Engineering Simple DC Circuit Analysis with Transistor

[Dan97]

Homework Statement

Homework Equations

The Attempt at a Solution

All loops are clockwise;

KVL Loop around 1V, 120kOhm and V_BE (I₁ is assigned to the loop):

-1V + 120kOhm * I₁ + V_BE = 0
I₁ = I_B = 0.3V / 120kOhm = 2.5 * 10^-6 A;

Loop around 10kOhm, v_CE (I₂ is assigned to the loop);

KVL Loop around 10kOhm, v_o, and 20V (I₃ is assigned to the loop):

10kOhm * I₃ + 20V - v_o = 0
v_o = 10kOhm * (I₂ - I₃)
10kOhm * I₃ +20V + (I₃ - I₂) * 10kOhm = 0 (1);

At the node connecting the collector part of the transistor, I_o and 10kOhm:

I₃ + (I₂ - I₃) + I_C = 0
since I_o = I₂ - I₃

I_C = B * I_B = 80 * 2.5 * 10_-6 = 2 * 10_-4 A

I₂ = -2 * 10_-4 A (2);

Subbing (2) into (1):

10kOhm * I₃ +20V + (I₃ + 2 * 10^-4 A) * 10kOhm = 0

20kOhm * I₃ = - 20V - 2 * 10^-4 A * 10kOhm

I₃ = -22 / 20k A = -1.1 * 10^-3

I₂ - I₃ = 9 * 10 ^-4 A ! = 6 * 10 ^-4 WRONG

And I've been stuck redoing the same equations for a day now... Would be nice if someone could help find the problem in my "solution" so I can finally move on haha.

 

[NascentOxygen]

Hi Dan97.

clockwise Σ of voltages: v_o + 10k × I₃ + –20 = 0

EDITED

 

[Dan97]

Hi @NascentOxygen ,

If loop 3 is clockwise, and it goes through the passive configuration, then it should be +20V - v_o +I₃ * 10kOhm?

I'm not sure what you're trying to say...

 

[cnh1995]

Hi @NascentOxygen ,

If loop 3 is clockwise, and it goes through the passive configuration, then it should be +20V - v_o +I₃ * 10kOhm?

I'm not sure what you're trying to say...

The given values of Vo and Io do not follow Ohm's law.
I am also getting Io=9*10^-4 A.

 

[Dan97]

Hi @cnh1995 ,

Could you show your working (if it's not too much of a hassle)? :)

 

[NascentOxygen]

Hi @NascentOxygen ,

If loop 3 is clockwise, and it goes through the passive configuration, then it should be +20V - v_o +I₃ * 10kOhm?

I'm not sure what you're trying to say...

Sorry, typo. Corrected.

Following your path, it will be +20V - v_o – I₃ * 10k = 0

Current through a resistor (in fact, through any element that's not a source) flows from its more-positive end to its less-positive end. If you draw an arrow showing this PD across the resistor, the arrow's tail is at the less-positive end and the arrow head is at the more-positive end.

 

[Dan97]

@NascentOxygen ,

Why would it be - I₃ * 10kOhm? Isn't that breaking Ohm's Law?

 

[cnh1995]

The given answers are 12V and 600uA, which are incorrect. They don't follow Ohm's law.

Hi @cnh1995 ,

Could you show your working (if it's not too much of a hassle)? :)

Collector current I_c=2x10^-4A.

Using KVL you can write,
10⁴*I_o=20-10⁴*(I_o+2*10^-4).

Solving this for I_o, we get I_o=9×10^-4 A.

 

[Dan97]

@cnh1995 ,

Oh actually I've made a big fuss for nothing! I've just look at the author's errata and the answer is indeed 9 * 10 ^-4!

 

[e0ne199]

what book is this anyway?

 

[Dan97]

what book is this anyway?

Alexander's Fundamentals of Electric Circuits