I am a newbie. I have a power bank which gives 5 v output from a USB. Now I need to step down the 5v into 1.5v to run a small motor. I have searched a lot to find any readymade IC... But unable to find. Can anyone help me, how to convert 5v to 1.5v ?

  1. Easy: use an adjustable 3-terminal voltage regulator and connect the "adjust" pin to ground. The downside to this is that for every watt that goes to the motor, you'll burn up around 2.3 watts in the regulator. This means you'll need to pay attention to heat sinking, and you'll get really low system efficiency (but it'll work).
  2. Medium-hard: Find a step-down converter, either a design to build or a module on ebay. If you find one that's current limited and nice and robust, and choose a current capability that's well above your motor's running current (because motors always consume more when starting than when running), it may work OK. You may have to do some experimenting, because by default step-down regulators don't play nicely with motors. Efficiency will be much better.
  3. Hard: Use a half-bridge driver with PWM, with a duty cycle of around 30%. One downside is that if you're a newbie, you have no clue what I just said. An upside is that you have all the bits you need to make a speed control; you just need to be able to adjust the duty cycle from 0% to 30%

Search for:

  • Step down converter
  • Buck converter
  • Switching converter
  • DC-DC converter

Check that the input can be set to 5 V, and the output can be set to 1.5 V. Some converters have fixed input/output voltages, others can be changed.

A ready made module also contains the necessary resistors/capacitors etc.


You could try something like this:


simulate this circuit – Schematic created using CircuitLab

The voltage divider allows us to have a 1.5v reference (200 kilo ohm resistor), I've chosen high value resistors in order to minimize as much the amount of power being lost. The comparator is there to quickly turn off and on the mosfet as soon as the load voltage increases over 1.5v and decreases to less than 1.5v. This very quick oscillation ensures that the voltage across the motor stays at 1.5v.

Perhaps this approach is a little bit hard to understand provided that you don't have much experience.

  • \$\begingroup\$ Using an NFET for high side switching feels wrong to me, but it does work here since the output voltage is so much lower than the input. Anyway, I would use a 200k resistor, not 201k, since 201k would require a harder-to-find resistor. Really, I'd use 47k and 20k, because high value resistors mean high Johnson-Nyquist noise. Though using a resistor divider like this isn't the best, really; a proper voltage reference would be preferable. Then again, I doubt the motor cares that much. \$\endgroup\$ – Hearth Feb 1 at 18:26
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    \$\begingroup\$ The power wasted in the resistor divider is going to be trivial compared to the power consumed by the MOSFET. \$\endgroup\$ – Elliot Alderson Feb 1 at 18:26
  • \$\begingroup\$ I'll also note that you're talking about the comparator as though it's switching fully on and fully off. This will not happen; comparators are just op amps optimized for something other than amplification. It's just going to work in a linear mode of operation. \$\endgroup\$ – Hearth Feb 1 at 18:31
  • \$\begingroup\$ ...and on top of that you might not be able to turn the NFET on hard enough to supply much current, depending on the threshold voltage of the FET in question. A PFET might be a better choice. A quick simulation (using FETs with 1.5V threshold voltages) shows that I can successfully pull about three times as much current out using a PFET than an NFET in this configuration (with the op amp inputs swapped, of course) \$\endgroup\$ – Hearth Feb 1 at 18:32
  • \$\begingroup\$ @Hearth Thanks for all of what you've said, I was already aware of most of the issues that you have mentioned. The reason why I decided to go with that circuit was due to its simplicity. The voltage across the load is a lot lower than the gate of the NFET therefore I've decided to use an NFET instead of a PFET, this is because they are usually harder to find and more expensive but since you've told me about how much more current you can pull using a PFET I've decided to change my schematic. Thanks for helping! \$\endgroup\$ – rr1303 Feb 1 at 18:56

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