Hello all,

I'm trying to detect the presence of a 24V AC signal source at 60Hz. Attached in the picture below is a schematic of what i'm trying to achieve.

Basically, I use a half-wave rectifier into a dual LED optoisolator. My CAD program doesn't have one with two diodes, so "D2" is there to represent said second diode.

My question is regarding the AC signal side with the resistors R1 and R2. These resistors are to limit the current from the 24VAC (~17Vrms) source, to a ~50mA current used by the LEDs. Figuring the needed resistance to be ~450 ohm to achieve a 38mA current, my question is:

Do you need R1 AND R2 in order to limit the current for BOTH diodes in the opto-isolator. My guess is that R2 is unnecessary because R1 will limit the AC current in both directions. Any ideas?

A quick second question:
is the power dissipated by the resistor equal to P=Vrms^2/R? (~.65W?)
If so, is a 1 watt resistor a good enough power handling ability for this resistor?

Thank you all,

CK

 

You only need either R1 or R2; not both (series circuit)

Since the current in the resistor only flows during half cycles, the power dissipation is half what you said.

 

Cool, thats what I was thinking. Good to know about the half power dissipation.

Thank you sir for your quick response. You've been very helpful.

C

 

If you use an opto-coupler with two inverse LEDs inside, you only need a single resistor. See attachment.

See also:-
http://uk.farnell.com/vishay/sfh620a-2/optocoupler-ac-input/dp/1045384

 

If you use an opto-coupler with two inverse LEDs inside, you only need a single resistor. See attachment.

See also:-
http://uk.farnell.com/vishay/sfh620a-2/optocoupler-ac-input/dp/1045384

He still needs the external diode to eliminate one of the two pulses he would get without it. If he doesn't mind 100 pulses per sec, then wire as shown int the link.