Electronic Help

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fastcharlie2

Joined Oct 16, 2014
4
I took the capacitor as an open and did a KVL from the 2V sourece to the 1.2kohm.
-Vb+Vbe+Ie(100+1.2k)=0
Ie=(2-.70/(100+1.2k)=1mA
Ie is approximate to Ic, Ic=1mA
Then a KVL at the output
-10V+1k x Ic+Vout=0
Vout=9V

The only thing I know for AC analysis is re=25mV/Ie=25ohm
 

t_n_k

Joined Mar 6, 2009
5,455
One would need to know the source frequency since there are frequency dependent components in series with the source.
 

t_n_k

Joined Mar 6, 2009
5,455
Are you able to draw the small signal AC equivalent circuit? This is essentially a grounded base configuration. How would you know that it is grounded base?
 
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