{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Professors HW3 Solutions

Professors HW3 Solutions - EE323 Homework 3 Solution...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 o s s s T d s T s s nV r nV I v v v v nV r R nV IR R I = = = + + + (b) (10points) Substitute given values into the above equation, we obtain 4 6 0.476 for 1 0.001 2 0.025 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 v I A I I A µ = = × = = = = × + = = (c) (5points) To have 0.5 o s v v = : 5 0.05 0.5 5 10 0.05 50 0.05 1000 s s v v I A mA A I µ = = × = = + 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 s s v V t t t ω ω ω = = = . Note 0.7 D V V = , so the peak value o V of the rectified voltage ( ) o v t across the load R is: 2 12 2 2 0.7 15.57 o s D V V V V = = × = (b) (6 points) For 1 2 and D D , conduction begins at an angle t θ ω = when sin 2 s s D v V V
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}