Formula sheet EE-434 Power Electronics
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Capacitor
Basic expressions
dv
CurrentiC = C
dt
Z
1
V oltagevc =
iC (dt) + V0−
C
1
Energyw(t) = Cv 2 (t)
2
1
τ =
RC
Average expression
Xavg
t
Z
1
=
T
x(t)dt
(1)
0
RMS expression
s
Xrms
1
T
Z
T
x2 (t)dt
(2)
v
u N
u X
2
=t
Xn,rms
(3)
Xrms =
0
n=1...
[an cos(nω0 t) + bn sin(nω0 t)]
(4)
T /2
1
f (t)dt
a0 =
T −T /2
Z T /2
2
an =
f (t)cos(nω0 t)dt
T −T /2
Z T /2
2
bn =
f (t)sin(nω0 t)dt
T −T /2
Z
(7)
Power and Energy
InstantaneousP owerp(t) = v(t)i(t)
Z t0 +T
W
1
p(t)dt =
AveragepowerP =
T t0
T
(8)
AveragepowerPdc = Vdc Iavg
(10)
ApparentP owerS = Vrms Irms
√
S = 3VL−L,rms IS,rms
(11)
ReactivepowerQ = Vrms Irms sin(θ − ϕ)
Z t2
Energy(W ) =
p(t)dt
(13)
p(t)dt
(14)
P
S
(26)
for 0 ≤ ωt ≤ β
∞
X
Vm
Vm
2Vm
+
sin(ωo t) −
cos(nωo t)
π
2
(n2 − 1)π
n=2,4,6...
Vo =
Vm
Vm
IL Xs
(1 + cosu) =
1−
2π
π
2Vm
(15)
(16)
Ripple voltage for C filter
∆Vo ≈
(18)
(19)
(20)
(28)
(29)
(30)
(31)
Full wave rectifier
Full wave rectified waveform fourier is
∞
X
2Vm
1
1
2Vm
+
−
cos(nωo t + π)
vo (t) =
π
π
n−1
n+1
n=2,4...
(17)
(27)
β ≤ ωt ≤ 2π
With source inductance
Inductor
di
V oltage =⇒ vL = L
dt
Z
1
CurrentiL =
vL (dt) + I0−
L
1
Energyw(t) = Li2 (t)
2
L
τ =
R
(25)
For controlled rectifier

 Vm
(α−ωt)/ωτ
sin(ωt
−
θ)
−
sin(α
−
θ)e
for α ≤ ωt ≤ β
Z
i(ωt) =

0
otherwise
Z β
1
Vm
(cosα − cosβ)
Vo =
Vm sin(ωt)d(ωt) =
2π α
2π
With free wheeling diode
(12)
0
P owerf actor(pf ) =
Z
sin(ωt − θ) + sin(θ)e−ωt/ωτ

0
v(t) =
T
W =
 Vm
(9)
t1
Z
i(ωt) =
(5)
(6)
(24)
q
Half wave rectifier
with
RL load

