Electronic – BJTs be saturated in switching circuits

bjtsaturationtransistors

Assume that I'm going to drive an LED with a bipolar transistor as a switch. According to books, when the LED is on, the transistor should be saturated, i.e. V_CE should be close to zero. But why do authors emphasize on saturation? I mean, in plots of collector current (I_C) as a function of V_CE, we can see that for a wide range of V_CE values, I_C is almost constant.

I can think of two possible motivations for using saturation:

  1. When a BJT is saturated, the calculations are simpler: no need to calculate V_CE and insert it in Kirchhoff's voltage law.
  2. When a BJT is saturated, all voltage provided by power supply can be given to the load (with no V_CE voltage drop)

Best Answer

Reason 2 is the answer.

Since P = VI we have two switch states:

  • Transistor off: I = 0 so P = 0. There is no power dissipated in the transistor.
  • Transistor on: I is high but V is low - typically 0.2 V or so for a BJT. That means that P = 0.2 × I. This may be as low as we can get with a BJT but will be a lot better than V/2 × I/2 which would be the half-way point on an analog arrangement.

Note that with high powered loads that this reduced power can still be high enough that serious looking heatsinks are required.

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