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The transistor as a switch

What will happen if in the example above, RB=100k instead of 1M?

IB=8.4/100k=84μA. You may expect that IC will be 84μA∙300=25.2mA, but that isn't possible since the voltage across RL would be 25.2V which is more than VS. IC,max in this circuit is VS/RL=9/1k=9mA. So even if IB=84μA, IC will be 9mA. IC/IB=107, which is less than hFE. In such a case, when IC/IB < hFE, we say that the transistor has become saturated and can be considered as a closed switch (between C and E).

Take a look at the diagram below.

You see a battery operated clock, operating at 3V. This clock has an alarm function: at a preset time, you hear a chime. Imagine that you don't want to hear a chime, but that you want to switch some other equipment on, for example a radio operating a 9V. This radio has an internal resistance of 100Ω. At alarm time, the output voltage of the clock is 3V. VBE=0.6V. hFE=100

What would be a proper value for RB? A large value may not saturate the transistor; a small resistor may overload the output stage of the alarm circuitry of the clock.

IC,max=9/100=90mA. IC/IB < hFE => IB > IC/hFE => IB > 90mA/100 = 0.9mA. The voltage across RB equals 3-0.6=2.4V. This means RB < 2.4V/0.9mA = 2.7k. To be on the safe side, 2.2k would be a good value. IB will then be 2.4/2k2 = 1.09mA.

Please note that this will only work if the "ground" of the clock (the minus terminal of its battery) is connected to the ground of our little switch (the emitter).