# Practical transistor problem

Discussion in 'General Electronics Chat' started by TheNight Silence, Apr 3, 2016.

1. ### TheNight Silence Thread Starter New Member

Feb 4, 2016
13
0
Pls I've seen this thread and i don't know how they get the answer

i think all available equations is 3 eqn,one at the input and two at the output
at i/p
Ib=(1-0.7)/120^3=2.5ua
and at the o/p
i assumed the transistor in the active region to apply the rule Ic=B*Ib,
then Ic=80*2.5u=200ua
1st eq
20=It*10^3+Vce
2nd eq
It : is the total current =Ic+Io
Vo=Io*10^3
then how to get the total current since Vce is unknown

2. ### Papabravo Expert

Feb 24, 2006
10,135
1,786
Vce = Vo for this problem. You write the sum of the currents into and out of the node at Vo equals 0 and it drops out. You have one equation in one unknown. The answer is 9V, not 12 as the problem stated.

$-\left (\frac{20 - v_o}{10\text K}\right )\;+\;200 \text \mu A\math\;+\;\left (\frac{v_o}{10\text K}\right )\; =\; 0$

Last edited: Apr 3, 2016
3. ### TheNight Silence Thread Starter New Member

Feb 4, 2016
13
0
oh my god,
it's very simple
Thanks very much

Feb 24, 2006
10,135
1,786
U R welcome