For the following question, how do we find \$I_B\$ and \$I_C\$?
Would it be
$$I_C = \frac{V_{in}-V_{BE}}{80k}$$
and
$$I_B = \frac{12-0.12}{5k}$$
0
\$\begingroup\$
\$\endgroup\$
-
1\$\begingroup\$ You have Ib and Ic the swaped. \$\endgroup\$ – RoyC Dec 16 '18 at 10:07
1
\$\begingroup\$
\$\endgroup\$
You have your IB and IC backwards in your calculations. Otherwise, you are on the right track.
