I've got an IC that requires 3.3V Vdd and 550uA.

My supply is at 5V.

I'm reading all over the internet that powering it via a voltage divider is not advised and I should get a voltage regulator/buck converter.

Nowhere did I find anything about a dropping resistor. Using this calculator I've calculated that a 3400 ohm resistor would create the required 1.7V drop and limit the current to the appropriate value.

I'm just here for a word of comfort, is my logic correct?

  • 3
    \$\begingroup\$ No, you can't drop it with single resistor if the load is not constant. Just use a regulator. What chip is it? \$\endgroup\$
    – Justme
    Feb 20, 2020 at 17:11
  • \$\begingroup\$ This will have all the same problems a voltage divider would have. \$\endgroup\$
    – Oskar Skog
    Feb 20, 2020 at 17:11
  • \$\begingroup\$ A seriers resistor is just the degenerate form of the voltage divider (lower resistor infinite) and will be worse even than a voltage divider. Use a regulator. \$\endgroup\$ Feb 20, 2020 at 17:27

2 Answers 2


This won't work as it required the current to be constant. In reality it is somewhere between 5nA and 550µA, depending on the state within the IC.

That's what the voltage regulator is for. It permanently adjusts the voltage drop over the regulator that way the output voltage is both independent from the input voltage as well as from the current drawn.


At this current level, if you have 5-20mA to spare, you could regulate it with a 3.3V zener and a resistor of about 160 ohms. The zener goes from 3.3V to ground, and the resistor goes from 5V to 3.3V. If you're running other things off the 3.3V supply, take those currents into account as well. When selecting a diode, make sure that over all current and temperature the zener will keep the supply voltage within the operating range.

Note that zener voltages are specified at different currents, so they're not interchangeable. Go from the curves, not the nominal specs, to find the tolerance on the supply.

  • \$\begingroup\$ Thank you but that's out of my competence. You should be aware given the question that I'm asking xD \$\endgroup\$
    – php_nub_qq
    Feb 20, 2020 at 17:30
  • 1
    \$\begingroup\$ It's almost identical to your proposal, you just change the resistor value and add a zener between 3.3V and ground. Unlike your IC, the zener takes enough current to maintain a fairly constant voltage. \$\endgroup\$ Feb 20, 2020 at 17:37
  • \$\begingroup\$ Note: 3.3V zeners aren't very good voltage references because the voltage changes depending on the current (but they are still better than resistors). In my hobby project I used a trim pot to adjust the current. \$\endgroup\$
    – user253751
    Feb 20, 2020 at 17:55
  • 3
    \$\begingroup\$ @user253751 If you set the zener current 10 times higher than the load current, (of course) the zener voltage will hardly change with load current. In OP's case, this would be 5 mA, wasting only 25 mW. For a mains powered project, this is perfectly acceptable, for projects running for ages on a small battery it isn't. \$\endgroup\$
    – Huisman
    Feb 20, 2020 at 18:44
  • \$\begingroup\$ Low voltage zener (less than ~6V) are very poor devices with a very soft knee. A regulator is a much better solution. \$\endgroup\$ Feb 21, 2020 at 2:43

Your Answer

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

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