Old Exam (2001)
1. The four quadrant DC-DC converter
a) As in your project reports!
b) Consider the blanking time circuit used in the laboratory converter
R1 open collector pull-down resistor and R2 potentiometer in parallel with diode
vC = VCC 1 e t
V
t = ln 1 C
VCC
The transfer at the buffer output occurs when VC =V
d) Fundamental component of the AC-side current
VLN,rms = 230 V, 50 Hz assume cos = 1
I1,line =
PLOAD,max
4000 W
=
= 17.4 A (rms, fund. comp.)
230 V
VLN,rms cos
e) Rectifier losses?
In the previous exercise the fundamental of the line side current was calculated.
However, The current is not sinusoidal but contains a large amount of harmonics.
From Figure 1 it is clear that if the fundamental has a peak level of 1 then the real
current has a peak value of 3.15. Therefore, the diode current has a peak value of
iD = 3.15 i1,line = 3.15 2 I1,line = 3.15 2 17.4 A = 77.5 A
The diode current pulses have duration tp=2.75 ms according to the same figure.
From Fig. 2 of the rectifier data sheets the forward characteristics are found to be
iD = 10 A v D = 0.78 V
iD = 100 A vD = 1.39 V
VD 0 = 0.712 V
RD = 6.78 m
The diode losses are calculated from the integral (fn=50 Hz Tn = 20 ms)
T
1 n
1 p
1 p
PD = p(t ) dt = p(t ) dt = VD 0 iD (t ) + RD iD2 (t ) dt =
Tn 0
Tn 0
Tn 0
2
iD
tp
tp
2
2
= VD 0 I avg + RD I rms
= VD 0 iD + RD
Tn
2 Tn
= 4.84 + 2.81 W = 7.65 W
The total rectifier losses are:
Prect = 4 PD = 4 7.65 W = 30.6 W
f)
DC link capacitor
The time derivative of the DC link voltage when no rectifier diode conducts is
found from Figure 1
dv DC
= 9789 V/s
dt
The average current drawn by the converter is given by
I DC,avg = DLOAD I C,max = {DLOAD = 0.95} I C,max = 16 A
For a capacitor
I DC,avg
dvC
iC = C
C DC =
= 1.6 mF
dt
dv DC dt
g) IGBT and freewheeling diode conduction characteristics
From Fig. 2 of the IGBT data sheets the forward characteristics of the IGBT are
found to be
I C = 1.0 A, T j = 150 o C VCE = 0.8 V
o
I C = 10 A, T j = 150 C VCE = 1.35 V
VCE 0 = 0.74 V
RCE = 0.06
VCE = 1.7 V at I C = 16 A, T j = 150 o C
From Fig. 13 of the IGBT data sheets the forward characteristics of the
freewheeling diode are found to be
I D = 1.0 A, T j = 150 o C VD = 0.9 V
o
I D = 10 A, T j = 150 C VD = 1.18 V
VD = 1.35 V at I C = 16 A, T j = 150 o C
VD 0 = 0.87 V
RD = 0.031
h) Switching losses of the IGBTs and freewheeling diodes
Note that the switching losses are calculated for the worst case DC link voltage,
i.e. with VDC = vLN = 325 V when the duty-cycle is close to zero and with the
average DC link voltage (290 V according to Figure 1) when the duty-cycle is
close to one.
IGBTs
From Fig. 10 of the IGBT data sheets the total switching losses of the IGBT are
found to be
ETS = 0.110 21.5mm 20 mm mJ = 1.19 mJ at VDC = 480 V, I C = 20 A, T j = 150 o C
In this application this means that the switching losses (for one IGBT) are
1.19 10 3
Psw = ETS f sw =
VDC I C f sw =
290 16 10 103 W = 5.75 W
VDC ,n I C ,n
480 20
ETS ,n
Freewheeling diodes
From the table Switching Characteristics:
tr=52 ns for IC = 20 A, RG = 10 and Tj = 150 C (for the IGBT)
diF
20
=
A/s = 385 A/s
9
dt 52 10
Fig. 15 gives (Note that log(385/100)=0.585 0.585 37.4 mm = 21.9 mm):
I RRM = 110 26 mm 25.5mm A = 10.46A
Fig. 17 gives (Note that log(385/100)=0.585 0.585 37.4 mm = 21.9 mm):
direc
= 100 10 24.75mm 50.5mm A /s = 309 A/s
dt
direc I RRM
I RRM
=
t rr 2 =
