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

  • \$\begingroup\$ What is the voltage at the inverting input of this op-amp? \$\endgroup\$ – Spehro Pefhany Jan 22 '14 at 1:13
  • \$\begingroup\$ the inverting input of this op-amp is 1 volt \$\endgroup\$ – Michael harris Jan 22 '14 at 1:43
  • \$\begingroup\$ Are you sure about that? Usually the non-inverting input is grounded. \$\endgroup\$ – Spehro Pefhany Jan 22 '14 at 2:02
  • \$\begingroup\$ @SpehroPefhany is it? What about op-amps in single supply applications? What if you want the amplifier to add a DC offset, as well as amplify? \$\endgroup\$ – Phil Frost Jan 22 '14 at 2:20
  • \$\begingroup\$ In textbooks, it usually is (to being with). In your 'typical' schematic below, it definitely is. In reality, not always, of course. I just wanted him to think about it. I see no mention of Avo or GBW product, so it seems it's introductory level. \$\endgroup\$ – Spehro Pefhany Jan 22 '14 at 2:46

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}\$.

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Assuming your circuit looks like the circuit here,


the input impedance is simply Rin. You are all set. The first link has a very in-depth explanation. Here is a bit easier read:


The input impedance of any sort of amplifier is simply the impedance detected by the input signal. Thus, Zin = Vin/Iin. Since there is a virtual ground at the inverting input of the op-amp, the input impedance is equivalent to Rin. For example, if the input signal is 5V, Iin will be (5V-0)/10kΩ = 0.5mA. Then Rin = Vin/Iin = 5V/0.5mA = 10kΩ.

For a non-inverting amplifier, the input impedance would be the impedance from the non-inverting terminal to ground.

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