The question incomplete! The complete question along with answer and explanation is provided below.
Question:
Augment the rectifier circuit of Problem 4.68 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% of the peak output and (ii) 1% of the peak output. In each case:
(a) What average output voltage results?
(b) What fraction of the cycle does the diode conduct?
(c) What is the average diode current?
(d) What is the peak diode current?
Problem 4.68:
A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through a 10-to-1 step-down transformer. It uses a silicon diode that can be modeled to have a 0.7-V drop for any current.
Given Information:
Input voltage = 120 Vrms
10 to 1 step-down transformer
Voltage drop at diode = 0.7 V
Load resistance = R = 1 kΩ
Required Information:
(i) 10% of the peak output and (ii) 1% of the peak output. In each case:
(a) What average output voltage results?
(b) What fraction of the cycle does the diode conduct?
(c) What is the average diode current?
(d) What is the peak diode current?
Case (i)
Vavg = 15.45 V
Conduction of diode = 7.11 %
Iavg = 0.232 A
Ip = 0.449 A
Case (ii)
Vavg = 16.18 V
Conduction of diode = 2.25 %
Iavg = 0.735 A
Ip = 1.453 A
Explanation:
Voltage at the secondary side of the transformer is
Vrms = Vpri/turn ratio
Vrms = 120/10 = 12 V
The relation between rms voltage and peak voltage is
Vp = Vrms/√2
Vp = 12√2 = 16.97 V
Vd = 0.7 V
First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.
case (i) 10% of the peak output:
(a) What average output voltage results?
Average output voltage = Vavg = Vp - Vd - 0.5Vr
Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp
Vavg = Vp - Vd - 0.5Vr
Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]
Vavg = 15.45 V
(b) What fraction of the cycle does the diode conduct?
ω = √2Vr/Vp - Vd
ω = √2*0.1(Vp-Vd)/Vp - Vd
ω = √2*0.1(16.97-0.7)/16.97 - 0.7
ω = 0.447 rad
Conduction of diode = (ω/2π)*100
Conduction of diode = (0.447/2π)*100
Conduction of diode = 7.11 %
(c) What is the average diode current?
Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]
Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]
Average current = Iavg = 0.232 A
(d) What is the peak diode current?
Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]
Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]
Peak current = Ip = 0.449 A
case (ii) 1% of the peak output:
(a) What average output voltage results?
Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]
Vavg = 16.18 V
(b) What fraction of the cycle does the diode conduct?
ω = √2*0.01(Vp-Vd)/Vp - Vd
ω = √2*0.01(16.97-0.7)/16.97 - 0.7
ω = 0.1417 rad
Conduction of diode = (0.1417/2π)*100
Conduction of diode = 2.25 %
(c) What is the average diode current?
Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]
Average current = Iavg = 0.735 A
(d) What is the peak diode current?
Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]
Peak current = Ip = 1.453 A
