Query on optocoupler with diode in series

[myatham]

Hello,

In our design, we used existing schematics for opto-coupler circuit.

It has one diode (D1) (whose Vr = 1500 V) in series to photo diode of opto-coupler. Photo diode current is limited to 1.1mA with current limiting resistor of 200K ohm. This circuit is to monitor 230Vac line voltage presence. Reverse break down voltage of photo diode on optocoupler is 6V.

My understanding is that, diode D1 will enahnce the reverse break down voltage of photo diode.

So, we expected that photo diode will get damaged if diode D1 is shorted as part of FMEA since Vr=6V for photo diode. But, there is no damage to the photo diode.


Is my understanding wrong i.e. diode is placed to enhance reverse break down of the photo diode?
Why opto-coupler is not damaged though its reverse breakdown voltage of phto diode is 6V? Is current limiting resistor helping?

 

[betwixt]

You have to understand the mechanism of breakdown. Without knowing the schematic, it sounds like the diode is there to prevent reverse polarity exceeding the reverse breakdown voltage of the LED in the opto-coupler not the photo diode which is at it's output side. The resistor may be helping but the damage isn't caused by excessive current (which the resistor would help with) but by structural damge to the LED material itself. A bit like flash over without the flash.

Please show a schematic for a more detailed explanation.

Brian.

 

[Ogu Reginald]

The resistor is helping because for every resistor with current flowing, there must be a voltage drop across the resistor.
Please attach diagram for easy analysis.

 

[WimRFP]

Hello,

Breakdown is not always a destructive process. If current is kept low enough, the heat generated by current*(breakdown voltage) will not damage the device. In your case the resistor (200 kOhms) limits the current. A 10V zener diode is always working in breakdown mode (when used as reference). So you are right, the series resistor limits the breakdown current and prevents it from damage due to heat.

However breakdown may result in device defects that may damage the device after some time (that can be days/months/years). Reverse BE breakdown in small signal transistors will not destroy them directly, but HFE drops over time making the transistor useless if high HFE is required at low current.

When looking to your series circuit of D, LED and R, I would add at least some capacitance (nF range) across the LED to protect it from mains transients (that can be in the kV range depending on the type of installation). I would not leave out the 1500V diode without serious testing. If you don't like the 1500V diode, leave it out and put a small signal diode (such as 1N4148) parallel to the LED in reverse direction. Note that the dissipation in the 200 kOhm resistor doubles.

 

[FvM]

Reading optocoupler or IR-LED datsheets strictly, there's no permitted reverse current specified, just a maximum reverse voltage. Some datasheets specify a typical and/or maximum reverse current,which gives an idea about acceptable reverse currents. The reverse leakage current of typical rectifier diodes will be probably higher, thus I think, a reverse protection is absolutely suggested for the said AC application.

I prefer to have the HV rectifier diode in addition rather than dissipating exessive power.

If you don't want the circuit to be polarity selective, an AC optocoupler with antiparallel LEDs might be the best solution.

 

[myatham]

One type error. Vr= 1000V. Circuit diagram is attached for your reference.

- - - Updated - - -

One type error, diode used is 1N4007 with Vr=1000V

 

[betwixt]

I see no diagram but the original question was about the "photo diode", given it's application, I think it was intended to be the emitter (LED) rather than the sensor side of the opto-coupler. Regardless, it isn't advisable to rely on a resistor to limit avalanche/ breakdown currents in components not designed to be used in that mode. Personally, I would add a second diode or even a resistor in parallel with an LED running off high AC voltage to drain away any leakage in the 1N4007 otherwise they share the reverse voltage between them without guarantee of how much each is holding back.

Brian.