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would be? is

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Martin Kennedy 2025-05-05 03:31:55 -04:00
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Final.tex
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@ -101,7 +101,7 @@ properties of an ideal op-amp, this is easy to see by inspection, as
when $v_p > v_n$, $v_{out}$ is driven as high as possible, and when $v_p > v_n$, $v_{out}$ is driven as high as possible, and
vice-versa when $v_p < v_n$. vice-versa when $v_p < v_n$.
At this point, it would be easy to derive the relationship between At this point, it is easy to derive the relationship between
$v_{out}$ and $v_{in}$: $v_n = v_p = \SI{0}{V}$; since all current $v_{out}$ and $v_{in}$: $v_n = v_p = \SI{0}{V}$; since all current
which flows from $v_{in}$ into $v_n$ then flows from $v_n$ into which flows from $v_{in}$ into $v_n$ then flows from $v_n$ into
$v_{out}$, the network of $R_1$ and $C_1$ into $R_2$ forms a voltage $v_{out}$, the network of $R_1$ and $C_1$ into $R_2$ forms a voltage