circuit would be
straight forward to calculate the on time (a percentage of the time
constant).

I'd prefer you stick to quite low voltages (6-12v), to minimize any
risk of shock. Who knows how this circuit might be physically
implemented!

Paul G.

D

#### David Nebenzahl

Jan 1, 1970
0
I think that a small sugar cube relay, diode, resistor and capacitor
will do the job !

Circuit, pleeze?

--
Made From Pears: Pretty good chance that the product is at least
mostly pears.
Made With Pears: Pretty good chance that pears will be detectable in
the product.
Contains Pears: One pear seed per multiple tons of product.

(with apologies to Dorothy L. Sayers)

D

#### David Nebenzahl

Jan 1, 1970
0
Why not take a crack at calculating the values of the
components and post your final circuit here. That way we
will get a good idea of how adept you are at circuit design.
I would still recommend putting R3 across C2 rather than C1
as per my original suggestion. That insures the drive to the
transistor truly goes to zero after some amount of time.

Hey, thanks for sticking with me. Check latest circuit incarnation at
http://www.geocities.com/bonezphoto/misc/One-shotBell.gif.

Component Value
-----------------------------
C1 100 uf/200 V.
C2 1 uf/100 V.
R1, R2 500 Î©, 2 W
R4, R5 10 kÎ©, 1/8 W
D1 1N4001
Q1 D1266 or equiv.

Rationale for values (assuming ~20 volt, 100 mA load):

C1: large enough to filter bulk of ripple
C2: sized for proper "on" time (WAG)
R1, R2: voltage divider to yield ~20 volts for load device;
calculations dictate 5 watt load, but since it's of short
duration, 2 w. should be sufficient
R3: bleeder (slow) for C1
R4, R5: this is also a WAG (wild-ass guess); see below
D1: 1A, 600 PIV
Q1: selected because I have these and have used them before.

OK, I admit that I do not know how to calculate those last 2 resistances
for proper biasing of Q1; so sue me. I do understanfd the general
principle of biasing; just never had the formal training to learn how to
I'd be curious to know how far off my guesses were.

BTW, I misunderstood where you wanted the bleeder resistor.

So give me a grade on my work.

--
Made From Pears: Pretty good chance that the product is at least
mostly pears.
Made With Pears: Pretty good chance that pears will be detectable in
the product.
Contains Pears: One pear seed per multiple tons of product.

(with apologies to Dorothy L. Sayers)

D

#### David Nebenzahl

Jan 1, 1970
0
Just one small question: earlier, you said:

So I thought I did that. But then you said:
R3 is still in the wrong place.

So what's the right place? I've put it across both caps and you still
don't seem to be happy.

--
Made From Pears: Pretty good chance that the product is at least
mostly pears.
Made With Pears: Pretty good chance that pears will be detectable in
the product.
Contains Pears: One pear seed per multiple tons of product.

(with apologies to Dorothy L. Sayers)

B

#### Baron

Jan 1, 1970
0
David said:
Circuit, pleeze?

Ok but excuse the ascii art.

live ----diode---relay---resistor---capacitor---neutral

Size the resistor to give the time constant with a particular value
capacitor. The relay contacts are isolated and can be connected to

The resistor should have sufficient wattage rating with respect to the
current and the capacitor should be rated for at least the maximum
voltage applied. ie 120v X 1.414. The relay can be almost anything
with a suitable coil voltage. ie 100 - 120.

The circuit works by utilising the charging current into the capacitor
to energise the relay. As the capacitor voltage increases the relay
will drop out. The time difference between energising and dropping out
is how long the annunciator will sound.

T

#### Tim

Jan 1, 1970
0

Unfortunately, your solution, while interesting, misses one of the
requirements of the whole deal: it needs to operate the
bell/chime/annunciator *momentarily*, not continuously.
Perhaps you are not inventive enough. My idea was to get a voltage low
enough to be safe. 5 volts could be used to fire a simple one-shot
circuit when the light comes on.

