Transistor Heat... ???

[lord loh.]

from
https://www.electro-tech-online.com/threads/improving-freq-counter-input-sensitivity.15672/

As you say, when transistors are fully ON or OFF, they dissipate very little power, but the switching from ON to OFF, and OFF to ON, isn't instantaneous, so during that short period of time they dissipate heat.

Does this mean that transistors dessipate p[ower only when they are in transit from one pont on the load line to another and do not dessipate heat when fixed at one point on the load line ??? :?: :?: :?:

I fail to figure out any explaination for this... If the transistor is on, it ought to let current pass through it. And it has a resistance and shall definitely get hot...

Am I wrong ?

 

[Miles Prower]

I fail to figure out any explaination for this... If the transistor is on, it ought to let current pass through it. And it has a resistance and shall definitely get hot...

Remember: P= IV

If you put a BJT into saturation, the V(CE) drops to a very low level (0.2V nominal for the small signal BJTs; 1.0V for power BJTs). Now, if it's pulling 5.0A through a 30V rail, that would be a dissipation of 5.0W. However, if it were operating in a linear mode, with the same 5.0A, and a V(CE) of 15V, the dissipation would be: 75W, considerably higher (and it had better be able to handle that much). I've built Class D (switches between cutoff and saturation) inverters that could burn up 10W resistors, and yet, the drive transistors would be just barely warm.

On the other end, there is a small leakage current, usually amounting to a few uA for small signal types, to several hunred for power transistors. Again, assuming a leakage current of 250uA, with a V(CE)= 30V, you'll get: 7.5mW of dissipation. Not much to worry about.

That's how you figure it: P(C)= V(CE) X I(C)

 

[Pommie]

I fail to figure out any explaination for this... If the transistor is on, it ought to let current pass through it. And it has a resistance and shall definitely get hot...
Am I wrong ?

No, you are not wrong. It is the amount of power dissipated/heat generated that is confusing you.

When a transistor is fully on, it drops a small voltage across the emitter and collector and a small voltage * a large current is a moderate amount of dissipated power and therefore heat.

When a transistor is off, then no (almost) current flows and therefore no power is dissipated and no heat generated.

If however, a transistor is half on, then a relatively large voltage is dropped across the collector and emitter and a relatively large current flows from collector to emitter and a large voltage * a large current = lots of power dissipated/heat generated.

HTH

Mike.

 

[lord loh.]

Oh....Thanks....It appears all so simple now that you have told me....

I was really bewildered....Thanks....