Help needed designing simple circuit

[PlainBill]

On Thu, 26 Mar 2009 17:12:00 -0800, David Nebenzahl
<nobody@but.us.chickens> wrote:

On 3/26/2009 9:21 AM PlainBill spake thus:

On Wed, 25 Mar 2009 18:01:12 -0800, David Nebenzahl
nobody@but.us.chickens> wrote:

On 3/25/2009 10:24 AM PlainBill spake thus:

Contrary to what several nay-sayers have expressed, this exercise has
a LOT to do with electronics repair. Any fool can keep replacing
components until he finds the bad one. The successful tech will be
able to analyse a circuit and make meaningful tests to identfy the
failing part.

Thank you. It's more than a little annoying being completely written off
as a clueless idiot simply because I reversed a diode's polarity *in a
schematic* (since corrected).

NOW your next exercise is to identify the values for these components.


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 ...

Several comments on this circuit. First of all, it could work, but
would require significant improvements, some of which have already
been discussed. Rather than using a doorbell or chime, a Sonalert
could be used. In part this will depend on the sound level desired.

So what improvements would you suggest to the current circuit (after I
made the changes recommended earlier)? (Circuit at
http://www.geocities.com/bonezphoto/misc/One-shotBell.gif)

You must add surge supression if you are using an inductive load.

Without it, the pulse generated when the transistor turns off will
eventually destroy it.

Second, some of the criticisms are valid. If you intend to do this as
a design exercise you can have fun designing it, but I would not
recommend building it. You are dealing with potentially lethal
voltages, and fairly expensive components to withstand those voltages.

To be honest, I think it's more fun designing things like this that will
operate "directly" off line voltage (i.e., without a transformer). One
of the criteria of this whole deal was to avoid the use of a transformer.

And in defense of that, there are tons of things in use every day that
operate just this way. Latest example I found were a bunch of electric
staple guns a neighbor gave me that operated directly off 120 volts,
firing a solenoid through a capacitor.

If I were to actually build this, I'm confident I could do it safely--at
least as safely as those staple guns, which have UL, CSA, etc., compliance.

And what components would be expensive? Seems to me the most expensive
part would be the transistor, or possibly the two power resistors, but
even those aren't terribly pricey.

Cost would depend on the nature of the load. If you are using a

Sonalert the load would be under .04A.

On the other hand, if you are using a chime the load will be higher,
and using a doorbell REALLY opens a can of worms. I think you had
better expect a load of .5A, and one heck of an inductive kick as the
transistor turns off. The typical doorbell is designed to operate on
about 15VAC. I would expect rater short contact life if you run it
off DC.

IMHO, you are using the wrong approach. Over 30 years ago I built
something similar using a few small caps, a few resistors, a diode, a
speaker, and a 74C914 hex schmidt trigger. I used 4 AA cells to power
it, your application should use a 'wall wart' putting out 6VDC at 100
ma. I'd have to look up the specs, but a 555 timer would also do the
job.

Could do, but see above.

PlainBill

 

[Tim]

In article <49c944ff$0$29984$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

(famous last words, "simple circuit" ...)

OK, so I'm trying to come up with a simple (maybe even elegant) solution
to a simple problem. Have an idea I want to run by y'all.

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.

Here's my idea for the circuit:
http://www.geocities.com/bonezphoto/misc/One-shotBell.gif

First of all, please don't laugh at this. I am *not* a double-E or in
any way an electronics expert. Also keep in mind that this is the
farthest thing from a mil-spec application. It's just for fun; no life
support medical devices will depend on it.

I'd like to know the following:

1. Will this circuit even work?
1a. Will it work but end up destroying one or more components?
2. Is there a simpler way of accomplishing this task?
3. If it'll work, what are the right component values?

Explanation:

D1 is a half-wave rectifier. C2 filters the DC to produce more-or-less
ripple-free current. C1 provides the momentary "on" signal, by charging,
then "shutting down" when charged (sized according to RC time constant
to provide the desired "on" time). R2 and R3 form a voltage divider to
supply the appropriate base voltage to Q1. R1 acts as a voltage divider
in series with the load to supply the appropriate output voltage. (I
chose 24 volts DC arbitrarily; it might be less, probably not more.)

