Electronic – arduino – Can you lower 12v supply voltage only through series resistor to power an Arduino of 5V

arduinoesp32power supplyresistorsvoltage

The more i learn about electronics the more questions i get XD

Today i learned that i could use a 3v led(20mA) in a 12V power supply by just adding 10k resistor… Which if i understand correctly makes the led power at 3V somehow, cool!

So if we just put aside all voltage regulators(don't have any, but do have alot of resistors :D)… What if i want to power an Arduino board(esp32 specifically) that normally demands 5V and can draw up to 150mA.

So if my calculations are correct(R=V/I -> 80=12/0.150) then this would mean that if i had an 80 ohm resistor in series with the Esp32, it would convert 12V power supply into 5V into the board and nothing would get fried??

The main question for asking this is simply because it sounds a little bit too good to be true XD. But if it is true which would be awesome! The board has a max current draw of 150mA when doing heavy tasks, so when it goes into normal tasks it drops to 40mA current draw. Is this a concern with my supplied 150mA? Does the voltage then maybe starts dropping/rising?

Thanks before hand!

Best Answer

The board has a max current draw of 150mA when doing heavy tasks, so when it goes into normal tasks it drops to 40mA current draw.

That's the problem. You'd size the resistor to give the required voltage drop for the minimum current and just give you the required 5 V. When the current draw increases the voltage would collapse and the circuit would work no longer.

Back to your calculations:

So if my calculations are correct(R = V / I -> 80 = 12 / 0.150) then this would mean that if I had an 80 ohm resistor in series with the ESP32, it would convert 12 V power supply into 5 V into the board and nothing would get fried?

You've made a mistake. The voltage across the resistor would be 12 - 5 = 7 V. R would then be V / I = 7 / 0.15 = 46.7 Ω. Now when the current drops to 40 mA you get V = IR = 0.04 × 46.7 = 1.87 V drop across the resistor giving you > 10 V into your microcontroller. Can you smell the smoke?

The LED, on the other hand, is having its current controlled by the series resistance.

For the microcontroller you need a voltage regulator.