NPN low transistor voltage in emitter despite high voltage collector [duplicate]

Question

I'm having a problem running fairly easy transistor case. I'd like to feed 12V to a short (9 diodes) LED strip using a signal from 3.3v controller. However after reading multiple articles on transistors, I can't wire this up correctly.

In this curcuit I'm using 2N3904 NPN transistor.

So I placed R1 so that there's 1mA current going to collector's base pin.

Collector is wirred directly to 12V and collector has:

• R2 as a pulldown resistor, so that it's grounded
• a short SMD5050 white strip (so that it doesn't exceed 200mA)

My intention here is to simply control 12v led stripe with a low voltage microcontroller.

^{simulate this circuit – Schematic created using CircuitLab}

The problem with this design is that I have only 2.7V on emitter when the button (SW1) is pressed - which looks like as if the base was used as a collector but I have checked the wiring with transistor datasheet multiple times.

I think I'm missing something very obvious, any ideas?

Answer 1

Here's a tip. Try not to put a load on the side of the BJT transistor's arrow unless you're doing something more weird.

Your circuit is not turning on because in order for the transistor to turn on, the current from the base to the emitter must be sufficient. However, in your circuit, the LED strip is in the way so current can't go through therefore it doesn't turn on.

If your LED strip is purely just LEDs and doesn't have a voltage rating but rather a current one, then you will need to include the resistor marked with the asterisk and looks up some LED ohms law calculator.

Answer 2

This is normal, you are using the transistor as an emitter follower.

Get rid of R2. Now, you can use a high-side switch:

...or a low side switch:

(source)

You can use MOSFETs or bipolar depending on current and what you have in stock.