# Thevenin Equivalent of an Amplifier with Negative Feedback

Discussion in 'Homework Help' started by mjakov, Sep 20, 2014.

1. ### mjakov Thread Starter New Member

Feb 13, 2014
20
5
Hi, everybody!

The question here is how to find the Thevenin equivalent circuit left of the load $R_l$.

$
k =10
a) R_2 = \infty
b) R_2 = 50 k \Omega
$

Applying KCL to nodes C and A of the circuit which is open at A-B:
$
\frac{10 - V_d}{10000} + \frac{v_A - v _d}{R_2}
\frac{v_d - v_A}{R_2} + \frac{-k v_d - v_A}{100}
$

Solving these equations we get:
a) $V_{Th} = V_a = - 100 V$
b) $V_{Th} = V_a = - 31.224 V$

To find the Thevenin resistance, the circuit is shorted at A-B and all the voltage sources:
$R_{Th} = (10k \Omega + R_2) \parallel (100 \Omega)$
a) $R_{Th} = 100 \Omega$
b) $R_{Th} = 99.834 \Omega$

According to the textbook, these results are correct except $R_{Th}$ in (b), which is given there as $37.48 \Omega$. So in (a), the circuit provides $- 1A$ current to the load resistor. Using the result from (b) here, the current should be $-0.313 A$, while according to the textbook the current should be $-0.833 A$. So according to the textbook, changing $R_2$ from $\infty$ to $50 k \Omega$ reduces the output current by only $\approx 17%$. Which solution to $R_{Th}$ in (b) is then correct? Should not using the same formula for both Thevenin resistances in (a) and (b) guarantee that if one is correct that the other is too?

2. ### gneill New Member

Feb 7, 2014
9
5
Hi mjakov.

You can't just suppress a controlled source when looking for the Thevenin resistance. Its behavior will influence the resistance perceived from the output. Your method worked for the first case because Vd was a constant (R2 being effectively an open circuit) turning the controlled source into a fixed source.

The classic way to handle this is to slap a test source onto AB and then use the voltage across AB and the current injected by the test source to determine the equivalent resistance.

mjakov likes this.
3. ### MrAl Distinguished Member

Jun 17, 2014
2,696
552
Hi,

Another interesting way is to simply analyze the circuit the way you would any circuit where you wanted to know what the output response was. For this circuit it would just be a regular amplifier circuit, nothing too special. You would get a result in a form like this:
Vout=f(Ein,R1,R2,R3,R4,A)

where f is the transfer function, Ein is the input voltage (10v in your case), R1 is 10k, R2 is R2, R3 is 100, and R4 is the load resistor, and A the gain (10 in your case).

Having that function really tells you everything about that circuit, and it comes from just a typical analysis. So the next step is to take the limit as R4 approaches infinity:

and doing this we end up with another function which does not contain R4 so it is in a form like this:

Next we equate this function F divided by 2 to the previous function:
f(Ein,R1,R2,R3,R4,A)=F(Ein,R1,R2,R3,A)/2

and now we end up with an equation in R4 only. Solving this last equation for R4 we get the Thevenin resistance.

Note that it may seem more complicated than it really is because of the notation, but it's really quite simple when we have all the values given to us beforehand like all the resistor values except the load resistor.

Also, if you dont want to have to find the limit above then just solve the circuit twice: once with R4 in it and once without R4 in it.

mjakov likes this.
4. ### mjakov Thread Starter New Member

Feb 13, 2014
20
5
Hi, the solution adds up for (b) nicely according to this method. Could you please explain why:
$v_{open\ circuit}^{out} = v_{unloaded}^{out} = 2 v_{loaded}^{out}$ ?

This method is relatively fast when all resistors except $R_l$ are numeric values. What method would in your opinion be best suited for solving algebraic equations when there are other unknown resistors in the circuit as well. Is it usual practice to just manipulate the equations intuitively or use some algorithm like Gaussian elimination?

5. ### MrAl Distinguished Member

Jun 17, 2014
2,696
552
Hi,

When you have a linear circuit with an output that has non zero output resistance and if you load it with a resistance Rx and that load causes the voltage to fall to one half the unloaded voltage then the equivalent output resistance must be equal to Rx also.

When there are other unknown resistances then you must solve the equation symbolically as you would any other equation while trying to isolate the resistance of interest (R4 in our case). It may not be that hard to solve really, just using your own judgment.

For example, starting with the function (however you care to obtain this):
VoutA=-(A*E1*R2*R4-E1*R3*R4)/(R1*(A*R4+R4+R3)+R2*(R4+R3)+R3*R4)

and then allowing R4 to go to infinity we get:
VoutB=(E1*R3-A*E1*R2)/(R3+R2+(A+1)*R1)

Now we form the solution equation VoutB=2*VoutA:
(E1*R3-A*E1*R2)/(R3+R2+(A+1)*R1)=-2*(A*E1*R2*R4-E1*R3*R4)/(R1*(A*R4+R4+R3)+R2*(R4+R3)+R3*R4)

and we can quickly see that E1 can be elimiated from both sides leaving:
(R3-A*R2)/(R3+R2+(A+1)*R1)=((2*R3-2*A*R2)*R4)/((R3+R2+(A+1)*R1)*R4+(R2+R1)*R3)

Factoring out the 2 on the right and expanding the denominators:
(R3-A*R2)/(R3+R2+A*R1+R1)=(2*R4*(R3-A*R2))/(R3*R4+R2*R4+A*R1*R4+R1*R4+R2*R3+R1*R3)

we see (R3-A*R2) on both sides so we get rid of that next:
1/(R3+R2+A*R1+R1)=(2*R4)/(R3*R4+R2*R4+A*R1*R4+R1*R4+R2*R3+R1*R3)

and now cross multiplying:
R3*R4+R2*R4+A*R1*R4+R1*R4+R2*R3+R1*R3=2*(R3+R2+A*R1+R1)*R4

subtract the right hand side from both sides:
-2*(R3+R2+A*R1+R1)*R4+R3*R4+R2*R4+A*R1*R4+R1*R4+R2*R3+R1*R3=0

factor the R4 out if possible (collect terms with R4):
(-R3-R2+(-A-1)*R1)*R4+(R2+R1)*R3=0

subtract any terms that do not contain R4:
(-R3-R2+(-A-1)*R1)*R4=-(R2+R1)*R3

multiply by -1 for convenience:
(R3+R2+(A+1)*R1)*R4=(R2+R1)*R3

divide both sides by the factor of R4:
R4=((R2+R1)*R3)/(R3+R2+(A+1)*R1)

Test this solution by inserting the values:
R4=((50000+10000)*100)/(100+50000+(10+1)*10000)

do the math:
R4=60000/1601=37.476577

Compare with the known result.

Last edited: Sep 21, 2014
mjakov likes this.