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Inductively Loaded Rectifiers
7
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
INDUCTIVELY RECTIFIERS
7.1
LOADED
CONTENTS 1. Current flow of an inductive circuit with a sinusoidal supply 2. Current and voltage waveforms of an inductively loaded single-phase rectifier 3. Approximations in the study of rectifiers 4. Half wave three-phase rectifiers 5. Three-phase bridge circuit 6. Table 7-3: single-phase rectifiers 7. Table 7-4: three-phase rectifiers 8. Evaluation
Uncontrolled rectifiers can be classified in various ways: according to the supply voltage, the type of load, the application, etc... Depending on the supply voltage, we can distinguish between single phase and three-phase rectifiers. As far as the load goes we distinguish resistive, capacitive and resistive inductive loads. If we look at applications we distinguish two large groups: 1. Linear and switch mode power supplies : mainly used for the supply to electronic equipment and circuits. 2. Industrial applications: this is mainly inductively loaded single phase rectifiers and three phase bridge rectifiers. This is the material discussed of this chapter. Examples of such industrial applications include: battery chargers, DC welding equipment, DC traction, electrical supply for electrochemical processes (electrolysis, chrome plating, nickel coating), rectifiers for automobile alternators, the supply for frequency converters, etc ... Power Electronics: Switches and Converters http://dx.doi.org/10.1016/B978-0-12-814643-9.50007-6, Copyright © 2018 Elsevier Inc. All rights reserved.
7.2
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
1. CURRENT FLOW OF AN INDUCTIVE CIRCUIT WITH A SINUSOIDAL SUPPLY 1.1 Mechanical switch At instantt1 we close the switch of the circuit with a resistive inductive load and a sinusoidal volt ^ · sin ωt. This is shown in fig. 7-1. The actual applied voltage for the series circuit age v = v of Rb and L b is shown to the right of the switch S. ______________
ω.Lb Rb 2 2 2 __ Assuming Z = Rb + ω . Lb ; tan Φ = ____ R ; cos Φ = Z ; b
√
L
b τ = R__ and α = ω . t1 . b
Calculating leads us to the following expression for the current flow through the load
[
(t − t)
]
(7-1)
_____ ^ 1 v __ i(t) = Z · sin (ωt − ϕ) + sin (ϕ − α) · e τ
i
S v
t1
v^
v v^
Rb
t 0
t1
t 0
t1 Lb
Fig. 7-1: R–L circuit connected to an AC voltage source
!
Remarks 1. In expression (7-1) we notice that the current has two components: (t1 − t) ______ ^ v __ Z · sin ( Φ − α ) · e τ . This is an exponentially decaying term which has practically disappeared after 5.τ. This is the transient term.
. .
^ v __ · sin ( ωt − Φ ) . After 5.τ only this term remains, this is referred to as the steady state
Z term. This is the nominal current which we have learned about in the classical electrical science of an RL-circuit when connected to a sinusoidal voltage. In expression (7-1) we have actually taken account of what happens immediately after closing the switch when connecting a sinusoidal voltage at an RL circuit, in other words we have taken the transient behaviour into account.
7.3
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD 2. If we close the switch in figure 7-1 at t1 = 0 then expression (7-1) becomes:
[
]
^ - t/τ v __ i(t) = Z · sin (ωt − ϕ) + sin ϕ · e
(7-2)
For this special case (t1 = 0) we have drawn the transient term and the nominal term together in fig. 7-2 (this is the total current i(t) !). After 5.τ the transient term is practically zero and i(t) = steady state term.
i
transient term
i(t) v^ z sin Φ
t 5.τ
0 steady state term v^ sin Φ z Fig. 7-2: Current flow of fig. 7-1 if t1 = 0.
1.2 Numeric examples: 1. Supply voltage: 230 V – 50 Hz. Load: 10 Ω – 100 mH. Switch closes at t1 = 0. Determine the current flow.
Solution:
b ^ = 325 V ; τ = ___ R = 10 ms ; Z = 32.96 Ω ; Φ = 72°20’ ; v
L b
i(t) = 9.85 [ sin ( ωt − 72°20’ ) + 0.952 · e
− 100.t
Fig. 7-3 shows the solution.
