OpAmp Input Resistance

When does the really high input resistance of the OpAmp go away? ..... I remember the 3 OpAmp Instrumentation amplifier.... and I believe I recall the final stage OpAmp being a differential OpAmp, and that it had a low input impedance, so there's the other 2 OpAmps as buffers feeding into the differential OpAmp....

is that right?

so, I was wondering... for example in a regular inverting amplifier with the gain of -Rf/Ri...... and the non-inverting input grounded..... is my input resistance of the op-amp still that really high input resistance... the Meg or Gig Ohm range that's associated with OpAmps? or have the resistors brought it down?

thanks!

Reply to
panfilero
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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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| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

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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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

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".

Reply to
krw

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Reply to
John Fields

it

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

Reply to
panfilero

Until the output rails. ;)

Cheers

Phil Hobbs

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Dr Philip C D Hobbs
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Reply to
Phil Hobbs

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"as long as the output isn't in the rails"

;-)

Reply to
keithw86

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