Fundamentals of Power Electronics
Week 10 Assignment Solutions
Q1. A Push pull converter shown in figure supplying a 20A constant load at 12V. The turns ratio
of the transformer is (Ns/Np) =0.5. The on-state voltage drops of diodes and transistors are 0.5v
and 1V respectively. What is the average input current (Iin) in Ampere?
(Note:- The transformer has zero losses and leakage. The load current and input current have
negligible ripple.)
Ans: 10A (9.8 to 10.2)
Q2. Refer to Q1, what is the percentage efficiency of the converter?
Ans: 92.3 ( 91.6 to 93)
Solution of Q1 and Q2:
Output power = 12*20 = 240W
V_s = 12 + 0.5 =12.5V (Diode voltage drop)
V_p = 12.5/0.5 = 25V
V_in = 25 +1 =26V (Switch voltage drop)
Input power = P_in = V_in*I_in = 1.I_sw + 1.I_sw + 0.5* I_d + 0.5*I_d + Po;
= 2*I_sw + I_d + P_o;
= 2*(I_in/2) +I_d + P_o;
= I_in + I_o/2 + P_o;
I_in(V_in-1) = Io/2 + Po;
I_in = 10A.
%Efficiency = (240/260)*100% = 92.3%
Q3. Refer to the figure of Q1, what will be PIV rating of the diode if the input to the converter is
supplied through a full bridge rectifier capacitor filter connected to 230V 50Hz supply with 10%
tolerance? (In this case consider Ns/Np = 0.25)
Ans: 178.89 (177 to 180 )
Solution:
2*(Ns/Np)*V_inmax = (1.1*230*sqrt(2))*2*0.25
Q4. Which of the following statement is/are true regarding the flux walking phenomenon?
It cannot be observed in Full bridge converter.
It can cause saturation of the converter transformer.
Flux walking occurs because of volt-sec unbalance across the winding.
Flux walking problem can be solved by connecting a series inductance with the primary winding.
Correcting the duty cycle in each PWM cycle can resolve the flux walking problem.
Ans: b, c, e
Q5. A TL494 PWM IC is driving a Push Pull Converter, which is supplying 40 V output from a
30 V source.
For the converter to be operated at 1000 kHz, Choose the correct pair of RT and CT values from
the given options below:
Note: Click Here to download the TL494 IC datasheet.
RT= 1kΩ, CT = 1nF
RT= 1kΩ, CT = 0.5nF
RT= 10kΩ, CT = 0.1nF
RT= 10kΩ, CT = 2nF
RT= 4kΩ, CT = 0.125nF
RT= 4kΩ, CT = 0.25nF
Ans: b, e
(Note: there was a miscalculation in the question, instead of 100 kHz switching frequency it will
be 1000kHz switching frequency)
Solution:
For Push Pull operation the switching frequency is:
f= 1/(2*RT*CT)
RT= 1 kΩ, CT= 0.5nF, f= 1000 kHz
RT= 4 kΩ, CT= 0.125nF, f= 1000 kHz
Q6. A full-bridge DC-DC converter is fed from DC link voltage of Vdc = 48V as shown in figure.
The ratio of transformer winding turns Ns/Np = 6 and converter operates at a switching frequency
of 50kHz with duty ratio of 0.5. The peak inverse voltage of the secondary side diodes D1 and D2
is _____ V.
Ans: 576 (Range: 570 to 580)
Solution:
PIV = 2*(Ns/Np)*Vdc = 576 V
Q7. Referring to question 6, the flux in the transformer core varies as
a. b.
c. d.
Ans: d
Q8. Referring to question 6, if the secondary of the full-bridge converter is modified as shown in
figure below, the peak inverse voltage of the diode is ____ V.
Ans: 288 (Range: 285 to 295)
Solution:
PIV = (Ns/Np)*Vdc = 288V
Q9: A half-bridge converter is designed to provide Vo = 10V where Q1 or Q2 of half bridge is
switched on for d*Ts period at an efficiency of 80%. If Vin=30V is applied to this converter with
d=0.4 at fsw=100kHz, calculate the rms current flowing through diode D1 in the secondary side if
input power supplied is 10W.
Note: 1. Ts is the time period for Q1/Q2 gate PWM cycle.
2. Take the upper integer value for number of turns(n)
3. Consider ideal components in the circuit
Ans:0.6197A (0.6 to 0.63)
Solution:
Pout = eff*Pin = 8W
Io = Pout/Vo = 0.8A
Id_rms = sqrt(Io^2 * d*Ts/Ts + ((Io^2)/4)*2*(1-d)*Ts/Ts) = 0.6197A
Q10: Referring to Q9, calculate the number of turns required to achieve the required specifications.
Ans: 4 (3.9 to 4.1)
Solution:
n= Vo/(Vin*d) = 3.33 => 4 turns
