Electronics > Beginners
Sensing (interfacing) high voltage with arduino or any Microcontroller
fourfathom:
--- Quote from: Laszlo on September 10, 2019, 02:27:28 pm ---I know that a single voltage divider would work in this case, but I would like to minimize the stresses on components, so I think multiple levels of attenuation with two voltage dividers+2 zeners will be something I will implement.
--- End quote ---
I don't think that multiple stages really buy you anything. The stress on the components can be calculated quite easily as long as your assumptions are correct. The first step is to check the power dissipation. In the simple divider the 150K resistor (or whatever value you select) will see the bulk of the power. With 150V input, the voltage across this resistor will be about 148V. That's essentially 150V, so the dissipation will be 1502 / 150,000 = 0.15W (simple Ohms Law stuff). A quarter-watt resistor will comfortably dissipate this amount of power, but you could use 1/2W part if you want even more margin or if you think the input voltage will greatly exceed 150V. The power dissipation in the rest of the components is trivial in comparison.
You then look at the current in the optodiode. This is being fed by the resistive voltage divider plus the series resistor, which has an equivalent output impedance of 1K + (150K || 2.7K) = 3.65K. At 150V input, the effective voltage driving this impedance will be 2.65V. Assuming the Optodiode forward voltage is 0.9V, the input current will be (2.65 - 0.9) / 3.65K = 0.48 mA. You could double the input voltage and still be nowhere near stressing the optodiode.
If you were to reverse the 150V polarity the reverse voltage at the optodiode would be -2.65V. Again, no stress on the opto. In fact, that zener or the 1K series resistor really don't add much in the way of protection -- this circuit is already pretty bulletproof.
You do want to consider the high-voltage input, and make sure that the 150K resistor and connections aren't going to leak or flash-over. A common 1/4W 5% resistor will have a voltage rating of around 350V (this is not related to power dissipation). Should be OK.
ledtester:
In this design how necessary is R2?
It seems that R2 is just an additional load on the Zener regulated power supply formed by R1 and D1.
Laszlo:
--- Quote from: mariush on September 10, 2019, 03:03:49 pm ---If you want fewer components, you could look into LR8 linear regulators: http://ww1.microchip.com/downloads/en/DeviceDoc/20005399B.pdf
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Tried it. Also tried the TL783, but the power dissipated was greater than I wanted it to be.
--- Quote from: fourfathom on September 10, 2019, 03:10:58 pm ---I don't think that multiple stages really buy you anything.
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Great answer. I've done the simulations, and you are right, the second voltage divider dissipates next to nothing, and the majority of the power gets "burned" at the first resistor. Built both circuits on a breadboard, and the only difference is the voltage levels in between the different stages at the divider. The first picture is the original, second is what you suggested. Ch1 is opto-led in, Ch2 is measured at the divider.
The other reasons why I was thinking of doing 2 stages of attenuation is to reduce the effect of EMI on the device. For example, the last Zener clamp would do nothing other than dissipating the occasional spikes. Am I wrong in there too?
Also, wouldn't an additional stage effect the ruggedness of the circuit (for example, the first resistor fails short.)?
Cheers,
Laszlo
fourfathom:
--- Quote from: ledtester on September 11, 2019, 12:33:28 pm ---In this design how necessary is R2?
(Attachment Link)
It seems that R2 is just an additional load on the Zener regulated power supply formed by R1 and D1.
--- End quote ---
R1 and r2 form the voltage divider that establishes the drive source for the opto. In normal operation the zener might as well not be there, it only serves to clamp transients that vastly exceed the positive or negative voltage input range. I doubt if you need it. You could also eliminate the series resistor going to the opto,
ledtester:
--- Quote from: fourfathom on September 11, 2019, 03:28:33 pm ---R1 and r2 form the voltage divider that establishes the drive source for the opto.
--- End quote ---
Ok, after thinking about it here's my reasoning on why R2 is desirable -- it effectively raises the trigger voltage of the opto. If we assume the opto's LED doesn't conduct until it has a forward voltage of, say, 1V, then the voltage divider requires the input voltage to be >= 1V * 153/3 = 50V before the opto LED starts to turn on. Without R2 the LED would start to turn on at a much lower voltage.
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