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47K in L and N lines
Resistor voltage rating & common mode noise immunity.Do you mean two 47K on both line and neutral sides? What is wrong with one 100K only on one side?
Well, at least I know who has the other half of the Worlds stock of 6N138s!
There is a trick you might wish to try but obviously no promises as to the outcome, if you add a resistor to ground from VB on the 6N139 it reduces the sensitivity a little but greatly speeds its switching speed. You might be able to recover a little more speed while still keeping it faster then a PC817. Try values between 10K to 20K and see what happens. I've got about 50 6N139s in stock here but a workbench on overload already so I can't help with experiments.
There is another completely different approach which might be worth looking at, use a small DC-DC converter to generate an isolated DC supply and use it to power a comparator driven from the AC line. The comparator output would drive an optocoupler LED to give the isolation you need.
Brian.
Hello Brian I always have lots of stock of things I don't use, so if you want anything useless I am your manYes I think your right the base connection is very handy for tuning the tradeoff so now I can't wait to get some in the new year to try
Yes a very interesting idea about the op-amp & isolation barrier that would decouple the sensing current from the opto-current and allow a normal transistor coupler to be used without high power dissipation, of course a step up in complexity!
I have looked everywhere for other alternatives, logic/gate-drive opto's came to mind but they all have threshold currents >=5mA.
Ohh I almost forgot I also thought of using a constant current regulator on the hot side but it did not cut the dissipation by a huge amount and again circuit complexity is my enemy!
I'm puzzled why you suggest that. The problem with opto-sensing the current is the need for the voltage to rise enough after crossing zero for the LED to light sufficiently. There will always be a short dead period between zero and LED Vf being reached. I'm not sure how a constant current source helps that, especially in an AC circuit. I agree the current waveform is unipolar sine and for most of the cycle it just wastes power but as long as peak cycle voltage doesn't result in excess LED current I don't see that as a problem. Rectifying first re-introduces the Vf of the bridge rectifiers again.Not complicated - A depletion fet and a resistor form a high voltage current source quite well (any depletion can replace this for a lower cost):
quoting asdf44:
I'm puzzled why you suggest that. The problem with opto-sensing the current is the need for the voltage to rise enough after crossing zero for the LED to light sufficiently. There will always be a short dead period between zero and LED Vf being reached. I'm not sure how a constant current source helps that, especially in an AC circuit. I agree the current waveform is unipolar sine and for most of the cycle it just wastes power but as long as peak cycle voltage doesn't result in excess LED current I don't see that as a problem. Rectifying first re-introduces the Vf of the bridge rectifiers again.
Brian.
quoting asdf44:
I'm puzzled why you suggest that. The problem with opto-sensing the current is the need for the voltage to rise enough after crossing zero for the LED to light sufficiently. There will always be a short dead period between zero and LED Vf being reached. I'm not sure how a constant current source helps that, especially in an AC circuit. I agree the current waveform is unipolar sine and for most of the cycle it just wastes power but as long as peak cycle voltage doesn't result in excess LED current I don't see that as a problem. Rectifying first re-introduces the Vf of the bridge rectifiers again.
Brian.
Am I missing something? The 100k resistor proposed is going to have a similar dead zone because it needs voltage before it reaches the required forward current of the opto. Fourtytwo cites 5V dead zone as a problem yet a 100k resistor will provide only 50uA at 5V. No likely to be a guaranteed ON through the opto I wouldn't think.
Now consider the nature of the linked current topology. The depletion is 'normally on' with a resistance <100ohm and the source resistor will be on the order of 1k. Before the depletion starts clamping off it's just a 1k resistor and will turn on the opto when a 1k resistor would. This is surely an improvement over 100k I think?
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