Res A unity feedback system has a loop transfer function
K_
cfw_s + m.- + 3m + 6)
where K = 2|]. Find the roots of the closed-loop sys-
tems characteristic equation.
Baa For the feedback system of Exercise 6.51. find the
value of K when two roots lie on t

Power electronics
Tutorial 3 Solution
Driving circuits for power switching devices
1.
a)
Given: VCE = 25 V, IC = 10 A, tr = 0.3s, tf = 0.7s, f = 100 kHz, D = 0.6
Since, turn-off energy loss (E) = energy stored in C,
VCE I C
1
2
(t r t f ) CVCE
2
2
C
I C

Power electronics
DC-DC Converter
1.
Tutorial 4 Solution
D = 1 E / VC = 1 30 / 120 = 1 0.25 = 0.75
If E =10V, possible output range of VC = 0 ~10V
It is D.C. Buck Converter
VC = D E
2.
Q VC = E /(1 D)
i L = I max I min = E / LxDT = 30 /(350 10 6 ) 0.75 /

Solution to Extra Tutorial 3
(A) Thyristor
1.
With reference to Figure 1, describe how does the firing pulse applied at the primary of the pulse
transformer turn on the SCR.
Figure 1
Ans: A gate firing (triggering) circuit consists of a transformer for i

Solution to Extra Tutorial 2
(A) Power Bipolar Transistor
1.
State how a transistor should be biased for it to function as a switch.
Ans: For a BJT transistor, when the base current is large enough to drive the VBE ,which is
greater than the threshold vo

Power electronics
Tutorial 1 Solution
Power semiconductor devices
Power Diode
1.
Power diodes have larger power-, voltage-, and current-handling capabilities. Frequency responses (or
switching speed) are slow compared to signal diodes.
2.
Diode continues

Suggested Solution to Extra Tutorial 5
Q1.
(a)
From Lecture 6, Page 16,
b
(b)
1
cos n
: b1 = 6.366
: b3 = 2.122
: b5 = 1.273
: b7 = 0.909
: b9 = 0.707
: b11 = 0.579
Fundamental mode
3rd order mode
5th order mode
7th order mode
9rd order mode
11th order m

Solution to Extra Tutorial 4
(A) Linear & Switching Mode Power Supplies
1.
Explain why the efficiency of linear power supply is usually quite low.
Ans: In linear power supply, a low-frequency, 50 Hz transformer is used to step down the ac
mains to a lowe

APPENDIX
SOME USEFUL FORMULAS
Half-wave rectifier
V
V dc m
O/P Vrms
Vm
2
Full-wave rectifier
2V
Vdc m
Output Vrms
Vm
2
Snubber circuit
Turn off energy loss E (unit: J) = 0.5 (Voff I ON (t r t f )
2
VCE I C (t r t f )
CVCE
Turn off energy loss E (unit: J

NOTE:
The answer of question (a) has been given by:
NOTE:
The answer of question (a) has been given by:
Note: (3.16) and (3.17) are the two equations of state-space model.

Power electronics
Tutorial 2 Solution
Power rectification with diodes and thyristors
1.
Calculate the current (IDC) flowing through a 3k resistor connected to a 110v single phase
half-wave rectifier as shown as following, and also the power consumed by th