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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 enter image description here

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    \$\begingroup\$ At least an architectural drawing would be helpful. \$\endgroup\$ – Peter Smith Feb 13 '16 at 17:45
  • \$\begingroup\$ Does it actually have to be resistive or are you looking for a voltage-controlled current source? \$\endgroup\$ – Oleksandr R. Feb 13 '16 at 17:47
  • \$\begingroup\$ The bottom line is a need a potentiometer. \$\endgroup\$ – coinmaster Feb 13 '16 at 17:56
  • \$\begingroup\$ Normally the anode is '+' and cathode is '-'. Is there a good reason this is reversed? \$\endgroup\$ – Transistor Feb 13 '16 at 18:29
  • \$\begingroup\$ Hmmm..Not sure. \$\endgroup\$ – coinmaster Feb 13 '16 at 18:31
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The linear opto-isolator

The IL300 Linear Optocoupler may be worth examining as a means of providing linear analog coupling with isolation.

IL300 linear opto-amplifier

Figure 1. IL300 isolated composite amplifier. (Source: datasheet linked above.)

The IL300 consists of a high-efficiency AlGaAs LED emitter coupled to two independent PIN photodiodes. The servo photodiode (pins 3, 4) provides a feedback signal which controls the current to the LED emitter (pins 1, 2). This photodiode provides a photocurrent, \$I_{P1}\$, that is directly proportional to the LED’s incident flux. This servo operation linearizes the LED’s output flux and eliminates the LED’s time and temperature. The galvanic isolation between the input and the output is provided by a second PIN photodiode (pins 5, 6) located on the output side of the coupler. The output current, \$I_{P2}\$, from this photodiode accurately tracks the photocurrent generated by the servo photodiode.

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:

The resistance (R) of an object is defined as the ratio of voltage across it (V) to current through it (I), while the conductance (G) is the inverse:

$$ R = {V\over I}, \qquad G = {I\over V} = \frac{1}{R} $$

The SI unit of electrical resistance is the ohm (Ω), while electrical conductance is measured in siemens (S).

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).

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. Programmable conductor.

  • Q is the variable 'resistor'. It will be a high-power, high-voltage device.
  • R1 / R2 form a voltage divider for the \$V_R/100\$ amplifier.
  • \$R_S\$ (shunt) monitors the current through Q. The signal is amplified to give \$-10 I_R\$.
  • The DIV box gain is set to give output \$\frac {10k \cdot I_R}{V_R} = \frac {10k}{R} \$ where R is the total resistance between V+ and V-.
  • All of the above forms a negative feedback circuit for OA which is controlling the resistance of Q set as by \$\frac {10k}{R} setpoint\$.
  • \$R_{SETPOINT}\$ is set by the micro-controller via the IL300 isolated composite amplifier shown in Figure 1.

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

schematic

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.

  • The measure the effective resistance of the circuit between V+ and V- we need to monitor both the voltage and the current. The OP says voltage could be up to 600 V and currents up to 100 mA. The analog circuits will operate between -10 V and +10 V.
  • R1 / R2 voltage divider is buffered by OA1 to give an output of \$ \frac {V_R}{100} \$. This will be 6 V at maximum voltage.
  • \$ R_{SHUNT}\$ is 10 Ω, giving 1 V at 100 mA. It is buffered an inverted by OA2 giving \$-10 I_R \$ volts per amp. This signal is inverted to facilitate the inverting input of the divider circuit (which will cancel the inversion).
  • OA3 and U1 form the divider circuit and is based directly on the application note above. OA3 virtual earth point (inverting input) will be 0 V when \$ W = 10 I_R\$, sourcing the current being sunk by OA2. This will happen when the following equation is true:

$$ 10 I_R = \frac{1}{10} V_{X1}V_{Y1} = \frac{1}{10} \frac {V_R}{100} V_{OA3} $$.

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.

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  • \$\begingroup\$ I'm not seeing how that could be used as a potentiometer in series with a circuit. \$\endgroup\$ – coinmaster Feb 13 '16 at 18:55
  • \$\begingroup\$ It gives you a means of controlling a transistor or FET as a variable resistor in your HV circuit. Am I correct in thinking you only want to control the 200k resistor in your other post? What's the voltage and current at max and min resistance? \$\endgroup\$ – Transistor Feb 13 '16 at 19:03
  • \$\begingroup\$ Max min is probably around zero to 3 Mohms. I know of no transistor or fet that can operate linearly under those conditions. \$\endgroup\$ – coinmaster Feb 13 '16 at 19:10
  • \$\begingroup\$ You'll require an amplifier with feedback to linearise the device. You didn't answer the question, "And what is the voltage and current at max and min resistance?". You really want to control the current in the whole circuit rather than resistance on the left? \$\endgroup\$ – Transistor Feb 13 '16 at 19:16
  • \$\begingroup\$ The current should be almost nothing, at least for the triode load design, I'm not sure about the variable zener design, my circuit analysis skills are still in development. Voltage range is -300v to +600v. \$\endgroup\$ – coinmaster Feb 13 '16 at 19:21
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Replace VR1 with 50kohm fixed, and use the LDR for R2 -- it only has 0.6 V across it.

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  • \$\begingroup\$ Not when I'm using it for a 600v supply. \$\endgroup\$ – coinmaster Feb 13 '16 at 19:48
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    \$\begingroup\$ Your circuit doesn't have 600 V across it -- at most about 7 V if cathode > anode. You may have 600 V between the circuit and GND, but not across the components themselves (else you'd have 16 W in R1). If you don't understand this, I suggest you stop before your lack of understanding of high voltages kills you. \$\endgroup\$ – jp314 Feb 13 '16 at 21:52
  • \$\begingroup\$ LTSpice disagrees. Also the method you stated limits how far down I can drag the voltage. \$\endgroup\$ – coinmaster Feb 13 '16 at 22:07
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    \$\begingroup\$ Then you have made a mistake with LTspice. Look at the voltage divider formed by R1, VR1, R2 and remember that Vbe on T1 is about 0.65V. The highest voltage across the circuit is about (((22k + 100k) / 12k) +1) * 0.65V ~= 7.25V \$\endgroup\$ – Dwayne Reid Feb 14 '16 at 13:44

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