I am using a TPS61235p to boost a 3.7V 18650 battery to 5.1v with a pure-resitive load of 2Ω. I expect it to draw 2.5A, but upon pulling the enable pin high the output current is limited to about ~1.75A due to the soft-start functionality described on page 10 of the datasheet.

I have confirmed this by increasing the load resistance higher than the Fig 8 Minimum Load Resistance at Startup, and then bypassing my increased resistance, confirming I now have the proper output. The booster never comes out of soft-start unless the load resistance is increased first, then reduced after the booster started switching.

The enable pin is controlled by an Arduino, so now I'm wondering how do I decrease the load resistance once I've enabled the booster.

My first thought was to put a resistor in-line like I am now, and then use a N-channel MOSFET to bypass the resistor after the booster has started. But is there a more robust / effecient way to do this?

Forgive my cramped circuit, I'm still learning.

  • SYS is VCC, battery voltage unless its being charged
  • VOUT goes straight to the load
  • EN and INACT go to the Arduino

Boost circuit

  • \$\begingroup\$ Soft start is basically always desirable. Why exactly is it a problem? \$\endgroup\$
    – Matt Young
    Commented Mar 1, 2017 at 2:15
  • \$\begingroup\$ When using such a low resistance load, the boost converter never comes out of soft-start. Its described on the top of page 11, and they recommend applying a small load or no load at all until the booster starts up. I just don't know the best way to do that yet. \$\endgroup\$
    – Jeff
    Commented Mar 1, 2017 at 2:22
  • \$\begingroup\$ The problem with a PFET is that Vout is 0V to start and it only stays off as long as Vout < Vgsth! Also want to make sure the PFET can handle the power dissipation during soft-start. What did you end up doing? \$\endgroup\$ Commented Jun 2, 2017 at 23:40

1 Answer 1


Costs a GPIO pin, but simple.


simulate this circuit – Schematic created using CircuitLab

Make sure the MOSFET you choose is rated for the 2.5A continuous,has a logic level gate and a low Rds on. Pull the gate low after Vout stabilizes. If that part had a power good output, you could even tie the power good signal to the gate.


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.