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What is the input impedance? I have tried to solve it, let me know whether it is right or not?

Thanks in advance.

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    \$\begingroup\$ no, it's not right \$\endgroup\$
    – Neil_UK
    Jan 26, 2021 at 13:46
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    \$\begingroup\$ Hint: assuming the amplifier is in normal operation (output not saturated), what is the voltage at the inverting input? (look up virtual ground). This might be considered a bit of a trick question but it is testing your knowledge of a key parameter of opamps. \$\endgroup\$ Jan 26, 2021 at 13:49
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    \$\begingroup\$ As another hint (in case you know the MILLER effect): Appears the "fictitoius" feedback resistor - referenced to the inverting input - larger or smaller if compared with its nominal value? By which factor? \$\endgroup\$
    – LvW
    Jan 26, 2021 at 15:34
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    \$\begingroup\$ One of those R's in each circuit does absolutely nothing. If you had numbered them, I could tell you which. \$\endgroup\$
    – user16324
    Jan 26, 2021 at 15:58
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    \$\begingroup\$ @BrianDrummond - Shhhhh :) That's what I am trying to get the OP to figure out. \$\endgroup\$ Jan 26, 2021 at 16:03

1 Answer 1

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I have illustrated the "magic" virtual short connection between the op-amp inputs by the conceptual picture below. I have extracted it from my answer to a recent similar question.

Inverting amplifier - input resistance

The op-amp output is represented by a variable voltage source producing a voltage VOA. It follows the voltage drop across the resistor R2; so VOA = VR2 (as magnitudes). Since it is connected in series to R2, it destroys the voltage drop across it (VR2 - VOA = VR2 - VR2 = 0)... and the combination of the two elements in series behaves just as a "piece of wire" ("virtual short connection" between the two op-amp inputs).

Now, if you connect the third resistor in parallel to this virtual short, it will be shunted... and nothing will change. The equivalent circuit input resistance will be determibed only by R1.

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