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Wien Bridge Oscillator

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Wien Bridge Oscillator Wien Bridge Oscillator Conventional analysis Consider the amplifier shown below. The potential at the noninverting input is equal to the input signal: v+ = vi . The potential at the invertiing input is related to the output via a voltage-divider: Rf1 v− = vo . Rf1 + Rf2 Using the ideal op-amp assumption that v+ = v– leads to vo = Gvi where Rf2 G ≡ 1 + . Rf1 This is the standard non-inverting amplifier configuration. Now let's ask if we can sustain a finite output if, instead of an external input, we feed the output back to the input through a frequency-dependent network. Using complex phasor notation such that v = ℝVê jωt (where ℝ means "real part of") , we write Vî = H(ω)Vô . We also had Vô = G(ω)Vi.̂ (Note: in the current case G actually does not have any frequency dependence. The notation is for purposes of generality.) Self-consistency requires Vî = G(ω)H(ω)Vi.̂ This requires that the loop gain G(ω)H(ω) = 1. This is the Barkhausen condition for oscillation, which implies both that the magnitude of the loop gain is unity and that the phase shift is zero or a multiple of 2π . Wien Bridge Consider the resistance - capacitance network shown above. Z1 i = o. Z1 Vî = Vô . Z1 + Z2 R 1+jωRC Vî = Vô . R R j 1 1+jωRC + ( − ωC ) 1 Vî = Vô . j ωRC 1 3 + ( − ωRC ) Thus 1 H(ω) = . j ωRC 1 3 + ( − ωRC ) Remember that Vô = GVî with G having the real value Rf2 G = 1 + Rf1 . So the loop gain requirement GH(ω) = 1 becomes G = 1. j ωRC 1 3 + ( − ωRC ) This is true provided G = 3. 1 ω = . RC These then are the conditions for oscillation with this network which is called the Wien bridge. Note that the above conditions then require Rf2 = 2. Rf1 In the practical circuit, we will select resistor Rf2 so that it can be varied until sufficient gain is reached to start oscillations. We consider Rf2 to be the bifurcation parameter of the system. Time-Domain Analysis The above criteria are useful in creating the oscillator design. However, insight into the oscillator dynamics - leading to an understanding of what controls the amplitude and shape of oscillation - requires a time-domain analysis. Consider the same circuit, re-drawn below with currents and voltages indicated at various points in the circuit. Linear circuit analysis We assume that all components have constant values. Later, the analysis will be generalized to allow the feedback resistor Rf1 to increase with potential drop across the resistor due to self-heating, introducing nonlinearity into the system and thereby limiting the amplitude of oscillation. Signal at the non-inverting input Let's first first sum potential drops across the R2C2 branch. 1 v+ + i2R2 + q2 = vo . C2 Differentiate this equation and use the fact that i2 = dq2 /dt: dv+ di2 1 dvo + R2 + i2 = . dt dt C2 dt Now from Kirchoff's Law for the sum of currents into a junction, using the ideal op amp behavior that zero current flows into the non-inverting (+) input: i2 = iR1 + iC1. Thus 1 dv+ i2 = v+ + C1 . R1 dt Inserting this into equ. [xx] we get: dv+ d 1 dv+ 1 1 dv+ dvo + R2 v+ + C1 + v+ + C1 = . dt dt ( R1 dt ) C2 ( R1 dt ) dt Collecting terms: d2 R C d dv R C 2 1 1 v o 2 1 2 + 1 + + + + = . [ dt ( R1 C2 ) dt R1C2 ] dt Signal at the inverting input At the inverting input we have Rf1 v– = v0 . Rf1 + Rf2 1 v– = v0 . Rf2 1 + Rf1 1 v = v , – G 0 where the gain is given by Rf2 G ≡ 1 + . Rf1 Combined results Now use again use ideal op-amp behavior to require that v+ = v– , so that in the above differential equation we can substitute for v+ using 1 v = v = v . + – G 0 Thus we re-write the original differential equation entirely in terms of v0 : 2 d R2 C1 d 1 1 dvo R C vo 2 1 2 + 1 + + + = . [ dt ( R1 C2 ) dt R1C2 ] G dt Multiply through by G and re-arrange terms: 2 d vo R2 C1 1 dvo 1 G vo 2 + 1 + + − + = 0. dt ( R1 C2 ) R2C1 dt (R1C2)(R2C1) If we match resistors so that R1 = R2 ≡ R and capacitors so that C1 = C2 ≡ C then this simplifies to: 2 d vo 1 dvo 1 + (3 − G) + vo = 0. dt2 RC dt (RC)2 Solving the linear equation st Again, assuming that all of the coefficients are constant, we can insert a soluton vo = Voe leads to the characteristic equation, 1 1 s2 + (3 − G) s + = 0. RC (RC)2 The solution is ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ (G − 3) ± √(G − 3)2 − 4 s = ω0. 