Why is the collector current at saturation less than the collector current in the forward active mode? Specifically,

Why is \$ I_C < \beta I_B \$ ?

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    \$\begingroup\$ It's not, so this question makes no sense. The collector current is less than the base current times the gain in saturation, which is one way to look at the definition of saturation. \$\endgroup\$ Dec 15, 2012 at 21:35
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    \$\begingroup\$ @TheP: You completely changed the question around with your edit. I don't think a edit of that magnitude is appropriate. You have unilaterally made the assumption that the OP just wrote the question wrong as apposed to having a deeper misunderstanding. Without him clarifying what he is really asking we don't know which it is. To be clear, he originally asked why saturated collector current as more than the base current times the gain. \$\endgroup\$ Dec 15, 2012 at 22:51
  • \$\begingroup\$ I'm with at @Olin on this one. The question now contains an internal contradiction -- the equation stating something orthogonal to the text, not aligned with it, but I like your answer so +1 for you. =P \$\endgroup\$ Dec 15, 2012 at 22:53
  • \$\begingroup\$ @DrFr: Now that you said this, I looked at the equation more closely, which I just glossed over before. It is now actually saying the same thing as the text after Photon's edit. So the original was a contradiction, and now we really don't know which way toiletfreak intended it to be. -1 for a very sloppily written question. \$\endgroup\$ Dec 15, 2012 at 22:59
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    \$\begingroup\$ If OP actually thinks the current increases in saturation, (s)he can always ask again. \$\endgroup\$
    – The Photon
    Dec 15, 2012 at 23:20

1 Answer 1


One way to look at this is, the BJT enters saturation when the supply is no longer able to provide enough current to keep it in forward active operation. For example, let's look at the basic common-emitter amplifier:

enter image description here

Let's say V273 is 5 V and R277 is 500 Ohm. Then if the base current were high enough that the forward active operating condition would give a collector current more than 10 mA, the collector would have to actually be pulled below ground (not indicated on the schematic, but let's say ground is the node connected to the emitter of the BJT).

To do this, the BJT would have to actually be delivering power to the rest of the circuit, which it simply can't do.

Of course, the BJT isn't even so close to ideal that it can pull the collector all the way down to ground, so saturation sets in at some higher collector voltage, which we normally take as about 0.2 V for typical silicon devices.


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