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Non-inverter OPAMP and Inverter OPAMP input impedance intuitive analysis

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eegchen

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Hi,

Thank you for watching this post.

For a simple feedback, like R1=R2=R, OPAMP's gain is A.

Non-inverting feedback amplifier, the input impedance is (1+AB)*Ri, B is feedback factor.
Inverting feed amplifier , the input impedance is R+R/(1+A)

I can do some math to get the result. But how to intuitive analysis this?
The Art of electronics gave the explanation like this,

"since voltage feedback tends to subtract signal from the input,
resulting in a smaller change (by the factor AB) across the amplifier's input resistance; it's a form of bootstrapping. Current feedback reduces the input signal by bucking it with an equal current."

But i am not quite understand, anyone can give more detailed explanation?

best
 

In case of inverting amp: lets take a practical example, your input has 5V R1=R2=1K, output is -5V.
Since you have the inverting inputs at virtual gnd, the input signal sees 1K load, simple is that.

In case of non-inverting configuration, the input sees only the opamps bias current which is negligible.
 

In case of inverting amp: lets take a practical example, your input has 5V R1=R2=1K, output is -5V.
Since you have the inverting inputs at virtual gnd, the input signal sees 1K load, simple is that.
In case of non-inverting configuration, the input sees only the opamps bias current which is negligible.

Sorry, that's not correct. The input IMPEDANCE is seen by the signal and has nothing to do with bias currents. You shouldn't mix up dc input resistance with signal input impedance. The inpuit impedance at the non-inverting input is enhanced by the feedback action and is (1+AB)*Ri (as mentioned above) with Ri=signal input impedance without any feedback.
 

Hi LvW,

I agree with you . The input impedance is a small signal parameter.

Do you know how to intuitively explain that ?
Why Ri*(1+AB) not Ri/(1+AB)?
 

Hi eegchen, remember that the opamp input stage is a differential pair.
Now, when you understand the operation principle of such a circuit and, in particular, the difference between differential and common mode input resistances, you have the answer.
Think of 100% feedback (unity gain configuration). In this case, both opamp inputs see the same signal: That's the common mode case with a very high input resistance.
Every other feedback factor is something between both extremes: No feedback (differential mode) and full feedback (common mode).
LvW
 

LvW,
So you are saying that the inputs (+/-) are at the same level only in case of the repetear ? Hmm I knew that the inputs are at the same level all the time (ideal opamp, no offsets).
Since you stated that i was wrong, and bias current has nothing to do with input resistance, i will get to the bottom of this :)
 

LaszloF, I agree with you. My explanation was a bit too short.
Of course, both inputs are always (nearly) on the same level, however, in case of low feedback (high gain) the amplifier must produce a larger voltage (if compared with greater feedback factor) in order to allow the n-input to follow the p-input voltage.
Thus, for low feedback factors the input transistors must provide more gain with the consequence of more input signal current resulting in a smaller input resistance. For 100% feedback the output is a low voltage (identical to the input voltage) and the input transistors draw only a small signal current.
Of course, exact calculations can reveal this effect more clearly - and this is only a more or less intuitive explanation for this effect.
 

Hi laszloF,

as I am not completely satisfied with my above explanation, let me try another (perhaps better and more exact) approach:
The key to understand the effect of resistance change is the fact that two sources drive a current, but the current is allocated to one single source only. That's the only way to define something like an input resistance at one port. In principle, we have a similar situation by explaining the MILLER effect (however, with a resistance decrease). And here we have something like a bootstrap effect.
I'll try to explain the non-inverting case:
The signal voltage Vp (at the p-terminal of the input differential pair) causes a certain current Ip1 flowing into the terminal. However, at the same time there is another voltage Vn (the feedback voltage) at the inverting terminal of the differential pair. Now, because Vn causes also a current In=Ip2 through the p-terminal in a direction opposite to Ip1 (working principle of diff. stage), we have a total current Ip=Ip1-Ip2 through the p-input that, of course, is much more smaller than Ip1.
Because we allocate this current Ip to the signal voltage Vp only (although also the "source" Vn is partly responsible), this leads to a drastic increase of the effective input resistance at the p-terminal.
I am convinced that this is the most correct explanation.
 
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Hi LvW,

Thanks. The explanation is quite clearly.
 

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