]
.
7.4
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
v (V) i (A)
v grid i(t)
325V 13,75A
9,85A
ωt 0 Φ 72°20’ -9,85A
-325V
i-transient term
t = 5.τ = 50 ms
i-steady state term Fig. 7-3: Current flow
2. Same details but now the switch closes at α = 72°20’ i(t) = 9.85 [ sin ( ωt − 72°20’ ) + 0 ] = 9.85 · sin ( ωt − 72°20’ ) .
v (V) i (A)
v - grid i(t)= i steady state
ωt 0 72°20'
Fig. 7-4: Current flow
Notice in fig. 7-4 there is no transient term, in other words we immediately arrive at the “nominal” state. The current lags the voltage by 72°20’ .
7.5
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
1.3 Calculating the current flow in fig. 7.1 S
v
i
^ v
^ v.sin ωt
t1
( for t > t1)
Rb
v
t
(b)
t
(c)
t1
0
v
Lb
^ v (a)
-t1 0
^ v.sin ω(t + t1)
Fig. 7-5: R-L circuit connected to an AC voltage source.
A resistive and inductive circuit is connected via a mechanical switch S at instant t1 to a sinusoidal ^ . sin ωt (fig. 7-5). Our goal is to determine the current flow in the circuit. voltage v = v To describe the applied voltage with Laplace notations we first consider fig. 7-5c. ^ . sin ω ( t + t ) = ^ . sin ( ωt + α ) with α = ωt . Here in: v = v v 1 1 This voltage only exists for t ≥ 0 + .
s · sin α + ω · cos α ^ . __________________ F(s) = L [ v ^ · sin (ωt + α) ] = v 2 2
s + ω
By moving the voltage waveform of fig. 7-5c over a distance t1 we arrive at fig. 7-5b. Application of the time delay rule gives:
s · sin α + ω · cos α ^ · e − t1 s · __________________ F(s) = v 2 2
s + ω ____________ ω b L ω Lb Rb __b L _____ _____ ___ 2 2 2 Set: τ = R ; Z = Rb + ω .Lb ; tan ϕ = R ; sin ϕ = Z ; cos ϕ = Z b
√
b
The current in fig. 7-5a can be written as:
s · sin α + ω cos α ^ · e − t1 s · _________________ v 2 2
s + ω
I(s) = _________________________ (R + s . L ) b
b
− t s ____________________ s · sin α + ω · cos α = v ^ · e 1 [ ] (Rb + s · Lb ) · (s 2 + ω 2)
Expressions between square brackets need to be split into partial fractions:
a
b · s + c s · sin α + ω · cos α _________ = ______________________ 2 2 2 (s + ω ) (Rb + s · Lb ) · (s + ω )
_________ + 2 R + s · L b
Solving gives:
b
Lb
a = Z__ · sin (ϕ − α) b = Z__1 · sin (α − ϕ) ω
c = Z__ · cos (ϕ − α)
7.6
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
[
]
So that:
Lb ω __ Z__ . sin (ϕ − α) Z__________________________ 1 . s . sin (α − ϕ) + Z__ . cos (ϕ − α) − t s ____________ ^ 1 I(s) = v · e · + 2 2
_______________________ I(s) = _______ · ________ + 2 1 __ Z
^ − t1 s v · e
[
(Rb + s . Lb )
(s + ω )
sin (ϕ − α)
s . sin (α − ϕ) + ω . cos (α − ϕ)
(s + τ )
(s 2 + ω )
]
Inverse Laplace transform: ^ v
[
]
(t − t )
1 − _____ τ
i(t) = Z__ · e
Z
· sin (ϕ − α) + sin (ωt − ωt + α − ϕ)
1 (t − t 1) _____ ^ − v i(t) = __ · e τ · sin (ϕ − α) + sin (ωt − ϕ)
[ ] i(t) = · [ sin (ωt − ϕ) + sin (ϕ − α) · e ] (t − t) _____ 1 τ
^ v __ Z
(7-1)
With α = ωt1 and ωτ = tg ϕ we can re-write (7-1):
2.
