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I have a voltage divider circuit producing a reference voltage as follows:

enter image description here

Vref is the control voltage for a regulator, and Vmeasure the output voltage of the regulator, with Vref being regulated to 1.21V.

Now, if RV1 were to fail open or become disconnected (it is connected on a pin header), Vref would fail to 0V. However, I need it to fail to Vmeasure, because otherwise a lot of circuitry may get damaged.

The simple solution would of course be to swap R1 and RV1 and adjust the values, but this cannot be done on this board.

I have tried adding a PNP transistor to bypass RV1 in case the voltage goes too low, but this didn't work as anticipated when RV1 is disconnected:

enter image description here

This results in the voltage being too high when RV1 is connected and no change when connected. The solution cannot be very complicated as it needs to be bodged onto the PCB.

Am I on the right track with this solution or what would you suggest?

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  • \$\begingroup\$ I see no apparent solution with what you've shown us. Show us more of the circuit and what can be "damaged" and we may be able to help. \$\endgroup\$ Jun 19, 2020 at 17:57
  • \$\begingroup\$ Use the simple solution, even if it means cutting traces and adding wires on the board. \$\endgroup\$
    – user16324
    Jun 19, 2020 at 18:04
  • \$\begingroup\$ @SpehroPefhany Vref controls an output voltage that is adjustable and damageable component(s) is what a user connects. This is part of a power supply for a stepper motor driver. \$\endgroup\$
    – namezero
    Jun 19, 2020 at 18:19
  • \$\begingroup\$ @BrianDrummond Ok, so given the situation you would recommend swapping R1 and RV1 yes? \$\endgroup\$
    – namezero
    Jun 19, 2020 at 18:21
  • \$\begingroup\$ @namezero yes... \$\endgroup\$
    – user16324
    Jun 19, 2020 at 18:23

2 Answers 2

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Add a zener diode parallel with RV1 (with cathode connected to Vmeasure and anode connected to Vref). Choose a proper zener with a voltage according to the minimum allowable/safe value of Vref and the value of Vmeasure: Vzener = (Vmeasure - Vref_minimum). In this way, Vref would never fall to zero, instead falls to Vmeasure - Vzener

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You could consider putting a fixed resistor in parallel with the RV1 connections to set a maximum value of the output voltage.

This will likely require increasing RV1 and will involve some nonlinearity, but it might be good enough.

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  • \$\begingroup\$ Mhh to fail to an acceptable level, the bypass resistor would have to be around 1k (to keep the voltage high enough with a 1/2 ratio). But that would preclude the resistance with RV1 connected from ever exceeding 1k, wouldn't it? \$\endgroup\$
    – namezero
    Jun 19, 2020 at 18:46
  • \$\begingroup\$ Well, pots have a tolerance of 10-20% often, so maybe 1.1K or 1.2K 1% would be okay and you could use a 5K or 10K pot. \$\endgroup\$ Jun 19, 2020 at 18:52
  • \$\begingroup\$ Ok, based on your last comment on the OP, I think there's a misunderstanding. Trimming the pot to 0 (or shorting the pins) is acceptable, the regulator will regulate the output voltage to the lowest level then. Which is what I want. In other words, if disconnected or failed, the pot should fail closed, not open. \$\endgroup\$
    – namezero
    Jun 20, 2020 at 5:59
  • \$\begingroup\$ With a 1k-ish resistor in parallel to the pot I could never achieve a Vmeasure/Vref ratio of > 2. \$\endgroup\$
    – namezero
    Jun 20, 2020 at 6:01

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