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I try to learn the theory of Transistors and there came up several questions, I can’t find an answer for.

  1. I’ve read that in order to use a transistor as a switch the base current should be high enough to fully saturate the transistor. If the base current is too low it could destroy the transistor. Why does the transistor get destroyed with less current?

  2. There was a sample where they switched a solenoid (5V, 1.1A) with a TIP120 transistor. When I check the datasheet the PC is 2W@TA=25°C, but isn’t 5V*1.1A already way too much for the TIP120? Does it have anything to do with VCE?

I’m sorry for these basic questions but I’m pretty confused right now.

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1) Power dissipation in the active region usually is much higher than in saturation. So most of the power is dissipated during switching time rather than in conduction (saturation) as you can see in the (last) graph below:

enter image description here

(Source Power Electronics Handbook)

This puzzle is best explained by looking at a (simplified) model of a [power] transistor (which ignores things like quasi-saturation) using a so-called (DC) load line:

  • in cutoff the transistor has high Vce (determined by the external circuit) but zero collector current. So power dissipation is zero.

  • in the active region both collector current and Vce can reach high values (moving toward the upper left on the line below during turn-on) so their product (=power) also reaches high values.

  • once saturation is reached, collector current still increases (a bit), but there's now a dramatic dropoff in Vce, so power dissipation falls off dramatically as well. In the more sophisticated model of a power transistor, this droppoff is not as sudden because there's a quasi-saturation region (not shown in the picture below).

enter image description here

2) You're confusing the voltage spec of the solenoid with the Vce of the transistor. At 1.1A you only have about 0.7V Vce in saturation for TIP120; see operating point (red X) I marked on the datasheet graph below. So power dissipated would be around 0.77W; easily dealt with a small heatsink. (Also beware that TIP120 is not a single transistor package, but a Darlington.)

enter image description here

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  • \$\begingroup\$ Thank you so much for your very detailed explanation. Now things are getting much clearer. \$\endgroup\$ – mt85 Oct 7 '15 at 9:52
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I’ve read that in order to use a transistor as a switch the base current should be high enough to fully saturate the transistor. If the base current is too low it could destroy the transistor. Why does the transistor get destroyed with less current?

A transistor can certainly get destroyed if the base current is too low - the transistor may not fully saturate and the voltage across it (collector to emitter) might be several volts (instead of a few hundred milli volts). Given that the current taken thru the collector might be (say) 10A will mean several tens of watts of power being dissipated in the transistor and it destroys. A bigger base current forces a lower saturating voltage for the same collector current and hence a much smaller power dissipation.

There was a sample where they switched a solenoid (5V, 1.1A) with a TIP120 transistor. When I check the datasheet the PC is 2W@TA=25°C, but isn’t 5V*1.1A already way too much for the TIP120? Does it have anything to do with VCE?

5V at 1.1A implies a power of 5.5 watts but, that is dissipated in the solenoid. If the transistor is saturating (to say 0.3 volts) then the power dissipated in the transistor is 330 mW.

Also note that the TIP120 can dissipate 65 watts if the tab (collector) is maintained at a temperature of 25degC.

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