I need to sense the voltage at some node in a circuit ranging from 0-200V DC. The sensing has to be done optically isolated and the input diode requires ~10mA for I_forward in order to operate in the linear current-transfer-ratio region. If I just use a limiting resistor this will make the sensing network consume up to 2W which is outrageous. Is there a more efficient way of doing this? I have thought of using a two stage circuit with a high-resistance voltage divider feeding into an op-amp and then feeding the op-amp's output to the optocoupler. Any other ideas? Simplicity and efficiency are a most! Thanks a lot.|||Yes, whatever you do, you are still going to have to find your 10 mA on the 'live' side of your optocoupler.
If you use an op-amp, you are going to have to provide a psu for that, again on the live side, so you can't take it from the 'safe' side. If you have a low voltage supply available, your idea is probably as good as any.|||What about isolating the 10mA source for your measurments from the source being messured creating an independent entity. Use ohms law to tear off the voltage drop of the intrusive equipment and calibrate your oscope to reflect that difference. that should give a linear image of whats going on. Honestly just geussing not a whole lot to go by|||You could use an op-amp to sense the node voltage. The op-amp has infinite input resistance, so it shouldn't effect the circuit. The output of the op-amp would pulse when the voltage is sensed. Connect the inverting side to your node, the non-inverting to the same reference ground as the circuit being sensed. You might have to attenuate on the gain, so that you don't slam into either supply rail of your op-amp. You'll get a high to low pulse when voltage is sensed above the differential voltage of the inputs, with reference to ground. You can apply caps and resistors to filter out noise and respond to a specific range of input values.
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