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This question already has an answer here:

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.

schematic

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?

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marked as duplicate by The Photon, Dmitry Grigoryev, Daniel Grillo, JRE, winny Aug 8 '17 at 7:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    \$\begingroup\$ Because you are using it as an emitter follower - Ve = Vb - 0.7V \$\endgroup\$ – JIm Dearden Jul 30 '17 at 14:20
  • \$\begingroup\$ It's working correctly. That's what this configuration (emitter follower) is meant to do. \$\endgroup\$ – Brian Drummond Jul 30 '17 at 15:46
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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.

enter image description hereYour 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.

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    \$\begingroup\$ (5050 LED x3 + R )/3 from 12V may draw >=150mA (ballpark) so Ib needs 5~10% of this thus Rb= 2.6V/7.5mA = 350 Ohms max. not 3k3.. The 22k is optional. \$\endgroup\$ – Sunnyskyguy EE75 Jul 30 '17 at 15:02
  • \$\begingroup\$ After applying Tony suggestion the circuit started to work (it wasn't working with 3.3k resistor. @Bradman175 if you could update the circuit picture that would be great. Thanks both of you! \$\endgroup\$ – Marek Lewandowski Jul 30 '17 at 15:21
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This is normal, you are using the transistor as an emitter follower.

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

enter image description here

...or a low side switch:enter image description here

(source)

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

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