^ v
[
(α − ωt) . cotg ϕ
i(t) = Z__ · sin (ωt − ϕ) + sin (ϕ − α) · e
(7-3)
]
CURRENT AND VOLTAGE FLOW OF AN INDUCTIVELY LOADED SINGLE PHASE RECTIFIER 2.1 Current flow
The diode in fig. 7-6 can be compared to a switch which is closed at t = 0. Expression (7-2) indicates the current flow: ^ →→ (7-2) − t/τ v i = Z__ · sin ϕ · e + sin (ωt − ϕ)
]
[
____________
Lb ωLb 2 2 2 __ Here in: Z = Rb + ω Lb ; Φ = arctan ____ ; τ = R . b R b We consider the same numerical example as that on p. 7-3. The current i(t) of fig. 7-3 can be redrawn as in fig. 7-7. There is however a fundamental difference, since in contrast with the mechanical switch the diode prevents the current becoming negative. The diode blocks at instant t2 , see fig. 7-7. Every period (T, 2 T...) the transient behaviour reoccurs again.
√
i vD
vL
b
v vR Fig. 7-6: Inductively loaded diode circuit.
b
Lb
v^
t
100mH
Rb
10 Ω
v
0
7.7
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
The 8.65 ms and the 13.75 A in fig. 7-7 are calculated in section 2.2. In section number 2-5 we find that t2 = 14.7 ms.
v (V) i (A)
i(t)
current
imax = 13.75A
supply voltage
v
325V
t2
tm
0
2T
T
t (ms)
3T
8.65ms -325V 14.7ms
Fig. 7-7: Current flow of fig. 7-6.
2.2 Time of maximum current strength By neglecting the voltage drop across the diode for every instant in time the voltage is: di
__ v = vR + vL = i · Rb + Lb . dt b
b
di The instant (tm ) of maximum current strength occurs at __ = 0. dt di
^ v
[
sin ϕ
so that:
[
]
− tm /τ
__ __ ____ From (7-2): dt = 0 = Z − τ · e
ωτ cos (ωtm − ϕ) − tm /τ _____________ =
+ ω · cos (ωtm − ϕ) or: e
]
ωτ cos (ωtm − ϕ) ______________ tm = − τ · ln sin ϕ
sin ϕ
(7-4)
^ . Expression (7-4) may be considered as exact insofar as v D << v di __ At the instant (tm ) of maximum current strength = 0 and it follows from dt di __ v = i · R b + L b · dt that v = imax · R b .
The time of imax and also of v R = imax · Rb is the intersection of v R with the v-curve. Function b
b
(7-4) is difficult to use since tm appears in the left and the right term. The easiest method is to
choose a few values for tm in the right-hand side and calculate if we get the same values on the left-hand side.
__
With the data from fig. 7-6 we find that tm = 8.65 ms. With v ^ = 230 · √2 = 325.26 V it follows from (7-2): imax = 13.75 A. This is indicated in fig. 7-7.
7.8
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD 2.3 Voltage waveform. Equal surface area criterion If we neglect the voltage drop across the diode in fig. 7-6 , then we find for every instant in time: v = vL + v R . Here in : v R = i · R b . b b b We reconsider the current waveform in fig. 7-8a that we already determined in fig. 7-7, but now for just one period of the supply voltage. The current reaches its maximum value imax at time tm . In fig. 7-8b we now draw vR = i · R b , together with the supply voltage v. b At every instant v L = v − v R is described. b b 2 1 From zero to tm the coil is energised with an energy 2__ · L b · imax . Betweentm and t2 the polarity of vL reverses (see fig. 7-8b) while i remains positive: the b
energy 2__ 1 · Lb · imax flows out of the coil. At t2 , i is zero. The energy in the coil has also become zero. In fig 7-8c we have only drawn v L . The shaded surface area indicates the flux in the coil. 2
b
[
tm
∫
]
From zero to tm the flux is built up ∆Φ = vL · dt (Wb) and between tm and t2 the flux b o will return to 0. The shaded area below the t-axis is exactly equal to the shaded area above the t-axis. The average voltage v L across the coil is zero. b For an ideal coil this equal surface area criterion is always valid. Note that the (perpendicular) shaded surface area to the left of tm in fig. 7-8b is equal to the shaded surface area to the right of tm . These are the same surface areas as shown in fig. 7-8c!