Component sizing:

R1 would obviously have to be large enough (in terms of power capacity)
to handle the load. The load would probably have a minimal current draw.
And since it would only be "on" momentarily, R1 could probably be a bit
undersized without worrying about damage.

Q1 would also need to be large enough to handle the load. I'm thinking a
common TO-220 type might work fine.

R2 & R3 could be small 1/8 watters.

OK, have at it. Rip 'er apart!



OK, so a lot of issues here deal with the AC power being detected and

transferred to something useful. I did a simple data logger for my well
pump (URL: http://personal.nbnet.nb.ca/snowowl/DataRecorder.html), that
had to deal 220V being sensed. I used a cheap little cell phone / palm
pilot charger that was universal 110/220 50/60hz and wired it parallel
with the motor leads. This gave me a nice 5 volts when the power was
applied to the pump, so I could log it's on cycles. You still have to
wire it up the the light, so that's a danger here as well. Dealing with
AC mains wiring, there always a need for extra safety.

- Tim -

 

[David Nebenzahl]

On 3/27/2009 3:55 PM Tim spake thus:

In article <49c944ff$0$29984$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

(famous last words, "simple circuit" ...)

OK, so I'm trying to come up with a simple (maybe even elegant) solution
to a simple problem. Have an idea I want to run by y'all.

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.

Here's my idea for the circuit:
http://www.geocities.com/bonezphoto/misc/One-shotBell.gif

OK, so a lot of issues here deal with the AC power being detected and
transferred to something useful. I did a simple data logger for my well
pump (URL: http://personal.nbnet.nb.ca/snowowl/DataRecorder.html), that
had to deal 220V being sensed. I used a cheap little cell phone / palm
pilot charger that was universal 110/220 50/60hz and wired it parallel
with the motor leads. This gave me a nice 5 volts when the power was
applied to the pump, so I could log it's on cycles. You still have to
wire it up the the light, so that's a danger here as well. Dealing with
AC mains wiring, there always a need for extra safety.

Thanks for your reply.

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)

 

[David]

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:49ce65a0$0$29745$822641b3@news.adtechcomputers.com...

On 3/27/2009 3:55 PM Tim spake thus:

OK, so I'm trying to come up with a simple (maybe even
elegant) solution to a simple problem. Have an idea I
want to run by y'all.

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.

Here's my idea for the circuit:
http://www.geocities.com/bonezphoto/misc/One-shotBell.gif

OK, so a lot of issues here deal with the AC power being
detected and transferred to something useful. I did a
simple data logger for my well pump (URL:
http://personal.nbnet.nb.ca/snowowl/DataRecorder.html),
that had to deal 220V being sensed. I used a cheap little
cell phone / palm pilot charger that was universal
110/220 50/60hz and wired it parallel with the motor
leads. This gave me a nice 5 volts when the power was
applied to the pump, so I could log it's on cycles. You
still have to wire it up the the light, so that's a
danger here as well. Dealing with AC mains wiring, there
always a need for extra safety.

Thanks for your reply.

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.

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

 

[Baron]

David Nebenzahl wrote:

On 3/27/2009 3:55 PM Tim spake thus:

In article <49c944ff$0$29984$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

(famous last words, "simple circuit" ...)

OK, so I'm trying to come up with a simple (maybe even elegant)
solution to a simple problem. Have an idea I want to run by y'all.

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.

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

--
Best Regards:
Baron.

 

[Paul G.]

On Sat, 28 Mar 2009 11:00:45 -0800, David Nebenzahl
<nobody@but.us.chickens> wrote:

On 3/27/2009 3:55 PM Tim spake thus:

In article <49c944ff$0$29984$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

(famous last words, "simple circuit" ...)

OK, so I'm trying to come up with a simple (maybe even elegant) solution
to a simple problem. Have an idea I want to run by y'all.

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.

