analysis_of_6_pulse_converter
TRANSCRIPT
Converter
Analysis
AC System
Dc line
Three phase 3-winding Transformer: one unit for a 12 pulse bridge
6-pulse Bridge
6-pulse Bridge
Converter configuration
ea
eb
ec
ia
Ib
Ic
V1
V2
V3 V5
V6V4
+
Vd
Id
a
b
c
Six pulse Graetz Bridge
Commutating inductance Lc
Smoothing inductance
Ls
ea= Em Cos ωt
eb= Em Cos ( ωt-120o)
ec= Em Cos (ωt +120o)
Valve pair 1-2 conducting
ea
eb
ec
ia
Ib
Ic
V1
V2
V3 V5
V6V4
Vd
Id
a
b
c
Six pulse Graetz Bridge
ea= Em Cos ωt
eb= Em Cos ( ωt-120o)
ec= Em Cos (ωt +120o)
Vd(t)1 3 5
24 6
a bc
1 3 5
24 6
a bc
6 ripples per cycle of ac fundamental
hence named as 6 pulse bridge
ea eb ec
time
Vd Average
Converter operation , with no delay and no overlap
1 3 5
24 6
a bc
ea eb ec
A1
Vd(t)1 3 5
24 6
a bc
6 ripples per cycle of ac fundamental hence named as 6
pulse bridge
eab
A1
time
Vd Average
Converter operation , with no delay and no overlap
Vd(t)1 3 5
24 6
a bc
6 ripples per cycle of ac fundamental hence named as 6
pulse bridge
eab
A1
time
Vd Average
Converter operation , with no delay and no overlap
Vd = 1/(π/3)[ ∫ (eab)dωt] - π/3
0
eab= √3 Em cos( ωt +30)
Vdo = 3√2 ELL/ π
Vdo = 1.35 ELL(rms)
-π/3
0
Vd(t)1 3 5
24 6
a bc
1 3 5
24 6
a bc
6 ripples per cycle of ac fundamental hence named as 6
pulse bridge
ea eb ec
time
Vd Average
Converter operation , with no delay and no overlap
eab
A1
A1
Valve currents
Instanteneous commutation
from valve 1 to 3
I1 I3 I5I5
I6 I2 I4 I6
ea eb ec
Firing delay α
Converter operation with delay and no overlap
Potential of negative terminal
wrt ground
Potential of positive terminal
wrt ground
Instant of natural
commutation
Firing delay angle α
ea eb ec
eb ec ea
With delay , No overlap
Vd (t)
Time, or ωt
Vd average
Vd(t)
1 3 5
24 6
a bc
Vd= Vdo Cos α
α
With delay and no overlap
1
2
3
4
5
6653
2
1
4
5
Valve currents
I1
I2
I3
I4
I5
I6
ea eb ec ea eb ec
α
120o
Id
ea eb ec ea eb ec
α
Valve currents
I1
I2
I3
I4
I5
I6
120o
Id
1
2
3
4
5
6653
2
1
4
5
2
53 3
4 6
Vd(t)
Phase voltages
α µ
Delay30o
Overlap
6 pulse bridge converter
Instant ofnatural commutation
Converter output with 30o delay.
Vx
Vy
30 deg delay
Cathode potential
Anode potential
Phase c voltage
60 deg delay
Cathode potential
anode potential
α=60o
90 deg
Vd(t)
90 deg delay
eab eac
1
4
1
Secondary side Line Currents
3
6
3
2
5
2
6
ILa
ILb
ILc
Line Voltages
eab eac
Vd
Inverter operation
α µ γπ
Valve pair 1-3 commutation
ea
eb
ec
ia
Ib
Ic
V1
V2
V3 V5
V6V4
Vd
Id
a
b
c
ea= Em Cos ωt
eb= Em Cos ( ωt-120o)
ec= Em Cos (ωt +120o)
αµ
I1 I3
Firing delay angle
Overlap
Valve pair 1-3 commutationCommutation loop driving voltage
eba=√2 ELL sin ωt
Firing delay angle
αµ
I1 I3
Overlap
ea
eb
ec
ia
Ib
Ic
V1
V2
V3
Id
a
b
c
Commutation loop i3
i1Lc
Lc
Commutation equation;
2 Lc (di3/dt )= eba =√2 ELL sin ωt
I3(t)=√2 ELLrms / (2 ωLc )[Cos α - Cos ωt]
ea eb
ec
eba
i3i1
Id Id
Commutation from valve 1 to 3
Commutation voltage
√2 ELL Sin(ωt)
αµ
Commutation voltage drop:ΔV= Vdo/2 [ Cos α - Cos (α + µ)] Valve 3 current during commutation:I3(t)=√2 ELLrms / (2 ωLc )[Cos α - Cos ωt]
Commutation summary
• No load direct voltage with α = 0; • Vdo = 3√2 ELLrms / π = 1.35 ELLrms
• Vd = Vdo/2 [ Cos α + Cos (α + µ)] • Vd = Vdo Cos α – Rc Id where Rc = (3ωLc / π)• Commuttion drop;• ΔV= Vdo/2 [ Cos α - Cos (α + µ)] • Incoming valve current;• i(t) = √2 ELLrms / (2 ωLc )[Cos α - Cos ωt]• Id = √2 ELLrms / (2 ωLc )[Cos α - Cos (α + µ)]
Commutation -ex
• 1. prove that the ratio of commutation drop to the direct voltage of a six pulse bridge is
• [ Cos α - Cos (α + µ)] / [ Cos α + Cos (α + µ)]
• 2.commutation overlap angle • µ = Cos-1[Cos α - √2ωLc Id / ELL] -α
Ex2.
• A six pulse converter when operating with a firing delay of 20o generates + 120 kV DC on no load. If the dc terminal voltage at 800A dc current is + 110 kV . Find
• (1) The commutation reactance• (2) Overlap angle• (3) Commutation time interval• (ans: 13.089 ohms, ?o , ? msec )• Extn; plot Vd_ Id characteristic for variation of
current from 1000A to o A .