= 33.9 s
dt
t rr 2
direc dt
1
1
QF = I RRM t rr 2 = 10.46 33.9 10 6 C = 177 nC at I F = 15 A
2
2
QF =
16
177 nC = 189 nC at I F = 16 A
15
The turn-off energy loss is thus
E Drr = QF VDC = 189 10 9 290 J = 54.8 J at I F = 16 A
The average switching power losses for one freewheeling diode are given by
Psw = E Drr Fsw = 54.8 10 6 10 103 W = 0.55 W
i)
Conduction losses
DLOAD = 0.95:
Pcond,IGBT=VCEICDIGBT=26.5 W/IGBT
Pcond,FWD=VDIDDD=0.54 W/FWD
DLOAD = 0.05:
Pcond,IGBT=VCEICDIGBT=0.68 W/IGBT
Pcond,FWD=VDIDDD=21.1 W/FWD
j)
Thermal design
DLOAD = 0.95
PIGBT = 32.3 W/IGBT, Rth,jh=0.77+0.24 C/W = 1.01 C/W
PFWD = 1.1 W/FWD, Rth,jh=1.7+0.24 C/W = 1.94 C/W
Prect = 30.6 W/FWD, Rth,jh=1.7+0.2 C/W = 1.9 C/W
Assume Tj = 125 C
Th,max,IGBT = 125-32.31.01 C = 92.4 C
Th,max,FWD = 125-1.1194 C = 122.9 C
Th,max,rect = 125-30.61.9 C = 66.9 C
PTOT = 2PIGBT +2PFWD +Prect = 97.4 W
Ta = 30 C
Rth,ha =
66.9 30 o
C W = 0.379 o C W 0.38 o C W
97.4
2. Snubbers
a) Draw a step down converter
Vdc
b) Draw a step down converter with an RCD
charge-discharge snubber
iC
iC
vCE
vCE Ds
Rs
Vdc +
Iload
Iload
Assume that the transistor current decreases linearly at turn-off (in principle true
for MOSFETs)
0 < t < t fi
iT = I 0 1
t
fi
This gives the snubber capacitor current
iCs
t
t
= I 0 iT = I 0 I 0 1 = I 0 0 < t < t fi
t
t fi
fi
The snubber capacitor voltage is calculated by integration of the capacitor current
vCs
I0 t 2
1 t
=
iCs dt =
CS 0
2 CS t fi
If it is assumed that the blocking voltage increases during tfi
vCs (t fi ) = VDC C S =
I 0 t fi
2 VDC
vCs
t
= VDC
t
fi
c) Stray inductance
vL = L
v
diL
di
70
= L T L = L =
H = 350 nH
9
dt
diT dt 26.1/130 10
dt
d) RCD clamp snubbers
Cs
Rs
Ds
CS
RS
e) Transistor technology?
Tail current IGBT
f)
Alter drive circuit?
Reduce over-voltage decrease the current derivative increase RG
VCC VGE
vGE = VCC + 1 - e
2 VCC t = ln
2 VCC
VCC VGE ,th
VCC VGE ,I 0
VCC VGE ,th
+ ln
= ln
t fi = ln
VCC VGE ,I
2 VCC
2 VCC
0
-t
= const
In the altered case
v L = 56 V v L = 30 V
diT
di
30
= 200 A/s T = 200 A/s = 107 A/s
dt
dt 56
This means that tfi should increase by a factor 200/107=1.87 since I0 is unaltered.
This is accomplished by increasing = RG(Cgs+Cgd). Note that Cgd is
approximately constant since the IGBT voltage is vCE 200 V during the current
fall. Therefore, RG should be increased by a factor 1.87.
3. The flyback converter
a) Draw the main circuit - As in your project reports!
b) Explain the need and operation of the snubbers - As in your project reports!
c) Determine the winding turns ratio and the air gap of the core
v1 = 300 V v2 = 30 V
N1 300
= 10
N 2 30
v2 = 30 V v1 = 300 V
30
di2 3.38 A
v2
H = 3.38 A/ms L2 =
=
=
H = 88.8 H
dt 10 s
di2 dt 3.38 10 -3
N 2 = 10 AL =
L2
N 22
88.8 H
10
= 887 nH 900 nH l = 200 m
d) Sketch the magnetizing current and calculate core losses
,max = 0.54 A, im
,min = 0.22 A, Triangle with f1 = 50 kHz and D 0.5
im
or
i
,min = 2.2 A, Triangle with f1 = 50 kHz and D 0.5
m,max = 5.4 A, im
= Lm
im
= B N1 AFe
2
9
m
L
i
(
)
10
10
900
10
0.54
m
B =
=
T = 281 mT
6
N1 AFe
10 10 173 10
NOT saturated!
= 0.38 +
im
0.1297
2
cos(n 2f1t ) A, f1 = 50 kHz
n
,1 = 0.1297 A peak
The fundamental component of the magnetizing current: im
n=1
= Lm
im
= B N1 AFe
2
9
(
)
L
i
10
10
900
10
0.1297
m m
=
B=
T = 67.5 mT
6
N1 AFe
10 10 173 10
Fig. 6 (67.5 mT and f=50 kHz): Pv=30 kW/m3
VFE=1780010-9 m3 PFe,50kHz=0.534 W
