elda minggu 3 [review basic consep]

1. Review of Basic Concepts
Pekik Argo Dahono

2. Average and RMS Concepts
• Periodic signals x(t ) = x(t + T )
• Average value
1 t o +T
x=
T to∫ x(t )dt
• RMS value
1 t o +T 2
X =
T to∫ x (t )dt

3. Single-Phase Power Concept
• Sinusoidal voltage and current
v = 2V cos(ωt )
i = 2 I cos(ωt − φ )
• Instantaneous power
p = vi = VI 444+ cos(2ω3] + VI 44 sin (2ωt )
1 cos φ [12444t ) 1sin φ244 3
resistive part reactive part

4. Single-Phase Power Concept
• Active or Average Power
t o +T
∫
1
P= pdt =VI cos φ
T to
• Reactive Power
Q = VI sin φ
• Apparent Power
S = VI

5. Single-Phase Power Concept
• Power triangle
S = P +Q
2 2 2
• Power factor
P
PF = = cos φ
S

6. Single-Phase Power Concept
E 2 = (V + ΔV )2 + δV 2
I P + jQ
E
R jX
Load V δV 2 << (V + ΔV )2
Thus
ΔV ≈ E-V = RI cos φ + XI sin φ
RP + XQ XQ
= ≈
E V V
δV
Losses = RI 2 = R(S / V )2
δ
φ V
I ΔV [
= R (P / V )2 + (Q / V )2 ]

7. Balanced Three-Phase Power
• Voltage and current
va = 2V cos(ωt )
( π
vb = 2V cos ωt − 23 )
vc = 2V cos(ωt + 23 )
π
ia = 2 I cos(ωt − φ )
( π
ib = 2 I cos ωt − 23 − φ )
( π
ic = 2 I cos ωt + 23 − φ )

8. Balanced Three-Phase Power
• Instantaneous power
p = va ia + vb ib + vc ic
p = 3VI cos φ
• Instantaneous power is a constant that is
equal to average power

9. Balanced Three-Phase Power
• Reactive power is defined as
Q = 3VI sin φ
• Apparent power is defined as
S = 3VI

10. Three-Phase Power System
Rs I Rs 3I
Rs I Rs
Rs I Rs
Ro Ro Ro
Ro / 3
Rs
Rs
Losses = 3RI 2 Losses = 18RI 2
PF = 1.0 PF = ?

11. Three-Phase Power System
Rs
Rs
Rs
− j1 j1 1
Rs
PF = ?

12. Three-Phase Four-Wire Systems
Series Losses Consideration : Three - Phase Three - Wire System
(
ΔP = rs I a + I b + I c + I n
2 2 2 2
) (unbalanced)
Ie = (I
2
)
+ Ib + Ic / 3
2 2
a
ΔP = 3rs I e
2
(V )
(balanced)
I a + Ib + Ic + I n
2 2 2 2 Ve = 2
an + Vbn + Vcn / 3
2 2
Thus, I e =
3
Shunt Losses Consideration : = (V 2
ab + Vbc + Vca
2 2
)/ 9
Vao + Vbo + Vco + Vno
2 2 2 2
ΔP = (unbalanced)
Rsh
Ve2
ΔP = 3 (balanced)
Rsh
Vao + Vbo + Vco + Vno
2 2 2 2
Thus, Ve =
3
or Ve = (V 2
an
2 2 2 2 2
)
+ Vbn + Vcn + Vab + Vbc + Vca / 12
Apparent Power S e = 3Ve I e
Power Factor = P / S e

13. Fourier Series Representation
• Fourier series
∞
x(t ) = co + 2 ∑C
n =1
n sin (nωt + θ n )
• Average value
x = co
• RMS Value
∞
X = co +
2
∑n =1
2
Cn

14. Total Harmonic Distortion (THD)
• Voltage signal
∞
v= 2 ∑V
n =1
n sin (nωt + θ n )
• Total Harmonic Distortion
1/ 2
⎛ ∞ ⎞
⎜ ∑
⎜ Vn2 ⎟
⎟
THD = ⎝ n=2 ⎠
V1