n=1
where,
(23)
R2 + (ωL)2
ωL
θ = tan−1 (
R
Impedance(Z) =
∞
X
(22)
Phasors
Fourier series
f (t) = a0 +
(21)
With LC filter
In CCM
Vm π
Vm
=
ωRC
2f RC
2Vm
π
2Vm
IL =
πR
3ωL
>1
R
Vo =
(32)
(33)
(34)
(35)
(36)
In DCM average inductor current.
Z
1 β 1
Vm (cosα − cosωt) − Vo (ωt − α) d(ωt)
iL =
π α ωL
RLE load CCM
D=
(37)
Imax = IL +
2Vm
π
−E
Vo − E
=
R
R
∞
X
vo (t) = Vo +
Vn cos(nω0 t + π)
Io =
DC-DC converters - CCM
(38)
(39)
∆iL
2
and Imin = IL −
where, V Co =
2Vm
π
and Vn =
2Vm
π
1
n−1
−
1
n+1
∆iL
.
2
Vo = DVs
Minimum inductance Lmin
(1 − D)R
2f
Vo (1 − D)
L=
∆iL fs
(54)
(55)
Output capacitor C
C=
Vm
io (ωt) =
[sin(ωt − θ) − sin(α − θ)
Z
e−(ωt−α)/ωτ ]f orα ≤ ωt ≤ β
(40)
β < α + π =⇒ DCM
(41)
For CCM
α ≤ tan−1
vo (ωt) = V0 +
∞
X
ωL
R
∆iL =
Vs −Vo
L
Vo =
1
π
Z
(42)
Vm sin(ωt)d(ωt) =
Vs
1−D
(57)
DT
Vo =
Minimum inductance Lmin
D(1 − D)2 R
2f
Vs D
L=
∆iL f
Lmin =
Vn cos(nω0 t + θn )
α
(56)
Boost converter
α+π
1−D
8L(∆Vo /Vo )f 2
(43)
n=1
2Vm
cosα
π
(44)
q
Vn = a2n + b2n
where,
2Vm cos(n + 1)α
cos(n − 1)α
−
π
n+1
n−1
2Vm sin(n + 1)α
sin(n − 1)α
−
bn =
π
n+1
n−1
2Vm
Vo =
cosα
π
an =
(46)
∆iL =
Vs
L
DT
Buck-Boost converter
Minimum inductance Lmin
D
1−D
(61)
(1 − D)2 R
2f
Vs D
L=
f ∆iL
(62)
Lmin =
(48)
Three phase full wave rectifier
Uncontrolled with R load
(60)
Vo = −Vs
(47)
(59)
D
R(∆Vo /Vo )f
C=
(45)
(58)
Output capacitor C
The harmonics are
(63)
Output capacitor C
∞
X
vo (t) = V0 +
(49)
∆iL =
where,
1
V0 =
π/3
Z
2π/3
π/3
D
R(∆Vo /Vo )f
C=
Vn cos(nωo t + π)
n=6,12,18,...
(64)
Vs DT
L
Flyback converter
3Vm,L−L
Vm,L−L sin(ωt)d(ωt) =
π
6Vm,L−L
Vn =
π(n2 − 1)
(50)
The amplitudes of the ac voltage terms are
for n=6,12,18,...
(53)
Lmin =
Full controlled with RL load
DCM
(52)
Buck converter
n=2,4..
ton
ton
=
= ton fs
ton + tof f
T
Vo = Vs
D
1−D
N1
N2
Minimum inductance Lmin
(51)
(1 − D)2 R
2f
Vs D
L=
f ∆iLm
(Lm )min =
N2
N1
(65)
2
(66)
(67)
Quasi Square wave inverter(RL series load)
Output capacitor C
∆Vo
D
=
Vo
RCf
∆iL m =
s
(68)
Vrms =
V sDT
Lm
Forward converter
Vo = Vs D
N2
N1
vn =
2
π
Z
(69)
Z
1
π
∆iL m =
π−α
Vdc sin(nω0 t)d(ω0 t) =
α
(70)
2α
π
4Vdc
cos(nα)
nπ
(76)
(77)
(78)
fcarrier
ftri
=
fref erence
fsine
(79)
Vm,ref erence
Vm,sine
=
Vm,carrier
Vm,tri
(80)
PWM inverter
mf =
DC-AC inverters
ma =
Square wave inverter(RL series load)
1−
4Vdc
cosα
π
V1 =
V sDT
Lm
Bipolar PWM inverter


Vdc
dc

+ Imin − VR
e−t/τ
 R
io (t) =

−V
Vdc

e−(t−T /2)/τ
 Rdc + Imax + R
Irms
2 d(ωt) = V
Vdc
dc
α
Output capacitor C
1−D
∆Vo
=
Vo
RLx Cf 2
r
π−α
for 0 ≤ t ≤
for
T
2
T
2
vo (t) =
≤ t≤T
(81)
2Vdc
nπ
n2π
2 + cos
− cos
f or n = 1, 5, 7, 11, 13...
3nπ
3
3
(82)
n=1
6 step inverter
−(T /2τ ) Vdc 1 − e
Imax = −Imin =
R 1 + e−(T /2τ )
s
s
Z T
Z T /2 Vdc −t/τ 2
1
2
Vdc
i2 (t)d(t) =
+ Imin −
e
dt
=
T 0
T 0
R
R
∞
X
4Vdc
vo (t) =
sinnω0 t
nπ
1,3,5..
q
pP∞
2
2
Vrms
− V1,rms
2
(V
)
n,rms
n=2
T HD =
=
V1,rms
V1,rms
∞
X
Vn sin(nω0 )t
(71)
(72)
(73)
Vn,L−L =
3 phase PWM inverter
Vn3 =
(74)
(75)
q
2
A2n3 + Bn3
(83)
where,



sin nπ
An3 = Vn sin nπ
2
3

nπ
nπ

Bn3 = Vn cos 2 sin 3
(84)