input impedance of op-amp circuits

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No sir, i am not too sure about the closed loop gain of the circuit. Anyway, please tell me if this is right:

Output impedance= R3/(1-(1/gain))= R3/(1+R1/R2).

Input impedance= R1 || R3/(1+R2/R1).

So, i think the result is Output impedance / input impedance.

Please correct me if i am wrong.

Thanks.
 

Hi rgamma,

1.) Sorry, but the output impedance is not correct. Did you calculate? Or guess?
Without any feedback the opamp output resistance is rather low (ro=50...100 ohms).
Due to negative feedback this resistance becomes much more smaller than ro!
Remember FvM's answer: Rout app. zero for ideal opamps.
Hint: For real oamps you have to consider the loop gain. (I suppose, you know what it is).

2.) Why do you think the gain would be the output-to-input impedance ratio ? Please, tell me: Where does this come from?
 
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I used miller effect to calculate the output impedance.
That would come along with the smalloutput impedance of opamps. So, this should be definitely bigger, so i ignored the one.

In common emitter amplifier, it is Rc/Re. So, i thought that would apply to this one too. Even in many feedback amplifiers, this is the case, but not for all.
 

I used miller effect to calculate the output impedance.
That would come along with the smalloutput impedance of opamps. So, this should be definitely bigger, so i ignored the one.


Sorry, it's the wrong way. I suggest to consult any reliable textbook to see how the output resistance is calculated in case of negative feedback.


In common emitter amplifier, it is Rc/Re. So, i thought that would apply to this one too. Even in many feedback amplifiers, this is the case, but not for all.

*At first, the ratio Rc/Re gives you a rough (very rough!) information on voltage gain of a common emitter stage (under certain conditions). What about Re=0? Gain infinite?
*Secondly, Re is NOT the output resistance of such a BJT stage.
Please tell me, in which of the "many feedback amplifiers" the gain is simply the ratio of the input to the output impedance.
 
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Rc||R-load is the output impedance of CE amplifer, but electronics is built on abstractions, so i simplified it. If Re=0, it would be Rc/hie?

There was a transresistance amplifier in which the gain was the ratio of impedances. Some examples in millman - integrated electronics. Could you suggest some good books where i can read op-amps? I just know basic circuit analysis and some amount of transistors, and other active devices.

Thanks.
 


Of course, in many cases the gain of an amplifier with feedback - in particular for opamps - can be with sufficiently accuracy expressed by the gain of 2 impedances (standard example: gain=-r2/r1 for an opamp wired as an inverter). However, this is NOT identical to a ratio output-to-input resistors.

A good book for opamps (among many others) is Sergio Franco: Design with operational amplifiers.
 
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So sir, could you tell me the output impedance of this circuit? I'll derive it and check with the answer you give.
 

could you tell me the output impedance of this circuit?
Click to expand...
Of which circuit, the one from post #15? It's showing an unspecified opamp, so assuming an ideal OP is the only reasonable answer. As already said, it has zero output impedance. Alternatively, if you're referring to uA741 from post #4, it has no output impedance specification in the datasheet, so nothing can be calculated. In any case, the output impedance will be low as long as the OP is operated within reasonable parameters.
 

rgamma said:
So sir, could you tell me the output impedance of this circuit? I'll derive it and check with the answer you give.
Click to expand...

OK, I will give you some introductory steps for calculating the ouput impedance for an opamp:
* opamp with non.inv. input grounded and resistive feedback k=R1/(R1+R2)
* Apply a test voltage u2 to the output
* Result: two currents: i2=i21 and i22
* i21=(u2-uo)/ro where ro=internal output resistor and uo=-Ao*un (internal output voltage, un=u2*k; Ao=openloop gain)
* i22=(u2-un)/R2
* Solve for Rout=u2/i2 (requires basic math only).
 

Oh, same method for any 2 port network?
 

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