Too many ohms??

I'm trying to make a 120VAC sensor... I'm using a full bridge rectifier and an optocoupler.
I'm having an issue when I connect the optocoupler into the circuit.

please see the schematic posted (left of the opto has separate ground from the right of the opto)

when I apply mains power with out the opto, LED A turns on.
when I apply mains power with the opto, neither LED A or B turn on.

(for the right hand side of the opto, i copied a schematic i found online)
(forgive me, i think i drew the recifier wrong)

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Please post the schematic that correctly represents the circuit you built.

The one you posted would destroy the optocoupler, because the output of the bridge is a negative voltage with respect to ground.

what do you mean "correct schematic" that's how my circuit is wired.

Then the optocoupler has probably been destroyed.

I would expect LED A to be destroyed as well, but perhaps you wired it backwards from what is shown in the circuit diagram.

1. You probably have drawn the bridge rectifier wrong (all diodes pointing the wrong way)

2. If you connect a ~2volt LED in parallel with a ~1.2volt opto LED, then all the current will go to the opto LED.
   Connect the indicator LED in series with the opto LED (between bridge+ and pin1). No resistor.

3. The opto transistor is connected with the wrong polarity.
   Pin4 goes to the battery and 3 to the LED/resistor.

4. The 150ohm resistor is like a short across the secondary LED, and will likely prevent the LED from coming on.
   Remove.
   Leo..

jremington:
Then the optocoupler has probably been destroyed.

I would expect LED A to be destroyed as well, but perhaps you wired it backwards from what is shown in the circuit diagram.

the rectifier is its own module. there is really no way to wire it wrong. it has two AC pins and +/- pins

Then the diagram is wrong, so it does not represent how you wired the circuit.

Wawa:

1. You probably have drawn the bridge rectifier wrong (all diodes pointing the wrong way)

2. If you connect a ~2volt LED in parallel with a ~1.2volt opto LED, then all the current will go to the opto LED.
   Connect the indicator LED in series with the opto LED (between bridge+ and pin1). No resistor.

3. The opto transistor is connected with the wrong polarity.
   Pin4 goes to the battery and 3 to the LED/resistor.

4. The 150ohm resistor is like a short across the secondary LED, and will likely prevent the LED from coming on.
   Remove.
   Leo..

1. the rectifier is its own ic
2. put the opto LED and LED A in series?
3. I will double check the pinout
4. noted.

jremington:
Then the diagram is wrong, so it does not represent how you wired the circuit.

being a bit nit-picky?

I already stated that the rectifiers were one module. Disregard the rectifiers in my circuit... as that is obviously not the problem. and if you read the original post with LED A working then that already shoots down the rectifiers being wrong.

if you read the original post with LED A working then that already shoots down the rectifiers being wrong.

I already pointed that out, and postulated in reply #3 that you wired the LED backwards.

If you want to get correct answers on this forum, please post the correct information describing the problem.

Yes, my apologies I suppose.

Posting the OP's schematic so I can see it without downloading it:

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Hi,

As you said you have the diodes reversed in your drawing.

I suggest:

1. Redraw the schematic with a square (or round) symbol for the bridge rectifier. Label the AC pins and + pin and - Pin

2. Get rid of the 220 ohms and put the LED (A) in series with the opto.

Your current is likely OK. The current will peak at ~ 3.3 ma but be below the useful on current of the opto input for much of the 120 hz rectified sine wave. Your output will go on and off at 120 hz as well.

I used a 120K in series with the input of a FOD814A300 opto to sense 120Vac and it worked fine.

Primary current needed depends on secondary current used (Current Transfer Ratio)
With non specified PC817s I would use twice the LED current that the opto transistor has to switch.

I had LEDs fail over time with circuits like that (weeks, months).
Could be wise to add a small cap (47uF/>=6.3v) to the + and - of the bridge rectifier.
The cap stops positive current spikes and negative spikes (slow diodes) from dirty mains power.
And also bridges the zero-crossing gaps (if you want that).
Leo..

Hi,

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Tom...

The 220ohm resistor for the primary LED is electrically connected in series with the two 25k resistors,
So is doing nothing useful there. Can be removed.

Still the other two problems left (3/4 swapped and 150ohm load).
Leo..

Thanks for the responses.
Wawa, I considered putting caps in the circuit after it was working on a basic level.
I will make changes and redraw my schematic as requested. However, my work day is about to start. So you guys should see it in about 10 hours

TomGeorge:
Hi,

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Tom...

What about using a 0.047uF 250V capacitor in place of the two resistors? In AC circuits, capacitors supply impedence (0.047uF at 60Hz = 56kohms) but they don't dissipate or generate heat like the resistors do.

Ok, the circuit in the attached is the new circuit... and it works. Thank you all for the help
The rectifier is a D3SBA20
The orange highlighter has not been added to the circuit. But I'm curious on what ratings should be used. (I haven't used caps a lot, I have some but not a huge variety)

See post#13 (47uF >=6.3v).
A 10-100uF, 6.3-35volt, whatever electrolytic cap will do. Just mind the polarity.
Max voltage on the cap is the two LEDs in series (~3.3volt).
Same for the rectifier. Not a problem if a bridge with a low voltage/current rating is used.
Leo..