Here's my idea for the circuit:
http://www.geocities.com/bonezphoto/misc/One-shotBell.gif

OK, so a lot of issues here deal with the AC power being detected and
transferred to something useful. I did a simple data logger for my well
pump (URL: http://personal.nbnet.nb.ca/snowowl/DataRecorder.html), that
had to deal 220V being sensed. I used a cheap little cell phone / palm
pilot charger that was universal 110/220 50/60hz and wired it parallel
with the motor leads. This gave me a nice 5 volts when the power was
applied to the pump, so I could log it's on cycles. You still have to
wire it up the the light, so that's a danger here as well. Dealing with
AC mains wiring, there always a need for extra safety.

Thanks for your reply.

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.

 

[David Nebenzahl]

On 3/28/2009 1:30 PM Baron spake thus:

David Nebenzahl wrote:

On 3/27/2009 3:55 PM Tim spake thus:

In article <49c944ff$0$29984$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

(famous last words, "simple circuit" ...)

OK, so I'm trying to come up with a simple (maybe even elegant)
solution to a simple problem. Have an idea I want to run by y'all.

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.

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)

 

[David Nebenzahl]

On 3/28/2009 12:26 PM David spake thus:

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:49ce65a0$0$29745$822641b3@news.adtechcomputers.com...

OK, so I'm trying to come up with a simple (maybe even
elegant) solution to a simple problem. Have an idea I
want to run by y'all.

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.

Here's my idea for the circuit:
http://www.geocities.com/bonezphoto/misc/One-shotBell.gif

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 Ί, 2 W
R3 220 kΊ, 1/8 W
R4, R5 10 kΊ, 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
calculate values to implement it. I'd appreciate your comments here, and
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)

 

[David]

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:49cec5b8$0$29736$822641b3@news.adtechcomputers.com...

On 3/28/2009 12:26 PM David spake thus:

"David Nebenzahl" <nobody@but.us.chickens> wrote in
message
news:49ce65a0$0$29745$822641b3@news.adtechcomputers.com...

OK, so I'm trying to come up with a simple (maybe even
elegant) solution to a simple problem. Have an idea I
want to run by y'all.

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.

Here's my idea for the circuit:
http://www.geocities.com/bonezphoto/misc/One-shotBell.gif

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 Ί, 2 W
R3 220 kΊ, 1/8 W
R4, R5 10 kΊ, 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 calculate values
to implement it. I'd appreciate your comments here, and
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.

So far the grade is not very good. The transistor is rated

at 60v. When the transistor is off it has to sustain the
full rectified voltage.of about 150v. The same is true of
C2. R3 is still in the wrong place. R5 should be much
smaller than R4 since Vbe will always be under a volt. R1,
R2 are close, but how sure are you that >20v will not
destroy the load? A better choice would be a smaller R1 and
larger R2 since the circuit as shown introduces a lot of C
to B feedback which will drastically slow down the turnoff
time and keep the transistor in a linear mode for longer
than necessary. Need load on time, and hfe of the transistor
you choose to calculate C2, R4, and R5.

Other prople replying have suggested either a transformer
isolated design or the use of a small relay and a simple
diode RC network to drive it. The relay is a good choice if
you can get the power to run the load elsewhere.

David

 

[David Nebenzahl]

Just one small question: earlier, you said:

... 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.

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)

 

[Baron]

David Nebenzahl wrote:

On 3/28/2009 1:30 PM Baron spake thus:

David Nebenzahl wrote:

On 3/27/2009 3:55 PM Tim spake thus:

In article <49c944ff$0$29984$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

(famous last words, "simple circuit" ...)

OK, so I'm trying to come up with a simple (maybe even elegant)
solution to a simple problem. Have an idea I want to run by y'all.

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.

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

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
your low voltage circuit.

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.

--
Best Regards:
Baron.

 

[Tim]

In article <49ce65a0$0$29745$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

On 3/27/2009 3:55 PM Tim spake thus:

In article <49c944ff$0$29984$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

(famous last words, "simple circuit" ...)

