Electronic Bell Construction and Circuit Analysis

Introduction

This circuit is based on the article Percussion Instrument Synthesizer by Forrest M. Mims, Popular Electronics, March, 1976, pgs. 90 - 102. which describes an audio amplifier with a twin-T notch filter in the feedback loop,

The R-C network gives a sharp peak at a single audio frequency at which feedback occurs. Disturbing the touch plate introduces a transient which starts a damped oscillation "ringing". I've added a Hall effect sensor on the input instead of a touch plate to inject the transient, and an audio power amplifer on the output.

Analyzing the RC Network

Kirchhoff's Laws

Kirchoff's Laws can be derived from Maxwell's equations for discrete circuit elements. Kirchoff's current law assumes that charge is constant in time at a given node: $\frac{ \partial Q(t) }{ \partial t} = 0$. Kirchoff's voltage law assumes magnetic field density is constant in time within a closed path: $\frac{ \partial B(t) }{ \partial t} = 0$. So they should be exact for DC and fairly accurate for lower frequency AC.

1. $i_1 - i_3 - i_6 = 0$
2. $i_4 - i_2 + i_7 = 0$
3. $i_3 - i_4 - i_5 = 0$
4. $i_6 - i_7 - i_8 = 0$

Recall the convention that currents flowing into the node are taken as positive.

First an aside, recall the Laplace transform of a function $f$ is defined to be $\mathscr{L}\{ f(t) \} = \int_0^{\infty} f(t) e^{-st} dt$ and has the property $\mathscr{L}\{ f'(t) \} = s \mathscr{L} \{ f(t) \} - f(0)$ and that the transform is linear.

Recall that the voltage across a resistor is $V = iR$. A capacitor has the equation $\frac{d V}{d t} = {1 \over C} i$ since capacitance is defined by $CV = q$. Then we have $$\mathscr{L} \{ V(t) \} = {1 \over s C} \mathscr{L} \{ i(t) \}$$ if we assume the initial voltage $V = 0$ at $t = 0$

Now we apply Kirchhoff's voltage law on the meshes 1-4. But first we apply the Laplace transform to all voltage terms in the loop using linearity and assuming initial currents and voltages are zero. For simplicity of notation, let's use $i(t)$ to denote $\mathscr{L} \{ i(t) \}$, and $V(t)$ to denote $\mathscr{L} \{ V(t) \}$.

1. $R_1 i_3 + \frac{1}{s C_1} i_5 - V_1 = 0$
2. $\frac{1}{s C_2} i_6 + R_2 i_8 - V_1 = 0$
3. $\frac{1}{s C_1} i_5 - V_2 - i_4 R_1 = 0$
4. $R_2 i_8 - V_2 - \frac{1}{s C_2} i_7 = 0$

Solving the Equations

Now we'll solve the equations by substitution. Mesh equation 1 gives $$i_5 = (V_1 - R_1 i_3) s C_1$$ Substituting $i_5$ into node equation 3 gives $$\text{ * } i_3 - i_4 - (V_1 - R_1 i_3) s C_1 = 0$$ Substituting $i_5$ into mesh equation 3 gives $$(V_1 - R_1 i_3) - V_2 - i_4 R_1 = 0$$ then $$i_4 = \frac{V_1 - R_1 i_3 - V_2}{R_1}$$ Substituting $i_4$ into equation * above gives, after rearrangement, $$i_3 = \frac{ \frac{ V_1 - V_2}{R_1} + s C_1 V_1 }{ 2 + s C_1 R_1}$$ Now solve mesh equation 4 for $i_8$ to get $$i_8 = \frac{V_2 + \frac{i_7}{s C_2} }{R_2}$$ Substitute $i_8$ into mesh equation 2 to get after rearrangement, $$i_7 = (V_1 - V_2 - \frac{i_6}{s C_2}) s C_2$$ Substitute $i_8$ into node equation 2 to get $$\text{ ** } i_6 - i_7 - (\frac{V_2 + \frac{i_7}{s C_2}}{R_2}) = 0$$ Substitute $i_7$ into equation ** above to get after some rearrangement, $$i_6 = \frac{ (V_1 - V_2) s C_2 + \frac{V_1}{R_2}}{2 + \frac{1}{s C_2 R_2}}$$ Finally, use node equation 1 to give $$i_1 = \frac{ \frac{ V_1 - V_2}{R_1} + s C_1 V_1 }{ 2 + s C_1 R_1} + \frac{ (V_1 - V_2) s C_2 + \frac{V_1}{R_2}}{2 + \frac{1}{s C_2 R_2}}$$ This is almost what we need because it relates input voltage $V_1$ to output voltage $V_2$ given only input current $i_1$.

Analyzing the Amplifier Plus RC Network

The Transfer Function H(s)

Returning to our complete circuit, we draw it using the twin-T notch filter block just analyzed, an ideal operational amplifer, and a voltage source. The ideal OpAmp has $\infty$ input impedance, $0$ output impedance, perfect linearity and gain $\mu \rightarrow \infty$ It's equation is $V_{o} = -\mu V_{i}$. The voltage source has internal resistance $r$. By Kirchhoff's voltage law, recalling we are using the Laplace transforms of voltage and current here, the mesh equation for the input is $$V_1 = -e + i_1 r = 0$$ Now lets eliminate $i_1$ and $V_1$ in the Twin-T equation using $$i_1 = \frac{e-V_1}{r}$$ $$V_1 = -\frac{V_2}{\mu}$$ giving $$\frac{e + \frac{V_2}{\mu} }{r} =$$ $$\frac{ \frac{ -\frac{V_2}{\mu} - V_2 }{R_1} + s C_1(-\frac{V_2}{\mu}) }{ 2 + s C_1 R_1} + \frac{ (-\frac{V_2}{\mu} - V_2) s C_2 + \frac{ -\frac{V_2}{\mu} }{R_2} }{2 + \frac{1}{s C_2 R_2}}$$ But the ideal op amp open loop gain $\mu$ is very high, so let $\mu \rightarrow \infty$ giving $$ \frac{e}{r} = \frac{ \frac{ -V_2 }{R_1} }{ 2 + s C_1 R_1 } + \frac{ (- V_2) s C_2 }{2 + \frac{1}{s C_2 R_2}}$$ Now let's put over a common denomiator, and invert to get $$ \frac{ V_2 }{ e } = -r \frac{ (2 + s C_1 R_1)(2 + \frac{1}{s C_2 R_2}) } { \frac{1}{R_1} (2 + \frac{1}{s C_2 R_2}) + s C_2 (2 + s C_