Onsider a half-wave rectifier circuit with a triangular-wave input of 5-V peak-to-peak amplitude and zero average, and with R = 1 k. Assume that the diode can be represented by the constant-voltage-drop model with VD = 0.7 V. Find the average value of vO.

Vo=0.324v

Explanation:Given: Area of shaded triangle is t1,t2 and T/4

Fprst find t1 and t2

(2.5/T/4)=(0.7/t1)

t1=0.07T

t2= (T/2)- t1

t2= 0.43T

Vo= 1/T × shaded triangle

Vo=(2.5-0.7)×(T/4-t)

Vo= 1/T × 1 .8×(1/4-0.07)

Vo=0.324v

6 cos 34, 6 sin 34 trig

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