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

Let's analyze the network in isolation and add the amplifier and input later.

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: Q(t)t=0. Kirchoff's voltage law assumes magnetic field density is constant in time within a closed path: B(t)t=0. So they should be exact for DC and fairly accurate for lower frequency AC.

First apply Kirchhoff's current law on nodes 1-4 gives

  1. i1i3i6=0
  2. i4i2+i7=0
  3. i3i4i5=0
  4. i6i7i8=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 L{f(t)}=0f(t)estdt and has the property L{f(t)}=sL{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 dVdt=1Ci since capacitance is defined by CV=q. Then we have L{V(t)}=1sCL{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 L{i(t)}, and V(t) to denote L{V(t)}.

  1. R1i3+1sC1i5V1=0
  2. 1sC2i6+R2i8V1=0
  3. 1sC1i5V2i4R1=0
  4. R2i8V21sC2i7=0

Solving the Equations

Now we'll solve the equations by substitution. Mesh equation 1 gives i5=(V1R1i3)sC1 Substituting i5 into node equation 3 gives  * i3i4(V1R1i3)sC1=0 Substituting i5 into mesh equation 3 gives (V1R1i3)V2i4R1=0 then i4=V1R1i3V2R1 Substituting i4 into equation * above gives, after rearrangement, i3=V1V2R1+sC1V12+sC1R1 Now solve mesh equation 4 for i8 to get i8=V2+i7sC2R2 Substitute i8 into mesh equation 2 to get after rearrangement, i7=(V1V2i6sC2)sC2 Substitute i8 into node equation 2 to get  ** i6i7(V2+i7sC2R2)=0 Substitute i7 into equation ** above to get after some rearrangement, i6=(V1V2)sC2+V1R22+1sC2R2 Finally, use node equation 1 to give i1=V1V2R1+sC1V12+sC1R1+(V1V2)sC2+V1R22+1sC2R2 This is almost what we need because it relates input voltage V1 to output voltage V2 given only input current i1.

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 input impedance, 0 output impedance, perfect linearity and gain μ It's equation is Vo=μVi. 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 V1=e+i1r=0 Now lets eliminate i1 and V1 in the Twin-T equation using i1=eV1r V1=V2μ giving e+V2μr= V2μV2R1+sC1(V2μ)2+sC1R1+(V2μV2)sC2+V2μR22+1sC2R2 But the ideal op amp open loop gain μ is very high, so let μ giving er=V2R12+sC1R1+(V2)sC22+1sC2R2 Now let's put over a common denomiator, and invert to get V2e=r(2+sC1R1)(2+1sC2R2)1R1(2+1sC2R2)+sC2(2+sC1R1) After more rearranging we get the transfer function of our circuit, H(s)=r(2+sC1R1)(2sC2R2+1)(R1R1C1R2C2R1C2)s3+(2R1C1)s2+(2R1C1R1C2)s+(1R1C1R2C2R1C2) where V2(s)=H(s)e(s)

Ringing the Bell

To ring the bell in operation, one applies a sharp impulse to the input. To approximate this, set e(t)=δ(t), the Dirac delta function. Then e(s)=L{δ(t)}=1 The solution will be V2(t)=L1{H(s)}

Time Domain Solution

Inverting the Laplace Transform

From complex analysis, for t0, the inverse Laplace transform for Re(s)>σ is given by h(t)=L1{H(s)}={ Residues of estH(s) at each of its poles } since H(s) is analytic on C except at a finite number of poles, in particular, analytic on the half plane {s|Re(s)>σ} and |H(s)|M|z||z|R for constants M>0 and R>0 Recall for a simple pole s0 the residue is Res(H(s),s=s0)=limss0(ss0)H(s) If we write H(s)=N(s)D(s), the roots of the cubic D(s)=s3+a2s2+a1s+a0=0 are the poles of H(s) where a2=2R1C1,a1=2R1C1R1C2,a0=1R1C1R1C2R2C2 and let the roots be s1,s2,s3 Assuming the roots are all distinct, we have the solution h(t)=limss1N(s)(ss1)est(ss1)(ss1)(ss1)+ limss2N(s)(ss2)est(ss2)(ss2)(ss2)+ limss3N(s)(ss3)est(ss3)(ss3)(ss3) or h(t)= N(s1)es1t(s1s2)(s1s3)+ N(s2)es2t(s2s1)(s2s3)+ N(s3)es3t(s3s1)(s3s2) Only the general form of the solution will concern us, V2(t)=k1es1t+k2es2t+k3es3t That is we don't need to know the exact amplitude of the voltage, but only the frequencies of its sinusoidal components, or the relative amplitiude of its exponential solutions, if any. Or problem is then to find the general roots of the cubic equation.