- Tim -

D

#### David Nebenzahl

Jan 1, 1970
0
Ok but excuse the ascii art.

live ----diode---relay---resistor---capacitor---neutral

You're excused. That illustrates the circuit perfectly.
Size the resistor to give the time constant with a particular value
capacitor. The relay contacts are isolated and can be connected to

The resistor should have sufficient wattage rating with respect to the
current and the capacitor should be rated for at least the maximum
voltage applied. ie 120v X 1.414. The relay can be almost anything
with a suitable coil voltage. ie 100 - 120.

The circuit works by utilising the charging current into the capacitor
to energise the relay. As the capacitor voltage increases the relay
will drop out. The time difference between energising and dropping out
is how long the annunciator will sound.

So I take it you'd like this circuit:
http://www.geocities.com/bonezphoto/misc/One-shotBell2.gif

And now can you give us the R and C values to give, say, a 1- or
2-second on time? I don't know how to calculate such things. (Understand
how they work, just never learned the actual math involved.)

--
Made From Pears: Pretty good chance that the product is at least
mostly pears.
Made With Pears: Pretty good chance that pears will be detectable in
the product.
Contains Pears: One pear seed per multiple tons of product.

(with apologies to Dorothy L. Sayers)

B

#### Bob Larter

Jan 1, 1970
0
David said:
I'd love to oblige you; unfortunately, I'm not yet at that level. I know
Ohm's law and some other basic stuff, but not enough about circuit
design to assign values with any confidence. I look forward to others
doing that. And someday, I intend to get a good basic electronics
textbook and seriously study it ...

"The Art of Electronics" by Horowitz & Hill. You'll save a fortune if
you buy a used student copy.

B

#### Baron

Jan 1, 1970
0
David said:
You're excused. That illustrates the circuit perfectly.

So I take it you'd like this circuit:
http://www.geocities.com/bonezphoto/misc/One-shotBell2.gif

That is just about it ! I hadn't considered connecting the relay
contacts back to the source to feed a transformer, but yes and the
transformer provides the safety isolation.
And now can you give us the R and C values to give, say, a 1- or
2-second on time? I don't know how to calculate such things.
(Understand how they work, just never learned the actual math
involved.)

Mmm, I tend to have trouble with decimal points.

Thats a little more difficult ! You need to know the characteristics of
the relay you are going to use.

For example: If the relay requires 0.01 amp (10ma)at 90 volts to pull
in. That would give a 90/0.01 = 9K Coil resistance.

Since the capacitor will be fully discharged initially the current to
charge it up will be limited by the resistance of the relay plus R. So
in this example there may be enough time constant due to the relay
itself if the capacitor value is well chosen.

The time is basically 63% of R X C Seconds.

Lets try 100uf @ 160vwg. 0.0001F X 9000 = 0.9 Seconds /100 X 63 = 0.56.
About half a second. So for this example a 220uf @ 160vwg would be

Since the current for the relay to drop out will be less than than 0.01a
the time will be less than this.

I would actually measure the relay I was going to use and adjust values
to suit. I'm sure you get the idea...

B

#### Baron

Jan 1, 1970
0
David said:
I thought you learned that there should be a discharge path
for the capacitor unless you want to wait hours or days
before it will work a second time.

David

Sorry my error I forgot to put one in :-(
You need to bleed about 1/10th of the relay current.
It should go directly across the cap.

D

#### David Nebenzahl

Jan 1, 1970
0
Following up on my own post, the circuit is poor for other
reasons. Think about what happens on the negative half
cycles of the AC input. You need at least a 'catch' diode in
there.

OK, I take your previous point about having a bleeder resistor to
discharge the cap.

I fail to understand why "negative half cycles of the AC input" are any
kind of a problem. Isn't that the very function of a rectifier? The
1N4001, f'rinstance, is rated at 600 PIV, so what's the problem? The
negative half cycles are simply blocked by the diode, right?

Just out of curiosity, what's a "catch diode"?

--
Made From Pears: Pretty good chance that the product is at least
mostly pears.
Made With Pears: Pretty good chance that pears will be detectable in
the product.
Contains Pears: One pear seed per multiple tons of product.

(with apologies to Dorothy L. Sayers)

B

#### Baron

Jan 1, 1970
0
David said:
I thought you learned that there should be a discharge path
for the capacitor unless you want to wait hours or days
before it will work a second time.

David

Yes you're right !
Sorry my error I forgot to put one in Â :-(
You need to bleed about 1/10th of the relay current.
It should go directly across the cap.

