Formula Sheet Final 2013 Power Electronics

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