Let us see how the feedback theory from my last week's post applies for a opamp in a typical non-inverting configuration:

In the diagram:

- $A$ is the gain of the amplifier with no feedback applied (its
)*open loop gain* - $x$ is the input signal
- $y$ is the output signal

The feedback network consists of two resistors $R_1$ and $R_2$. The gain $B$ of the feedback network is that of a voltage divider: $$B=-{R_1 \over R_1+R_2}$$The minus sign here accounts for the fact that the feedback is applied to the inverting input of the opamp.

Using the formulas from last week's post:

$$STF={A \over (1-A*B)}={A \over (1+A*{R_1 \over R_1+R_2})}$$

$$ETF={1 \over (1-A*B)}={1 \over (1+A*{R_1 \over R_1+R_2})}$$

$$LG=A*B=A*{R_1 \over R_1+R_2}$$

As $A$ (and hence $LG$) increases, $STF$ approaches the familiar ${R_1+R_2 \over R_1}=1+{R_2 \over R_1}$.

One special case here is an opamp connected as a unity gain buffer:

This is equivalent to the general non-inverting connection with $R_1$ open and $R_2$ shorted. In this case, $B=-1$ and

$$STF={A \over (1-A*B)}={A \over (1+A)}$$

$$ETF={1 \over (1-A*B)}={1 \over (1+A)}$$

$$LG=A*B=A$$

$LG$ equals $A$ - in this configuration, all available open loop gain is applied to reduce the output error. As $A$ increases, $STF$ approaches unity.

In the next post, I will look at the opamp in the inverting configuration.

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