Professors HW3 Solutions

Professors HW3 Solutions - EE323 Homework 3- Solution,...

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EE323 Homework 3--- Solution, Au/2004 3.54 (25points) (a) (10points) To find the small-signal response, open the dc current source I and short the capacitors 1 C and 2 C . Also, replace the diode with its small signal resistance: T d nV r I = . s R and the diode form a voltage divider, such that T d T os s s T ds T s s nV r nV I vv v v nV rR n VI R R I == = ++ + (b) (10points) Substitute given values into the above equation, we obtain 4 6 0.476 for 1 0.001 20 . 0 2 5 0.01 3.333 for 0.1 10 2 0.025 1000 9.804 for 1 10 mV mA mV mA o mV A I A vI A I I A µ × = = ×+ (c) (5points) To have 0.5 = : 5 0.05 0.5 5 10 0.05 50 0.05 1000 ss I A m A A I = = = + 3.78 (25points) (a) (6 points) Refer to the full wave rectifier of fig (3.27) on Page 176. The ac voltage between the two terminals of the secondary winding is: sin 120 2 sin (1/10) 12 2 sin vV t t t ωω ω = . Note 0.7 D VV = , so the peak value o V of the rectified voltage ( ) o vt across the load R is: 2 12 2 2 0.7 15.57 D V V =− = −× = (b) (6 points) For 12 and DD , conduction begins at an angle t θω = when sin 2 s sD V θ , i.e.
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This note was uploaded on 02/21/2011 for the course ECE 205 taught by Professor Hemami during the Spring '08 term at Ohio State.

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Professors HW3 Solutions - EE323 Homework 3- Solution,...

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