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It is known that in the saturation region of the BJTs the pn junction is reverse biased. The reason for being reversed biased could be explained as follow: Vce=Vcb+Vbe Therefore Vcb=Vce-Vbe We can see that if the Vce<Vbe (saturation region) then Vcb would be negative, which means Vc<Vb and therefore the base-collector junction is forward biased. Since we consider positive direction of collector's current while the base-collector junction is reversed biased, in case of forward-biasing we need to consider the current in opposite direction (negative).

enter image description here

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  • \$\begingroup\$ Are you saying that the graph in your question is wrong? \$\endgroup\$
    – Andy aka
    Sep 13, 2022 at 13:28
  • \$\begingroup\$ You should cite where that graphic comes from. I have problems with the 0.7V saturation region and linear breakdown region. \$\endgroup\$ Sep 13, 2022 at 14:46

1 Answer 1

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The behavior in the deep saturation region is not correctly shown in that diagram.

Here is a simulation for Ib = 10mA, 0 < Vc < 100mV and a 2N4401:

enter image description here

If the collector voltage is less than about 20mV the net current flow is out of the collector.

Note also that at Vc = 0, the majority of the base current flows out of the collector rather than the emitter. This is a hint as to how you can usually identify collector from emitter on an unknown transistor.

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