I have some high voltage (300v+) analog circuits that I want to control digitally which requires the use of a voltage controlled linear resistor that can withstand the high voltages. I don't expect the current levels to be that high.
I originally settled on LDR optocouplers but it turns out they can't handle big voltages so that leaves me with transistor or diode optocouplers.
As far as I understand it, photodiodes are light controlled zeners and phototransistors are light controlled transistors.
Which one should I choose for a high voltage linear variable resistor meant for a voltage divider?
Here is an example of how I would use the variable resistor
Electronic – High voltage voltage controlled linear variable resistor
high voltagephotoresistorvariablevoltage divider
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Best Answer
The linear opto-isolator
The IL300 Linear Optocoupler may be worth examining as a means of providing linear analog coupling with isolation.
Figure 1. IL300 isolated composite amplifier. (Source: datasheet linked above.)
This could be a good start to a solution.
You could have one complete Figure 1 circuit feed the control signal to the HV side and another giving feedback to the LV side, if required.
The variable resistor
Making a voltage controlled resistor to go up towards infinity presents a problem in that our DAC isn't able to output infinite control voltage. If, instead, we control conductance the problem becomes simpler. First, some definitions from Wikipedia:
$$ R = {V\over I}, \qquad G = {I\over V} = \frac{1}{R} $$
Controlling conductance makes this a little easier. A value of zero conductance (control voltage at 0 V) means infinite resistance. We can set 100% control voltage to give any chosen maximum conductance (minimum resistance). In this circuit I will set the minimum resistance to 1 kΩ (= 1 mS). So full range is 0 S to 1 mS (∞ to 1 kΩ).
simulate this circuit – Schematic created using CircuitLab
Figure 2. Programmable conductor.
So, for setpoint = 0, R = 10k / 0 = ∞. For setpoint = 10 V, R = 10k / 10 = 1k. For setpoint = 2 V, R = 10 kΩ, etc.
A separate isolate PSU is shown. This will generate a dual 15 V supply with the common floating at V- potential.
The (almost) full circuit
simulate this circuit
Figure 3. A conductance control circuit. All chips require decoupling capacitors from +Vs and -Vs to PSU common (and are not shown to reduce clutter).
The circuit is based on Analog Devices' AD633, page 10, Figure 16, "Connections for division". The AD633 is a four-quadrant multiplier but when installed in the op-amp feedback loop in this configuration it becomes a four-quadrant divider.
$$ 10 I_R = \frac{1}{10} V_{X1}V_{Y1} = \frac{1}{10} \frac {V_R}{100} V_{OA3} $$.
Solving for the output of OA3,
$$ V_{OA3} = 10k \frac {I_R}{V_R} = \frac {10k}{R} $$
Feeding this value to the inverting input of OA4 completes a feedback loop required for OA4 which drives Q1 to control the resistance of the circuit. Control is achieved by setting the required conductance on OA4's non-inverting input.
The arrangement shown in Figure 1 can be used to control the conductance circuit of Figure 3.
I have not tested either circuit. It might be worth simulation.