Ibam using a Lipo battery to boost the voltage from 3.7V to 10V using a MT3608 module.

Since the MT3608 module always boosts the voltage no matter if I connect the load at 10V. Is there a way where module starts boosting automatically if a load is connected using any transistor?

If there is a possibility using a transistor, where should I connect the transistor and what all connections should be changed?

P.S : I am trying to boost the voltage from 3.7V to 10V to control 3 LEDs (each LED with a forward voltage of 3.3V and forward current of 30mA.?

Circuit Diagram of MT3608 module

  • \$\begingroup\$ The trouble is that if you are not boosting voltage, what do you use as an indicator that you do need to boost the voltage? The three series LEDs sat with around 3.7 volts across them will not draw any appreciable current so that's probably out of the question. Be also aware that in your previous question you stated that the LEDs ran at 3 volt each and the extra 1 volt I assumed you would be using as a current limiter circuit. You can't just connect LEDs across a voltage supply without some form of current limiting device because they may take excessive current and fail. \$\endgroup\$
    – Andy aka
    Jul 30, 2020 at 15:20
  • \$\begingroup\$ If you worry about the life of the battery then don't, because the IC is capable of decreasing the consumption at light (and no-) load conditions. \$\endgroup\$ Jul 30, 2020 at 15:28
  • \$\begingroup\$ I would be using a resistor along with those 3 series LEDs when connecting them with 10V boosted voltage. @Andyaka \$\endgroup\$ Jul 30, 2020 at 15:35

1 Answer 1


You want to regulate LED string current rather than output voltage. So place a resistor at the bottom of your LED string, use an amplifier to gain up the signal, and send that back to the FB pin. That's the general idea. You could just pick a resistor large enough that the LED string current created the regulation feedback voltage, but I suspect the power lost on the sense resistor would be too high. (P = I^2*R). An op amp lets you break the problem in half, pick a resistor for lowest power dissipation, pick an op amp gain to increase that sensed voltage. Good luck.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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