Cubic Equation Solution

From classical algebra, the roots of the cubic s3+a2s2+a1s+a0=0 are given by the following formula, Let Q3a1a229 and R9a2a127a02a2354 Let the discriminant be D=Q3+R2 Let SR+D3 and TRD3 Then the roots are s1=13a2+(S+T) s2=13a212(S+T)+12i3(ST) s3=13a212(S+T)12i3(ST)

If D=0 all roots are real and at least two are equal, and if D<0 all roots are real and unequal. The solutions in this case would be of the form ke±ut, either growing exponential solutions (output increasing until limited by the circuit?) or decaying exponential solutions (maybe shows up as a click?). I'm not sure about the existence of these.

If D>0 we have one real root and a complex conjugate pair of roots s1=γ, s2=u+vi, s3=uvi which would give us one solution of the form keγt as above and two solutions ±keutsin(vt).

Conditions for Damped Oscillations

Since our cubic has real coefficients, if Q>0 then D=Q3+R2>0. But Q=3(2R12C1C2)(2R1C1)2= 6C14C29R12C12C2>0 iff 3C1>2C2 since the denominator > 0.

Resonant Frequency at the Edge of Oscillation

Recall Newton's formulas for the coefficients in terms of the roots, s1+s2+s3=a2 s1s2+s1s3+s2s3=a1 s1s2s1s3=a0

Then multiplying out the roots gives a2=γ+u+iv+uiv=γ+2u a1=γ(u+iv)+γ(uiv)+(u+iv)(uiv) =2γu+(u2+v2) a0=γ(u+iv)(uiv)=γ(u2+v2)

On the cusp of steady oscillation, u=0 and we can simplfy the equations to a2=γ a1=v2 a0=γv2

Combining the equations gives a0=a2a1 or 1R2C2=4R1C1

The frequency of the oscillating solution is f=v2π=a12π=12πR12C1C2 and the transient term decays as ke2R1C1t

For this design, pick the capacitances to satisfy 3C1>2C2 pick a small value for R1C1 pick a frequency f which gives R1 as R1=12πf2C1C2 finally R2 is R2=R1(C14C2)

Circuit Schematic

NOTES:

  • The dual 100K resistors connected to pin 3 of LM308 provide a floating ground for this op amp.
  • You'll get excessive current drain unless the 330pF capacitor is connected to pin 2 of LM380N. The chip will actually heat up! Parasitic oscillation in the ultrasound range, perhaps?
  • The 150K resistor at the output of the LM308 isolates the two amplifiers. Again, hissing and high current drain happened when the two amplifiers were directly connected.
  • High current drain also occurred when pin 12 of the LM380N was grounded. I don't know the reason.
  • Quiescent power drain is about 7mA

Operation

  1. Turn the coarse and fine duration adjustment potentiometers clockwise as far as they will go. There will be a continuous tone.
  2. Adjust the pitch.
  3. Adjust the loudness. Not too loud or the mini speaker will saturate and the tone will sound harsh.
  4. Rotate the coarse adjust clockwise until oscillation stops, the clockwise until it barely begins again.
  5. Rotate the fine adjust counterclockwise to set the duration of the bell tone, while ringing the bell with the push button switch.
  6. You might have to tweak all the pots for best results.
  7. Use a fresh 9V battery.

Construction

The electronic bell simulator is enclosed in a small aluminum chassis box 4cm X 5cm X 10cm to fit comfortably in the hand.

Mount the circuit on a single-sided fiberglass-epoxy circuit board. Make the ground paths wide for circuit stability and ground to the chassis. Wiring is #26 stranded Cu wire. I plotted speaker holes on polar coordinate graph paper which I used as a template for drilling.

The moving parts of the Hall effect sensor switch came from brass stock and a disassebled push button switch. Lubricate with vacuum grease or plumber's grease.

Lay out the PC board pattern on a sheet of graph paper in colored pencil. Tape to the copper foil side of the circuit board and use an awl to make indentations at the positions of the holes. Draw circuit layout by hand on the Cu foil with a resist pen. Correct excess resist with a dissecting needle.

Electronic Bell Layout.

Keep all connections short as the circuit is very sensitive.

Lettering is rub-on stencil, protected by clear-seal transparent plastic adhesive film. The box is polished with fine sandpaper and steel wool.