If we assume does not turn on then the circuit

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If we assume does not turn on, then the circuit resembles that in Fig. 3.54(a), requiring that rise and turn on.

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BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 108 (1) 108 Chap. 3 Diode Models and Circuits Solution As rises from zero, attempting to place positive charge on the left plate of and hence draw negative charge from , the diode turns off. As a result, directly transfers the input change to the output for the entire positive half cycle. After , the input tends to push negative charge into , turning on and forcing . Thus, the voltage across remains equal to until , at which point the direction of the current through and must change, turning off. Now, carries a voltage equal to and transfers the input change to the output; i.e., the output tracks the input but with a level shift of , reaching a peak value of . Exercise Repeat the above example if the right plate of is 1 V more positive than its left plate at . We have thus far developed circuits that generate a periodic output with a peak value of or for an input sinusoid varying between and . We surmise that if these circuits are followed by a peak detector [e.g., Fig. 3.30(a)], then a constant output equal to or may be produced. Figure 3.57(a) exemplifies this concept, combining the circuit of Fig. 3.56 with the peak detector of Fig. 3.30(a). Of course, since the peak detector “loads” the first stage when turns on, we must still analyze this circuit carefully and determine whether it indeed operates as a voltage doubler. V in 1 D t D C 2 out V C 1 X t in V t in V C 1 C 2 2 C in V 1 C 2 in V C 1 C 2 V p in V C 1 C 2 2 t 3 t 4 in C 1 C 2 V t 1 t 5 V p V out V 2 p Figure 3.57 Voltage doubler circuit and its waveforms. We assume ideal diodes, zero initial conditions across and , and . In this case, the analysis is simplified if we begin with a negative cycle. As falls below zero, turns on,
BR Wiley/Razavi/ Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 109 (1) Sec. 3.5 Applications of Diodes 109 pinning node to zero. Thus, for , remains off and . At , the voltage across reaches . For , the input begins to rise and tends to deposit positive charge on the left plate of , turning off and yielding the circuit shown in Fig. 3.57. How does behave in this regime? Since is now rising, we postulate that also tends to increase (from zero), turning on. (If remains off, then simply transfers the change in to node , raising and hence turning on.) As a result, the circuit reduces to a simple capacitive divider that follows Eq. (3.111): (3.112) because . In other words, and begin from zero, remain equal, and vary sinu- soidally but with an amplitude equal to . Thus, from to , a change of in appears as a change equal to in and . Note at , the voltage across is zero because both and are equal to . What happens after ? Since begins to fall and tends to draw charge from , turns off, maintaining at . The reader may wonder if something is wrong here; our objective was to generate an output equal to rather than . But patience is a virtue and we must continue the transient analysis. For , both and are off, and each capacitor holds a constant voltage. Since the voltage across

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