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I want to modify a DC-DC step-up converter that uses a single potentiometer that sets the voltage to use a single SPDT switch where I can select the output voltage during operation without turning the module off.

I think that I need a resistor in parallel to the switch so when there is switchover there is some resistance and not open circuit to not blow the chip up.

The main chip is SX1308/MT3608.

Vout to be set at 10.6V and 10.8V, because these values are so close I want to use poteniometers to set the voltage manually instead of trying to string resistors. My only limtation are the potentiometers, as I only have 10K and 100K on hand, the R1 and R2 are not a problem if these would be E12 standard values.

I've recreated the PCB (I think it's correct) of the module with my attempt at making what I want to achieve.

Edit: instead of trying to switch from one or the other resistor, have a switch in parallel to the resistor and switch it.

enter image description here

I'm not sure if the circuit is correct, as the original circuit has the 3 pins of potentiometer connected, it connects the wiper to the feedback pin.

enter image description here

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    \$\begingroup\$ You might need to consider what might happen during the transfer of SW1 from one side to the other when neither side is connected. This could be solved by having the higher value pot in circuit all the time and switching the other one into circuit in parallel with the first. \$\endgroup\$ – Transistor Apr 20 at 17:30
  • \$\begingroup\$ Expanding on @Transistor correct point, your switch will have a travel time and a contact bounce time, both typically many milliseconds. You need to look at the spec of that part. Far better is to keep the first resistance (pot' etc) there all the time and switch a higher value second resistance in parallel with a MOSFET. Fast, clean switching. Use a sharp debounced switch level to drive the MOSFET gate. Be mindful of the MOSFET off-state capacitance effect on that circuit. Plenty on the internet on all that. (If I wasn't snowed under, I'd post an answer.) \$\endgroup\$ – TonyM Apr 20 at 17:39
  • \$\begingroup\$ @TonyM I like the idea of having an additional resistor in parallel and switch it instead of switching between 2 resistors. I've updated the circuit, I'm only not sure about the connections of the potentiometers, as the original circuit uses all 3 pins of the potentiometer. I'm not concerned about the contact bounce etc as the voltage difference is small and the voltage is not critical, + 2v is acceptable during the switching. \$\endgroup\$ – Oxmaster Apr 20 at 19:40

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