How Do You Calculate Thevenin Equivalent Circuits?

[orangeincup]

Homework Statement

Determine the Thevenin equivalent with respect to the terminals a,b

Homework Equations

Rth=Voc/Isc

The Attempt at a Solution

I'm doing this practice problem and I calculated the the short circuit current below:

500uA*100ohm/(100ohm+1310ohm)= 35.460 uA

since 80*ib=current in v2, 35.460 A * 80=2836.87 uA

So Isc=2836.87

Now for the open circuit voltage..
80ib* 50kohm= 40.0*10^5 ib
Now to solve for ib is where I'm lost.
In the solution they have 4*10^-5v2=-160ib, but I have no idea where it came from?

 

[NascentOxygen]

A fraction of v₂ is fed back to influence the input. That fraction is given as 4x10^-5v₂, so substitute in this your value for v₂ (which you have in terms of i_b).

 

[orangeincup]

Thank you. I have a question about calcuating Isc vs. Voc, why isn't Voc simply Isc/50kohms?

I know the circuit is suppose to be "open" but isn't it still the same thing?

 

[NascentOxygen]

Thank you. I have a question about calcuating Isc vs. Voc, why isn't Voc simply Isc/50kohms?

It would be if the Thèvenin resistance were 50kΩ, but it isn't. That's what you are trying to find! The feedback changes the output impedance. And it changes the input impedance, and it changes the gain.

 

[orangeincup]

So is the Vth or Voc I'm calculating simply the voltage across the 80ib dependent current source?

In the open circuit voltage I thought the 50kohm resistor would still be part of the circuit, and the voltage across that would equal the voltage across a-b?

 

[NascentOxygen]

So is the Vth or Voc I'm calculating simply the voltage across the 80ib dependent current source?

Yes, but the 50k remains in place, it's part of the model. (Without some resistive load, the voltage across a current source would be infinite.)

In the open circuit voltage I thought the 50kohm resistor would still be part of the circuit, and the voltage across that would equal the voltage across a-b?

Yes, and yes.