detecting 18vac?

[JimDod]

I am working on a little xbee project where I detect a door bell, which is 18vac ( I expect this can vary from 15vac to 24vac). I'm a software guy and have made smoke when I fiddle with hardware so I thought I'd ask for some advice.

the current plan is to use a H11AA4M which is a AC Optoisolator simular to a 4N35 but with two LED's.

What I *think* I need to do is simply have a resistor in series with the LED side to limit the current.

My calculations are :

18V - 1.2V = 16.8V voltage from transformer minus voltage required by Optoisolator

16.8v / .010 = 1680 ohm voltage to be dissipated divided by current (10ma)



So will a 1.8k ohm resistor in series work with no smoke?

also could I do the same thing with a cheaper 4n35 and a 1n4001 diode?

Any other solutions to the problem?

 

[Reloadron]

Yes, I would use the H11AA4M with a 1.8 KΩ resistor in series with the emitter. That is the sort of application it is designed for and your math is correct.

Ron

 

[JimDod]

thanks Ron,

I was thinking a 1/4 watt would work ... just a guess

how would I calculate wattage for this circuit?

 

[Diver300]

The voltage across the diodes is so small that you can ignore it.

The maximum power is 24x24/1800 = 0.32 so 1/4 W is marginal. I would use at least a 0.5 W resistor.

You should be aware that the response time of the opto isolator is fast enough that you will see a 100 or 120 Hz signal on the output when the AC signal is present.

 

[Reloadron]

Power is equal to the current times the voltage.
Power is equal to the voltage squared divided by the resistance.
Power is equal to the current squared times the resistance.

Let's use the last formula. .010 * .010 = .0001 so .0001 * 1800 = 0.18 or 180 mW

Absolutely a 1/4 watt (250 mW) will be fine.

Ron

 

[Diver300]

The power depends on the square of the voltage. If you are going to get 24V, the power will be around 0.3 W. If the voltage is 18 V, the power will be 0.18W.

If you are not sure about the voltage, and the signal is on for longer than a minute or so, a 1/4W resistor could be too small.

 

[JimDod]

Thanks fo the help,
My further research indicates that the standard voltage for US door chimes is 16VAC, my voltmeter measurement was 18VAC, I think this is a typical variance for something unregulated.

I like things to be rugged, so I may just use a 1/2 watt.

One other thing I am concerned about is overloading and frying the H11AA4M given that I have little or no control over the unregulated input. Is this a valid worry?

 

[Diver300]

There is no problem there.

You are aiming for 10 mA, and the LEDs on the H11AA4M have an absolute maximum of 60 mA, so you have a good safety margin.

 

[Reloadron]

What tends to happen is the door chime uses a transformer which is a ratio type device. So in reality the output is a function of the input. Meaning if mains voltage is a little high or low the output will be a little high or low. Not quite an exacting science. As Diver mentioned I doubt you will hurt the opto coupler with 1.8 K in there.

Ron

 

[Diver300]

The other thing that happens is that the transformer output goes down when it is loaded. The transformer will be made to give its rated output (is that 16V?) when it is at full load. At no load it could rise to around 20 V, so the voltage will depend on what bell you are using.