I have a 12v led strip and have tested the current draw at ~500mA per colour.

I want to drive this via my Wemos GPIOs which are 3.3V @ 12mA max.

For testing I have been using 2N2222 transistors with 330Ω resistors and it seems to work well, in that the current draw is similar as via 12V directly. Ideally I should aim for 12mA base current right, so 220Ω resistor?

Although measuring the base current only gives 4mA which is less than I would have expected given (3.3 - 0.7)/330 = ~8mA, any explanation for this?

Also with a transistor gain of 500/8 = 62.5 (or 500/4 = 125?) I think the transistor is not fully switched on, and it does start to get hot quite quickly so this is probably true.

I suspect that gain is too high so I need to cascade 2 transistors to increase the gain in two steps. Like 5mA -> 50mA -> 500mA. So it would be something like GPIO 3.3V via 510Ω to give (3.3 - 0.7)/510 = 5mA, then 12V via 220Ω to give (12 - 0.7)/220 = 51mA

I also have some IRL540N mosfets which I tried but these don't fully switch on at 3.3v so would need a transistor to switch these anyway.

edit: Looking back at my original order for these transistors and these are what I bought https://www.aliexpress.com/item/2N2222A-Free-shipping-100pcs-in-line-triode-transistor-NPN-switching-transistors-TO-92-0-6A-30V/32816748799.html they say 2N2222A on the packaging but are not the old style metal case ones, so not entirely sure what they are now MSP2222 maybe?

This is what I had in mind for the circuit

  • 1
    \$\begingroup\$ At half voltage and half current --- 6v and 0.25 amp---- any 2N2222 wiii burn up. Unless you use duty cycle to modulate the average intensity. \$\endgroup\$ – analogsystemsrf Mar 25 at 3:43
  • \$\begingroup\$ What do you mean by half voltage, half current, half of what? \$\endgroup\$ – trapper Mar 25 at 4:18
  • \$\begingroup\$ For a switch (or linear element, such as bipolar wires as linear amplifier) in series with a resistive load (we'll assume the LEDs behave like this), the maximum power dissipation is at half power and half voltage. \$\endgroup\$ – analogsystemsrf Mar 25 at 10:49
  • \$\begingroup\$ I don't understand where you got 6V & 0.25A from though, half of what exactly? \$\endgroup\$ – trapper Mar 25 at 10:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.