I'm designing a circuit that uses the power MUX TPS2116. The design is normally powered by a 4.2 V battery, but as soon as the battery is removed it should be powered by a secondary power source, a 3 V battery. The TPS2116 allows configuring at what voltage level the primary source should drop to before the switch should occur, and internally, it uses a comparator against a voltage reference.

My problem is that a voltage divider introduces current consumption, and I want to ideally minimize it down to below 1 µA for battery life. So given 4.2 V and Ohm’s law, it gives me a total resistance of 4.2 MΩ, which to me sounds very large, and might be susceptible to noise etc., but how can I know if this is too large or not?

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  • 3
    \$\begingroup\$ What’s the input impedance or leakage current into PR1? \$\endgroup\$
    – winny
    Commented Jul 5, 2023 at 22:00
  • \$\begingroup\$ @winny PR1 pin leakage is 0.1uA according to datasheet. \$\endgroup\$ Commented Jul 5, 2023 at 22:07
  • 1
    \$\begingroup\$ Can you tolerate 0.42 V offset on you voltage measurement? If not, you need to decrease your resistance. \$\endgroup\$
    – winny
    Commented Jul 5, 2023 at 22:27
  • \$\begingroup\$ Since the datasheet only gives a 'max' rating for PR1 leakage, I'm not sure it's safe to assume that as the 'typical' value and use it to calculate a specific error offset value. PR1 leakage could easily be much less than that too... \$\endgroup\$
    – brhans
    Commented Jul 5, 2023 at 23:28
  • \$\begingroup\$ @brhans, could be, but isn't it best to assume the worst case? \$\endgroup\$ Commented Jul 6, 2023 at 6:11

2 Answers 2


The PR1 pin has a leakage current of up to 0.1 µA, which means that the voltage divider must work for any leakage between 0 and 0.1 µA. In other words, the normal current through the divider should be larger than that.

A current of 1 µA can work, if you ensure that the worst case still results in the desired voltage. If you fear noise, you can put a capacitor in parallel with R2; R1 and C then form a low-pass filter.

The voltage divider is necessary to allow configuring the switch-over threshold. If you want to avoid the voltage divider current, you need to use a different device that determines the threshold in another way. As it happens, the TPS2116 has a variant that just switches to the higher of the two inputs, the intuitively named LM66200. This will work if your primary battery never goes below 3 V, or if you consider 3 V low enough for the switchover. (If the primary battery actually goes down to the secondary's voltage, the LM66200 will regularly switch between them whenever the current one goes down a little, i.e., both will go further down together.)



The PR1 pin has a nominal threshold of 1V ± 8%.

If the switchover point is 4.2V, and you want the divider current to be 1µA then the total divider resistance is 4.2V / 1µA = 4.2M ohms. The individual resistors would be R1 1V / 1µA = 1M ohms, and R2 = 4.2M ohms - R1 = 3.2M ohms.

Per the datasheet PR1 has an input leakage of 0.1µA max. That leakage will cause an error in your switchover threshold. The amount of added error will be the pin current into the divider Thevenin resistance, or \$I_\text{PR1} \frac{R_1 R_2}{R_1 + R_2} = 0.1\,\mu A \frac{(1\,M\Omega)(3.2\,M\Omega)}{(1\,M\Omega)+(3.2\,M\Omega)} = \pm 0.076\,V\$.

If ±0.076V of added error is acceptable, then you can use 3.2M ohms and 1M ohms. If not, reduce the resistor values accordingly.

Note that the ±0.076V is an additional 1.8% error on top of the 8% tolerance in the PR1 pin threshold. If using 1% resistors, then your overall worst-case tolerance will be 8% + 1.8% + 2% = 11.8% (about ±0.5V).

  • \$\begingroup\$ Is it just a matter of voltage offset tolerance? I don't need to consider noise susceptibility of weak resistors? I don't really care how low the voltage goes on primary supply, I simply want it to switch when the primary battery is physically removed. So I basically want the switchover point to be as low as possible, and I think the lowest possible is internal vref. If so, for simplicity, can I make R1 = 0 (remove it) and R2 = 4.2Ohm? \$\endgroup\$ Commented Jul 5, 2023 at 22:43
  • \$\begingroup\$ @user2251965 What is the minimum voltage that your load can tolerate? 1 V sounds much too low. And what is the secondary voltage? (If R1 = 0, you can remove R2.) \$\endgroup\$
    – CL.
    Commented Jul 6, 2023 at 5:24
  • \$\begingroup\$ @CL 2V is the lowest it can tolerate, but primary supply is never going to fall that low except when removed. Secondary voltage is a 3V battery. Why can I remove R2? I was thinking about removing R1, that is, to short PR1 to primary supply, and then let R2 act as a weak pulldown (R2) to ground to just make sure PR1 goes low when primary supply is removed. \$\endgroup\$ Commented Jul 6, 2023 at 6:09
  • \$\begingroup\$ @user2251965 A voltage divider with R1=0 is not a voltage divider. You do not need a pulldown; the load itself will pull down its supply, and if not, it does not matter anyway. \$\endgroup\$
    – CL.
    Commented Jul 6, 2023 at 9:18
  • \$\begingroup\$ @user2251965 If you are concerned with noise affecting the measurement then simply add in a small value ceramic capacitor between PR1 and ground. Even like 1nF should be fine here. Also, unless its happening at least every few minutes, it doesn't seem like the consequences for switching over due to noise are actually that bad. Worst case scenario you draw from your backup for a few microseconds and then switch back to primary power. \$\endgroup\$
    – user4574
    Commented Jul 6, 2023 at 17:09

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