# Practical transistor problem

#### TheNight Silence

Joined Feb 4, 2016
13
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

#### Papabravo

Joined Feb 24, 2006
14,855
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
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.

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

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#### TheNight Silence

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

#### Papabravo

Joined Feb 24, 2006
14,855
U R welcome