Formula sheet – Final Exam – IGEE 401 / ELEC 6411 Average and rms values The instantaneous power is:
p(t ) = v i
PAV =
1 T
T
T
∫ p(t ) dt = o
Sinusoidal steady-state and phasors v = 2 V sin ωt
Power, reactive power and power factor S = VI* = Vej0 IejΦ = VI ejΦ = SejΦ S = VI cos φ + jVI sin φ
i = 2 I sin (ωt − φ )
S = P + jQ,
I = V Z , Z = R 2 + X 2 L −1 X L = ω L, φ = tan ( X L R )
V = Ve j 0 and I =
T
1 1 2 v i dt I = I rms = i dt ∫ To T o∫
S 2 = P2 + Q2
Apparent Power (VA) = S = VI Real Power (W) = P = Re[S] = VI cosΦ Reactive Power (VAr) = Q = Im [S] = VIsinΦ PF = P = cosΦ S
V Ve j 0 V − jφ = jφ = e = Ie − jφ Z Ze Z
In phase (IP) and out-of-phase (IQ) components: I P = I cosφ → P = VI P , IQ = I sin φ → Q = VIQ Three-phase systems j0
Va = Ve , Vb = Ve
−j
2π 3
j
, Vc = Ve
2π 3
π
Vab = Va − Vb =
j( ) 3Ve 6
π
= VLL
j( ) e 6
S 3φ = 3 VL I L , P3φ = S 3φ cosθ , Q3φ = S 3φ sin θ = 3VL2ωC
Transformers (ideal):
NpIp = NsIs Vp/Np = Vs/Ns VpIp = VsIs
Line current distortion 12
1T is (t ) = is1 (t ) + ∑ ish (t ), I s = ∫ is2 (t ) dt T h ≠1 0
=
I
∑ I sh2 , THD(%) = 100 Idis , where I dis =
h =1
I s2 − I s21 =
∑ I sh2 , Crest factor = h ≠1
s1
Average power, total power factor, and displacement power for a nonlinear load: PAV =
1 T
T
∫
2Vs sinωt is dt = Vs I s1 cos φ1 , S = Vs I s , PF =
o
Thyristor based ac controllers
I n = V0 _ n Z n
Phase controlled Fundamental current for resistive load:
Vs I s1 cos φ1 I s1 I 1 = cos φ1 = s1 DPF , PF = DPF Vs I s Is Is 1 + THDi2
where Z n = R 2 + (nωL) 2
Iˆ1 = a12 + b12 ,
Fundamental component for inductive load:
I L1 =
VS
πωL
a1 =
2 VS (cos 2α − 1), 2π R
(sin 2α + 2π − 2α )
On-off or integral half-cycle control: n cycles on and m cycles off.
(rms)
b1 =
2 VS (sin 2α + 2π − 2α ) 2π R
I s , peak Is
Single-phase diode bridge rectifier (considering a highly inductive load):
Vd 0 = 0.9V s , I s = I d and I s1 = 0.9 I d , DPF = cos φ1 = 1, PF = DPF Line voltage distortion: di di di vPCC = vs − Ls1 s , is = is1 + ∑ ish , vPCC _ 1 = vs − Ls1 s1 , vPCC _ dist = Ls1 ∑ sh = dt dt h ≠1 h ≠1 dt Three-phase full bridge rectifiers:
v d = v Pn − v Nn , Vd 0 =
I s1 = 0 .9 Is
∞
∑ (I sh X Lsh )2 h ≠1
1 π /6 2V LL cos ωt dωt , Vd 0 = 1.35V LL = 2.34V s π / 3 ∫−π / 6
Input current components considering a highly inductive load: I s = 0.816 I d
I s1 = 0.78 I d
I sh =
I s1 h
DPF = 1
PF = 0.955
Single-phase thyristor ac-dc converters: 2 2 V dα = V S cos α = 0 . 9V S cos α , Pdc = I dc V d α
π
S = V s I s , S 1 = V s I s 1 , P1 = V s I s 1 cos α , Q 1 = V s I s 1 sin α , I S 1 = 0 . 9 I dc Three-phase thyristor ac-dc converters considering a highly inductive load: V dα = 1.35V LL cos α , I s =
