Measuring battery voltage without consuming current

[MikeK]

I was pretty pleased with the performance of my constant current dummy load battery tester.  I used it a couple weeks ago to test some Chinese camera batteries (cells).  The battery voltage was monitored at a PIC ADC pin.

Now I'd like to test another battery, but this one is 7.2V.  So I need to divide the voltage into what the PIC can handle (less than 5V).  This got me to wondering...the voltage divider will consume current and thus affect the results.

Can I measure voltage without consuming too much current?  And if not, at what point do the resistors in the voltage divider cause problems?  Can I just use two 100k resistors and grab half the voltage, and ignore the 36uA?

[hlavac]

You could buffer it with low bias current opamp

[MikeK]

I'd still need to use a voltage divider at the op-amp input, no?  Unless I run the op-amp from a higher supply.

[T4P]

the key is low bias ... voltage divider or not, you can get away with low bias ...

[Psi]

I've used a 100-200K voltage divider on an AVR ADC input before.
If you only want 8bits resolution it works fine.
The ADC is really intended for a 10K load but 100K doesn't seem to cause any issues for 8bit sampling.
You can also clock the ADC slowly so the sampling logic has time to charge up the internal cap through the 200K resistance.

200K would consume 36 micro amps at 7.2V

Another option could be to put some diodes in series with the 7.2V to bring it under 5V.
If you know the Vdrop on the diodes is reliably X.X volts then you can add it to the voltage reading you sample on the ADC.

It means you cannot read voltages below X but for a 7.2V battery you only need to drop say 3V and if your 7.2V battery is at 3V it's 100% flat anyway so there's no real need to know the exact voltage.

[digsys]

It's a catch 22. The lower the sense current, the worse the linearity you create. The ADC will have an
effective impedance, not in the real sense, but bad enough. This may be 100K or 1M. So if you have a
100K voltage divider, you'll de-linearize the sense voltage range. If you don't have a choice, then you'll
need to make up a LookUp table (or maths function) to correct the errors.
If you use a BUFFER op-amp, as suggested, that'll fix that problem, but now you have an Input offset
to worry about (still quite doable). BUT -
Unless you REALLY need to see the unloaded CELL voltage, then having a 0.mA, 1.0mA or GREATER
load is no issue. You're only applying it during the "read" cycle anyway. And even left on, it's a TINY
percentage of the CELLs energy storage. If you need the know the total energy precisely, then just add
the ~1mA in firmware/software. When testing batteries, designers often have an output to drive a
programmable (set) load, to see how the CELL behaves under load, as the CELL will have an
internal leakage current ANYWAY. For NiMH (non LSD) this can be quite (relatively) high !!

[MikeK]

Thanks for pointing that out.  I completely missed that the internal cap has to charge through the top voltage divider resistance.

Thanks all.  I think I will go for the op-amp buffer.  Especially since I've found that when I shut off the PIC power supply, the battery being tested powers the circuit through the ADC input.

[Psi]

It's a pretty tiny cap. but yeah, if you stuck a 10Meg in series its not going to work properly.

The best thing is to try it and see. I found 200K didn't really cause an issue for my application.

[MikeK]

Even if I go with 10k divider resistors, we're only talking 0.3mA.  I'm testing the battery at 50mA and 100mA, so it's not much to be worried about.

Are op-amps stable at <1 gain?  I'd rather have the division through the op-amp so that shutting off the power doesn't drain the battery through the resistor divider.

[george graves]

Attache a voltage divider to the positive supply rail you want to measure, the center tap to a ADC input on the micro, then attach the "ground" end of the divider to a I/O pin.  Keep the I/O pin as in input(or an output pulled high) until you want to measure the voltage, then switch it to a output, and pull it low to ground. Take a measurement, then switch it back.  Bingo.

[jimmc]

The current pulses caused by the ADC caps have a very low duty cycle and so a relatively large capacitor (say 10nF or more) across the ADC input will average these pulses to a very low value and allow the use of large value resistors in the divider.
Of course this will limit the frequency response but that is not an issue when measuring a battery discharge voltage.

Jim