Answer 15.1

_psets/12.md

@@ -9,22 +9,152 @@ title: Problem Set 12
 {:.question}
 Show that the circuits in Figures 15.1 and 15.2 differentiate, integrate, sum, and difference.
 
+I'll take as given that op amps with negative feedback keep their inputs at the same voltage. I also
+assume ideal op-amps that draw no current from their inputs. All the proofs follow from [Kirchhoff's
+current law](https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws) (aka conservation of charge).
+I write $$R_f$$ to refer to the resistor in the feedback path.
+
+For the integrator,
+
+$$
+\begin{align*}
+\frac{V_i}{R} &= - C \frac{dV_o}{dt} \\
+\frac{dV_o}{dt} &= \frac{-V_i}{R C} \\
+V_o &= - \frac{1}{RC} \int V_i \mathrm{d}t
+\end{align*}
+$$
+
+For the differentiator,
+
+$$
+\begin{align*}
+\frac{-V_o}{R} &= C \frac{dV_i}{dt} \\
+V_o &= -RC \frac{dV_i}{dt}
+\end{align*}
+$$
+
+For the summing amplifier,
+
+$$
+\begin{align*}
+\frac{V_1 + V_2 + V_3}{R_i} &= -\frac{V_o}{R_f} \\
+V_o &= -\frac{R_f}{R_i} (V_1 + V_2 + V_3)
+\end{align*}
+$$
+
+For the difference amplifier, neither input terminal is grounded. Call their voltage $$V_i$$.
+Applying Kirchhoff's current law to the non-inverting input,
+
+$$
+\begin{align*}
+\frac{V_2 - V_i}{R_i} &= \frac{V_i}{R_f} \\
+\frac{V_2}{R_i} &= V_i \left( \frac{1}{R_i} + \frac{1}{R_f} \right)
+\end{align*}
+$$
+
+Then looking at the inverting input,
+
+$$
+\begin{align*}
+\frac{V_1 - V_i}{R_i} &= \frac{V_i - V_o}{R_f} \\
+\frac{V_1}{R_i} - \frac{V_o}{R_f} &= V_i \left( \frac{1}{R_i} + \frac{1}{R_f} \right) \\
+\frac{V_1}{R_i} - \frac{V_o}{R_f} &= \frac{V_2}{R_i} \\
+V_o &= \frac{R_f}{R_i} (V_2 - V_1)
+\end{align*}
+$$
+
 ### (b)
 
 {:.question}
 Design a non-inverting op-amp amplifier. Why are they used less commonly than inverting ones?
 
+Connect the signal you want to amplify to the non-inverting input. Then create a voltage divider,
+connected to ground on one end, the inverting input in the middle, and the output on the other end.
+I'll refer to the resistor connected to ground as $$R_g$$, and the feedback resistor $$R_f$$ as
+before.
+
+$$
+\begin{align*}
+\frac{V_o - V_i}{R_f} &= \frac{V_i}{R_g} \\
+\frac{V_o}{R_f} &= V_i \frac{R_f + R_g}{R_f R_g} \\
+V_o &= V_i \left( 1 + \frac{R_f}{R_g} \right)
+\end{align*}
+$$
+
+A potential downside is that the gain must be at least unity. A potential upside is that the input
+impedance is very high. And, of course, the output isn't inverted.
+
 ### (c)
 
 {:.question}
 Design a transimpedance (voltage out proportional to current in) and a transconductance (current out
 proportional to voltage in) op-amp circuit.
 
+For a transimpedance amplifier, replace the input resistor in an inverting amplifier with a wire.
+Then the current flowing through the feedback resistor is equal to the current at the input. So $$I
+= V_o / R_f$$, and $$V_o = R_f I$$.
+
+For a transconductance amplifier, replace the feedback resistor with a wire. Then the current out
+is $$I = -V_i / R_i$$. If you want a non-inverting transconductance amplifier, connect the input
+straight to the non-inverting terminal and connect the inverting terminal to ground via a resistor
+$$R_g$$. Then $$I = V_i / R_g$$.
+
 ### (d)
 
 {:.question}
 Derive equation (15.16).
 
+The currents flowing through $$R_1$$, $$R_o$$, and $$C$$ must be equal. Call this current $$I$$
+(I'll take left to right to be positive for all currents). Because of the negative feedback, the
+inverting input of the op-amp is a virtual ground. Recall that $$Q = CV$$, so $$dQ/dt = I = C
+dV/dt$$. Thus we can express $$I$$ three ways:
+
+$$
+\begin{align*}
+I &= \frac{V_{PD}}{R_1} \\
+&= -\frac{V_{RC}}{R_0} \\
+&= C \left( \frac{dV_{RC}}{dt} - \frac{dV_F}{dt} \right)
+\end{align*}
+$$
+
+I'm using $$V_{RC}$$ to denote the voltage at the junction between $$R_O$$ and $$C$$.
+Differentiating the first two lines yields
+
+$$
+\begin{align*}
+\frac{dI}{dt} &= \frac{1}{R_1} \frac{dV_{PD}}{dt} \\
+&= -\frac{1}{R_0} \frac{dV_{RC}}{dt}
+\end{align*}
+$$
+
+Thus
+
+$$
+\frac{dV_{RC}}{dt} = -\frac{R_0}{R_1} \frac{dV_{PD}}{dt}
+$$
+
+As such we can write the current through the capacitor as
+
+$$
+I = C \left( -\frac{R_0}{R_1} \frac{dV_{PD}}{dt} - \frac{dV_F}{dt} \right)
+$$
+
+This can still be equated
+
+$$
+C \left( -\frac{R_0}{R_1} \frac{dV_{PD}}{dt} - \frac{dV_F}{dt} \right) = \frac{V_{PD}}{R_1}
+$$
+
+So all that remains is to solve for $$dV_F/dt$$.
+
+$$
+\frac{dV_F}{dt} = \frac{V_{PD}}{R_1 C} + \frac{R_0}{R_1} \frac{dV_{PD}}{dt}
+$$
+
+This differs in sign from (15.16) because the book makes the positive current direction right to
+left.
+
+
 ## (15.2)
 
 {:.question}