Reverse momentary N.O. Switch behavior using PNP Transistor?

Question

I want to install an led board with a switch in my car. The switch would be pressed down by the lid of the arm rest when closed. I only have N.O. tact switches to use for this, since I need momentary action. I think this would work:

^{simulate this circuit – Schematic created using CircuitLab}

When the switch is pushed in, the lid is closed and the light should be off. RSwitch should keep the 2n3906 off. When it is released, the 2n3906's base is connected only to ground, so the lights should be on. Is this right?

The LED light current is roughly 64mA, 100mA at worst case, so the 2n3906 I have should work fine I would think.

Additionally, how do I calculate the right values for RBase and RSwitch? Is 1kΩ each correct or should they be different? Can Rswitch be omitted, or should it be included for safety reasons?

Answer 1

You need to omit Rswitch so that the transistor base is pulled completely up to the emitter voltage....that's how you turn off a PNP transistor. 1k for Rbase will give you about 12mA of base current which seems like a reasonable choice. Without knowing anything about your LEDs it's hard to say whether the 150 ohm resistors have the right value. You might think about using two transistors, one for each string. You can just connect both bases to the switch and both emitters to 12V. That will reduce the power dissipated in the each transistor although it's probably not necessary.