2.4 Average output voltage of a single phase rectifier The average output voltage in fig. 7-8d (and in fig. 7-6!) depends upon the conduction angle β . β
β
∫ o
∫ o
1 ___ 1 ___ Vdi = 2. π v · dωt = v ^ · sin ωt · dωt 2.π
^ v ___ Vdi = 2. π · ( 1 − cos β )
di V : d = average value (= DC voltage) i = ideal. We consider a rectifier without losses in the transformer or the diode. Since v L
di V ___ = 0, the output DC current may be determined from: Id = R .
b (AV)
The value of the conduction angle β will be determined in section 2.5.
b
(7-5)
7.9
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
i imax (a)
ωt ωt2
t π β ωm
0
2π
v v
v^ vL
(b)
vR
b
b
vR
b
0
t2 t m
b
(c)
T
b
tm
∆Φ = 1
vL
T 2
vL
t
vL .dt b
0
∆Φ 1 + ∆Φ2 = ZERO 0
t
t2
tm
t2
∆Φ = 2
v0
tm
T
vL .dt b
v^ (d)
t 0
T 2
Fig. 7-8: Current and voltage waveforms from the configuration in fig. 7-6
t2
T
7.10
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD 2.5 Conduction angle The angle ωt2 = β (fig. 7-8a) is called the conduction angle. We can determine β by making the current equal to zero in expression (7-2): ^
[
]
− t2 /τ
v __ 0 = Z sin ( ωt2 − Φ ) + sin Φ . e
− β/tgΦ With ωt2 = β and ω·τ = tan Φ we find: sin ( Φ − β ) = sin Φ · e .
The value of β as a function of Φ can only be determined by iteration. The solution is graphically presented in fig. 7-9.
β°
(1-cos β )
360
0
340
0,06
320
0,23
300
0,5
280
0,82
265° 260
1,17
240
1,5
220
1,76
200
1,93
180 0
2 10
20
30
40
50
60
70 72°20'
Fig. 7-9: Relationship between conduction angle and inductivity in the configuration of fig. 7-6.
With Φ = 72°20’ we find: β = 265°. From ωt2 = β = 265° it follows: t2 = 14.7 ms. This is shown in figure 7-7.
80
90
Φ°
7.11
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
3. APPROXIMATIONS IN THE STUDY OF RECTIFIERS For simplicity in the study of rectifiers a number of ideal properties are assumed, namely: 1. Switches without voltage drop or leakage current. 2. Switching occurs without time delay. 3. The supply voltage is perfectly sinusoidal. 4. There is a constant DC output current. These ideal properties are closely approximated by industrial configurations. In the following cases we make use of these approximations.
4. HALF WAVE THREE-PHASE RECTIFIER - RESISTIVE LOAD 4.1 Operation From t1 (fig. 7-10) v1 is the most positive secondary voltage and D1 conducts. Via D 1 , v1 is applied to the cathodes of D 2 and D3 so that these block. At t2 , v2 is larger than v 1 and D2 conducts. At the cathode of D 1 , v2 appears so that D 1 blocks. In the same fashion D3 conducts from t3 .
!
Remarks 1. The phase order of the supply determines which (diode-) switch conduct and which switches turn off (commutate). Since the supply network dictates this sequence we talk about net commutation or natural commutation. The points 1, 2, 3, 4,... (fig. 7-10) are the natural commutation points. 2. As will become clear later, in this circuit pre-magnetisation of the transformer core occurs.
L1
L2
v
L3
1
2
v1
v2 3
v34
v2
v1
v3
t 0
v1
N
v2
v3
Rb
v0 D1
D2
v0
D3
P
0 t1
D1
t2
D2
t3
Fig. 7-10: Half wave three-phase rectifier. Configuration and voltage waveform.
