I am doing a project in which I have a component that outputs a 5 Volt signal. Since the Pi is only able to handle 3.3 V I connected the signal to a voltage divider. I used a 4.7 kiloOhm and a 2.2 kiloOhm resistor.

5V * 4700Ω/(4700Ω+2200Ω)= 3.4 V

Can the Pi handle that current or do I have to reduce it even further?


4 Answers 4


You will still be exceeding the maximum voltage of the GPIO (admittedly not by much).

There is no allowance for tolerances; ±5% for resistors, ±2.5% for external power. Good design always makes allowance for worst case values.

There is also NO NEED; GPIO will be low if <0.8V; high if >2.2V so I suggest you design for 2.3V

The ACTUAL threshold for GPIO is not specified; my tests indicate 1.25V, although there is some variation between individual Pi; 1.3V is probably the threshold, but a higher voltage gives a greater noise margin.

See Electrical Specifications of GPIO

  • Ok. So I changed my setup output 1.6 V by switching both resistors, but if i used two resistors with the same resistance, that should give me 2.5 Volts, which is greater than 2.3 V. Thank you for your help Jan 2, 2018 at 23:20

I believe the Pi will be fine with that arrangement. The voltage (3.4V) is slightly larger than the official 3.3V limit. The excess current will be easily handled by the GPIO protection circuitry (excess voltage 0.1V, resistance 4700 ohms, current 21 microamps).

Remember the resistance values may be +/- 20% anyhow.


The Raspberry Pi's GPIO ports are designed to be used with voltages at most around 3.3V. To be safe, I would reduce your voltage further.


The voltage divider isn't causing a current of any significant, the input resistance (impedance) is 45-65 k ohm. The total current is approx 0.77 mA and a 1/10 will be passing through the internal resistance (77 uA).

So taking in the internal resistance in the calculation the divider will be 1.7v vs 3.3v and that's within the Raspberry Pi's tolerance


simulate this circuit – Schematic created using CircuitLab

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