S =
3V LL I s , S 1 =
3V LL I s 1 , P1 =
3V LL I s 1 cos α , Q 1 =
Q SVC = QTCR − Q FC =
SVC (TCR+FC): Voltage regulation
I 2 3 3 I d = 0.816 I d , I s1 = 0.78 I d = I s , DPF = cos φ1 = cos α , PF = DPF s1 = cos α 3 Is π π V s2 X L1 _ eff (α )
−
3V LL I s 1 sin α
V s2 , XC
I Th _ rms =
2I 2
VS = jX S I + VT , VS = V S ∠δ , VT = VT ∠0, I = I∠ − φ , VT = VS − jX S I
Tuned harmonic filters
f h = h ⋅ f grid =
Buck dc-dc converter:
(Vd − Vo )t on = (− Vo )t off , Vo Vd
=
1 2π C h Lh
t on v I V 1 = D = ctrl , o = d = TS Vˆst I d Vo D
DTS TV 1 I L , peak = (Vd − Vo ) = S o (1 − D ) = I oB 2 2L 2L 2 2 2 fc ∆Vo 1 TS (1 − D) π 1 = = (1 − D) , f c = Vo 8 LC 2 2π LC fs
I LB =
Boost dc-dc converter:
(Vd )t on = (Vd I LB =
− Vo )t off ,
Vo I 1 = , o = (1 − D ) Vd 1 − D I d
V t VT 1 I L , peak = d on = o S D (1 − D ) 2 2L 2L
I oB =
V o TS D (1 − D ) 2 2L
∆Vo DTS T = =D S Vo RC τ
Single-phase inverters:
id (t ) =
Vo I o V I cos φ − o o cos(2ωt − φ ) = I d + i d 2 Vd Vd
SPWM (half-bridge)
ma =
Vˆcontrol f , m f = tri f control Vˆtri
(VˆAo )1 = m a
Vd Vˆ , m a = control ≤ 1 2 Vˆtri
f h = ( j m f ± k ) f 1 , j integer k odd/even h = j mf ±k
Square Wave (half-bridge) (VˆAo )1 =
4 Vd V = 1.273 d π 2 2
(VˆAo ) h =
(VˆAo )1 , h odd integer h
Voltage cancellation: 4V (180 − α ) , half of a pulse width. Vˆoh = d sin(hβ ), β = πh 2 3-phase SPWM
(V LL )1 _ rms =
3-Phase square-wave:
3 2 2
V LL 1 _ rms =
m aVd , m a ≤ 1 6
π
Vd , V LL h _ rms =
6 Vd , h = 6n ± 1 πh
For SPWM controlled dc-ac converters it should be noted that the values in the table are peak values normalized with respect to the voltage that appears at the output of the converter: Vd for full bridge and 0.5 Vd for half bridge. Also, the principle of voltage cancellation
Design of a second-order LPF: Gain =
Vout _ h , Gain(dB) = 20 log(Gain), fres = Vin _ h
fh 10
Gain( dB ) − 40 dB / Dec
UPF diode rectifier: V I V I cos 2ωt pin (t ) = Vˆs sin ωt Iˆs sin ωt = Vs I s − Vs I s cos 2ωt , p d (t ) = Vd id (t ), id (t ) = s s − s s Vd Vd v d ,ripple (t ) ≈
I sin 2ωt Vd 1 ic dt = − d , For constant frequency control : I rip = ∫ Cd 2 ω Cd 4 f s Ld
Switch-mode bi-directional grid interface:
P = V s I s1 cos θ =
[
V s2 ω Ls
Vconv1 V2 sin δ , Q = V s I s1 sin θ = s ω Ls Vs
Vconv1 = V s2 + (ω Ls I s1 )
]
2 0.5
=
ma 2
Vconv1 1 − cos δ Vs
V dc
HVDC:
V dc _ B = −1.35 V LL cos α B , γ B = 180 o − α B , extinction angle, V dc _ B = 1.35 V LL cos γ B PdB = 1.35 V LL I d cos γ B , Q dB = 1.35 V LL I d sin γ B
