Opto coupler biasing

[Suraj143]

HI Mike thanks for the support.

I think you missed my point.My load varies from 0A to 5A. It can be a any electronic household equipment.Water heater,TV,Refrigerator etc....

I just want to sense if the load exceeds 800mA. Thats it.The output doesn't needs to be logic high continuously. Pulse output will be fine.Even pulse timing not critical.

Even If I can sense the load exceeds 600mA then it is much better.The trip point in this case 600mA. It also doesn't needs to be perfect it can be 500mA or 700mA.

 

[ronv]

You can use bigger diodes. Probably still cheaper than a huge resistor.

https://www.mouser.com/ds/2/302/BYC15X-600-116142.pdf

 

[MrAl]

Hi MrAL I like your idea & Rons Idea.It doesn't need to be accurate.

Here I have drawn what you mentioned.Is this what you mean?

Hi,

Yes that is it. I dont know about the resistor value, you have to find that experimentally i think. As an approximation the voltage across the LED will be:

vLED=230*1.4*(R1+R2)/(R1+R2+R4)

which should come out to around 0.7 volts or something like that. With R4=470k it wont be high enough to bias the LED enough so you'd have to come down on that value. Of course you have to test it to make sure it works over time.

Most likely you'd have to try a value for R4 of around 20k, which would mean it would have to be a power resistor too of around 5 watts as it would dissipate around 2.7 watts of power. So you'd have to think about that too.
For less power we'd have to use a capacitor and resistor combination.

 

[alec_t]

...... or, you could avoid the hassle and power wastage of a sense resistor by using a small current-transformer, instead of the opto, to give mains isolation. Something like as shown below.
L1 is one of the mains wires (line or neutral) carrying the load current. The second opamp has hysteresis and is used as a comparator. The first opamp has adjustable gain to set the trip point.

Edit: The indicated inductance values for the CT are speculative, for the sim. Some experimenting with the number of turns would probably be needed, depending on the core available.

 

[MikeMl]

Alec, I think there should be a burden resistor tied directly across the secondary of the CT. If there is ever a sharp rise-time pulse in the primary with the opamp not powered, there is no virtual ground to terminate the secondary. The step-up action of the CT could easily create a voltage high enough to blow the opamp inputs.
The opamp would be seeing the voltage developed across the burden resistor as its input.

 

[MrAl]

Hello,

I like the current transformer idea too.

I forgot to mention one detail in my last post however. That is, if the 47 ohm resistor is raised that relaxes the lower Ohmic requirement of the resistor R4, so R4 can be raised.
For example, if with 47 ohms (which is close to 50 ohms) if a 20k resistor is required for R4, then with 47 ohms changed to 100 ohms that means roughly a 40k resistor for R4 would be required. That reduces the power requirement for R4 to 1.3 watts so a 2 watt resistor might be ok. Following that line of thought, raising the 47 ohm to 200 ohms means roughly an 80k resistor for R4 might work, with the power requirement being 0.65 watts. Similarly, a 400 ohm resistor and 160k resistor for R4 brings us down to around 0.3 watts and maybe a little higher like a 500 ohm resistor and 200k for R4 brings the power requirement down to 0.27 watts so a 1/2 watt resistor would work.

This should be tested carefully however because the accuracy is always in question.

 

[alec_t]

Alec, I think there should be a burden resistor tied directly across the secondary of the CT.

Good point, Mike. What value would you suggest? My gut feeling says ~100 Ohm would do.