Help needed designing simple circuit

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.

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.

PlainBill

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:

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 -

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

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.

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(with apologies to Dorothy L. Sayers)

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

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

--
Best Regards:
                     Baron.

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.

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

Circuit, pleeze?

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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)

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

Hey, thanks for sticking with me. Check latest circuit incarnation at

Component Value

----------------------------- C1 100 uf/200 V. C2 1 uf/100 V. R1, R2 500 ?, 2 W R3 220 k?, 1/8 W R4, R5 10 k?, 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.

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Made From Pears: Pretty good chance that the product is at least
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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)

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

So I thought I did that. But then you said:

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

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Made From Pears: Pretty good chance that the product is at least
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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)

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.

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 -

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

You're excused. That illustrates the circuit perfectly.

So I take it you'd like this circuit:

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.)

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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)

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

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

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.

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.

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

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.

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

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)