15. Power concept under nonsinusoidal waveforms
∞
• Voltage v = Vo + 2 ∑V
n =1
n cos(nωt + α n )
∞
• Current i = Io + 2 ∑I
n =1
n cos(nωt + β n )
• Instantaneous power
p = vi

16. Power concept under nonsinusoidal waveforms
• Average power
∞
P = Vo I o + ∑V I
n =1
n n cos(α n − β n )
• Apparent power
S = Vrms I rms
• Power factor
∞
P
Vo I o + ∑V I
n =1
n n cos(α n − β n )
PF = =
S Vrms I rms

17. Example
• Voltage
v = 1 + 210 cos(100πt ) + 2 2 cos(300πt )
• Current
( )
i = 2 5 cos 100πt − 60 − 2 cos(300πt )
o
• Average power
( )
P = 50 cos 60 o + 2 cos(π ) = 25
• RMS Voltage and Current
V = 12 + 10 2 + 2 2 = 105 I = 5 2 + 12 = 26
• Power factor
P 25
PF = = = 0,478
S 105 6

18. Sinusoidal voltage case
Average power : P = V1 I1 cos(α1 − β1 )
Power factor :
P I I1
PF = = 1 cos(α1 − β1 ) = cos φ1
S I rms ⎡ 2 ∞ 2⎤
1/ 2
⎢ I1 +
⎢
∑In ⎥
⎥
⎣ ⎦
φ1 = α1 − β1
n=2
where :
Relationship between power factor and THD:
cos φ1
PF =
1 + THD 2

19. Transformer and inductor
• Inductor is used to store temporary energy and
also used to smoothing the current
• Transformer is used for voltage conversion,
galvanic isolation, and also used to store
temporary energy.
• Transformer and inductor are the heaviest
component in power electronics system.
• In power electronics applications, transformer
sometimes has to operate with both dc and ac
voltages or currents.

20. Electromagnetism
Hukum Ampere :
∫ H.dl = ℑ = NI At
Hlc = NI
I Φ H=
NI
At/m
lc
A μNI
B = μH = Wb/m 2
N lilit lc
μA
Φ = BA = NI Wb
lc
μAN 2
λ = NΦ = I
lc
λ μAN 2
L= =
I lc

21. Airgap Influence
ℑ = NI = H c lc + H g g
Bc Bg ⎛ l g ⎞
NI = lc + g = Φ⎜ c + ⎟
μ μo ⎜ μAc μ o Ag ⎟
⎝ ⎠
Ac ≈ Ag = A
I
Φ μ = μ r μo
NI
Φ=
N lilit g lc g
+
μr μo A μo A
N 2I
λ = NΦ =
lc g
+
μr μo A μo A
N2
L= ≈ μ o AN 2 / g
lc g
+
μr μo A μo A

22. Transformer
i1 i2
v1 N1 N2 v2

23. Ideal Transformer
i1 i2
• • v1 N1
= =−
i2
v1 N1 N 2 v2 v2 N 2 i1
Ideal transformer neither
dissipates nor stores energy

24. Practical Transformer
i1 i2
Ll1 • • Ll 2
v1 Lm N1 N2 v2

25. Symmetrical Components
Any unbalanced three-phase quantities can be
composed into three symmetrical components:
- Positive sequence components
- Negative sequence components
- Zero sequence components
Ib2
I c1
I ao = I bo = I co
I a1 I a2
I b1 I c2

26. Symmetrical Components
⎡Vao ⎤ ⎡1 1 1 ⎤ ⎡Va ⎤ ⎡Va ⎤ ⎡ 1 1 1⎤ ⎡Va1 ⎤
⎢V ⎥ = 1 ⎢1 a ⎢V ⎥ = ⎢a 2 1⎥ ⎢Va 2 ⎥
⎢ a1 ⎥ 3 ⎢ a 2 ⎥ ⎢Vb ⎥
⎥⎢ ⎥ ⎢ b⎥ ⎢ a ⎥⎢ ⎥
⎢Va 2 ⎥ ⎢1 a 2 a ⎥ ⎢Vc ⎥ ⎢Vc ⎥ ⎢ a
⎣ ⎦ ⎣ a2 1⎥ ⎢Vao ⎥
⎦⎣ ⎦
⎣ ⎦ ⎣ ⎦⎣ ⎦
2π
j
a= e 3 I co
I c2
I c1
Ia
Ic
I ao
Ib
I a1 I a2
I bo
I b1
I b2

