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I have a power supply that regulates its output by turning a potentiometer. It supplies the full 90V to the two leads of a 200k potentiometer. By sending a voltage in-between 12-90V to the wiper input on the power-supply, it will adjust its output. I tried to use a mcp41010 digital potentiometer, however it promptly let the smoke out since you can only apply 0 to VDD across its leads. Is there a way to add additional circuitry to use a digital potentiometer, or is there a better solution to divide 90V using a microcontroller?

Edit: The powersupply I'm trying to dynamically adjust it this one:

enter image description here Found here:

https://www.banggood.com/DC10-60V-30A-1500W-To-12-90V-Boost-Converter-Step-Up-Power-Supply-Module-p-1076169.html

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    \$\begingroup\$ I recommend you reverse engineer the power supply and come up with a schematic to really know how the voltage control works. This is a high power and relatively high output voltage converter. Tinkering with it without knowing what you're really doing can be dangerous. \$\endgroup\$ – joribama Jul 16 at 2:22
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If you have a bit more detail about your feedback circuit (where the pot wiper goes) that would be helpful, as well as the resistance values used for the pot and its load resistor.

I'm thinking an optoisolator like this one: https://www.digikey.com/product-detail/en/on-semiconductor/H11G1SM/H11G1SM-ND/1793958 could be used to modify the voltage divider feedback using low-voltage drive. You'd drive the LED side with a filtered PWM signal to vary its resistance, or you could use a DAC.

If the feedback works the way I think it does, using a voltage-controlled current sink on the low side would provide a way to change the voltage too. You'd use your digital pot to control it, and have the current-sink open-collector or drain go to the main voltage divider.

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  • \$\begingroup\$ If the pot wiper is actively providing regulating feedback (seems highly likely) then you can't just just generate a voltage and feed it to the terminal that used to connect to the pot wiper. I'm interested int his voltage-controlled current sink though, what are you visualizing? \$\endgroup\$ – Kent Altobelli Jul 16 at 1:31
  • \$\begingroup\$ He says the voltage varies between 12 and 60V, and uses a pot to control it. If the feedback is coming from a voltage divider, then I assume that the pot ties between the output and feedback nodes, and a low-side resistor sets the max voltage. So when the pot is zero, output = feedback, so Vf =12V. By replacing the low-side resistor with a low-side current source, you vary the current through the high side to change the Vo to Vf voltage. Less sink = lower voltage, more sink = higher voltage. \$\endgroup\$ – hacktastical Jul 16 at 1:36
  • \$\begingroup\$ The way you're describing it sounds like a low side resistor and only using two terminals of the pot to create a variable resistor for the upper half of the voltage divider. OP said 90V is applied across the pot though, which makes me think the wiper itself is providing the reference voltage that is compared by the "brain" inside. If the feedback from the wiper (probably reduced with voltage divider) is too low it increases the output voltage and visa versa. We need to know the nitty gritty details to understand how to hacktastically hack it. \$\endgroup\$ – Kent Altobelli Jul 16 at 1:53
  • \$\begingroup\$ Yes, kind of need to know the values. If the divider values are low then it would be difficult to pull enough current out of Vfb to move the supply to the max. So the high-side optocoupler approach would work, with the transistor across where the pot is. The more the LED drive, the lower the voltage. \$\endgroup\$ – hacktastical Jul 16 at 2:15
  • \$\begingroup\$ It's a booster supply, so my theory about Vfb being 12V is... all wet. \$\endgroup\$ – hacktastical Jul 16 at 2:49
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All regulators use a linear feedback ratio to a fixed voltage.

You can adjust the ratio on the low side by connecting an NPN small signal transistor to the FB pin of the IC but then the pot has to be adjusted to Vmin and then it may overheat with power dissipation when shunted on the low side to raise the output voltage. IT might work, but more details are needed. The grounds must be shared with the transistor emitter next to the Vfb on IC.

Choose the R Value by trial and error starting from 10k on base to input and 1k on Emitter to ground with collector on Vfb with Pot on a mid scale . Then apply 0.6 to 1V for trial tests and record and plot results. A small cap 0.1uF across Vbe will reduce noise injected.

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