feedback resistor

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.

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.

• I do use 10 times instead of double. Commented Oct 11, 2012 at 6:29
• could you post somekind of application note on selecting this? This would be very useful for me. Commented Oct 11, 2012 at 6:31
• @sandundhammika Googling for 'introduction to operational amplifiers' will yield what you're looking for. Commented Oct 11, 2012 at 9:25