I'm trying to power my ESP32 which requires an operating voltage of 3.3 V. I have a 5 V, 5 A DC power supply so I set up a voltage divider circuit with a 5 kΩ resistor and a 10 kΩ resistor, therefore Vout = 5 V · (10k / (10k + 5k)) = ~3.33 V.

When measuring the voltage in the middle when just a wire is there, I get a nice reading of 3.2 - 3.4 V which is perfect (see image below).

fig 1

Here I measured the DC voltage at the white wire and put the COM probe to GND. I get the excepted readings.

THE PROBLEM: When I hook up my ESP-32 to where the white wire was, and connect the ground of the ESP to the ground rail directly connected to the PSU, the board only receives a voltage of 0.6 V (I measured the voltage at the 3.3 V & GND pins on the board with my DMM). It's really weird since the voltage at the node should be 3.2 - 3.4 V. I don't know if I'm wiring something wrong, I'm still very new to electronics.

ESP32 Wiring

Additional info:
This board is the ESP32 WROVER CAM Board by Freenove. The PSU in fact does output 5 VDC, 5 A (measured both with a DMM).


3 Answers 3


Voltage dividers aren't good power supplies/voltage sources.

The voltage you want will only be there when no current flows (no load). As soon as you connect a load, current will flow, and voltage will drop across your voltage divider resistors, so the voltage supplied to the load will vary with the load current.

What your ESP32 actually "sees" is a 3.3 V voltage source with a resistor in series (several kΩ in your case).

You should use a voltage regulator.

Also see:

  • \$\begingroup\$ Thanks I didn't know that, could you give me an example of where voltage and current dividers can be useful in electronics? We learn about them so much in Circuit Analysis class, surely they are used somehow? \$\endgroup\$
    – jasonmzx
    Commented Dec 26, 2022 at 0:48
  • 4
    \$\begingroup\$ @jasonmzx that is a totally separate question. But one example would be in sensing a battery voltage much higher than the processor voltage. You can use a divider to reduce a 12 V battery voltage by 6:1 so it is now about 2 V. This 2 V from the divider can go to the ADC input pin of a microprocessor. Inside the microprocessor, code multiplies by 6 to calculate the battery voltage. ADC inputs are usually designed to have very low current consumption so the ADC does not "load" the divider enough to make a difference (usually). \$\endgroup\$
    – user57037
    Commented Dec 26, 2022 at 0:50

If you're using the ESP32 WROVER-CAM board specifically, check to see if yours has a "VCC" pin, as this pin has a 5 V to 3.3 V converter built onto the pin (this is the pin that is also attached to the power from the micro USB slot)

I attached my 5 VDC PSU to this pin and everything is working perfectly!


The ESP32 draws a considerable amount of current, probably at least 5 mA, which would be a load of about 660 Ω at 3.3 V. A 660 Ω resistor across your 10 kΩ resistor will drop the voltage to about 0.6 V as you see.

You could add an NPN emitter follower to get closer to the 3.3 V you want, but a better option would be a 3.3 V LDO regulator.


simulate this circuit – Schematic created using CircuitLab


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