How to determine the feedback resistor in negative feedback op-amp configuration ? How does the feedback resistor affect the signal output of op-amp ? What is the consequence of using high resistance or low resistance in feedback circuit ?


The transfer function of the inverting amplifier is

\$ V_{OUT} = -\dfrac{R_{feedback}}{R_{in}} \cdot V_{IN} \$

For the non-inverting amplifier it is

\$ V_{OUT} = \left(1 + \dfrac{R_{feedback}}{R_{g}}\right) \cdot V_{IN} \$

where \$R_g\$ is the resistor to ground.

So the gain is determined by a resistance ratio, where a higher feedback resistance gives a higher gain. About the choice for higher or lower resistance values: lower is better, because at higher resistances the input bias current may begin to play a role. But don't overdo it: if your inverting amplifier would have a 1 kΩ feedback resistor and you want a gain of 10 \$\times\$, then the input resistance should be 100 Ω, and that may be a bit too little for the signal source. So see how much current the source can supply, and calculate the feedback resistor from that.

The non-inverting amplifier doesn't have that problem: the input signal feeds directly to the high impedance of the non-inverting input. To minimize offset error you'll have to make the input impedances for both inputs equal, that means on the signal input a series resistance equal to \$R_{feedback}\$ and \$R_g\$ in parallel. Example: if the feedback resistor is 10 kΩ and \$R_g\$ 1 kΩ then place a 9.1 kΩ in series with the source.

By the way, that equal impedance rule also goes for the inverting amplifier. You'll often see the non-inverting input directly connected to ground, but again placing a resistor between the input and ground will reduce offset error. Again the resistance is the parallel of the other two resistors.

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The feedback resistance divided by the input resistance gives the closed-loop gain of the circuit. Don't set it lower than double what the op-amp can tolerate as a load: in the typical inverting configuration the inverting input is a virtual earth so Rf is in parallel with the external load. On the other hand setting it too high means a proportionally higher input resistor, which increases input noise. Lowering the C/L gain lowers the AC distortion and increases the DC accuracy.

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  • \$\begingroup\$ I do use 10 times instead of double. \$\endgroup\$ – Standard Sandun Oct 11 '12 at 6:29
  • \$\begingroup\$ could you post somekind of application note on selecting this? This would be very useful for me. \$\endgroup\$ – Standard Sandun Oct 11 '12 at 6:31
  • \$\begingroup\$ @sandundhammika Googling for 'introduction to operational amplifiers' will yield what you're looking for. \$\endgroup\$ – Marquis of Lorne Oct 11 '12 at 9:25

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