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ECE 427
Homework #3
Solutions
1. Question 12.6 on page 632.
What is a static VAR compensator (SVC)?
An answer is on pages 612-613. Important points to address are the circuit diagram, the
combination of a TSC and a TCR, what it does for the power system (stabilize voltage levels in a fast
and continuously controllable fashion), and the role of reactive power in its operation.
2. Problem 2.1 on page 73.
2.1 The reverse recoery time of a diode is trr=5μs, and the rate of fall of the diode current is
di/dt=80A/μs. If the softness factor is SF=0.5, determine
a. the storage charge QRR and
b. the peak reverse recovery current IRR.
A
trr  5  μs
rate  80
SF  0.5
For SF=0,
1
2
QRR   rate trr  1000 μC
2
μs
But SF=0.5. Deriving a value for ta ,
tb = SF  ta
ta  tb = trr
ta  SF  t a = trr
ta 
trr
1  SF
 3.333  μs
Solving for IRR from the di/dt value given,
IRR  ta rate  266.667 A
QRR is the area under the discharge curve,
1
QRR   IRR trr  666.667  μC
2
IRR  trr rate  400 A
3. Problem 2.6 on page 74.
Two diodes are connected in series as shown in Figure 2.11 and the voltage across each diode is
maintained the same by connecting a voltage-sharing resistor, such that V D1 =VD2 =2000V and
R1 =100kOhms. The v-i characteristics of the diodes are shown in Figure P2.6. determine the
leakage currents of each diode and the resistance R2 across diode D2 .
VD1  2  kV
VD2  2 kV
R1  100  kΩ
From Figure P2.6, at 2kV,
Is1  8  mA
Is2  16 mA
From Figure 2.11,
VD1
IR1 
 20 mA
R1
By KCL, we find the current in R2 ,
IR2  IR1  Is1  Is2  12 mA
R2 
VD2
IR2
R2  166.667  kΩ
It is possible to interpret the problem as having
Is1  16 mA Is2  8  mA
From Figure 2.11,
VD1
IR1 
 20 mA
R1
By KCL, we find the current in R2 ,
IR2  IR1  Is1  Is2  28 mA
R2 
VD2
IR2
R2  71.429 kΩ
The direction of currents shown in
Figure 2.11 can make this problem
tedious. Be careful.
4. Problem 2.26 on page 76
A diode circuit is shown in Figure P2.26 with R=10Ω, L=5mH, and Vs=220V. If a load current of 10A
is flowing thriough freewheeling diode Dm and switch S1 is closed at t=0, determine the expression
for the current i through the switch.
R5  10
L5  0.005
Vs  220
IDm  10
With the switch open initially, the inductor current is equal to the diode current.
IL0  IDm  10
Closing the switch, where u(t) is a unit step function,
d 
L5   iL   R5  iL = Vs u ( t)
 dt 
Take a LaPlace transform of the differential equation,

Vs
L5  s  R5  IL( s)  L5  IL0 =
s

Partial fraction expansion,
Vs  L5  IL0  s
IL( s) =
s L5  s  R5


Vs  L5  IL0  s

s L5  s  R5
K2
K1
IL( s) =

L5  s  R 5
s

=


K1  L5  s  R5  K2  s

s L5  s  R5
Vs
K1 =
R5
Vs = K1  R5
L5  IL0 = K1  L5  K2
L5  IL0 =
Write the solution ,
 Vs 
 
 R5  
IL( s) =
s

 Vs

 R  IL0
 5

s
R5
L5
Vs L5
R5
 K2
K2 = L5  IL0 
Vs L5
R5
Back to time domain,
t
 L5 
Vs  Vs
  R5 
iL( t) 

 IL0  e  
R5

L5
4
τ5 
 5  10
R5

R5
Substituting into the expression for iL (t)
Vs
R5
 22
1
IL0  10
τ5

3
 2  10
iL( t)  22  12 e
A
 2000 t
20
iL( t)
10
0
0
4
5 10
3
1 10
3
1.5 10
3
2 10
t
6. Question 9.7 on page 500.
9.7 What are the means of turning on thyristors?
An answer is on page 449 of the textbook. This explains the five important ways to turn on a
thyristor:
a. Gate pulse
b. Light pulse
c. Excessive forward voltage
d. Excessive dv/dt between anode and cathode
e. Too high temperature
6. How does a thyristor turn off? Show a diagram revealing charge concentrations to explain your answer.
To turn off, we must evaucate the gate region of excess charge carriers. We can do this by natural
commutation, reverse bias turnoff, or gate turnoff. In a sequence of snapshots of the excess charge carrier
concentrations,
np
pn
pn
np
Anode
Cathode
np
pn
pn
np
Anode
Cathode
np
pn
pn
np
Anode
Cathode
np
pn
pn
Anode
np
Cathode
7. Questions 9.16, 9.18, and 9.21 on page 500.
9.16 What is the difference between an SC and a TRIAC?
The SCR has a unidirectional current flow. The TRIAC is bidirectional and can be turned on with a gate
signal in either polarity.
9.18 What are the advantages and disadvantages of a GTO Thyristors?
The main advantage is the ability to turn off. The main disadvantage is the low reverse current gain during
turnoff.
Other advantages are the same advantages as an SCR. Other disadvantages are the long tail curent and
the minimum sustaining voltage and current. The GTO is more expensive and does not have quite the peak
voltage and current range overall as the SCR.
9.21 What are the advantages and disadvantages of an LASCR Thyristor?
The main advantage is the optical triggerring and the isolation capability that is brings. The main
disadvantage is the speed and lack of an ability to turn off optically. The other characteristics are the same
as an SCR.
7. Question 9.28 on page 500.
9.28 What are the common techniques for voltage sharing of series connected thyristors? An answer is
on pages 475 and 476. The resistors provide steady state voltage balancing when the thyristor block while
the interaction of resistors and capacitors provide dynamic voltage balancing.
Another problem not in the homework set provided for your study examples.
Problem 9.8 on page 502. Two thyristors are connected in parallel to share a total load current of
IL =500A. The on-state voltage drop of one thyristor is VT1 =1.0V at 300A and that of the other
thyristor is VT2 =1.5V at 300A. Determine the values of series resistances to force current sharing
with 10% difference. Total voltage v=2.5V.
VT1  1.0 V
VT2  1.5 V
IT  500  A
IT
share 
ITmax 
  1 
  262.5 A
2 
2 
v  2.5 V
share  10 %
IT
share 
ITmin 
  1 
  237.5 A
2 
2 
If the thyristors have the give currents and the given voltages, we want to make the sum of voltage
across them to be equal.
We use a model of the thyristor as a constant forward voltage. Therefore, series resistors much
make up the difference in voltage to get an overallvoltage of 2.5 Volts.
3
mΩ  10 Ω
VT1  ITmax RT1 = v
RT1 
v  VT1
ITmax
VT2  ITmax RT2 = v
 5.714  mΩ
RT2 
v  VT2
ITmax
 3.81 mΩ