# Output impedance of an active circuit

Discussion in 'Homework Help' started by atagar, Oct 21, 2009.

1. ### atagar Thread Starter New Member

Oct 21, 2009
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Hey everybody, I have an homework due tomorow morning, and I keep arguing with my friends at school about this question. With the circuit below, the question is "What is the output impedance value".

1)So my point of view is that since this is a ideal amp-op, we can eliminate the part on the Vin side, since we know that there are no current going out of the V- (the negative of the amp-op). That way, we can simplify the circuit as if the 8R was connected to a source.

When simplifying, we kill the source, then we connect the 8R directly to ground, and we calculate the equivalent resistance at the end.

2) The other way of thinking is that, by definition, the V+ and V- are equal (ideal amp-op) and that way, when killing the Vin, everything on the left goes to 0V, and that the V- is also at 0V and it also short out the R beneath the 11R. And then we proceed the same way as in the first method.

Answer : 609/259R (not sure... I did it in my head but you get the point...)

So which on is good ? I'm pretty sure we are missing something in one of the case that make it impossible. Thank you! have fun File size:
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2. ### steveb Senior Member

Jul 3, 2008
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469
Neither method is correct. There are at least three ways to approach this.

1. Use equations to calculate output resistance Rout as Vout/Iout, by driving the output with a current source Iout.

2. Calculate output conductance Gout as Iout/Vout, by driving the output with a voltage source Vout. Then Rout=1/Gout

3. The simpler way is to realize that an ideal opamp makes the non-inverting amplifier stage look like it has zero output impedance. Then calulate the effect of the remaining resistors.

Note that the output of the noninverting amplifier is not at the output of the opamp itself. It's important to identify the output is at the node where the three resistors connect together.

Last edited: Oct 21, 2009
3. ### atagar Thread Starter New Member

Oct 21, 2009
8
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I'm having trouble to understand why the 8R will not be part of the impedance; as I see it, the current that come from the op-amp will go through that resistor and there will be a voltage drop right?

So from what you are saying, I should take the equivalent of the four resistance to the right? (I just ignore the 8R, and suppose the op-amp is not connected to the two R at the left)

4. ### steveb Senior Member

Jul 3, 2008
2,432
469
The best thing to do is derive the equations for output impedance of simple opamp circuits. Then you will see why the 8R is not a factor. Basically, the feedback completely masks the resistor value.

Not four resistance, but only two are relevant. The resistance looking back into the non-inverting opamp stage includes the feedback resistors. Hence they do not matter either.

It's a difficult concept at first, so sometimes it's better to brute force it with equations until you are comfortable.

5. ### atagar Thread Starter New Member

Oct 21, 2009
8
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Thank you, I will see what I can do. I will try with the equations method.

6. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,666
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steveb is giving you good advice; I'll try to motivate you a little further.

If you have gotten this far in your studies, you should have learned several methods for solving networks. In particular, do you know how to formulate the nodal equations for a network?

If you designate the minus input of the opamp as V1, the junction of the 8R, 11R and 6R resistors as V2, and the output as V3, and the gain of the opamp as A, can you write three equations satisfying Kirchoff's current law for those three nodes, with a current source of 1 amp connected to V3?

To get you going, the equation at V3 would be:

(V3-V2)/6R + V3/3R = 1

7. ### atagar Thread Starter New Member

Oct 21, 2009
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I do know how to use the nodal equation, however I was stuck on the fact that I have no idea what is Iout. But I will try with your suggestion of putting a 1 amp source at Vout and trying to get the Vout. If I get Vout, I will be able to do Vout/1amp = R = my output impedance right?

8. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,666
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That is exactly right.

As a further hint, I would leave that opamp gain as an explicit variable, A. Then, when you get your expression for Vout, it will involve A, which you can then allow to increase toward ∞.

Just let the voltage at the opamp output be -A*V1 and it should all work out.

Let us know what you get.

9. ### atagar Thread Starter New Member

Oct 21, 2009
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ok well I did what you told me, and it gives me that V3 = 1. So my output impedance = 1? that is strange!

10. ### atagar Thread Starter New Member

Oct 21, 2009
8
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I'm actually getting 6...

Last edited: Oct 21, 2009
11. ### The Electrician AAC Fanatic!

Oct 9, 2007
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When you get your final answer, show your equations so if there's a mistake, we can help you find it.

12. ### atagar Thread Starter New Member

Oct 21, 2009
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Ok here's my 3 equations : 1) V3+V2 = 6*R
2) 3*(-A*V1-V2) + 2*V2 = 4*(V2-V3)
3) 12*V1 = V2

so V3 = 6R and it doesn't make any sense!

My friend just made me notice that since we calculated the transfer function for a previous question and also the input impedance, we can use those two values to get the output impedance.