W

#### whit3rd

Jan 1, 1970
0
Function:  person has a motion-detector light installed in their home.
They want a buzzer/bell/annunciator of some kind to go off *momentarily*
whenever the light is activated.

There are motion detectors in the X10 lineup that wirelessly
communicate with a variety of modules; ring a chime, turn on a light,
whatever floats your boat. OK, I realize this is a learning
hobby-style project....

A

#### Andy

Jan 1, 1970
0

Unfortunately, your solution, while interesting, misses one of the
requirements of the whole deal: it needs to operate the
bell/chime/annunciator *momentarily*, not continuously.

--
Made From Pears: Pretty good chance that the product is at least
mostly pears.
Made With Pears: Pretty good chance that pears will be detectable in
the product.
Contains Pears:  One pear seed per multiple tons of product.

(with apologies to Dorothy L. Sayers)

First of all, I think that figuring things out is exactly what these
forums are for. I think your circuit is too complicated. I do this
kind of circuit all the time, but I usually use digital components and/
or comparators, microprocessors, etc. But I like your idea of using
the RC charge cycle. My version of your circuit includes a small DC
relay. Let me try to explain it, since I'm not educated enough to
figure out how to post a drawing. The 120V input voltage is dropped
across two resistors. The hot leg of the 120V is tied to a 33k
resistor, which is tied to a 1k resistor, and the 1k is tied to the
neutral. Now, from the connection point of the two resistors, place a
diode which is in series with the contact switch for the light
sensor. So one leg of the light sensor contact will be on the diode.
The other leg of the light sensor contact will go to the high side of
a 5V relay coil. On the other side of the relay coil, place a cap and
resistor in parallel. The other legs of the resistor and capacitor
will be tied to neutral. The light will turn on until the cap is
charged to the point that it doesn't allow current through the relay
coil. This timing is dependent on the cap size and on the 33k
resistor (which is of course adjustable). The resistor which is in
parallel with the cap allows a discharge path. But the resistor has
to be large enough to limit the current through the relay coil. The
coil I'm talking about is a little Omron 5G series (about \$2.00) and
it requires somewhere around 5mA to operate. Anyway, this will work.
But you'll have to mess around with the values. Good luck.

R

#### Ron D.

Jan 1, 1970
0
Why not take a crack at calculating the values of the
components and post your final circuit here. That way we
will get a good idea of how adept you are at circuit design.
I would still recommend putting R3 across C2 rather than C1
as per my original suggestion. That insures the drive to the
transistor truly goes to zero after some amount of time.

David

What you can do is purchase a Delay-On-Make block from www.ssac.com.

I already have a wireless motion detector that has that option.
Buzzer/and or outlet.
I don't use it that way. I have the dector mounted inside and the
receiver connected to a 7 W bulb.

D

#### David Nebenzahl

Jan 1, 1970
0
First of all, I think that figuring things out is exactly what these
forums are for.

Thank you. That's what I think they're for as well.
I think your circuit is too complicated. I do this kind of circuit
all the time, but I usually use digital components and/ or
comparators, microprocessors, etc. But I like your idea of using the
RC charge cycle.

Thanks again.
My version of your circuit includes a small DC
relay. Let me try to explain it, since I'm not educated enough to
figure out how to post a drawing. The 120V input voltage is dropped
across two resistors. The hot leg of the 120V is tied to a 33k
resistor, which is tied to a 1k resistor, and the 1k is tied to the
neutral. Now, from the connection point of the two resistors, place a
diode which is in series with the contact switch for the light
sensor. So one leg of the light sensor contact will be on the diode.
The other leg of the light sensor contact will go to the high side of
a 5V relay coil. On the other side of the relay coil, place a cap and
resistor in parallel. The other legs of the resistor and capacitor
will be tied to neutral.

Great--another entry in the contest!

I've drawn what I think you just described:
http://www.geocities.com/bonezphoto/misc/One-shotBell3.gif

Is that what you had in mind? Notice I added the snubber diode across

So now *I* get to critique *your* circuit:

1. While the diode is needed to provide (pulsating) DC to charge the
capacitor, I don't think the motion sensor is going to like DC; I think
it wants AC. So that may make this not work at all.

