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Please see attached image. How does the voltage gain end up being equivalent to Av = 1-error?

Thanks Question Solution

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    \$\begingroup\$ It is hardly readable. Please, make an effort and rewrite the question using the built-in schematic and formula engines.. \$\endgroup\$ – Eugene Sh. May 12 '16 at 21:04
  • \$\begingroup\$ Oh. I'm not too sure how to use them, especially the formula engines. Apologies \$\endgroup\$ – Arsenal123 May 12 '16 at 21:11
  • \$\begingroup\$ MathJax Tutorial \$\endgroup\$ – Tom Carpenter May 12 '16 at 21:12
  • \$\begingroup\$ Are you asking why \$\frac{1}{1 + \epsilon} \approx 1 - \epsilon \:\$? \$\endgroup\$ – Alfred Centauri May 12 '16 at 22:49
  • \$\begingroup\$ Nope, I was confused on how to actually derive the error term \$\endgroup\$ – Arsenal123 May 29 '16 at 10:09
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Assuming that your question is how do you get from:

$$A_v = \frac{1}{1+\frac{R_{out}+R_L}{A_oR_L}} \tag{1}$$

To:

$$A_v \approx 1 - \frac{R_{out}+R_L}{A_oR_L}$$

It is an approximation rather than an equality. Let's examine it.

First we will say:

$$x = \frac{R_{out}+R_L}{A_oR_L} \tag{2}$$

Substituting (2) into (1), we get:

$$\frac{1}{1+x} = (1+x)^{-1}$$

Using the Binomial expansion of this (see here), we basically get:

$$\frac{1}{1+x} = 1 - x + x^2 - x^3 + ... $$

This is where we make an approximation. If \$x\$ is very small (i.e. \$x<<1\$), then it follows \$x^2\$ and \$x^3\$ and so on, are going to be very very small. So small that we make an approximation that they are negligible. As a result, we can say:

$$\frac{1}{1+x} \approx 1 - x \tag{3}$$

Substituting (1) and (2) into (3), we then get:

$$A_v \approx 1 - \frac{R_{out}+R_L}{A_oR_L}$$

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