27. Balanced nonsinusoidal quantities
Let assume:
∞
∞ ∞
∑ [( )] ∑ [( )]
∑
π
Vn cos n ωt + 23
va = Vn cos[n(ωt )] vb = Vn cos n ωt − 2π
3
vc =
n =1
n =1 n =1
For n=1:
va1 = V1 cos(ωt ) (π
vb1 = V1 cos ωt − 23 ) π
vc1 = V1 cos ωt + 23 ( )
For n=2:
va 2 = V2 cos(2ωt ) ( π
vb 2 = V2 cos 2ωt + 23 ) π
vc 2 = V2 cos 2ωt − 23 ( )
For n=3:
va 3 = V3 cos(3ωt ) vb3 = V3 cos(3ωt ) vc3 = V3 cos(3ωt )

28. Balanced nonsinusoidal quantities
For n=3k-2, The harmonics are similar to
positive sequence quantities.
For n=3k-1, the harmonics are similar to
negative sequence quantities.
For n=3k, the harmonics are similar to zero
quantities.

29. Symmetrical components
• Symmetrical components theory can be
applied to both steady-state phasor
quantities and instantaneous quantities.
• Symmetrical components theory can be
derived also by using linear algebra and
treated as variable transformation

30. Voltage and Current across the inductor
diL 1 to + t
vL = L iL = ∫ vL dt + iL (to )
dt L to
Steady-state:
iL (t ) = iL (t + T ) vL (t ) = vL (t + T )
Thus t +T
∫t vL dt = 0
Average voltage across the inductor under
steady-state condition is zero.

31. Voltage and current across the capacitor
dvC 1 to + t
iC = C vC = ∫ iC dt + vC (to )
dt C to
Steady-state:
iC (t ) = iC (t + T ) vC (t ) = vC (t + T )
Thus t +T
∫t iC dt = 0
Average current through the capacitor under
steady-state conditions is zero.

32. Batasan Topologi
• Sumber tegangan hanya boleh diparalel jika sama
besar
• Sumber arus hanya boleh diseri jika sama besar
• Sumber tegangan tidak boleh dihubungsingkat
• Sumber arus tidak boleh dibuka
• Sumber tegangan bisa berupa kapasitor
• Sumber arus bisa berupa induktor

33. Hubungan Berikut Harus Dihindari
+
+

34. Dualitas
Sumber tegangan Sumber arus
Hubungan paralel Hubungan seri
Induktor Kapasitor
Resistor Konduktor
Reverse conducting switch Reverse blocking switch
Variabel tegangan Variabel arus

35. Contoh Dualitas
R L C
+
Vs is Ip vp G C L
dis 1 t dv p1 t
Vs = Ris + L + ∫ is dt + vC (0) I p = Gv p + C + ∫ v p dt + iL (0)
dt C 0 dt L 0

36. Contoh Dualitas
+ +

37. Penggunaan Komputer
• PSIM, MATLAB, EMTP, PSPICE, etc.
• Switching Concept
• Averaging Concept

38. Switching Concept
ii io
vo=SwEd
S1
R Vo(ω)=Sw(ω)Ed
Ed
S2 vo Io(ω)=Vo(ω)/Z(ω)
L Ii(ω)=Sw(ω)Io(ω)
Sw=1 IF S1 ON and S2 OFF
Sw=0 IF S1 OFF and S2 ON

39. Switching Concept
ii S1 io
R
Ed
S 2 vo IF vref > car THEN S w = 1 ELSE S w = 0
L dio
= (S w E d − Rio ) / L
dt
ii = S wio
vref
+
car −

