I received some NPN transistors (2N2222). I hope to use them to power on / off some sensors, requiring too much energy to be supplied by the Arduino card (next step!).

But before doing that, I need to understand how use them.

These bugs have 3 pins:

Collector
Base
Emitter

I understood that they can be used to let current pass or block it.
Even better, the power source to switch off / switch on the transistor as not to be the same as the one controlled.

By the way, this french video is great to understand how the BJT transistors work: https://www.youtube.com/watch?v=Zud1Y9S7Jd0

The four transistor operation modes are:

Saturation – The transistor acts like a short circuit. Current freely flows from collector to emitter.
Cut-off – The transistor acts like an open circuit. No current flows from collector to emitter.
Active – The current from collector to emitter is proportional to the current flowing into the base.
Reverse-Active – Like active mode, the current is proportional to the base current, but it flows in reverse. Current flows from emitter to collector (not, exactly, the purpose transistors were designed for).

Today, I want to do something simple, and just use it in saturation mode and cut-off mode.

The current passing through the collector (I_C) is linked to the current passing through the base (I_B), knowing that I_C >> I_B

Each NPN transistor have a "current gain", noted h_FE or β.

h_FE varies with temperature and even between transistors of the same sort

h_FE min/max can be found in the datasheet of the transistor, and some multimeters are able to mesure it.

h_FE links I_C and I_B with the following equation:

I_C = h_FE * I_B

And

I_E = I_B + I_C

I use my multimeter to check the value of the one I want to use for my circuit.
After doing my circuit, I learned that it is a bad practice. In reality, you have to look in your datasheet, and take the lowest value of h_FE for your calculation.

Another thing good to know :

For silicon transistors V_BE ≅ 0.7V

I made a little circuit to use my NPN transistor:

Saturation mode

How compute R_B to be in saturation mode ?

Please note that in reality, I used the same power source for P₁ and P₂

V_P1 = V_P2 = 5V

For my transistor:

35 < h_FE < 300

Here, I want to have

Ic = 18mA

So the sub-question is :

Which current have to pass through I_B to be in saturation mode ?

As I'm dumb, and I use the real value of h_FE for my transistor (instead of h_{FE_min})

h_FE = 290 (should be h_FE = 35)

I_C = 18mA = h_FE * I_B

I_B = I_C / h_FE = 18 * 10^-3 / 290 = 62µA

Now, we can compute the value of V_R :

V_R = V_P2 - V_BE = 5V - 0.7V = 4.3V

And finally, the value of R_B!

R_B = V_R / I_B = 4.3 / (62 * 10^-6) = 69kΩ

As I don't have a 69kΩ resistor, I will use a 64kΩ resistor.

I didn't know when I made my circuit, but it's better to multiply I_B by a factor of 1.5 to be sure that the transistor will be in saturation mode.

I_{B_sat} = I_B * 1.5 = 93µA

R_B = V_R / I_{B_sat} = 4.3 / (93 * 10^-6) = 46kΩ

Cut-off mode

It's way easier :)

As we saw earlier :

I_C = h_FE * I_B

So,

If I_B = 0, I_C = 0

To be in cut-off mode, you just have to turn off the power supply

V_P2 = 0