I'm developing a small project that contains one AAA NiMH battery that is charged via USB with a MAX DS2711 IC. The output of the battery is connected to a TPS61006 to boost the ~0.9 - 1.3v cell voltage to a 3.3 usable output. The problem is that while charging, the MAX chip outputs up to 5V into the NiMH battery, which would be too high of an input voltage to the TPS61006. I'd like to be able to add a switch between the battery charger and boost converter to turn off the boost converter while the battery is charging, and just take the 5V down to 3.3V via an LDO to run the rest of the electronics off the input charger power.

The rough schematic: Battery Charger Schematic

I'm finding out, however, that switching voltages down to 1v isn't exactly trivial.

First thought: Use a PNP BJT. The problem is that the CE junction drops too much voltage between the collector and emitter for such small voltages as these. An NiMH battery is fully charged at about 1.3v and completely dead at roughly 0.9v. With a Vce sat of 0.3 or 0.4V, If the battery is full at 1.3v, the input of the boost converter is seeing 0.9v, which is below it's recommended minimum Vbat in. One thing I found was these: I'm not sure what the catch is but the Vce for the PNP looks like 60mV at a 100mV output.

Second thought: Use depletion-mode NMOSs. Unfortunately, finding a surface mount, depletion-mode NMOS, with a low enough Rds(on) is expensive, and these are the only ones that fit the bill, I think. $10 is too much for this little hobby project.

Third thought: As a noob electrical engineer I saw that there are relays that come in small packages such as these normally-closed solid-state relays. However once I got my hands on them to play with, there's a drop across the output because the "on-state resistance" (essentially the Rds(on) stackup, I believe) is 2.5ohms. Too much to use because the voltage drop is too high.

Fourth thought: A coworker recommended just a PFET since the Rds(on) is what dictates the voltage drop across the FET, which is usually milliohms. This is true, but at a 1V source input, the gate threshold voltage is usually negative unless I'm misunderstanding something, or there's a configuration I'm not aware of. So at 1V at the source and 0V at the gate, the PNP is already turned off.

Fifth thought: The same coworker pointed me in the direction of JFETs. However if you just look at the selection of JFETs on Digikey, it seems like there are small SOT packages that handle up to 50mA Id, then it jumps to big TO-247 packages that handle 26A, and nothing really in between.

Other than that, I'm all out of ideas. To boil the post down, I need a switch with the following requirements, and I have yet to find anything that works:

  1. It is normally on.
  2. It can't drop more than, say, 50mV cross its input to output.
  3. It is controlled by a 0 - 5V input.
  4. It has to handle ~500mA of current.
  5. A small SOT SMD package, but not those tiny, lead-less QFN packages.
  6. I don't want to add additional batteries in series. That would make the design too easy, and it wouldn't be fun if it was easy.
  • \$\begingroup\$ How do you intend to turn the device on/off when using battery power (not connected to USB)? \$\endgroup\$ Commented Jun 7, 2020 at 4:45
  • \$\begingroup\$ @BruceAbbott There's a microcontroller on the board. After idle time, it shuts off all the loads, and desserts the enable pin on the boost converter, then goes into a low power mode, waiting for interrupts. There's no physical switches on the board, just push buttons going to the controller. \$\endgroup\$
    – justynnuff
    Commented Jun 7, 2020 at 4:55
  • \$\begingroup\$ So the only real problem is the possibility of over-voltage at the TPS61006 battery input if the cell doesn't hold its voltage below 3.6V? (ie. cell becomes disconnected or goes open circuit internally) \$\endgroup\$ Commented Jun 7, 2020 at 5:30
  • \$\begingroup\$ One more question. How do you intend to regulate the charging current? \$\endgroup\$ Commented Jun 7, 2020 at 5:39
  • \$\begingroup\$ 1. PFET or P Channel Fet - NOT PNP FET. || 2. Thought the 4th - P Channel FET can work. You need a PFET with Vgsth << 1V - ie so 1V negative wrt input on gate has it well enhanced at desired current. Such devices exist - see eg \$\endgroup\$
    – Russell McMahon
    Commented Jun 7, 2020 at 7:42

1 Answer 1


The problem is that while charging, the MAX chip outputs up to 5V into the NiMH battery, which would be too high of an input voltage to the TPS61006.

