Essentially I am getting confused trying to do the sums for an op amp with a gain of 10dB and an input impedance of 1kohm.
I worked out that $\frac{V_{out}}{V_{in}}=-\frac{R_{2}}{R_{1}}$ because $V_{+}$was going to ground, $=>V_{-}=0$.
I know that the output impedance of the amplifier itself is very high.
I know that the compensation resistance $R_{3}=\frac{R_1R_2}{(R_1+R_2)}$ but I am not certain why.
I had thought the input impedance would be the $R_1||R_2$ (or whatever else would go to the node for $V_-$ which in this case is just $R_1$ and $R_2$) but I am doubting myself.
Can anyone clarify what this input impedance is actually referring to?
I should also perhaps add that I am going to construct this for real out of a 741 amplifier so I am trying to figure out what resistances to pick to get my 1000 $\Omega$. I can't believe that $R_2$ wouldn't matter in this, so if anyone can clarify that, it would be useful.
Answer
@DaveTweed wrote a good verbal proof for $R_{3}=\dfrac{R_1R_2}{R_1+R_2}=R_1||R_2$.
Here's an algebraic version.
Let's drop the ideal OpAmp assumption that OpAmp input impedances are infinite. Then input bias currents are nonzero.
$I_b=I_{b+}=I_{b-}\neq0$
In practice, Ib can vary between different batches of ICs. Ib isn't known. Let's assume that it's fixed.
First, consider the case without compensation resistor, $R_3=0$.
$\dfrac{V_{in}}{R_1}+\dfrac{V_{out}}{R_2}+I_b =0$,
$V_{out}=-V_{in}\dfrac{R_2}{R_1}-I_bR_2$
notice the $I_bR_2$ nuisance.
Second, consider $R_3\neq0$. Let's find $R_3$ such that $V_{out}$ is closest to $-V_{in}\dfrac{R_2}{R_1}$
Voltage at the positive input: $V_{(+)}=I_bR_3$
$\dfrac{V_{in}-I_bR_3}{R_1}+\dfrac{V_{out}-I_bR_3}{R_2}+I_b=0$
$\dfrac{V_{in}}{R_1}+\dfrac{V_{out}}{R_2}+I_b\left(\dfrac{R_3}{R_1}+\dfrac{R_3}{R_2}-1 \right)=0$
$I_b\left(\dfrac{R_3}{R_1}+\dfrac{R_3}{R_2}-1 \right)=0$, when $\dfrac{R_3}{R_1}+\dfrac{R_3}{R_2}=1$
which can be solved for $R_{3}=\dfrac{R_1R_2}{R_1+R_2}=R_1||R_2$
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