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)
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    \$\begingroup\$ It's also important to know the main reason not to have BJTs saturated in switching circuits: it is much slower to switch off a transistor that has been driven to saturation. So it is an efficiency vs. speed consideration. \$\endgroup\$ – jpa Apr 2 at 12:31
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    \$\begingroup\$ Also worth mentioning (though maybe taking your application a bit too literally) that there are several reasonable BJT LED driver circuits with the transistors operating in the forward-active region. \$\endgroup\$ – W5VO Apr 2 at 16:52

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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    \$\begingroup\$ I guess there is also another reason: when BJT is saturated, I_C is almost independent of I_B. To get a specific I_C from a transistor in "active" mode, a specific value for I_B is needed. \$\endgroup\$ – apadana Apr 2 at 8:01

There is a third reason, related to your reason #2: Because the voltage across the BJT is low, its power dissipation is also low.

This point was driven home for me many years ago when I was building a variable DC power supply1 with current limit. A transistor switched on a lamp2 when the current limit was engaged. When I was testing it, I deliberately overloaded it, the lamp started to light up, and the transistor promptly caught on fire!3 I soon realized that I had failed to saturate the transistor, and its power dissipation was far too high as a result.

1 723-based, which gives you an idea of just how many years we're talking about...

2 28-volt incandescent, not a wimpy LED!

3 I still have the scorch mark on my breadboard socket.


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