HW5soln - PHYS3360/AEP3630 Electronic Circuits, Spring...

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Unformatted text preview: PHYS3360/AEP3630 Electronic Circuits, Spring 2011, HW5 Solutions 1 Diodes Solutions 1. Remember that the charge on a capacitor is proportional to the voltage across it, Q = C V, and that a diode is on when the voltage across it is nonnegative. Let Q = C Vmax and define all capacitor charge to be positive. Whenever a diode is on, charge will flow until the voltage on either side of it is equal. Thus, we require V1 0, V2 V1, V3 V2, and Vout V3, for all possible charge configurations. This implies positive charge will only accumulate on the right hand plate of each capacitor (as drawn). For the following argument, refer to the diodes from left to right as D1, D2, D3, and D4. When Vin = - Vmax, D1 is on, and C1 accumulates charge Q. Then, when Vin = 0, initially V1=Vmax, so D1 is now off and D2 is on, allowing some of the charge to move to C2. Charge will flow until the potential difference across D2 is zero; it will distribute itself evenly, giving Q1=Q/2 and Q2=Q/2 (if you feel that some charge should also move across D3, note that the instant a single charge crosses to C3, V3=V1+Q3/C>V2 and D3 switches off). Continuing this line of reasoning, one finds that the following is true in general. When Vin = 0, Q1+Q2 distributes evenly between C1 and C2, while Q3+Q4 distributes evenly between C3 and C4. When Vin = - Vmax, Q1 = Q, Q2+Q3 distributes evenly between C2 and C3, and Q4 does not change. Working through the math gives the following table of values (in charge units of C Vmax and voltage units of Vmax): a- d....
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HW5soln - PHYS3360/AEP3630 Electronic Circuits, Spring...

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