2. Not sure why you have a voltage divider (the 33k and 1k resistors)
across the line, rather than just using the 33k resistor in series with
the diode/relay/capacitor chain; why the 1K resistor? You realize that
those resistors are going to draw power from the line all the time,
right? And where did you some up with those values?

3. And of course this still leaves the question of the values for what
I've labeled as "R" and "C", the components that provide the delay. I
wonder if someone could be so good as to take a guess at them, and,
better yet, tell us how to calculate them (yes, I know, you need to know
the inductance and resistance of the relay coil).

Not knocking your design, mind you (which, now that you see it, is
really *not* all that simple); just trying to learn something, which is
the object of this whole exercise.

--
Made From Pears: Pretty good chance that the product is at least
mostly pears.
Made With Pears: Pretty good chance that pears will be detectable in
the product.
Contains Pears: One pear seed per multiple tons of product.

(with apologies to Dorothy L. Sayers)

B

#### Baron

Jan 1, 1970
0
David Nebenzahl Inscribed thus:
Thank you. That's what I think they're for as well.

Thanks again.

Great--another entry in the contest!

I've drawn what I think you just described:
http://www.geocities.com/bonezphoto/misc/One-shotBell3.gif

Is that what you had in mind? Notice I added the snubber diode across

So now *I* get to critique *your* circuit:

1. While the diode is needed to provide (pulsating) DC to charge the
capacitor, I don't think the motion sensor is going to like DC; I
think it wants AC. So that may make this not work at all.

2. Not sure why you have a voltage divider (the 33k and 1k resistors)
across the line, rather than just using the 33k resistor in series
with the diode/relay/capacitor chain; why the 1K resistor? You realize
that those resistors are going to draw power from the line all the
time, right? And where did you some up with those values?

3. And of course this still leaves the question of the values for what
I've labeled as "R" and "C", the components that provide the delay. I
wonder if someone could be so good as to take a guess at them, and,
better yet, tell us how to calculate them (yes, I know, you need to
know the inductance and resistance of the relay coil).

The value of the inductance of the relay coil is not needed for
calculating the delay time. Just the coil resistance !

A

#### Andy

Jan 1, 1970
0
Thank you. That's what I think they're for as well.

Thanks again.

Great--another entry in the contest!

I've drawn what I think you just described:http://www.geocities.com/bonezphoto/misc/One-shotBell3.gif

Is that what you had in mind? Notice I added the snubber diode across

So now *I* get to critique *your* circuit:

1. While the diode is needed to provide (pulsating) DC to charge the
capacitor, I don't think the motion sensor is going to like DC; I think
it wants AC. So that may make this not work at all.

2. Not sure why you have a voltage divider (the 33k and 1k resistors)
across the line, rather than just using the 33k resistor in series with
the diode/relay/capacitor chain; why the 1K resistor? You realize that
those resistors are going to draw power from the line all the time,
right? And where did you some up with those values?

3. And of course this still leaves the question of the values for what
I've labeled as "R" and "C", the components that provide the delay. I
wonder if someone could be so good as to take a guess at them, and,
better yet, tell us how to calculate them (yes, I know, you need to know
the inductance and resistance of the relay coil).

Not knocking your design, mind you (which, now that you see it, is
really *not* all that simple); just trying to learn something, which is
the object of this whole exercise.

--
Made From Pears: Pretty good chance that the product is at least
mostly pears.
Made With Pears: Pretty good chance that pears will be detectable in
the product.
Contains Pears:  One pear seed per multiple tons of product.

(with apologies to Dorothy L. Sayers)

I have used this circuit many times (without the relay) because it
will output a pretty stable DC signal at right around 5-6V. It works
because the charge cycle is much faster than the discharge cycle. The
resistor values are chosen to make sure that your DC output is
compatible with TTL, or a 5V logic input. This is a quick,
inexpensive circuit which I have used for monitoring motor states.
But, in this application, I have to say that it probably won't work.
It will turn on the light, but it will happen so fast as to be
invisible. The only way to make this work is to have a HUGE
capacitor.

B

#### Baron

Jan 1, 1970
0
David said:
(famous last words, "simple circuit" ...)

Did you have a go with any of the suggestions ?

D
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