input impedance of op-amp circuits

[rgamma]

Hi,

What should i do if i have to find the input impedance of an op amp circuit. For example, a inverting amplifier with load resistance Rf.

Also, what is meant by dynamic range?

---------- Post added at 16:34 ---------- Previous post was at 16:33 ----------

Also tell how to measure output impedance also.

 

[emresel]

Hi;
Since internal impedance of the opamp is infinite (theoretically) then your input imp -for an inverter opamp amplifier- is determined by the input resistor (Ri) (which also define part of the gain).
Dynamic range can be considered as the range;
- possible maximum input signal amplitude, where output signal shouldn't have any saturation and distortion
(ie for an amplifier after some certain input level, amplifier cannot amplify anymore because output should exceed the power supply, then it is saturated)
- possible minimum input signal which can be detectable from the noise at the output of the circuit.
(ie for an amplifier some small input signals cannot detected at the output of the amplifer becasue of the amplifer internal noises)
hope helps

 

[LvW]

Hi;
Since internal impedance of the opamp is infinite (theoretically) then your input imp -for an inverter opamp amplifier- is determined by the input resistor (Ri) (which also define part of the gain).

The reasoning is not correct.
The effectice input impedance is identical to the resistor R1 (between the signal source and the inv. input) because of the feedback: Principle of virtual ground.

 

[rgamma]

Thank you, How about the output impedance?

also, what will happen if i short the two inputs of an op-amp?

In this circuit, if i change the value of R3, the gain is not changing. Is it due to some circuit error or is there any reason?

 

[LvW]

* After a short circuit acroos both inputs the opamp cannot work anymore. Surprising?
* The next question is worth to give a detailed answer:

(1) Assuming an ideal opamp it does not matter because R3 has no influence (because the voltage across the input tends anyway toward zero)
(2) However, for a real opamp R3 can play a very important role: it can stabilize the circuit! That means, you can use even uncompensated or partly compensated opamps for each selected closed-loop gain value
However, R3 does not influence the closed loop gain because it reduces the feed-forward factor in the same amount as the feedback factor. Thus, the closed-loop gain remains constant.
But due to the reduced feedback the loop gain is larger (correction: smaller) leading to a better stability. The price is a reduced closed-loop bandwidth.

---------- Post added at 13:55 ---------- Previous post was at 13:33 ----------

Regarding the input and output impedances there are simple relations to the gain and feedback parameters:

Assuming resistive feedback k=R1/(R1+R2)

*Inverter: Zin=R1+R2/(1+|Ao|) where Ao=opamp open loop gain (frequency-dependent)
Good approximation: Zin=R1

*Non-inverter: Zin=Zd*(1+T) where Zd= diff. input resistance and T=k|Ao| (loop gain)
Approximation: infinite

* Output impedance: Zout=Ro/(1+T) where Ro=opamp output impedance (without feedback) and T=k|Ao|.
Approximation: zero

 

[FvM]

However, for a real opamp R3 can play a very important role: it can stabilize the circuit!

Yes., it's used for this reason in special cases. But referring to 741 or other universal compensated OPs, R3 will only detoriate the amplifier''s performance: lower bandwidth, more noise, higher offset voltage.

 

[LvW]

Yes., it's used for this reason in special cases. But referring to 741 or other universal compensated OPs, R3 will only detoriate the amplifier''s performance: lower bandwidth, more noise, higher offset voltage.

Yes, that's the price to be paid for this simple stabilization method.
For my opinion, another example for the known fact that no improvement is possible (in this case: more stabilization margin) without performance degradation of some other parameters. Everything is a trade-off (as in the whole life!).
However, I doubt if someone will use this method for opamps that are already unity-gain stable.

 

[emresel]

What you mean while saying "uncompensated, partly compensated" opamps?
Could you please explain alittle bit?
Thanks

 

[rgamma]

one more doubt, i connected an op amp (741c) from output to positive terminal and it started smoking. Why does it happen?

 

[LvW]

one more doubt, i connected an op amp (741c) from output to positive terminal and it started smoking. Why does it happen?

First two questions from my side:
- what "pos. terminal"? Power or signal input?
- Why did you do that?

---------- Post added at 16:38 ---------- Previous post was at 16:35 ----------

What you mean while saying "uncompensated, partly compensated" opamps?
Could you please explain alittle bit?
Thanks

Explanations would involve stability issues - and this is to complex and time-consuming to explain it here.
Please, consult a suitable textbook. However, specific questions (like: what is the purpose of R3) can and should be placed here in the forum. Sorry.

 

[rgamma]

i meant the Noninvertinng terminal.
i was building an instrumentation amplifier. instead of giving feedback of the input buffer to the inverting, i gave to non inverting. It started smoking.

 

[FvM]

I can hardly imagine how a wrong feedback connection should cause overheating or damage of a 741. This may possibly happen with a GHz OP. There must be some serious violation of device specifications added in your circuit. You should post the complete circuit to make the issue understandable.

Regarding the significance of loop gain reducing "R3", you'll find OPs compensated for G >= 5 or G >= 10. If you want use them with a lower gain for some reason, adding "R3" can help.

 

[rgamma]

its just a normal instrumentation amplifier in which i connected both the input buffers from output terminal to the non inverting terminal.

 

[FvM]

Both buffer outputs will go near to either positive or negative supply voltage. But it's still a legal 741 operation condition.

 

[rgamma]

oh, but, is there a possible condition where it can burn?

btw, could you tell the input and output impedance of this circuit?

 

[2010ee179]

btw, could you tell the input and output impedance of this circuit?

View attachment 59566

You should do circuit analysis to find out input impedance (Vin/Iin) and output impedance (Vout/Iout)...

 

[rgamma]

i won't be asking for help if i knew it well. So, please help if you know.

Thanks.

 

[LvW]

I recommend to split the task into 2 parts:

* Without R3 the input resistance Rin1 is known to be simply R1 (opamp ideal)
* path with R3: As for all classical amplifiers with negative feedback: Rin2=R3/(1-gain).
In your case: gain=-R2/R1 (note the minus sign!).
* Final result: Rin=Rin1||Rin2
______________

 

[FvM]

oh, but, is there a possible condition where it can burn?

E.g. by reversing the supply voltage. Output short circuit should be tolerated.

could you tell the input and output impedance of this circuit?

Restricting the discussion to ideal OPs, you get zero output impedance. A regular inverting amplifier has simply an input impedance of R1. In this case, you have R3 as additional input load, multiplied with 1/(1+R2/R1). I don't see a reasonable purpose of R3, however.

 

[LvW]

I don't see a reasonable purpose of R3, however.

In this context, a question to rgamma: Do you know the closed-loop gain of the given circuit (opamp ideal)?