Is there an easy circuit or IC that will convert negative voltage to positive voltage and retain the initial value? Context below.

I am trying to add a volt meter to my power supply using some old parts I have lying around. My power supply is two 2-12V channels, one positive and one negative. I have an adc that can take anywhere from 0 to 5v. To start, I have a voltage divider on my channel output to my ADC so that my max voltage on both channels is just under 5v. However, I don't think my ADC can take negative voltage. I only have one 7 seg display, so I am using a switch to decide which voltage I am measuring.

On my ADC there is a VCC, +VREF, and a -VREF. I have VCC coming from my rectifier and my VREFs are coming from my regulated output. Initially I was using the switch to swap both VREFs depending on the channel. For example, my positive channel would have +VREF on my regulated positive output, and -VREF would be ground. On my negative channel it would switch so my +VREF was ground and -VREF was my negative output. This would still provide up to a 5v drop to my ADC for measurement. However, it doesn't seem to work that way.

My theory is that the 5V VCC doesn't change, so when I switch to my negative channel, it becomes much more than 5V relative to my VREF. So I want to put a negative to positive voltage converter on my negative output so that the ADC will receive a 5v drop relative to the initial ground. I just have no idea how to do that and all google seems to give me are positive to negative converters. Any help is appreciated! Thanks!


How about a simple inverting amplifier:

$$ Vout = - \frac{Rf}{Ri} Vin $$


simulate this circuit – Schematic created using CircuitLab

Ri at around 10Kohm can be a good starting value. Choose Rf to scale the output voltage such that it is always within the output voltage limit of the opamp and the input limit of the ADC.

  • \$\begingroup\$ Wow this works very well! I have simulated it in ltspice with an LM741 and it seems to work nicely. Thanks a bunch! \$\endgroup\$
    – Gigaxalus
    Nov 26 '15 at 0:09
  • 1
    \$\begingroup\$ LM741 is not a good selection in the circuit as given. It requires more than 10V of supply voltage and output voltage swing cannot get right near the supply rails. I searched for a rail-to-rail opamp that can operate with 10V, the cheapest one I found is TLV271. \$\endgroup\$
    – rioraxe
    Nov 26 '15 at 1:02

I'd use an LTC1043: -

enter image description here

Basically you can sample a voltage that is negative on pin 13 and 0V on pin 7 then, when you toggle the device the charged up capacitor gets rearranged so that the neg terminal of the cap is connected (via pin 14) to measurement 0V and the 0V terminal of the cap is connected (via pin 8) to the input of the ADC.

In the diagram above it might be a good idea to keep the LTC1013 to act as buffer to your ADC.

Another issue is that you'll need to provide +5V and -5V supplies to the chip AND your signals will need to be scaled down to 5V before the differential input.

Do you also see that you can use two LTC1043s to measure +V and -V voltages - operate them inverted - whilst one is presenting a voltage to thE OP-AMP the other can be sampling.

  • \$\begingroup\$ Hmm this might work. A few questions: When you say "toggle the device", is that a physical switch or a pin that needs to be set high? It looks like the cap is connected to pins 11 and 12 but I can't visualize how it is swapped. The 5v and -5v are not a problem since I can get both values from a voltage divider on my rectifier. I am not quite sure what you mean in your last paragraph. \$\endgroup\$
    – Gigaxalus
    Nov 25 '15 at 22:24
  • \$\begingroup\$ There is a pin that shifts the internal capacitor between the two pairs of pins. You can use two of these chips feeding the same instrumentation amplifier to minimize some errors. \$\endgroup\$
    – Andy aka
    Nov 25 '15 at 22:37
  • \$\begingroup\$ I am confused as to how switching the pin would provide voltage for more than a split second. Wouldn't you need to be constantly flipping the pin on and off the switch the cap? \$\endgroup\$
    – Gigaxalus
    Nov 25 '15 at 22:43
  • \$\begingroup\$ The cap across the input terminals of the LT1013 can be removed. This means that when you reposition the sampling cap it gets placed across the input impedance of the chip and this will start to discharge it very very slowly so, yes, you need to regularly toggle the cap between measurement nodes and sampling nodes. Input resistance might be 1 giga ohm so you do the math! It's of no consequence in reality because you'd be switching between positive and negative measurements at least once per second anyway. \$\endgroup\$
    – Andy aka
    Nov 25 '15 at 23:00

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