D3
t4
D1
t5
D2
t6
t
7.12
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD 4.2 Commutation Up to now we have assumed that the diodes commutate immediately in the natural commutation points. From practical point of view the leakage reactance of the supply transformer prevents the currents in the conducting diode from suddenly decreasing to zero so that at instant t2 (fig. 7-10) D2 conducts and D1 cannot yet block. Fig. 7-11 indicates an equivalent circuit for this situation. The phases 1 and 2 are temporarily short-circuited. The short-circuited voltage v K at the cathodes of D 1 and D2 is indicated by: v1 + v2 v K = ______ 2 After t2 , v2 becomes more positive and v 1 becomes less positive so that id2 increases and id1 decreases. The difference v K − v1 together with X L1 determines the time at which id1 becomes zero, while v 2 − v K and X L2 determine the increase of id2 . v1
leakage inductance
P
phase 1
XL 1
v2
id 1
phase 2
XL 2
v3
id 2
phase 3
vK
Rb
N
Fig. 7-11: Short-circuited voltage v K during commutation
The commutation time can also be expressed as the overlap angle µ. The shaded surface area (fig. 7-12) shows the voltage loss as a result of commutation. µ
v v1
v2
v3
v2
v1
v3
v1
v2
vK =
v3
vK =
t
v1 + v2 2
v2 + v 3 2
v + v3 vK = 1 2
i
id 1
(DC)
id 2
id
id 2
id 1
id 3
3
id 1
id
2
id
3
t t1
t2
t3
t4
t5
Fig. 7-12: Voltage and current waveform during commutation
t6
7.13
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
The gradual slopes of the flanks during commutation (fig. 7-12) results in less harmonics in comparison to the theoretical square wave pattern of the diode current.
4.3 Average rectified voltage v
v^ v3
v1
-� 0 3
v2
v3
v2
ωt
+� 3
v0
v^
v1
vdi Vdi =
2. �
0
.v^
ωt
Fig. 7-13: Rectified voltage over half wave rectified
We select the zero point of ωt in fig. 7-13 in such a way that it corresponds with the maximum of the phase voltage v 1 .We can then write v 1 = v ^ · cos ωt . The average rectified voltage in the case of an ideal rectifier is then easy to calculate:
[
+π/3
]
|
__
+π/3
^ ^ 3·√ 3 ___ ___ ___ ^ Vdi = 3 · 2.π 1 v · cos ωt · dωt = 2.π 3 · v · ^ sin ωt = 2·π 3· v · 2 · sin 3_ π = ____ 2·π · v −π/3
∫
−π/3
__
3 · √ 3 ______ ^ Vdi = 2 · π · v
(7-6)
___
If Vf is the secondary voltage of the transformer, then v ^ = √ 2 . V f and we find for Vdi : __
3 · √ 6 ______ · Vf = 1.17 · Vf Vdi = 2 · π
(7-7)
Vdi The key output DC current has the same time based waveform as vdi and : Id = ____ R b
7.14
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
5. THREE-PHASE BRIDGE CIRCUIT 5.1 Three-phase bridge circuit. Resistive load 5.1.1 Configuration The diode bridge of fig. 7-14 can in principle be immediately connected to a three-phase network. To adjust the value of the output voltage a transformer is placed at the input of the bridge rectifier. This provides galvanic isolation between the supply network and the DC voltage output. In contrast to normal supply transformers the star point of the secondary may not be earthed. If it is earthed then we can only work with a “floating” output. P
i0 D1
L1 three phase grid
v1 L2
v2 L3
v12
D3
D5
v31
Rb
v0
v23
v3
D4
D6
D2
Fig. 7-14: Three-phase bridge rectifier with diodes
N
5.1.2 Line voltages Between the three line conductors there are three AC voltages present. In relation to each other these voltages are separated by one third of a period. The notation used for the instantaneous line voltages in fig. 7-14 has the following meaning: the order of the subscript letters of v 12 means that v12 is positive when the potential of line L 1 is higher than the potential of line L 2 . If v12 is the largest instantaneous line-voltage, then current will flow from L 1 , through D1,Rb , D6 , to L2 .
