# Help with Thevenin Equivalent

Discussion in 'Homework Help' started by nb2pc, May 12, 2014.

1. ### nb2pc Thread Starter New Member

May 12, 2014
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This is one of the problems for my circuits 1 review packet.

http://imgur.com/YrfMo9f

I am supposed to find the equivalent Thevenin circuit from the perspective of the $R_c$ resistor. I have deduced that the currents $i_a$ and $i_b$ are both zero. I am not really sure what is to be done to get the answer.

Any help would be appreciated. Thanks.

2. ### WBahn Moderator

Mar 31, 2012
17,455
4,701
Any work would also be appreciated.

How have you deduced that those currents are zero?

3. ### nb2pc Thread Starter New Member

May 12, 2014
7
0
First I used KCL at the middle left node where $i_b$ converges with $\beta i_b$.
Then I performed KCL at the node where $i_a$ and $\beta i_a$ meet with the current through $R_e$.
I also used KCL on the top and bottom nodes, getting everything in terms of $i_b$.
With some substitution, I came to this equation:
$(\beta+1)^2 i_b - \beta i_b (\beta + 2) = i_b$

Since $\beta$ is a fixed parameter, I concluded that $i_b$ must be zero. So $i_a$ is also zero due to KCL at the middle left node.

I just really don't know where to go from here. I tried to find the short circuit current and realized that it must be zero since the two dependent sources are providing no current.
Does this mean that the solution is trivial or did I make a mistake somewhere?

Last edited: May 12, 2014
4. ### WBahn Moderator

Mar 31, 2012
17,455
4,701
And got ... what? I'm not a mind reader.

And got ... what? I'm not a mind reader.

And got ... what? I'm not a mind reader.

And I just really don't know how to help you because the equation you came to is wrong, but since you haven't shown how you got that equation, I have no clue what you might have done wrong. Why? Because I'm not a mind reader.

That you made a mistake somewhere. But I have no way to even guess where since I'm not a mind reader.

Do you notice the theme here?

5. ### nb2pc Thread Starter New Member

May 12, 2014
7
0
Here's more of my work...

KCL around $R_b$ and $R_a$:
$i_a = (\beta + 1) i_b$ (equation 1)

KCL at top node ($i_c$ is current through $R_c$):
$i_c = -\beta (i_b + i_a)$
$i_c = -\beta i_b (\beta + 2)$ (using equation 1)

KCL at $i_a$, $\beta i_a$, and $i_e$:
$i_e = (\beta + 1) i_a$
$i_e = (\beta + 1)^2 i_b$ (using equation 1)

KCL at bottom node:
$i_c + i_e = i_b$

$-\beta i_b (\beta + 2) + (\beta + 1)^2 i_b = i_b$
$0 = i_b (\beta + 1)^2$

That's how I got $i_b$ to be zero.

6. ### WBahn Moderator

Mar 31, 2012
17,455
4,701
Looks good.

You don't indicate what direction i_c is defined in and force me to guess based on your equation. Don't do that -- engineering is not about guessing. Annotate your drawing with the direction if all of your currents.

From your equation, I am assuming that i_c is flowing downward through Rc.

What direction have you defined i_e to be in? Again, don't rely on people making the same assignment that you did.

From your equation, I am assuming that i_e is flowing downward through Rc.

Looks good.

Huh????

Do this one a bit more carefully.

7. ### nb2pc Thread Starter New Member

May 12, 2014
7
0
I evaluated it more carefully this time and realized that it didn't give me any information at all (0=0). So I tried a different approach....

I used KVL in the lower left loop:

$-v_s + i_b R_b + (\beta + 1) i_b R_a + (\beta + 1)^2 i_b R_e = 0$

I used the results of KCL in the two middle nodes to get it all in terms of $i_b$.

Solving for $i_b$:
$i_b = \frac{v_s}{R_b + (\beta + 1) R_a + (\beta + 1)^2 R_e}$

Using KCL results to get $i_a$:
$i_a = \frac{(\beta + 1) v_s}{(R_b + (\beta + 1) R_a + (\beta + 1)^2 R_e}$

Using KCL on the top node ($i_c$ going upwards this time):
$i_c = \frac{\beta (\beta + 2) v_s}{R_b + (\beta + 1) R_a + (\beta + 1)^2 R_e}$

Is my work correct? How can I use this to solve for the equivalent circuit?

Last edited: May 12, 2014
8. ### WBahn Moderator

Mar 31, 2012
17,455
4,701
Correct. The basic problem is that all you really did was confirm that KCL was, indeed, satisfied by all of the various currents and that your equations were self-consistent. The red flag that was staring at you was that none of your equations involved any voltages (such as the input voltage, Vs).

Hopefully you also see the value of showing your work in detail. It allowed me to tell you what was right and where you went off the rails and, from that, you were able to quickly resolve the problem and move on.

Now, consider the following questions:

Q1) Does the current in Rc depend on the value of Rc?

Q2) If not, what kind of circuit produces a current in a resistor that is independent of the resistance?

Q3) Looking at the schematic, does it make sense that the answers to the above two equations are what they are?

9. ### nb2pc Thread Starter New Member

May 12, 2014
7
0
Thanks for your help! I was a long way down the wrong path.

Answer 2) An independent current source would do that. What does this mean for the Thevenin Resistance though? Is it zero?
Answer 3) It is starting to make sense. Would it be wrong to say that it behaves like a voltage controlled current source?

10. ### WBahn Moderator

Mar 31, 2012
17,455
4,701
Correct for the most part. It is, indeed, a voltage-controlled current source. Since the current is completely independent of the load, it is further an ideal current source. Keeping in mind that the output resistance of a current source is in parallel with the source, what is the effective resistance? Hint: it is NOT zero.

It might help to note that ideal voltage sources do not have Norton equivalents and ideal current sources do not have Thevenin equivalents.

FYI: It appears to be the small signal model for a pair of Darlington-connected NPN BJT transistors.

11. ### nb2pc Thread Starter New Member

May 12, 2014
7
0
Would the resistance be infinite? A lower resistance would cause the current through the load to droop as the load resistance increased, right?

12. ### WBahn Moderator

Mar 31, 2012
17,455
4,701
Correct -- and your reasoning is correct as well. Only with an infinite resistance will the current NOT going through the load be zero regardless of load resistance and, hence, all of the current will always go through the load. Of course, this ignores the case of when the load resistance is also infinite, but that is a case of the immovable object and the irresistible force and has no easy answer.

13. ### nb2pc Thread Starter New Member

May 12, 2014
7
0
I have one last question; the prompt for the question requested a Thevenin equivalent w.r.t. R_c. I don't really see how that could be given. Is the question ill-formed?

14. ### WBahn Moderator

Mar 31, 2012
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4,701
I think the question is ill-formed if it only gives you the option of providing a Thevenin equivalent.