Engineering Solve for Vo in Transistor Circuit

AI Thread Summary
The discussion focuses on solving for the output voltage (Vo) in a transistor circuit using Kirchhoff's Voltage Law (KVL). The user initially calculates Vo as 4.7V but is expected to find it as 5V. They identify a mistake in their calculations regarding the direction of voltage drops in the loops, specifically how they handle the 2V and 0.7V values. A correction suggests that the base current (Ib) should be 25μA, leading to a collector current (Ic) of 2.5mA and confirming that Vo equals 5V when recalculated correctly. The conversation emphasizes the importance of accurately applying KVL and considering voltage polarities.
Tekneek
Messages
70
Reaction score
0

Homework Statement


So I am supposed to find Vo in this circuit. Given, Vb-Ve=0.7, ic=100ib, ie=ic+ib

The Attempt at a Solution


In order to solve for Vout i did KVL for the two loops:

Loop 1 (left side, clockwise):

Vb-Ve-ie(416)+2-ib(10*103)=0
0.7-ic(416)-ib(416)+2-ib(10*103)=0
0.7-ib(100)(416)-ib(416)+2-ib(10*103)=0
2.7-52016ib=0
ib=5.19*10-5A

ic=5.19mA
ie=ic+ib=5.2mA
Ve=ie(416)=2.18V (this one i am not sure)

Loop 2 (Right side, anti-clockwise)

10-ic(2*103)-Vo-Ve-ie(416)=0
Subbing in everything i get
Vo=4.7V

But the answer is supposed to be 5 Volts, i am not sure where i made a mistake.

xlwnqr.png
 
Physics news on Phys.org
The 2V and the 0.7V in the left hand loop are in opposite directions, so they should subtract, not add, so you should get 1.3-52016*Ib=0.
 
Last edited:
phyzguy said:
The 2V and the 0.7V in the left hand loop are in opposite directions, so they should subtract, not add, so you should get 1.3-52016*Ib=0.

I still don't get the right answer, this time around i get 7.87
 
You've made another mistake in the right hand loop, but you don't really need that. With what I gave you, you should get Ib = 25μA. Ic = 100* Ib = 2.5 mA. Vout = 10.0 - 2.5mA * 2.0 kΩ = 5.0V
 

Similar threads

Back
Top