2 where ω0 ≡ 1/RC. The values are complex for 1 < G < 5 and the solution grows exponentially when G > 3. Of course, indefinite exponential growth is not possible; eventually the power supply limits are reached. However, if an introduction of nonlinearity causes the gain to diminish with growing amplitude the oscillations will saturate. In such a case we have a bifurcation to an oscillatory solution at G = 3. This occurs when Rf2 = 2, Rf1 for which case s = ±jω0. Note that we can re-write the differential equation (flipping the middle term to conform better with the form of the Van der Pol equation - see below) as: 2 Rf d vo 2 dvo 2 ω ω vo 2 − − 2 0 + 0 = 0. dt ( Rf1 ) dt In [2]: # TO BE ADDED Code to calculate eigenvalues as a function of feedback resistor ratio Amplitude-dependent gain We will now model the feedback resistor Rf1 as a lamp filament that heats up as the output v0 increases from zero. As the filament temperature T increases from a reference temperature T0 , Rf1 (T) = Rf10[1 + α(T − T0 )]. We need to model how the temperature of the lamp filament varies with power P dissipated within the filament, which in turn depends on the potential drop across the filament and temperature-dependent resistance of the filament. 2 Rf T 2 vo 1 ( ) 2 P = (if1) Rf1 (T) = Rf1 (T) = vo . ( Rf (T) + Rf ) 2 1 2 [Rf1 (T) + Rf2 ] Suppose the filament has heat capacity Cfil . Let us assume that the heat transport away from the filament is proportional to the difference between the filament temperature T and the ambient temperature Ta : 1 Q˙ = (T − Ta ), transfer ℜ where ℜ is the effective thermal resistance of the bulb. Even radiative transport is modeled this way near onset when the temperature difference is still modest relative to the initial absolute temperature (about 300K): 4 4 4 4 3 εσ(T − Ta ) = εσ[Ta + (T − Ta )] − Ta ≈ 4εσTa (T − Ta ). (Here ϵ is the emissivity of the filament and σ is the Stefan-Boltzman constant.) So the heat balance equation is dT f = P − Q˙ . C dt transfer Using above results: Rf T dT 1 ( ) 2 1 Cf = vo − (T − Ta ). dt 2 ℜ [Rf1 (T) + Rf2 ] Inserting the temperature dependence of the resistance: Rf α T T dT 10[1 + ( − 0 )] 2 1 Cf = vo − (T − Ta ). dt 2 ℜ {Rf10[1 + α(T − T0 )] + Rf2 } Re-writing: 2 dT 1 + α(T − T0 ) vo 1 f T Ta C = 2 − ( − ). dt Rf2 Rf10 ℜ 1 + + α(T − T0 ) [ Rf10 ] If we assume that the oscillation period is much shorter than the time to reach thermal equilibrium, we suppose that the system reaches a steady state that everages over the heating per cycle by the signal at a given amplitude. Then we say that dT f ≈ 0. C dt This requires that 2 − (T − a ) = 0. 2 1 + α(T − T0 ) vo 1 T Ta 2 − ( − ) = 0. Rf2 Rf10 ℜ 1 + + α(T − T0 ) [ Rf10 ] 2 As we will see, the temperature difference T − T0 ∼ vo , so the equation above can be simplified to lowest order in vo2 : 1 vo2 1 T Ta 2 − ( − ) = 0. Rf2 Rf ℜ 1 + 10 [ Rf10 ] We'll also assume that the ambient temperature and reference temperature for the filament are the same: Ta = T0 . Then we have: 1 ℜ 1 ℜ T T vo2 vo2 ( − 0 ) ≈ 2 ≈ , Rf2 Rf 9 Rf 1 + 10 10 [ Rf10 ] where we have used the critical value of the resistance ratio Rf2c = 2. Rf10 The dynamical equation including temperature dependence of the resistance Rf1 is: 2 Rf d vo 2 dvo 2 ω ω vo 2 − − 2 0 + 0 = 0. dt ( Rf10[1 + α(T − T0 )] ) dt Assuming, at least near the onset of oscillations, that the self-heating is modest and thus the correction is small compared to 1: 2 Rf d vo 2 dvo 2 α T T ω ω vo 2 − [1 − ( − 0 )] − 2 0 + 0 ≈ 0. dt ( Rf10 ) dt Substituting for T − T0 : 2 Rf d vo 2 1 ℜ 2 dvo 2 α vo ω ω vo 2 − 1 − − 2 0 + 0 = 0. dt ( Rf10 [ 9 Rf10 ] ) dt 2 Rf Rf d vo 2 2 1 αℜ 2 dvo 2 vo ω ω vo 2 − − 2 − 0 + 0 = 0. dt ( Rf10 Rf10 9 Rf10 ) dt 2 Rf d vo 2 2 αℜ 2 dvo 2 vo ω ω vo 2 − − 2 − 0 + 0 = 0. dt ( Rf10 9 Rf10 ) dt where we used in the last term Rf2 ≈ 2 Rf2 ≈ 2 Rf10 near the onset of oscillations. Scaled equation The model equation is 2 Rf d vo 2 2 αℜ 2 dvo 2 vo ω ω vo 2 − − 2 − 0 + 0 = 0.
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