Circuit to save one battery when another battery is present

Question

I have a battery powered system and I want to preserve the test batteries during setup and calibration. I want the system to use the setup batteries if they are present and use the test batteries if not.

The test batteries are either LiOn (nominally 7.2V) or NiMh (nominally 6.0V) and the setup batteries are NiMh (nominally 6.0V). I'd prefer a simple system, and also one that doesn't necessarily depend on the exact voltages and charge levels (as it's possible the setup batteries could be higher voltage than the test batteries in some cases).

I'm guessing the following PMOS circuit would work, provided that the Vgs(th) is higher than the max difference between Vtest and Vsetup. Getting a PMOS FET with Vgs(th) of about 4V seems about right.

^{simulate this circuit – Schematic created using CircuitLab}

There are a few other SO topics about undervoltage and undervoltage cutoff that use voltage dividers, opamps and PMOS FETs, and another using an SCR, but I'm hoping for something simpler.

It seems like a JFET circuit might work here too (but I've never used a JFET yet).

Will my circuit work? Is there a better way?

Answer 1

The problem with your diagram is that the Test battery will also drive the MOSFET gate high. You'll end up dropping V_GS(th) across the MOSFET. Vsetup and the gate need to be isolated from Vpwr by another diode.

The simplest solution is to use a SPDT relay:

^{simulate this circuit – Schematic created using CircuitLab}

But if you want to create the same function using transistors, you'll need a few more:

^{simulate this circuit}

Note that I've shown the MOSFET body diodes explicitly to emphasize the fact that source and drain on M2 have been swapped.¹

When Vsetup is present, Q1, Q2 and M2 are all on. Q2 keeps M1+M3 off.

When Vsetup is not present, R1 turns M1+M3 on and R3 keeps M2 off. R5 makes sure that any leakage current from Vpwr does not turn on Q1.

---

¹ The first version of this schematic had M1 reversed, too. That's incorrect, because we need its body diode to block Vtest current unless M1 is explicitly turned on.² As it is, there will be a constant drain from Vtest through R1 and R2 as long as Vsetup is connected. Increase their value as needed.

² Actually, it's worse than that. If Vsetup is greater than Vtest, then it will try to charge Vtest through M1's diode. Therefore, we need a third MOSFET M3 to protect against that. Now you can see why specialized ICs have been developed to deal with this kind of application.