impedanceinput-impedanceinverting-amplifieroperational-amplifier
I want to Design an op-amp inverting amplifier circuit with
Closed-loop voltage gain = -10 and Input impedance = 10k
Since the it's an inverting amplifier then the Closed-loop voltage gain =-Rf/Rin
So if Rf = 100k and Rin = 10k then the Closed-loop voltage gain = -10
But I don't know how to set the input impedance of an inverting amplifier.
the input of this op-amp is 1 volt
When we talk about the input resistance of a circuit, we're describing how it affect the other circuit that is providing the input signal.
Specifically, we want to know, in order to change the input current by i amps, how much do we need to change the input voltage? That's why the input resistance is, by definition, \$ \dfrac{\mathrm{d}v_i}{\mathrm{d}i_i}\$.
So what's the input resistance of this circuit?
The key point is that in this configuration, as long as we avoid saturating the op-amp output, the inverting input of the op-amp is a virtual ground. The feedback in the circuit operates to keep that node at 0 V. So whatever input current we want, by Ohm's Law, the required input voltage is \$\mathrm{R_{in}}\times{}i_{i}\$. Therefore the input resistance is Rin.
Then there are answers like this that mention that input impedance is infinity.
That answer was talking about the input resistance of the op-amp, which was assumed to be ideal, not the input resistance of the whole circuit.
Your hyperphysics link is correct, and so is your conclusion about the input impedance of the voltage follower. The math follows from the basic control system diagram. You can see a nice presentation on it here:
http://slideplayer.com/slide/1496732/
Best Answer
Here's a typical op-amp inverting amplifier:
The input impedance is simply \$R_{in}\$, so for your requirements, \$R_{in} = 10k\Omega\$. \$R_f\$ is then whatever it needs to be to realize the desired gain. You want a gain of -10, so:
$$ -\frac{R_f}{10k\Omega} = -10 \\ R_f = 10 \cdot 10k\Omega = 100k\Omega $$
Why does the input impedance depend only on \$R_{in}\$? So long as the op-amp is not saturated, the inverting input is held at the same potential as the non-inverting input. Here, that's just ground, though any DC voltage works. So, you might as well consider \$R_{in}\$ as connected to ground, because the voltage is the same as it was. Once you realize the inverting input is effectively ground, it's easy to see the input impedance is just \$R_{in}\$.