I’m working on a low power sensor and would like to measure its battery voltage.

The battery voltage is too high my ADC input so I have been dividing it with a pair of 1M resistors and a 100n cap so that the ADC’s input capacitance doesn’t load it too much while sampling and affect the results.

The trouble is that the resistor divider drains a few uA all the time and this is far from ideal. Is there a common solution with a low BOM cost to this problem?

I thought of switching an N channel mosfet to the base of the divider but it would allow the ADC input to float too high. Putting a P channel on the +ve rail would be difficult to drive.

Do special ADCs exist that can measure beyond the +ve supply rail?

Any suggestions welcome.

  • \$\begingroup\$ Maybe a voltage divider will solve your problem? \$\endgroup\$ – Harry Svensson Oct 30 '17 at 7:04
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    \$\begingroup\$ @HarrySvensson I think the OP has described his voltage divider! \$\endgroup\$ – Neil_UK Oct 30 '17 at 7:09
  • \$\begingroup\$ @Neil_UK Oh, I didn't see the word "pair", I thought he just used a "voltage divider" by using one 1 MΩ resistor and one 100 nF capacitor. Whoops. \$\endgroup\$ – Harry Svensson Oct 30 '17 at 7:11
  • \$\begingroup\$ what's your system? If it's a CMOS circuit which draws little power, powered from the battery, then a logic '1' output will be very close to the battery voltage, use the '1' to power your divider, set it to '0' output when you don't want the drain. It will cost you use of a logic output, which you might be able to multi-use with other functions, and it only measures the IC rail, not the battery, which of course could be different. Instead of a P-type, use a single gate CMOS device if that difference is important. \$\endgroup\$ – Neil_UK Oct 30 '17 at 7:12
  • \$\begingroup\$ Use a p-fet on the high side of the divider instead of an n-fet at the bottom. You may also need a transistor to drive the gate of the p-fet. \$\endgroup\$ – Dean Franks Oct 30 '17 at 7:33

Use a simple circuit like this. The P-Channel MOSFET can be some other low current device - The one shown is just the one that showed up in the schematic tool.


simulate this circuit – Schematic created using CircuitLab

The idea is to connect the GP_OUT signal to your MCU and have the software drive that high at the time an A/D measurement is to be made. I show using 100K resistors for the divider but these may need to be scaled down depending on the input impedance of the A/D input pin. It's been my experience that 1Meg resistors used for this purpose lead to too much measurement error.

Some delay from setting the GP_OUT high will be needed till the A/D conversion is commanded to allow for settling of the A/D input.

Note that when the NPN transistor is off the only load on the battery is leakage current.

  • \$\begingroup\$ Thanks, I think this will do the trick if I can find a spare GPIO. Why an NPN and not an N channel on the bottom? Cost? \$\endgroup\$ – Michael Oct 31 '17 at 9:00
  • \$\begingroup\$ The NPN solution can be the lowest cost in high volume considerations. You can use an N-Channel MOSFET for the bottom device as long as your GP_OUT has enough voltage swing to accommodate it's threshold voltage. There are some very nifty parts that contain both an N and a P channel MOSFET in the same package and would be minimal size for this application. Take a look at something like the Diodes Inc BSS8402DW. mouser.com/ProductDetail/Diodes-Incorporated/BSS8402DW-7-F/… \$\endgroup\$ – Michael Karas Oct 31 '17 at 12:32

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