Simple NPN Switch

The transistor based switch is probably the "hello world" of electronics. It works based based a few simple "known" facts.

1. Most transistors have a very high impedance between collector and emitter when connected to nothing (except your multimeter of course, which is also a very high impedance load when measuring resistance

2. Most transistors will bring the high impedance between C-E when a small (~0.6 V relative to emitter) is applied to its base

The circuit below was built and tested by me. It was drawn in fritzing

Choosing the values of components

Internet tells me that the current through an LED is 20mA. This may have been true 30 years ago, but today any ordinary LED will light up nice and bright by the time it has about 5mA flowing through it. The typical V_CE of this transistor is about 0.2 volts and the drop across the leads of the LED is about 1.2 volts. So,

I_C = (12 - (0.2 + 1.2)) / 2.2*1000

    = 0.0048 Amps or just 5mA

And I tested this by removing the transistor from the circuit (yes that offsets the 0.2 V C-E drop). Led shines bright no problem

Myth: Typical LED current is 20mA - Busted, it is only 5mA could be less

Now let us focus on the left hand side.

We want a nice 0 (or near zero) to 1 volt swing on the wiper of our pot. That means we want 1 volt or less drop across our pot (1K). That gives us a current of

1V/1000Ohm = 1mA. The remaining 11 volts must drop across our R2. Therefore

R2 = 11v/1mA = 11kOhm. Hard to find, so used 10k. But let us recalculate that current now

which is = 12V / 10k Ohm + 1 kOhm = 0.00109 Amp

Thus making the drop across the pot to be 1.09V. Not too bad. We can still resolve fine voltages across its wiper.

Connecting a multimeter in the base-emitter we see that the LED starts to glow real faint by the time we have about 0.45 volts. By the time we have about 0.6 volts the LED is fully ON and by going any further has no effect.

The total current drawn by the circuit it about 6 mA (0.0048 + 0.00109) which is confirmed by the multimeter.