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I have attached the problem and the solution I have for the problem.

Is the solution ok ?

The current through 1 kOhm resistor will be 9.2 mA?

And the remaining current 9.2mA - 2.1 mA = 7.1 mA will come from the base right?

Isn't that too much of current as Ib > 3Ic ?

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  • \$\begingroup\$ The solution is correct. In saturation region, Ic = beta*Ib do not hold anymore. \$\endgroup\$
    – G36
    Commented Jan 27, 2018 at 10:50

1 Answer 1

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The solution is correct. Once \$ V_{ce} \approx 0.2V\$, the transistor is pushed to the saturation region and a maximum current, \$ I_{csat} \$ will flow in the collector circuit. When it is in the saturation region, increase in \$ I_{B} \$ will have no effect on \$I_C \$ anymore. It will only push the transistor deeper into saturation. So \$ I_B \$ can be greater than \$ I_C\$ in such circuits.

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