# Converting negative DC voltage to positive equivalent

Requesting some input on this topic which has already been discussed (partially for my requirement) here: Negative to Positive Voltage Converter

I also have a positive-only display and I need to measure up to -32v DC. I need to convert the minus voltage to the positive equivalent in a simple manner and I am currently thinking of doing this in TWO op-amp stages as illustrated.

Questions: 1. Is there a more elegant & simple way to do this? Using only one op amp perhaps? The limit on +/- supply is a limiting factor hence my 2-stage design. 2. Does anyone see a problem doing it this way? This is a basic block diagram only.

• Exactly what is your 'positive-only display', and what is producing the -32V? Mar 7, 2018 at 0:55
• Basic block diagram or not - Take the first learning and drop the idea of trying to use an ancient relic of an 741 opamp. They are basic garbage compared to most modern types. Mar 7, 2018 at 1:23
• Fair comment Michael. I will in fact use 2 x LM358s if I go with this current idea. Mar 8, 2018 at 1:33
• Bruce, a link to the meter is here ebay.com.au/itm/…. I'm using 2 of them: one each for the pos & neg rails. The 32 v rails derive from this LM317 PSU: ebay.com.au/itm/… Mar 8, 2018 at 1:44
• One more thing Bruce, these meters have the peculiarity of measuring current in the return 0v rail so I can't just "switch wires around" which would be oh so easy. And they need (+ve) power, too. Mar 8, 2018 at 2:37

If your input ranges from say 0 volts down to -32 volts then you can use a simple inverting op-amp configuration: -

Consider the op-amp chosen will work well with -in and +in close to 0 volts. This is not unusual with rail-to-rail op-amps. Next, as a simple mind-experiment imagine that R1 is 32 x R2. So if Vin is -32 volts then Vout has to be 1 volt to drive -in to 0 volts. I use the word "drive" because this is what happens in an op-amp with negative feedback - the output drives the -in input to match the voltage at +in.

No negative voltages are required to be produced by this op-amp configuration. So, if Vcc is (say) 20 volts, you could make R1 twice as big as R2 and this would convert -32 volts at Vin to +16 volts at Vout. -16 volts at Vin would produce +8 volts at Vout.

There are a few op-amps that work very close to 0 volts but, as always, to widen your scope of choice, it might be better to make a small negative rail (-3 volts is all that is needed) to power the op-amp's negative supply pin: -

• That's very interesting Andy. Perhaps I was wrong in thinking that the rail supply voltage needs to be in excess of the + & - voltages on both the input and output as seems to be the case you've proposed - no? Hence thought it impractical to use (an expensive) opamp with +/- 35v supplies. Mar 8, 2018 at 1:57
• I think I get what you're saying now: the actual voltage on the neg VIN pin will only ever be say 0v or a +ve voltage hence a neg supply rail to it isn't actually needed or as you suggest apply say a tiny neg supply "just in case". Right? Mar 8, 2018 at 2:02
• One more thing Andy, must not lose sight that although your idea deals with converting -32v in to a +ve out, it's not achieving the exact equivalent +ve voltage out for the digital display to actually measure. Mar 8, 2018 at 2:16
• @StevePeck oh, how do you work that out? What mapping of a negative voltage to a positive voltage do you need that cannot be accomplished by the right values of resistors R1 and R2? Maybe you should provide a link to your display or explain that a little more. Mar 8, 2018 at 9:48
• Hey Andy, put simply: I need a positive equivalent of the negative voltage for the display to measure. 100mv tolerance acceptable. How to do this without needing +/- 32v power to the opamp...in 1 stage? LM358 is my preferred choice, ideally. Mar 8, 2018 at 20:11