I need Ideas/Suggestions

[drkidd22]

How do you get 2xVf?

I found the LED drop voltage of 2Vf using a resistor in series with the 15V power supply. Voltage across led was 2V.

The diodes are in anti-parallel. When the line polarity is in one direction, one diode conducts aad drops ~1.2V. When the line polarity reverses, the other diode conducts and has a drop of Vf=~1.2v, but with the opposite polarity. One set of 3-series diodes functions with one LED and the other set of 3-series diodes functions with the other LED. Since the resistor is not polarity sensitive, it functions half the time with one LED, and half the time with the other...on opposite polarity line cycles.

Yes, I noticed this. I used the DMM and it read 2.07V across the diodes just like you say. Or just 2V. 0.66Vf per diode drop. This was on "bench test" not Proteus.

The Max rating is what they "can" dissipate, not what they "will". So, If your lamp draws ~100mA, and each diode drops ~0.7V that's 0.7Vx0.1A=0.07W (70mW) per diode. And since as you say, they only conduct half the time, that's an average dissipation of 35mW per diode.

That makes sense. I will put in 120AC to this thing and see if I don't burn the house down.

 

[KMoffett]

I found the LED drop voltage of 2Vf using a resistor in series with the 15V power supply. Voltage across led was 2V.

...

Vf for an LED is not 2Vf, it is just Vf for an LED. The LED Vf varies some with current and a lot with color: Technical LED Color Chart.
Vf for diodes also vary with current and type: **broken link removed**

So you measured the Vf of the LED on your bench as 2.0v. Measure the Vf of the optocoupler when you get it. Spec says @10mA it has a typical of 1.17v and a max of 1.4v.

Ken

 

[drkidd22]

I guess I just didn't word it right.
Vf = 2V not 2Vf. Got it.

Nice info. Thanks