TSTE19 Power Electronics · Half-wave rectifier output voltage ... • In the ideal three-phase...

TSTE19 Power ElectronicsLecture 3

Tomas JonssonICS/ISY

Outline• Rectifiers

– Current commutation

• Rectifiers, cont.

– Three phase

2015-11-09 2TSTE19/Tomas Jonsson

Effect of Ls on current commutation• Current commutation =

current path changed fromone diode to another

– Commutation notinstantaneous when Lsnonzero

– Magnetic energy change

• Use simplified example

– Output represented byconstant dc current source


vs>0

Effect of Ls on current commutation• Current commutation =

current path changed fromone diode to another

– Commutation notinstantaneous when Lsnonzero

– Magnetic energy change

• Use simplified example

– Output represented byconstant dc current source


vs<0

Source inductance effects, cont• Waveform if L=0

• Prior to ωt = 0, vs is negative, current flow through D2

– vd = 0, is = 0


Current commutation• During commutation (ωt > 0)

– vs positive, D1 turns on

– iD1 = is

– iD2 = Id – is

– iD1 +iD2 = Id

– D2 stops conductingwhen iD2 = 0


0 5 10 15 20 25 300

2

4

6

8

10

12

wt [deg]

[A]

iD1

iD2

After commutation completedValid for 0 < is < Id

Commutation current

• Commutationcurrent– Temporary

currentcontributionrelated toenergy transfer

Current commutation waveforms• Large Ls used

to clearly showeffect

• Time forcommutationdepend onLs size andcurrent changein Ls


Current commutation time• is through inductor starts at zero, end at Id when ωt=u

• Integrate both sides, left is area Au (voltage * angle)

• Commutation angle can be calculated


= 2 sinω = = ω ω 0 < ω <

2 sinω ω = ω

= ∫ 2 sinω ω = 2 1 − cos = ω ∫ = ω

cos = 1 −ω

2

Half-wave rectifier output voltage• Vdo = Ideal average voltage of half-wave rectified voltage

(effect of the commutation inductance Ls neglected)

• = ∫ 2 sin = −cos =

• ≈ 0.45


0 30 60 90 120 150 180 210 240 270 300 330 360-1.5

-1

-0.5

0

0.5

1

1.5vd

vsVd

Output voltage incl commutation voltage drop

• Vd = Vd0 - DVd = Vd0 – = 0.45Vs -

• Commutation voltage drop appears as a resistance to the dc-side current. Rcomm =


Commutation conclusions• Conduction:

– Magnetic energy isstored related to theinductance of theconduction path

• Commutation– Transfer of current

between two paths:– ÞStored magnetic

energy needs to betransfered!

– Output voltagereduction proportionalto Id and Ls

Exercise 5-5Consider the basic commutation circuit of Fig. 5-11a with Id = 10 A.

a) With Vs=120 V at 60 Hz and Ls = 0, calculate Vd and theaverage power Pd

b) With Vs=120 V at 60 Hz and Ls = 5 mH, calculate u, Vd , and Pd

c) With data as i b) calculate u, Vd , and Pd with Id = 20 A


Current commutation in full-bridge• Same principle for area Au due to Ls


Rectifier during current commutation• vs negative before t = 0

– D3 and D4 conducting

– is = -Id

• vs positive

– D1 and D2 startsconducting(Short circuit paththrough D3 and D4)

• iu are commutationcurrents

• vd = 0 during commutation


Valid for -Id < is < Id

Current commutation angle• is change is double that of previous example (from -Id to Id)


cos = 1 −2ω

2

Full-bridge rectifier output voltage• Vdo = average voltage of full wave rectified voltage

(effect of the commutation inductance Ls neglected)

– = ∫ 2 sin = −cos =

– ≈ 0.9


0 30 60 90 120 150 180 210 240 270 300 330 360-1.5

-1

-0.5

0

0.5

1

1.5vd

vs

Vd

In the single-phase rectifier circuit shown in Fig. 5-14a, Vs = 120 Vat 60 Hz, Ls = 1 mH, and Id = 10 A.