Input impedance : 2R
transfer function : 2

so we have 2R = Vin
2R+ x = 2Vin
x = 2R.

Is that right?

Last edited: Oct 21, 2009
13. ### steveb Senior Member

Jul 3, 2008
2,432
469
That is correct.

Deriving the equations directly is not too bad if you assume a perfect OPAMP.

The following equations are relevant. Note that the input voltage becomes a source of zero since we are looking for the current response only due to the hypothetical current source that we place at the output node.

Vout=I1*3R where I1 is the current in the 3R resistor
I1=Iout-I2 where I2 is the current in the 6R resistor
I2=(Vout-Vn)/6R where Vn is the node where 3 resistors connect together
Vn=0 since the input source is shorted for this calculation

Do the algebra and you get Vout/Iout=2R

You get the same answer if you use the simple method I suggested earlier. If you already know that an ideal opamp provides zero output resistance, then you just short the node voltage Vn to ground directly, and then it is obvious that you simply have a 3R resistor in parallel with a 6R resistor.

The derivation is more cumbersome if you do the calculation for a nonideal opamp, as The Electrician suggested. But, this is very good practice, and yields information critical to understanding practical limitations of opamp circuits, particularly if you note that compensated opamps have gain that rolls off at high frequency.

Last edited: Oct 21, 2009
14. ### atagar Thread Starter New Member

Oct 21, 2009
8
0
I finally did it, and I worked. Thank you guys, I you will probably see me around.

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Interesting problem.....

Taking up Electrician's challenge of the non-ideal amp case ....

If the gain is a finite value Av and input resistances are still infinite.

I have

Rout=2R{1-[1/(1+3Av/92)]}/{1-[1/(1+Av/36)]}

For Av=100 then Rout=2.0816R ohms

For Av=1000 then Rout=2.0103R ohms

It seems Rout is fairly insensitive to Av, given Rout=2R for Av=∞
(Av for an op-amp would typically be of the order of tens of thousands at least.)

Any confirmation on this outcome?

16. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,666
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Sorry I had to be gone for about half the day.

I don't think you can determine the output impedance from just the input impedance and the transfer function.

For example, change the 3R resistor to 4R, and change the 11R resistor to 9R. Then the transfer function remains unchanged, but the output impedance is different.

So, it's true that the output impedance of your given circuit is 2R, but not for the reason you gave.

The three equations you need are:

(V1-V2)/11R + V1/R = 0

(V2+A*V1)/8R + (V2-V3)/6R + (V2-V1)/11R = 0

(V3-V2)/6R + V3/3R = 1

If you solve this system, you will get V3 = R*(6A + 216)/(3A+92). This is the voltage at V3 when you inject 1 amp into the output node, so it is also the output impedance. If you let A -> ∞, then you get 2R.

17. ### The Electrician AAC Fanatic!

Oct 9, 2007
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That's the same result I get.

18. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Thanks Electrician - you always keep us "on the ball" with the extended challenge for a problem!

Rgds,

t_n_k

19. ### steveb Senior Member

Jul 3, 2008
2,432
469
Good catch! Sorry, but I didn't read his post closely enough. I just saw that the final answer was right, and figured he had found a valid method. It's amazing how he got the right answer with the wrong method, but it's clearly just coincidence. Not only is the basic idea of using input impedance and transfer functions wrong, but the equations don't even have correct matching of units. Resistance can not be equated to voltage, as was done in this case. Further, the algebra was not even correct because, once Vin was set to zero, he should have x=-2R: a negative output impedance!

Aside from the flaws revealed, the input impedance and transfer function don't allow you calculate output impedance by any method. In the example given, the input voltage divider is completely irrelevant for calculating Rout, since the input voltage source is shorted in the calculation of Rout. However, gain and input impedance are highly dependent on the input voltage divider.

It is very dangerous to develop ad hoc methods based on intuition. Shortcuts need to be developed from rigorous calculations and then applied very carefully after that. The shortcut I originally mentioned was based on my previous experience calculating the output impedance of many opamp circuits. The general rule is that ideal opamps generate zero output impedance in circuits with negative feedback, even if there is a large open loop source resistance on the output of the opamp (i.e. 8R in this case).

20. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,666
473
I knew you must have done something like that. :-(

My reaction when I tried to understand his algebra was: huh?

He definitely has a serious misunderstanding here, but I think there's hope. He did catch on to the notion of injecting 1 amp at the output as evidenced by his comment:

"But I will try with your suggestion of putting a 1 amp source at Vout and trying to get the Vout. If I get Vout, I will be able to do Vout/1amp = R = my output impedance right?"

More practice should help.

Your commentary is right on. I fear the OP (and his friend) is indeed relying on intuition, as you say. He said he was familiar with the node method, but the equations he derived indicate he needs more practice there as well.

I hope he has a look back here before he turns in his homework!