# Low power LiPo cell ADC with an nMOS switch and a resistor or capacitor divider circuit?

This is my first question on this platform and am looking for advice on doing a very low power LiPo battery measurement for a wearable device. My current circuit looks like this: (See image below)

Previously, I have implemented resistor dividers of 10Mohm resistors (for MCU that can take only a max. of 3.7V as ADC input which doesn't work with a fully charged 4.2 V LiPo cell), but I didn't like the idea of them continuously draining the battery. So I am adding an nMOS which will be turned on by MCU only when it needs to measure (like every 60s or 60 min). The on resistance on nMOS can be calibrated in the MCU firmware.

However, I am considering further using a low ESR 2.2uF capacitor divider instead of resistor divider here, to minimize current consumed even further when nMOS is ON. Will capacitor divider provide similar ADC results (considering the charge time constant doesn't factor in if I wait 1 second to measure ADC). I couldn't find a similar circuit online and was wondering maybe its so because of a reason. Can you guys advice, suggest or provide feedback on using C vs R divider network for this particular idea?

Any suggestion would be appreciated here.

Thanks! RQ

Edit:

Based on suggestions from @DKNguyen and @BruceAbbott, I have edited my circuit below, realizing only nMOS won't work in either high or low sides. This is because, as @DKNguyen mentioned, that the source of nMOS is floating for nMOS to initially turn ON (Vgs > 0), on the lower side, the ADC pin will pull current directly from the battery (which is why I was using switch in the first place). The new circuit follows the same example solution highlighted by @stevenvh in the following link: Measure Lithium ion battery voltage (thus remaining capacity)

I am using 0.01% resistors for higher accuracy, using enhancement type MOS to minimize energy consumption when they are OFF when not in use. Also, thinking at this point, the cost of this setup and size almost equal, if not more, than cheap and small battery fuel gauges they have out there like MAX17301 which is only \$3.63/unit and 1.7mm x 2.5mm only which I am seriously considering at this point. But its always great to learn more about basics.

Now my question remains, can we still use capacitor divider instead of resistor divider in the new circuit to minimize current consumption even further?

• You have an NMOS on the high-side. Are you actually able to turn it on without a floating gate drive? As it stands, it looks like a source follower to me. – DKNguyen Jan 30 at 23:56
• It will only turn on with an active high signal, which is what I want... – Raza Jan 30 at 23:57
• You don't understand my concern which means you do not realize your circuit has a problem. The gate voltage (implied to be relative to GND) does not control a NMOS. Your NMOS cannot see GND and does not care what GND is. It only cares about the voltage between gate and source. What happens when you try and apply 5V to the gate (relative to ground) to turn it on? Current flows through resistors and the source voltage rises up to Vbat. You get Vgs $\leq$ 0. It's a contradiction. You MOSFET can never turn on all the way. It can only conduct enough to be a source follower, not to be a switch. – DKNguyen Jan 31 at 0:04
• Now it's a different issue. NMOS on low side will never work because when it is open, the ADC will be exposed to the full Vbat. You could use a PMOS on high side, but if your Vbat is too high, have will have the issue where pulling the gate to GND to turn it on will exceeds the maximum Vgs. You will also have trouble turning it off since your IO pin can't drive up to Vbat to turn off the PMOS. But you can drive PMOS gate with an open-drain output pin (or a NMOS/NPN in open drain/collector) and pull-up resistors to fix that. Use zeners to limit Vgs when gate is pulled LO. – DKNguyen Jan 31 at 0:20
• electronics.stackexchange.com/questions/39412/… – Bruce Abbott Jan 31 at 0:45