The diodes keep the output waveform from getting too large.

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1 Wien Bridge Oscillat CIRCUIT: The Wien bridge oscillat, see Fig., consists of two voltage dividers. It oscillates (approximately) sinusoidally at the frequency that produces the same voltage out of both voltage dividers. When the two wavefms from the voltage divider are the same, the op-amp sees two wavefms that differ by the offset voltage of the op-amp, which is nonzero. This causes the output wavefm at that frequency to be large, since changing the size of the output signal cannot eliminate the offset. This renders the negative feedback ineffective, and the open-loop gain of the op-amp applies to the offset voltage. The Wien-bridge oscillat circuit is followed by a voltage follower with a DCblocking capacit. Fig.. Wien-bridge oscillat. The diodes keep the output wavefm from getting too large.

2 Wien Bridge Oscillat (cont.) In terms of poles, the circuit has poles just to the right of the real axis. The diodes ensure that the poles are just to the right of the real axis at all times, preventing the output wavefm from dying out if the poles were to drift into the left half-plane. Analysis of circuit (skip to "To adjust the oscillat" if desired): We equate the voltages out of the two voltage dividers and perfm many steps of algebra to find the poles. + s + + R s s + + sc + s + sc s + sc + s + + sc s

3 Wien Bridge Oscillat (cont.) + sc s + sc sc + s + s C R sc + s + s s C + s( C + ) + s C s + s + C + R s 3 C s + s + C C + C s + s + τ τ + τ s + τ τ R 4 ± + τ τ R 4 τ

4 Wien Bridge Oscillat (cont.) F purely imaginary roots, we want the real part (in front of the square root), to be zero. In actuality, we want the real part to be just slightly positive so our poles will be in the right half-plane. + τ τ 0 + τ τ + C C C + C + C + C F the roots to be in the right half-plane, / must be larger than the right side of the equation. The diodes in the circuit, move the poles toward the left half-plane by lowering the value of when the amplitude of the output signal gets too large. s ± j τ ± j C If C nf, we get poles at audio frequencies using R's that draw only modest current from the op-amp.

5 Wien Bridge Oscillat (cont.) s ± j C ± jgr/s F example, if we want an oscillation frequency at khz, we have the following calculations: s ± jgr/s ± jgr/s ± jk Hz π k Hz π Gr/s Gr/s k Hz π ( 60kΩ) F we have a solution with a standard 5% resist value: 60 kωω F this solution, we need f poles on the imaginary axis. Now we turn to the question of how to pick so that the peak current through will result in a voltage across the diodes that will conduct enough current to move poles to the left half-plane. That is, the current in the diodes should be a significant proption of the current in. We assume a desired magnitude V sinusoidal output. TABLE I: N94 DIODE I-V DATA (FAIRCHILD) [] i d v d µa 0.73 V 5 µa 0.35 V 0 µa V 00 µa V

6 Wien Bridge Oscillat (cont.) Since and the output voltage is across, we will have V across and V across. Say the value of 0 kωω. i R3 V 00kΩ 0µA Say we wish to have 5% of the current in, i.e., 5 µa, flowing in the diodes. This effectively reduces the resistance to an equivalent of 80 kωω. Thus, we adjust to 40 kωω so it will be half the value of the equivalent. From the data f the N94 diode, the voltage drop across the diodes will be 0.35 V. The part of that drops 0.35 V when the current is 0 µa is given by Ohm's law: top.35v 0µA 35.kΩ To adjust the oscillat, first adjust until the circuit oscillates. Then adjust until the amplitude of the oscillation is as desired. REF: [] (accessed 3 July 07)

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