# Simple Op-Amp analysis (find R eq.)

Discussion in 'Homework Help' started by tAllann, Oct 26, 2013.

1. ### tAllann Thread Starter New Member

Oct 26, 2013
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0

Note that the two terminals are the one shown and the ground reference.

I am supposed to find Req.

First I made a label on the upper terminal Vs, and labeled Vn and Vp, and Vo, and a current I from right to left across the 1 k ohm resistor.

I then made the assumption this is an approximately ideal op-amp, and set

1) Vp = Vn, and Ip = In = 0.

Then I decided that

2) Req = Vs / I

Then,

3) I = (Vo - Vp) / 2k
4) I = Vp / 1k

1, 3, 4 => Vo = 3Vp = 3Vn

Then I did a nodal analysis at the node with Voltage Vn.

(Vn - Vs) / 10k + (Vn - Vo)/10k = 0

Then subbing 3Vn for Vo,

Vs = -Vn

Then combining equ. 4 and 2,

Vs/Vp = Req K ohms

and then subbing Vn for Vp and then -Vs for Vn,

Vs/-Vs = Req K ohms

Req = -1 k ohms

But the answer is supposed to be 5 k ohms. Did I make a mistake or interpret the question incorrectly?

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2. ### WBahn Moderator

Mar 31, 2012
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You mistake is in defining your test current. The equivalent resistance is the ratio of the voltage applies to the input (Vs) and the current that flows out of Vs into the input resistor, which is the current into the leftmost 10kΩ resistor.

The current in the 1kΩ resistor is NOT the same current as in the input resistor. Remember that the opamp interacts with ground, too.

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3. ### tAllann Thread Starter New Member

Oct 26, 2013
11
0
I see. That makes sense now. Thank you very much. I think I can solve it now.

4. ### WBahn Moderator

Mar 31, 2012
23,109
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Good. If you have any other problems, just come on back!

5. ### anhnha Well-Known Member

Apr 19, 2012
866
61
I just can't figure out why the two current are not equal. If I apply a voltage source, Vs, between the terminal and ground, then why are 10K(leftmost resistance) and 1K not in series?

To me it seems that Vs, the leftmost resistance 10K and 1K are in series.

6. ### WBahn Moderator

Mar 31, 2012
23,109
6,954
Really? To be in series, it means that (conceptually) a charge carrier that enters the leftmost 10kΩ resistor has NO CHOICE but to work its way eventually through the 1kΩ resistor. But what happens after this carrier gets through the rightmost 10kΩ resistor and makes it to the opamp output? Can't it hang a right and go into the opamp instead of going straight? But it only goes through the 1kΩ resistor if it goes straight.

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7. ### anhnha Well-Known Member

Apr 19, 2012
866
61
I failed to see that in the opamp is an active device here. To operate, it has to be provided with E+ and E- DC voltage supplies. If I draw those voltage sources into the circuit, the two resistance 10K and 1K are not in series but when these voltage source are left out, the opamp really looks like a two terminal device and 10K and 1K also looks as if they are in series.

8. ### WBahn Moderator

Mar 31, 2012
23,109
6,954
It looks like a two-terminal device? It sure looks like it has three terminals.

But the key point is that in this series chain you think you are seeing is a junction that has THREE (i.e., more than two) devices connected to it and we have never put ANY constraint saying that current cannot flow into or out of the opamp's output terminal. Quite the opposite, we have assumed that it will source or sink any amount of current necessary to enforce the assumption that the input terminals are at the same voltage.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,497
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I solve for Rin and I get very interesting solution
Req = R1 - (R2*R4)/R3

10. ### anhnha Well-Known Member

Apr 19, 2012
866
61
Jony, what are R1, R2, R3, R4?

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Here you have

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12. ### anhnha Well-Known Member

Apr 19, 2012
866
61

In this circuit, I get Vn = 3Vs.

Is= (Vs -Vn)/10K = -Vs/5K
(Is: current flowing from left to right through the leftmost resistance, 10K)
If the minus sign is taken into account then:

Rin = Vs/Is =-5K

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,497
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Req will be positive because for positive Vin op amp output voltage will be negative Vout.
So Vin source will deliver the current, so Rin is positive.

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14. ### LvW Active Member

Jun 13, 2013
674
100
Jony`s formula is correct.
The gain of the circuit is G=-3.
For an output Vo=0 the input resistance, of course is 20kohms.
For Vo=G*Vin and according to the Miller theorem the equivalent input resistance is
Req=20k/(1-G)=20k/4=5kohms.

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15. ### anhnha Well-Known Member

Apr 19, 2012
866
61
Thanks. I made a mistake. The voltage at the inverting, Vn = -Vs.
Is= (Vs -Vn)/10K = Vs/5K
(Is: current flowing from left to right through the leftmost resistance, 10K)
Rin = Vs/Is = 5K