# Finding Zin of non ideal non-inverting amplifier

I would like to know how to solve for Zin of the non ideal non-inverting amplifier shown below: I wonder how to get to this expression: $$\beta = \frac{R_1}{R_1 + R_2}$$ $$Z_{in} \approx 2R_{icm} || (R_{id} + \beta A(s) R_{id})$$

Where A(s) is the gain of the op-amp in the s-domain (again, it's non-ideal). Obviously there are some approximations I do not see. I know $R_{icm} > R_{id}$, and probably $R_0 >> R_1 \: or \: R_2$, but otherwise I'm in the dark, totally in the dark in fact.

Thanks!

EDIT: Maybe I should have been more precise regarding my question. I'm dealing with a textbook not really showing anything. Of course I can use shortcuts regarding negative feedback, I can also use google (altough I haven't found any links properly explaining what I'm seeing here), and I understand the basics of negative feedback.

I want to solve it simply using standard analysis method, $Z_{in} = V_{i}/I_{i}$. The author does the following steps:

1. $V_x = V_o \frac{2R_{icm} || R_1}{2R_{icm} || R_1 + R_2} \approx \beta V_o = \beta A(s) V_{id}$
2. $I_x = V_{id}/R_{id}$

3. $Z_x = \frac{V_x}{I_{x}} = \frac{\beta A(s) V_{id}}{V_{id}/R_{id}} = \beta A(s) R_{id}$

4. $\implies Z_{in} = 2R_{icm} || (R_{id} + Z_x) = 2R_{icm} || (R_{id} + \beta A(s) R_{id})$

What I don't understand is:

1. I don't even know where is $V_x$ because no schematics show it. I guess it's the voltage at $V_{IN-}$, and I also guess they approximated the current entering $V_{IN-}$ to be zero in order to get that relation (but again, why is it a valid approx considering we assume finite resistances, because $R_1 \: or \: R_2 << R_{icm} \: or \: R_{id}$?). For the rest I understand, including the simplification to $\beta$.

2. What is $I_x$? The current flowing in V_{IN-} seems to be approximated to $V_{id}/R_{id}$, correct? But why? Is it because $R_icm >> R_{id}$ (so that the current in the icm resistors are negligible)?

3. I don't have any idea what is $Z_{x}$ at this point? What is this equivalent impedance, or what is the equivalent circuit with $Z_x$?

4. I have no idea why $Z_x$ is in series with $R_{id}$. And why $2R_{icm}$ is in parallel with those 2.

If someone could explain these steps that would be really useful to me so I can get on with my reading. I'm more interested into the analytical solution than why I get the final result.

Thanks!

It should be noted that $Z_{in}$ means the input impedance at IN+ (and not IN-).

Due to negative feedback, the IN- voltage will be pulled towards IN+. The voltage across $R_{id}$ is therefore kept very small, causing only a very small current to flow from IN+ to IN-. So because of feedback, the resistor $R_{id}$ looks bigger. We are dealing with series-shunt feedback!

Basic feedback theory will tell you that in this case $R_{id}$ needs to be multiplied by $1 + \beta A$ (one plus the loopgain). So what you see is a resistor to ground ($2R_{icm}$) in parallel with a resistor to IN- modified by feedback ($R_{id}(1 + \beta A)$).

To be correct, the equivalent resistance to ground at IN- still has to be added in series to the modified $R_{id}(1+\beta A)$, but this contribution will probably be much smaller.

So you end up with a total input impedance of approximately

$Z_{in} \approx 2R_{icm}||\left[R_{id}(1 + \beta A(s))\right]$

EDIT:

1. In order to get the same formulas as the book for $V_x$ (the voltage at IN-), you need to assume that $R_{id} \gg R_{icm} \gg R_1,R_2$ and $R_o\approx 0$. So this means that $R_{id}||R_{icm}\approx R_{icm}$. By blocking the path via $R_{id}$, $V_i$ cannot influence IN- directly.

2. $I_x$ is the current flowing from IN+ to IN-, and so is also equal to the total current seen flowing to "ground" at IN- as far as IN+ is concerned. So if we were to replace everything at IN- by a single impedance to ground (which is the $Z_x$ we're trying to calculate), then the current through that impedance would be the same as through $R_{id}$.

3. $Z_x$ is the total equivalent impedance to ground at IN-. This is the voltage $V_x$ divided by the current to ground $I_x$.

4. And so finally, you can find the total equivalent impedance to ground as $R_{id}+Z_x$ in parallel with the shown $2R_{icm}$.

As you can see, the formulas reflect again that the current is artificially made small by the feedback. $V_x$ is brought closer to $V_i$ by $V_o$, and this reduces the current through $R_{id}$ compared to an $R_{id}$ to ground.

• Hi, what I don't really understand is actually why we multiply $R_{id}$ by $1+\beta A$ from a circuit point of view (see my updated answer). Which is precisely what I want to get to. Is it possible to explain further? Either way, much thanks for your time. – Yannick Feb 5 '18 at 18:56
• I have edited my answer to match your question. – Sven B Feb 6 '18 at 9:18
• Hi Sven sorry for the late answer, love your answer. Thank you very much for your time. May I ask a last question? If Z_x is the total impedance from IN- to ground, why in 4) do we consider R_id in serial with Z_x? Shouldn't be R_id be somewhat "included" in Zx? Or is it because Z_x is the impedance to ground seen by IN- (and thus would explain why IN+ sees it as in serial with R_id)? – Yannick Feb 12 '18 at 20:43
• I think your second part is correct! – Sven B Feb 12 '18 at 22:29

Here are some hints: -

Do you know how negative feedback works - for a perfect amplifier (with A or A(s) equalling infinity), the voltage at the inverting input must be exactly the same as the voltage at the non-inverting input. If those voltages were not equal, the output form the amplifier would be infinity and we all know that this can't happen.

Take that on board and you should be able to see that under these circumstances, the input impedance is purely $2R_{icm}$. It is precisely this figure because, with both inputs at the same level, there can be no current flowing through $R_{id}$.

When A is finite (a real amplifier) the voltage difference between both inputs drives the amplifier to produce the output voltage. So, there is a voltage difference across the inputs in a real amplifier and this voltage is a fraction of $V_i$. Now the input impedance becomes more complex to calculate because you have $2R_{icm}$ in parallel with some factor x $R_{id}$ and that factor is based on the open-loop gain (A) being non infinite.

If you still can't figure this yourself there are various proofs on google for op-amps with finite values for A.