v v31
v12
v13
v23
v31
v12
v23 t
0
t 0
t2
Fig. 7-15: Line voltages of the three-phase diode bridge of fig. 7-14
7.15
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
When the largest instantaneous voltage is v 23 then D3 and D2 conduct. If v 31 is the largest instantaneous value then there is conduction via L 3 through D5 , Rb ,D4 to L1 . We now look at what happens when v 31 has the largest negative value of all the line voltages. This is shown in fig. 7-15 from the instant t2 . Now L1 is positive in relation to L 3 and a current will flow from L1 through D1, Rb , D2 to L3 . Instead of working with the negative part of v 31 it is simpler to reason with the positive part of v 13 . We find the same scenario with the negative parts of the other line voltages. In future we work only with the six (positive) voltages as shown in fig. 7-16 while realizing that in reality we have three-line voltages (v12 , v 23 , v 31 ) and their inverse forms (v21 ,v32 , v 13 ). In fig. 7-16 we have only drawn the part of the line voltage that has the largest instantaneous value of all the six line voltages. The line voltage v 13 has been drawn completely for a didactic reason that will later becomes obvious.
v32
v0
1
2 3 4 5 6 1 2 3 v12 v13 v23 v21 v31 v32 v12 v13
ω v12
v13 v32
t
0
t1 t2
-
D6
�/ 6
D3 D2
+�/ 6
D5 D4
v13
v2
v3 v31
t3
D1
v1
v23
v21
D1 D6
D2
2. �
ωt
0
Fig. 7-16: Instantaneous value of the largest line voltage. Commutation points.
5.1.3 Natural commutation points From instant t1 in fig. 7-16 v 12 becomes the largest instantaneous voltage of the three phase net-
work . In other words L1 is positive with respect to L 2 . The current flows then from L 1 , through D1 , R b , D6 to L2 . Diodes D1 and D6 therefore conduct . At instant t2 , v13 becomes the largest voltage andD1 and D2 will conduct. We say that D 2 has taken over from D 6 or that there is commutation from D 6 to D2 . At t3 , v 23 becomes larger than v 13 and D3 together with D 2 will conduct, there is commutation from D1 toD3 . The points 1, 2, 3 etc… in fig. 7-16 are the points where the diodes commutate in a natural manner. These points are referred to as points of natural commutation.
7.16
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD We see that there is always one diode from the top half of the bridge that conducts together with one diode from the bottom half of the bridge. If we ignore the voltage loss across the conducting diode pair, then the output voltage v o is formed by all the peak values of v 12 , v 13 , v 23 , as is shown in fig. 7-16. The output voltage v o of the configuration in fig. 7-14 is again drawn in fig. 7-17, now together with the output current io of the bridge rectifier.
v0 i0
v0
i0 =
v0 Rb t
0 Fig. 7-17: Output voltage and output current of a bridge rectifier
5.1.4 Reference line voltages The natural commutation points are formed at the intersection of two line voltages. The commutation point 1 is the instant that diode D1 conducts for the first time (together with D6 !). To electronically determine the commutation points, it is easier to use the zero crossover of a voltage rather than the intersection point of two voltages of several hundred volts. Luck is on our side because in fig. 7-16 we see that the commutation point 1 (intersection of v32 and v12 ) corresponds with the zero crossover of v13 . We call v13 the reference voltage for D1 . For every diode in the circuit shown in fig. 7-14 we can determine the reference line voltage. The result is presented in table 7-1. A diode conducts for the first time at the zero crossover of its reference line voltage.
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
7.17
A simple rule of thumb allows us to quickly determine the reference line voltage: 1. for a diode from the top half of the bridge we consider the line that is connected to the anode and after that the previous supply line. For example diode D 1 : anode (line 1), previous supply line (3): the zero crossover of the line voltage v 13 is the natural commutation point of diode D1 . 2. for a diode from the bottom half of the bridge : the inverse voltage of the diode above it. Example: diode D6 : Reference line voltage for D 3 is v 21 so that the inverse v 12 reference is for D6 .
Table 7-1
Table 7-2
reference line voltage
diode
driving voltage
Pair of diodes
v13
D1
v12
D6 + D1
D3
v23
D5
v31
v23 v21 v31 v32
v12
D2
v13
D1 + D2
D4
v21
D3 + D4
D6
v32
D2 + D3
D4 + D5 D5 + D6
5.1.5 Driving voltage
We have seen that the diodes D 6 and D1 conduct together with the voltage v 12 . We refer to v 12 as the driving voltage for D6 + D1 . In this way in fig. 7-14 we can determine the driving voltage for every pair of diodes that conduct simultaneously. These driving voltages are indicated in table 7-2.
