OpAmp Input Resistance

Depends on your definition of "input resistance" :-)

The input impedance of the OpAmp itself is still quite high... most modern-day OpAmp inputs just look like small capacitances.

In the inverting case you mention... the "system" input impedance is Ri (viewed from the source driving that node). ...Jim Thompson

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Search around until you find a page that describes the "virtual short" model of op-amp circuits.

In short, if the circuit is stable and the op-amp's gain is infinite, then the op-amp will do whatever it needs to do to hold the negative input voltage equal to the positive input voltage. So, to anything connected to the negative input, it "looks like" it's shorted to the positive input.

So what do _you_ think the input impedance of the _circuit_ would be in that case?

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Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
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The two input OpAmps are in the non-inverting mode, so the inputs are fed directly into the '+' inputs. In that configuration the OpAmp is high impedance. The input OpAmps aren't just "buffers" though. There is a feedback resistor (from outputs to '-' inputs) and a resistor between inputs to set the gain. You're correct, the output amplifier is a differential amplifier, typically with unity gain.

No, it's Ri. The '-' input is at a "virtual ground" so current flows in Ri. The '-' input is a virtual ground ("zero" volts from '-' to '+' inputs) as long as the output isn't in the rails (think: infinite gain).

The input impedance *is* the resistor. The other end of it is "grounded".

---

View in Courier:

+V>--------------+ | +--|--[R2]--+ | | | Vin>-----[R1]-+-|-\ | | >------+-->Vout GND>------------|+/U1 |

-V>--------------+

Since the non-inverting (+) input of the opamp is at GND (0 volts) and the opamp has to do whatever it must to make the voltage on the inverting input equal to the voltage on the non-inverting input, it'll force Vout to whatever value is required to make that happen.

For example, assume that a gain of -1 is needed and that Vin is at 1V.

Then, in order to force U1- to 0V, R1 and R2 must be equal and Vout must go to -1V, like this:

+V>--------------+ | R +--|--[R2]--+ R | | | 1V>------[R1]-+-|-\ | | >------+-->-1V GND>------------|+/U1 |

-V>--------------+

Notice that R1 and R2 form a voltage divider:

1V | [R1] | +----0V | [R2] | -1V

and that since the junction of R1 and R2 is forced to 0V by the opamp, Vin will be driving R1, which will be the input resistance of the circuit.

Notice also that as long as R1 and R2 are equal, the gain of the circuit will always be -1, so the input resistance of the circuit can be made as high or as low as desired just by changing the values of the resistor pair, while keeping them equal.

That is, if R1 = R2 = 1k, then the input resistance of the circuit will be 1k.

Likewise, if R1 = R2 = 10K, then the input resistance of the circuit will be 10k.

There are limits however, and you can't run the thing into the ground. (Unintended but gleefully accepted pun :-)

--- JF

it

Thanks for all the answers, it makes some sense now!

Until the output rails. ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058
hobbs at electrooptical dot net
http://electrooptical.net

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