I'm using a 3.7V LiPo to power a device that normally runs on 3V (and draws about 100-200mA). I've determined that my device can run properly if I can achieve an output voltage that scales the input by a constant factor of about 0.7 (I could also use a diode to simply drop the voltage rather than scale, but I'd like something less affected by current and temperature). Please note that I do not want a constant output voltage, or I’d just use a regulator. I need to track the voltage curve of the LiPo somewhat.

So the desired transfer function is:

Vout = 0.7 * Vin

I think a differential amplifier circuit of the following configuration should work (see schematic), since it gives:

Vout = Vin * (R2/R1)

(according to this article)


simulate this circuit – Schematic created using CircuitLab

To achieve a scale close to 0.7 and have a low parasitic current (since the battery will always be connected), I'd like to use R1 = 1M and R2 = 680k. You can see a simulation here.

I don't know a whole lot about op-amps in practice, so I have a few questions:

  • When applying a load, does that inadvertently affect the output voltage, due to current draw from the output of the op-amp? In simulating, it didn't seem to have an effect, which is nice. I'm avoiding a simple voltage divider for the very reason that the load ends up affecting the desired scaling.

  • Is it practical to use that high of resistances? I typically see ohms in the hundreds, or maybe 10k, but I don't often see circuits using 680k or 1M. Just want to know if that's a smell of some sort. Side note: for a 1000mAh battery, it would take about 30yrs to drain, according to a quick calculation with these resistances.

  • Is it unwise to use an op-amp in a DC application? I typically see examples with AC circuits.

  • Are there any blatant bugs in my circuit design that I might be missing?

  • While I'm not asking for part recommendations, I am considering an LM158 or similar. If there is a good reason to avoid this for my application, letting me know would be much appreciated.

Thanks so much for your help and advice!


I'm using Rload to represent the device being powered. I don't know its exact resistance, but I estimate it to be about 20ohms on average.

  • \$\begingroup\$ How much current do you need from the \$\mathrm{0.7\,\cdot\,3 = 2.1\,V}\$ supply? Do you intended the output to be at the constant \$\mathrm{2.1\,V}\$ or dynamic \$\mathrm{0.7\,V_{in}}\$? \$\endgroup\$ – Unknown123 Apr 11 '19 at 6:35
  • \$\begingroup\$ Vin is the battery, so it will range from 3.0V to 4.2V (3.7V nominal). The output would be 4.2 * 0.7 = 2.94V at its highest and 3V * 0.7V = 2.1V at its lowest. The output voltage being dynamic is desired (otherwise I.d use a regulator). As mentioned I’d need to be able to support up to 100-200mA from the output to support the load (device). \$\endgroup\$ – Andrew M. Apr 11 '19 at 11:59

You've left out a few things for us to answer fully. What is Rload? Is it to power some other part of the circuitry or just to provide a voltage to another IC? What is powering the op-amp?

So my guess here is that you want to measure the LiPo voltage but the measuring circuit has a maximum input voltage of 3V. In this case I would check what that circuit's input impedance or input leakage current is.

A simple voltage divider should do the trick. If R-load is too high for reasonably high divider resistors put an op-amp in follower mode between the voltage divider and the measuring pin.

With your chosen resistors (1M and 680k) the current flowing through the divider would be ~2.2uA so if the measuring pin leakage current is much less than 1/20th (100nA or less) of that current the amplifier isn't needed. If you intend on powering circuitry from the scaled version of the LiPo voltage then the first question is why?

EDIT Based on the OP's clarifications.

I would not use an op-amp to act as a power source. The other issue with your proposed circuit is the powering of the op-amp from the same voltage as the one you are scaling. This means that the op-amp needs to accept an input voltage up to it's positive supply voltage.

I would simply use a diode in series with the LiPo. This would drop ~0.7V, consumes no current on it's own and at 150mA going through it would dissipate ~73mW in heat.

Alternatively, if you can gain access to the circuit that detects the low battery voltage, break the sense pin out. Then you can use a DC-DC converter to power your device and a resistive divider linked to the broken out pin for the supply detection.

  • \$\begingroup\$ Thanks for your answer. I've updated the question with a note on Rload. I'm not actually trying to measure the voltage per se. Rather, I'm powering a device that normally runs on 2xAA batteries. The device has a native low-battery indicator that comes on at about 2.4V (at its input voltage, not the original battery). I'm trying to scale down the LiPo voltage so it will power the device and still allow the device's low-battery indicator to work (which is why I can't use a 3V regulator). \$\endgroup\$ – Andrew M. Apr 11 '19 at 13:13
  • \$\begingroup\$ Also, I don't have any other power source except for the battery. Can I use it to power the op-amp? I'm not shooting for a gain > 1, so maybe it's ok in this case? \$\endgroup\$ – Andrew M. Apr 11 '19 at 13:21

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