OK, so I'm trying to come up with a simple (maybe even elegant) solution
to a simple problem. Have an idea I want to run by y'all.

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.

Here's my idea for the circuit:
http://www.geocities.com/bonezphoto/misc/One-shotBell.gif

OK, so a lot of issues here deal with the AC power being detected and
transferred to something useful. I did a simple data logger for my well
pump (URL: http://personal.nbnet.nb.ca/snowowl/DataRecorder.html), that
had to deal 220V being sensed. I used a cheap little cell phone / palm
pilot charger that was universal 110/220 50/60hz and wired it parallel
with the motor leads. This gave me a nice 5 volts when the power was
applied to the pump, so I could log it's on cycles. You still have to
wire it up the the light, so that's a danger here as well. Dealing with
AC mains wiring, there always a need for extra safety.

Thanks for your reply.

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 -

 

[David Nebenzahl]

On 3/29/2009 6:43 AM Baron spake thus:

David Nebenzahl wrote:

On 3/28/2009 1:30 PM Baron spake thus:

David Nebenzahl wrote:

On 3/27/2009 3:55 PM Tim spake thus:

In article <49c944ff$0$29984$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

(famous last words, "simple circuit" ...)

OK, so I'm trying to come up with a simple (maybe even elegant)
solution to a simple problem. Have an idea I want to run by y'all.

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.

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

Circuit, pleeze?

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
your low voltage circuit.

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)

 

[Bob Larter]

David Nebenzahl wrote:

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.

--
W
. | ,. w , "Some people are alive only because
\|/ \|/ it is illegal to kill them." Perna condita delenda est
---^----^---------------------------------------------------------------

 

[Baron]

David Nebenzahl wrote:

On 3/29/2009 6:43 AM Baron spake thus:

David Nebenzahl wrote:

On 3/28/2009 1:30 PM Baron spake thus:

David Nebenzahl wrote:

On 3/27/2009 3:55 PM Tim spake thus:

In article <49c944ff$0$29984$822641b3@news.adtechcomputers.com>,
nobody@but.us.chickens says...

(famous last words, "simple circuit" ...)

OK, so I'm trying to come up with a simple (maybe even elegant)
solution to a simple problem. Have an idea I want to run by
y'all.

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.

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

Circuit, pleeze?

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
your low voltage circuit.

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

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
about 1 second.

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...

--
Best Regards:
Baron.

 

[David]

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:49d0228d$0$29738$822641b3@news.adtechcomputers.com...

On 3/29/2009 6:43 AM Baron spake thus:

David Nebenzahl wrote:
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
your low voltage circuit.

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.)

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

 

[David]

"David" <someone@somewhere.com> wrote in message
news:u73Al.15608$D32.7304@flpi146.ffdc.sbc.com...

"David Nebenzahl" <nobody@but.us.chickens> wrote in
message
news:49d0228d$0$29738$822641b3@news.adtechcomputers.com...
On 3/29/2009 6:43 AM Baron spake thus:

David Nebenzahl wrote:
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
your low voltage circuit.

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.)

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

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.

David

 

[David Nebenzahl]

On 3/30/2009 5:01 AM David spake thus:

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)

 

[Baron]

David wrote:

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:49d0228d$0$29738$822641b3@news.adtechcomputers.com...
On 3/29/2009 6:43 AM Baron spake thus:

David Nebenzahl wrote:
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
your low voltage circuit.

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.)

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  :-(
You need to bleed about 1/10th of the relay current.
It should go directly across the cap.

--
Best Regards:
Baron.

 

[David]

"David Nebenzahl" <nobody@but.us.chickens> wrote in message
news:49d10b60$0$29750$822641b3@news.adtechcomputers.com...

On 3/30/2009 5:01 AM David spake thus:

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"?

This is getting a bit old, but a DC relay coil is an
inductor. What happens when a current is flowing in an
inductor and the current source is cut off as will be the
case in the negative half cycles? I am NOT going to design
the circuit for you since you need to learn some basics.
While you are learning, what coil voltage would you choose?

David