http://www.allaboutcircuits.com/vol_3/chpt_8/9.html

Stated: "Another limitation of this amplifier design is the fact that its input impedances are rather low...Each input voltage source has to drive current through a resistance, which constitutes far less impedance than the bare input of an op-amp alone. The solution to this problem, fortunately, is quite simple. All we need to do is "buffer" each input voltage signal through a voltage follower"

Why does the addition of a resistor lessen the impedance? Shouldn't it be the other way around?

Thanks for your help!

 

When you have two paralleled resistors, the equivalent resistance tends towards the smaller one. Exemple, if you have R1||R2 and R1=50ohms; R2=1k, then:
\(R_{eq} = \left( \frac{1}{R_1}+\frac{1}{R_2}\right)^{-1} = 47.6\Omega\)
Same thing apply here. Even if you parallel a high impedance with something smaller, then the resistive effect is inevitably reduces.

 

Understood, but wouldnt their equivalent resistance add to the impedance of the op amp input since the resistors are in series with its leads?

 

Impedance is the sum total of resistance, inductance, and capacitance. Usually the overwhelming number is the resistance.

 

Maybe I should be asking the question of how input impedance is measured in an op amp.

 

Usually, both inputs of an opamp is directly connected to the gate of a FET, or the base of a BJT in a differential pair. On a MOSFET, the resistance seen at the gate is almost infinite (theoretically). See http://en.wikipedia.org/wiki/File:Long-tailed-pair.gif

What you can do to mesure input impedance of the opamp is putting a voltage source in series with a known resistor (preferably near the expected input impedance), and those components in series with the opamp's input. You then mesure the voltage at the opamp's input. Use the voltage divider equation to find the input impedance.

\(V_{IN} = \frac{V_{INPUT} R_{input}}{R_{input} + R_{known}}\)

 

The input impedance at the two inputs to the first circuit on the page you referenced:

http://www.allaboutcircuits.com/vol_3/chpt_8/9.html

can be determined by inspection.

If the + and - inputs of the opamp itself are assumed to have such a high impedance that it can be taken as infinite, then the impedance at the V2 input is clearly just 2R. The impedance at the V1 input is easy to determine, because the + input is grounded through a resistance of R which is so low compared to the (essentially) infinite input impedance of the + input of the opamp, that the + input is essentially just grounded. That means that the - input is also a (virtual) ground. Then the input impedance at V1 is just R.

Both these impedances (R at the V1 node, 2R at the V2 node) are very much lower than the very high input impedance of the opamp + and - inputs.

 

Well Ive got it figured out.

Thank you all very much for your input.

Sincerely
User