For the following question, how do we find \$I_B\$ and \$I_C\$?

Would it be

$$I_C = \frac{V_{in}-V_{BE}}{80k}$$


$$I_B = \frac{12-0.12}{5k}$$

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  • 1
    \$\begingroup\$ You have Ib and Ic the swaped. \$\endgroup\$ – RoyC Dec 16 '18 at 10:07

You have your IB and IC backwards in your calculations. Otherwise, you are on the right track.


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