40. Averaging Concept
0 < t < TON
ii S1 io
dio
Rio + L = Ed
dt
R TON < t < Ts
S 2 vo dio
Ed Rio + L =0
dt
L
Averaging
dio
Rio (TON + TOFF ) + L (TON + TOFF ) = Ed TON
dt
vref Divided by Ts results in
+ dio T
Rio + L = E d ON = vo
car − dt Ts

41. D-Q Transform
f qdo = Kf abc f abc = K -1f qdo
(f qdo )T = [ f q fd fo ] ⎡ cos θ sin θ 1⎤
(f abc )
T
= [ fa fc ]
⎢
(π
K = ⎢cos θ − 23
-1
) ( π
sin θ − 23 ) ⎥
1⎥
fb
(
⎢cos θ + 2π ) ( π
sin θ + 23 ) 1⎥
⎡cosθ ( π
cos θ - 23 ) ( π)
cos θ + 23 ⎤
⎣ 3 ⎦
2⎢
K = ⎢ sinθ
3⎢ 1
( π
sin θ - 23 ) ( )
π ⎥
sin θ + 23 ⎥
1 1 ⎥
⎣ 2 2 2 ⎦
t
θ = ∫ ω (ς )dς + θ o
0

42. D-Q Transform
fb
fq
θ
fa
fd
fc

43. Example
ia = 2 I cos ω s t
( 23π )
ib = 2 I cos ω s t +
ic = 2 I cos(ω s t − 23 )
π
iq =
2 2I
3
[ ( π) (
cos ω s t cos ωt + cos ω s t + 23 cos ωt + 2π
3
)+ cos(ω s t − 23π )cos(ωt − 23π )]
iq = 2 I cos(ω s − ω )t
id =
2 2I
3
[ ( ) (
π π ) ( π ) ( π
sin ω s t sin ωt + sin ω s t + 23 sin ωt + 23 + sin ω s t − 23 sin ωt − 23 )]
id = 2 I sin (ω s − ω )t
io = 0

44. D-Q Transform
Power Invariance
Pabc = va ia + vb ib + vc ic
3
(
Pabc = Pqdo = vq iq + vd id + 2vo io
2
)
Instantaneous Reactive Power :
3
(
q = v q id − v d iq
2
)

45. DQ Transform of Stationary Elements
v abc = pλabc
[ ] [ ]
v qdo = Kp K -1λabc = Kp K -1 λqdo + KK -1 pλqdo
⎡ − sin θ cos θ 0⎤
v abc = ri abc [ ]
p K -1
⎢
( π
= ω ⎢ − sin θ − 23 ) ( π
cos θ − 23 ) ⎥
0⎥
v qdo = KrK i qdo-1 ⎣ (
⎢− sin θ + 2π
3
) ( π
cos θ + 23 ) 0⎥
⎦
Thus
KrK -1 = r ⎡ 0 1 0⎤
[ ]
Kp K -1 = ω ⎢ − 1 0 0⎥
⎢ ⎥
⎢ 0 0 0⎥
⎣ ⎦

46. DQ Transform of Stationary Elements
v qdo = ωλdq + pλqdo
(λdq ) = [λd
T
− λq ]
vq = ωλd + pλq
vd = −ωλq + pλd
v o = pλ o

47. DQ Transform of Stationary Elements
iq R L
+
ia R L vq ωLid
a id L
R
ib
n vd ωLiq
b +
ic io R L
c
vo

48. DQ Transform
• If the speed of reference frame ω is equal to
the supply frequency, it is called
synchronous reference frame.
• If the speed of reference frame is zero, it is
called stationary reference frame.
• If the speed of reference frame is not equal
to the supply frequency, it is called
asynchronous reference frame.

