Help needed designing simple circuit

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

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....

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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.

What you can do is purchase a Delay-On-Make block from

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.

On 3/31/2009 4:57 AM Andy spake 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:

Is that what you had in mind? Notice I added the snubber diode across the relay coil that the other David had asked for earlier.

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.

1. 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?

2. 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.

David Nebenzahl Inscribed thus:

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

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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.

Did you have a go with any of the suggestions ?