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enter image description hereI understand base-collector (current) linear relationship is lost in saturation region and collector current will be lesser than \$ \beta * Ib \$ . Hence,at max, we can call such a region as non-linear region.

Saturation literally means something remaining constant and not increasing further (upon variation of another quantity). Which quantity is getting saturated in so called 'saturation region' of BJT ?

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  • \$\begingroup\$ related: electronics.stackexchange.com/questions/51405 \$\endgroup\$ – The Photon Jun 14 '16 at 5:55
  • \$\begingroup\$ I would like to say you are running out of carriers to conduct any more current, hence saturated, and I would say it's majority carriers, but it's been a long time since school. google "saturation region majority carriers" and see what pops up? \$\endgroup\$ – winny Jun 14 '16 at 7:42
  • \$\begingroup\$ Reading this question and comments might also help: electronics.stackexchange.com/questions/198064/… It is the base region of the BJT which gets saturated with carriers. \$\endgroup\$ – Bimpelrekkie Jun 14 '16 at 8:01
  • \$\begingroup\$ It's just the region where the transistor is "saturated" with current, it's the point at which the transistor will not allow more current to flow through Collector-Emitter junction no matter what load is connected. The current hits a limit and Vce increases instead with an increasing load. The only way out is to apply more base current (which works but only up tto a point) \$\endgroup\$ – Sam Jun 15 '16 at 23:36
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The NPN transistor consists of two diodes the B-E diode and the B-C diode.

Assuming a B-E voltage of 0.7V the B-E diode is forward biased and electrons are injected from the emitter into the base region.

For a collector voltage of, say, 1V the B-C diode is reverse biased, a field exists in the depletion zone that pulls the electrons into the collector. If the voltage at the collector terminal is reduced fewer carriers are removed from the base and for some voltage the collector is unable to remove all the carriers from the base region. The base region is saturated with carriers.

The transistor is said to be saturated.

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Which quantity is getting saturated in so called 'saturation region' of BJT ?

Obviously the collector current. It can be seen very clearly from the output characteristic graph that as you decrease the collector to emitter voltage, the corresponding current increases with reference to the DC load line. When the voltage becomes very less the collector current starts to saturates (not increasing any more for irrespective of the variation in voltage between collector and emitter.)

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  • \$\begingroup\$ Actually, I believe "saturation" refers to the minority carriers in the base region. There are more of them available than can be "used". \$\endgroup\$ – Olin Lathrop Jun 15 '16 at 11:09
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enter image description here

I think this can be explained only after considering the effect of load line. As it is evident from the figure shown, a change in base current causes a negligible change in collector current in the saturation region. Whereas in active region , it causes a linear change.

Hence collector current is the one which gets saturated upon changes in base current.

Learning(sounds silly!): It is a bad idea to understand transistor operation from characteristics itself. Always understand the effect of load line (external circuitry) for any any realistic operation.

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