CS代写 ENGN 1218 Introduction to Electronics Week 09 Problems with Solutions – cscodehelp代写

AUSTRALIAN NATIONAL UNIVERSITY Research School of Engineering
ENGN 1218 Introduction to Electronics Week 09 Problems with Solutions
Design a power supply with the following parameters: Si diode model, f = 60 Hz, Vo(DC) = 9 V, Vr(pp) = 2 V, Io(DC) = 100 mA. Use half-wave rectifier circuit with smoothing capacitor shown in Figure 1.
vin(t) C RL vo(t)
Figure 1: The circuit for Question 1.
Consider the diode-bridge full-wave rectifier circuit with smoothing capacitor shown in Figure 2. Assume Si
diode model, RL = 600 Ω, C = 2200 μF, f = 60 Hz, and vin(t) = ( (a) Find Vo(DC), Io(DC) and ripple factor r.
(b) Find diode ratings.
2)240sin(ωt).
Figure 2: The circuit for Question 2.
Design a power supply with the following parameters: f = 50 Hz, Vo(DC) = 3.7 V, r = 1% and Io(DC) = 350 mA. Use diode-bridge full-wave rectifier circuit with smoothing capacitor.
Week 09 Problems page 1
ANU ENGN 1218
Q4 (Challenge Design Problem)
Design a power supply with the following parameters: f = 50 Hz, Vo(DC) = 14 V, r = 20% and Io(DC) = 0.5 A. Use the circuit shown in Figure 3.
N D2 D4 C RL vo(t)
Figure 3: The circuit for Question 4.
(a) Prove that for a diode half-wave rectifier circuit with smoothing capacitor Vr(pp) = Vo(DC) . fRLC
(b) Prove that for a diode-bridge full-wave rectifier circuit with smoothing capacitor Vr(pp) = Vo(DC) . 2fRLC
Week 09 Problems
Complete Solution
The design specifications are:-
AUSTRALIAN NATIONAL UNIVERSITY Research School of Engineering
ENGN 1218 Introduction to Electronics Week 09 Problems with Solutions
vin(t) C RL
half-wave rectifier circuit with smoothing capacitor, Si diode model,
f = 60 Hz, Vr(pp) = 2 V, Vo(DC) = 9 V, Io(DC) = 100 mA.
The design steps are:-
Step1: Find the load resistance
RL =Vo(DC)=9=90Ω Io(DC) 0.1
Step2: Find the peak output voltage
Vo(DC) = Vo(p) − 12Vr(pp) Vo(p) = 10V
Step3: Find the peak input voltage Vin(p) = Vo(p) + 0.7 = 10.7V
Hence the input voltage is vin(t) = Vin(p) sin(ωt) = 10.7sin(2π60t). Step4: Find the smoothing capacitor
C = Vo(DC) = 833.33 μF fRLVr(pp)
Step5: Find diode ratings
PIV = ID(max) =
2Vin( p) − 0.7 = 20.7V Vo( p) = 111.111mA
0.7 × ID(max) = 77.77mW
Using factor of safety of 50%, the diode ratings are
Peak Inverse Voltage (PIV) rating = 1.5× PIV = 31.05 V. Diode current rating = 1.5 × ID(max) = 166.66 mA.
Diode power rating = 1.5 × PD(max) = 116.66 mW.
Week 09 Problems
Complete Solution
The analysis specifications are:-
full-wave rectifier circuit with smoothing capacitor Si diode model,
RL = 600 Ω,
C = 2200 μF,
f =60Hz, √
vin(t) = ( 2)240sin(ωt)
The analysis steps are:-
Step1: Find the peak input voltage
Vin(p) = ( 2)240 = 339.41V Step2: Find the peak output voltage
Vo(p) = Vin(p) −1.4 = 338.01V Step3: Find the peak to peak ripple voltage
Vr(pp) = Vo(DC) ∼= Vo(p) = 2.134V 2fRLC 2fRLC
The assumption Vo(DC) ∼= Vo(p) can be made in analysis, assuming the rectifier circuit given is properly de- signed.
Step4: Find the average load voltage
Vo(DC) = Vo(p) − 21Vr(pp) = 336.94V Step5: Find the average load current
Io(DC) = Vo(DC) = 0.5615A RL
Step6: Find the ripple factor
r = Vr(pp) = 6.333 × 10−3 = 0.6333%
Week 09 Problems
Complete Solution
The design specifications are:-
full-wave rectifier circuit with smoothing capacitor, Si diode model, f =50Hz,r=1%,Vo(DC) =3.7V,Io(DC) =350mA.
The design steps are:-
Step1: Find the load resistance
RL = Vo(DC) = 3.7 =10.57Ω Io(DC) 0.350
Step2: Find the peak to peak ripple voltage
Step3: Find the peak output voltage
Vo(DC) = Vo(p) − 12Vr(pp) Vo( p) = 3.7185V
Step4: Find the peak input voltage
Vin(p) = Vo(p) +1.4 = 5.1185V
Hence the input voltage is vin(t) = Vin(p) sin(ωt) = 5.1185sin(2π50t). Step5: Find the smoothing capacitor
C = Vo(DC) = 94.607mF 2fRLVr(pp)
Step6: Find diode ratings
PIV = Vin(p)−0.7=4.4185V
ID(max) = Vo(p) = 351.79mA RL
PD(max) = 0.7 × ID(max) = 246.26mW
Using factor of safety of 50%, the diode ratings are
Peak Inverse Voltage (PIV) rating = 1.5× PIV = 6.628 V. Diode current rating = 1.5 × ID(max) = 527.68 mA.
Diode power rating = 1.5 × PD(max) = 369.38 mW.
Verify design using PSPICE. Week 09 Problems
Vr(pp) Vo(DC)
Vr(pp) = rVo(DC) = (0.01)(3.7) = 37mV
Q4 (Challenge Design Problem) Partial Solution
The design specifications are:-
full-wave rectifier circuit with smoothing capacitor, Si diode model,
Transformer input (240V rms),
f =50Hz,r=20%,
Vo(DC) = 14 V, Io(DC) = 0.5 A.
The design steps are:-
Step1: Find the load resistance RL
Step2: Find the peak to peak ripple voltage Vr(pp)
Step3: Find the peak output voltage Vo(p)c
Step4: Find the peak input voltage Vsec(p) (in Q3, this was denoted by Vin(p)) Step5: Find the smoothing capacitor C
Step6: Find diode ratings
Step7: Find the transformer turn ratio n
Vsec(p) Vpri(p)
Note: Vpri(p) = ( 2)240V.
Q5 Solution
See Hambly textbook pages 504-505 OR see Lecture 32 doc cam notes. Hints:
• Capacitor discharges for nearly the entire cycle T for half-wave rectifier circuit with smoothing capac- itor.
• Capacitor discharges for nearly the entire half-cycle T2 for full-wave rectifier circuit with smoothing capacitor.
Week 09 Problems page 6