The DS2711 uses an external PNP transistor to regulate charging current. In your circuit this transistor's Emitter is connected to +5 V USB, which is why the battery voltage could potentially go up to +5 V (if the battery became disconnected or went open circuit internally). So you just need a way to prevent the charging voltage from going over 3.6 V (maximum allowed input voltage of the TPS61006).

The simplest solution is to power the current regulating transistor from a voltage regulator that puts out 3.6 V or less. A 3.3 V regulator would provide ~3.3-1.7 = 1.6 V of 'headroom' between the charging input and battery voltages.

The circuit looks like this:-


simulate this circuit – Schematic created using CircuitLab

At 500 mA (a realistic maximum charging current for a AAA NiMH cell) the 3.3 V regulator would have to dissipate ~(5-3.3)*0.5 = 0.85 W. Without the voltage regulator Q1 would have to dissipate this power, so total dissipation is the same but it is now spread over two devices. The FCX718 has a maximum recommended power dissipation of 1 W on a 15 mm2 PCB area. If used alone it might have to dissipate up to ~(5-1.2)*0.5 = 1.9 W at 500 mA, whereas powered through the 3.3 V regulator it would only need to dissipate a maximum of ~(5-3.3)*0.5 = 1.05 W.

Another potential issue is the charging current sense resistor (R5 in my circuit), which drops up to 0.125 V. You could use a standard 2.5 V Gate drive FET to bypass this resistor in battery mode. The FET would become effective once the booster had started up, permitting higher battery current draw without extra voltage drop across the sense resistor.

The circuit for that could look like this:-


simulate this circuit

Q1 keeps the FET turned off in USB power mode. In battery mode the FET is operating in reverse, with Drain and Source functions swapped.

  • \$\begingroup\$ This is very very good stuff. I don't know why I didn't think to put the LDO as the voltage source to charge the battery itself! This redesign avoids all the headache I've had trying to think of a switch that works down to under a volt. And yes, you caught the voltage drop across Rsns as well. I noticed when I prototyped the circuit on perfboard that voltage being dropped. I thought there was no good way to prevent that, but you offered a solution for even that! Unfortunately, I'm in for a large redesign, but I love it. Thank you for your awesome input! \$\endgroup\$
    – justynnuff
    Commented Jun 9, 2020 at 4:19
  • \$\begingroup\$ Two small things: 1) Would having the voltage source into the battery be 3.3 instead of 5V affect the charge time (the headroom you spoke of)? I can look into that myself. 2) "The TPS61006 uses an external PNP transistor to regulate..." I think you meant the DS2711. Just a small typo, but I know what you're saying. \$\endgroup\$
    – justynnuff
    Commented Jun 9, 2020 at 4:21
  • 1
    \$\begingroup\$ Yes I meant the DS2711 (corrected). It regulates charging current by varying the current supplied to the transistor's Base. This will work if there is enough voltage difference to keep the transistor out of saturation. FCX718 has saturation voltage of ~0.15V at 500mA and current gain of 100, so 1.6V is plenty. \$\endgroup\$ Commented Jun 9, 2020 at 6:21
  • \$\begingroup\$ Not sure if you're still following this section, but as far as selecting which 3.3v source to use when plugged in vs not plugged in, would you recommend an ORing diode setup, just taking the output of the boost converter (I don't think this is efficient) or something else? \$\endgroup\$
    – justynnuff
    Commented Jun 10, 2020 at 6:26
  • \$\begingroup\$ The boost converter already has a diode on its output. The 3.3V USB regulator may have problems with backfeeding, depending on the particular chip used. You might need a diode on the regulator input, to make sure the USB +5V line goes low enough to detect lack of USB power. \$\endgroup\$ Commented Jun 11, 2020 at 4:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.