49. Space Vector
N N e jθ + e − jθ
na = cos θ =
2 2 2
( π
j θ − 23 ) + e − j (θ − 23π )
nb =
N
2
( π
cos θ − 23 =)2
N e
2
( π
j θ + 23 ) + e − j (θ + 23π )
nc =
N
2
( π
cos θ + 23 = ) N e
2 2
Fa = na ia + nb ib + nc ic
N jθ ⎛ π
− j 23 j 23 ⎞
π N − jθ ⎛ π
j 23 − j 23 ⎞
π
= e ⎜ ia + ib e + ic e ⎟ + e ⎜ ia + ib e + ic e ⎟
4 ⎝ ⎠ 4 ⎝ ⎠
3 N jθ r 3 N − jθ r *
= e i+ e i
8 8
r 2⎡ π
− j 23 j 23 ⎤
π
i = ia + ib e + ic e
3⎢
⎣ ⎥
⎦
is called space vector of current.
The same definitions apply to voltage and current.

50. Example of Space Vector
ia = 2 I cos ω s t =
2
2
[
I e jω s t + e − jω s t ]
(
ib = 2 I cos ω s t + 2π
) =
2
I ⎡e j (ω s t + 23π ) + e − j (ω s t + 23π ) ⎤
3
2 ⎢
⎣ ⎥
⎦
2 ⎡ j (ω s t − 23π ) − j (ω s t − 23π ) ⎤
( π
ic = 2 I cos ω s t − 23 = ) 2
I e
⎢
⎣
+e
⎥
⎦
r
i =
3
2
⎨
⎩
[ ] ⎢
⎣
j (ω t + 2π )
I ⎧ e j ω s t + e − jω s t + ⎡ e s 3 + e
− j (ω s t + 23 ) ⎤ − j 23
π
⎥
⎦
e
π
⎢
⎣
j (ω t − 2π )
+ ⎡e s 3 + e
− j (ω s t − 23 ) ⎤ j 23 ⎫
π
⎥
⎦
e ⎬
π
⎭
r
i = 2 Ie jω s t

51. Transformation of Space Vector
r ω r − j ωt
x = xe
Previous example
rω j (ω s −ω )t
i = 2 Ie
It should be noted :
r
xa = Re[x ]
r
⎡ xe j 23π ⎤
xb = Re
⎢
⎣ ⎥
⎦
r
⎡ xe − j 23π ⎤
xc = Re
⎢
⎣ ⎥
⎦

52. Numerical Methods To Solve Differential Equations
Let
f ' ( x, t ); x (t o ) = xo ; h
Euler Method :
x(t o + h ) = x(t o ) + hf ' ( xo , t o )
Fourth - Order Runge - Kutta Method :
k1 = f (xo , t o )
⎛ k1 h⎞
k 2 = f ⎜ xo + , t o + ⎟
⎝ 2 2⎠
⎛ k h⎞
k 3 = f ⎜ xo + 2 , t o + ⎟
⎝ 2 2⎠
k 4 = f ( xo + k 3 , t o + h )
h
x(t o + h ) = x(t o ) + (k1 + 2k 2 + 2k3 + k 4 )
6

53. Example
di
= 10 − 5i
dt
i0 = 0
h = 0 .1
i (0.1) = 0 + 0.1× (10 − 5 × 0 ) = 1
i (0.2 ) = 1 + 0.1× (10 − 5 × 1) = 1.5
i (0.3) = 1.5 + 0.1× (10 − 5 × 1.5) = 1.75
i (0.4 ) = 1.75 + 0.1× (10 − 5 × 1.75) = 1.875

54. Numerical Integration
We have N data for the function y in the interval from a to b :
b−a
Let h =
N
b N −1
∑
∫a ydt = h n=0 yi (the simplest rule)
b h⎡ N −1 ⎤
∫ ∑
ydt = ⎢ y0 + y N + 2 yi ⎥
2⎢ n =1 ⎥
a
⎣ ⎦

55. Example
T 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
i 1 2 3 1 2 3 1 2 3 1
h = 0.1
N = 10
N −1
∫0idt = h ∑ i = 1.9
0.9
0
N −1
∑ i 2 = 4.3
0.9 2
∫0 i dt = h
0