1. Calculate u, Vd, and Pd

2. What is the percentage voltage drop in Vd due to Ls?

Exercise 5-8


3-phase full-bridge rectifier, general view

• Less ripple on output

• Handles higher power

• No current in neutral wire


3-phase full bridge rectifiers, L = 0• One diode in each group is conducting at any time


6-pulse Graetzrectifier bridge

To load

From load

+

-0

D1

ua - ub

ua

ub

uc

D6

D3 D5

D4 D2

Diode rectifier


To load

From load

0

+

-

ua

ub

uc

D1

D6

D3 D5

D4 D2

ua - uc

Diode rectifier


To load

From load

0+

-

ua

ub

uc

D1

D6

D3 D5

D4 D2

ub - uc

Diode rectifier


To load

From load

0

-

+

ua

ub

uc

D1

D6

D3 D5

D4 D2

ub - ua

Diode rectifier


To load

From load

0

-

+

ua

ub

uc

D1

D6

D3 D5

D4 D2

uc - ua

Diode rectifier


To load

From load

0-

+

ua

ub

uc

D1

D6

D3 D5

D4 D2

uc - ub

Diode rectifier

3-phase full bridge rectifier waveforms• Every diode conducts 1/3

of the cycle

• Output waveformcontains 6 segments

• Instantaneouscurrent commutationdue to L = 0


= −

= 2

=1π 3⁄ 2

⁄

⁄

cosω ω

= 1.35

Principles of AC/DC conversion, 6-pulse bridgeId

1 3 5

Ud

4 6 2

Ua

Ub

Uc

Ia

Ib

a

c

b

Ic

Ua Ub Uc

16

12

32

34

54

56

Uac Ubc

tw

Exercise 3-100• In the ideal three-phase rectifier circuit, construct the wave

forms of diode D1 and D2 voltages and currents.


Input line current3ph rectifier• No 3rd harmonic

• Compare with single phasePF = 0.9


: =23

: =1π 6 = 0.78

= cosϕ = 1.0

= =cosϕ

= cosϕ =3π = 0.955

§ Fourier analysis gives additional harmonic components§ Remember calculation uses RMS of Is, Is1 and Id

=2π 2 = 0.9 = 0 ℎ

= ℎ ℎ

2015-11-09TSTE19/Tomas Jonsson 31

Single phase rectifier, input current

Source inductance effects, DC current load

• Source L not 0

• Only one current commutation at a time

– 6 commutations during oneline-frequency cycle


Transfer of current from valve 1 to valve 3

1

I

d a b

d

I I1 3

3

U

Ua

b

Xc

cX

t

tu

I

I

1

3

U = 0.5 ( U +U )

2015-11-09TSTE19/Tomas Jonsson

Current commutation• Current commutation

phase c -> phase a (D5 -> D1)

• Au indicates the currentcommutation voltage drop

• Au only half of areabetween va and vcbecause of twoinductances


+ =

= ω ∫ = ∫ ω =

=∫ ( ) ω = ( )

Effect of commutation inductance


= ω ∫ = ω =

=∫ ω =∫ ( ) ω = ( )

D =ω

3=

3ω

= - D = 1.35 − ω

A 3-ph rectifier feeding a constant current load has the following data:VLL = 400 V at 50 Hz, Ls = 7 mH.

The maximum ac-side rms current Is = 10 A.

Exercise 3-101


Calculate

a) Max dc-side current

b) Average dc-side voltageat max current

c) Max active power

d) Diode average current

e) Diode rms current

Exercise 3-102Using results from exercise 3-101, calculate

• Diode conduction losses (pD1=∫ uD1 iD1)using the diode BYW29E withV0=0.79V, Rs = 0.013 ohm (Tj=25C)

• Use the diode on-state model belowwhere iD can be expressed with itsaverage and rms current as calculatedabove


Inrush current• LC-circuit fed by voltage step

– Worst case when input voltage at maximum when applied

– Peek voltage twice the input voltage step

• DC circuit needs to support twice the peak input voltage!

• Alternative: limit current, using resistor. Short resistor afterstart using thyristor


= 2 2 (singlephase)= 2 2 (threephase)

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TSTE19 Power Electronics · Half-wave rectifier output voltage ... • In the ideal three-phase...

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