I have a voltage source I'd like to measure, it is from 0V to 1kV, any value in that range. The accuracy isn't that important, but measuring with an accuracy with 1-5% would be perfect. This has to be then optically isolated, so I could read the voltage on a microcontroller. The second part is not hard at all, when using regular optoisolators. The problem is scaling the voltage down so that I can light up the diode of the optoisolator, and keeping the results semi-linear or semi-logarithmic so I could calculate the output voltage. Is using voltage dividers with a diode the only available option? With such a wide input voltage range getting good results with that method is not possible, because the voltage defines what resistor types and values to use. Any advice?
Electronic – Very wide range high voltage measurement with optical isolation
measurementmicrocontrolleropto-isolatorvoltage
Related Solutions
A couple of thoughts:
- To make an opto's output transistor switch on hard, quickly, you need to put a lot of current into the LED briefly, but I don't know that you'd have to keep it on that hard. I would think that driving the LED briefly with a higher current and then using a lower current to hold it might be helpful; if the time between cutting back on the holding current and switching it off altogether was sufficient, that reduced holding current could help the transistor switch off more quickly.
- It might be helpful to add some circuitry on the output side of your opto. You may have to switch 15mA, but that doesn't mean the opto has to do so. If you will only have to start asserting the output when it's high (meaning you won't have to do things like stretch pulses which originate with other devices) and won't have to hold it too long, you could perhaps use an NFET, a diode, a small cap, and two optos. One opto would connect the gate of the NFET to the drain via diode. The other would connect it to the source. The drain would have a small cap to the source to maintain the gate voltage while the NFET is on. To start asserting the output, one would pulse the opto connecting the gate to the drain. To stop asserting it, one would pulse the opto connecting the gate to the source. When the device is idle, one should drive the latter opto just enough to overcome any possible leakage through the former. There are many possible variations on this circuit; having a "supply" capacitor may be better than having a capacitor on the gate, but putting the capacitor on the gate would probably yield more consistent behavior (since the capacitor would always start discharged when one started to assert the output, and since releasing the output would not require charging the cap).
Depending upon how much voltage drop you can tolerate, other approaches may be workable as well. For example, you might be able to use an opto together with a BJT to form a Darlington arrangement, but use a second opto to accelerate the turn-off behavior. Such an approach would have a much higher voltage drop than the MOSFET-based approach, but would eliminate any need for a capacitor.
I think you missed that optically coupled elements are potential free therefore you can place them in any arm of the corresponding series within the circuit (from vcc to gnd in your case) consructing the logic behavior you want. For buttons (switches) the same is actual.
As you can see at the picture above, it is possible to use the same schematic to control both switch or photo transistor.
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Best Answer
If you're REALLY careful with your choice of parts, you can resistively divide the 1kV input into something you can read directly. It won't be isolated that way, so you'd need to figure out some kind of protection in case a resistor comes loose or you drop a wire across it or something like that. Maybe a sacrificial buffer amp?
If you use an active buffer between the source and the ADC, you can also apply some gain and offset so that the expected range to measure fills up the entire range of the ADC (with some headroom for out-of-range detection). That might be a good idea by itself.