The crude triangle-to-sine converter of can be improved considerably by rounding the sides of the triangular wave input, besides clipping it at the top and bottom. The circuit of provides a VTC with a slope of 1 V/V near the origin, where all diodes are off. As the magnitude of vT rises and approaches a diode drop, either D1 or D2 goes on, in effect switching R2 into the circuit. At this point, the slope of the VTC decreases to about R1y(R1 1 R2). As the magnitude of vT rises further and vS approaches two diode drops, either the D3-D4 pair goes on, clipping the top of the waveform, or the D5-D6 pair goes on, clipping the bottom. Let us arbitrarily impose Vsm 5 2 3 0.7 5 y2)1.4 5 2.2 V, and let us assume diodes with Is 5 2 fA and nVT 5 26 mV, so that at 0.7 V they draw 1 mA. To fi nd suitable values for R1 and R2, arbitrarily impose the following pair of constraints: (1) When vT reaches its positive peak Vtm, let the current through the D3-D4 pair be 1 mA; (2) When vT reaches half its positive peak, or Vtmy2, let the slope of the VTC match that of the sine function there, which one can readily prove to be cos 458, or 0.707 V/V.
(a)Guided by the above constraints, fi nd suitable values for R1 and R2. (b) Simulate the circuit via PSpice. Use a 1-kHz triangular wave with peak values of 62.2 V, and display both vT and vS versus time. Make multiple runs, each time changing the values of R1 and R2 a bit until you come up with a set that gives what you think is the best sine wave. (c) Using the optimized wave shaper of part (b) as basis, design a circuit that accepts a triangular wave with peak values of 65 V and yields a sine wave also with peak values of 65 V. Hint: at the input, replace R1 by a suitable voltage divider to accommodate the increased triangular wave while still meeting the aforementioned constraints. At the output, use a suitable amplifi er implemented with a 741-type op amp.