5.1.6 Pulse number Important factors that determines the design of a rectifier are amongst others the value and the frequency of the ripple voltage. We call pulse number p:
ripple frequency fundamental harmonic of the output voltage p = ______________________________________________________ frequency mains supply
(7-8)
The majority of single phase rectifiers are two pulse and the majority of three phase bridge rectifiers are six pulse. The indexes of E 1 , B2 , B6 refer to the pulse number. The lower the pulse number and the larger the permitted output ripple, the cheaper the output filter will be. In fig. 7-16 we see that the output voltage has a frequency six times larger than the grid frequency: p = 6.
7.18
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD 5.1.7. Average rectified voltage and current From fig. 7-16 we can calculateV di :
[
+π/6
∫
]
|
^ 3. v 1 ___ ^ Vdi = 6 · 2.π v · cos ωt · dωt = ___ π · sin ωt −π/6
^ 3· v
= ___ π
−π/6
__
3 · √ 2 · Vline ^ ___________ 3 · v ____ Vdi = π = π
^ v __ With Vline = ___ becomes: √2
Vdi :
+π/6
(7-9)
d = average value ( = DC voltage ) i = ideal rectifier (transformer and diode losses are ignored)
If the line voltage is 3 x 230V then we find Vdi = 310 V. At 3 x 400V becomesV di = 540V . Vdi ___ The average DC current through Rb is given by: Id = R b
5.2 Three-phase bridge. Resistive-inductive load 5.2.1 Voltage, current Fig. 7-18 shows the configuration. Without the coil the current flow would be as in fig. 7-17. With a coil Lb in series with Rb (fig. 7-18) the ripple in io is even less so that as a good approximation io may be considered constant (fig. 7-18). We call this constant DC current Id . In table 7-2 we see that v12 and v13 are driving voltages for diode D1 while v21 and v31 are driving voltages for diode D4 . We can easily sketch iD andi D in fig. 7-18. 1 4 These currents ( iD and iD ) together form the current iS . 1 4 1 In an identical manner iS and iS can be drawn. If the turns ratio is 1:1, then i p ( = i S ), 1 2 3 1 ip ( = iS ) and ip ( = i S ) can easily been drawn. 2
3
2
3
From the primary phase currents we can derive the line currents iL , iL ,i L ( see evaluation 1 2 3 question no. 7.2). __
The output voltage is the same as for a resistive load:
3 · √2 · V
line ___________ = V π
di
(7-9)
The larger the self inductance Lb , the more constant the output current. Vdi ___ With the average voltage across the coil zero, the average current is: Id = R b
(7-10)
In fig. 7-18 note that iS is practically a block shaped current for 2/3 of a period. 1
The effective value of the secondary line current follows from (5-6): __ ___ 2 = √ δ · I = __ · I or:
IRMS = Iline
d
√3
d
___
√ 32
__
Iline = · I d
(7-11)
7.19
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD 5.2.2 Transformer power
___
The secondary apparent power of the transformer is: S sec. = √ 3 · Vline · Iline __ __ π · V 2 di ______ __ Ssec. = √ 3 · · __3 · I = 1.05 · P d di 3 · √ 2
√
Ssec. = 1.05 · Pdi
D1
^ (v12 = v.coswt)
iL 1 L1
iL 2
L2
3
1
iL 3
L3
iS2
ip
ip
ip
v0
v32
iS1
D3
D5
(7-12)
i0 +
Lb
1
v12
iS3
v0
2 3
2 D4
D6
D2
Rb -
v13 v23 v21 v31 v32 v12 v13 v23 v21 v31
v12
ωt
i0
0
Id
0
iD 1 iD 4
0
Id Id
0
1
iS 2
iS 3
0
ωt ωt
Id
iS
ωt
ωt
Id ωt
0
ωt 0
Fig. 7-18: Inductively loaded three-phase bridge
7.20
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD The solution of evaluation question 7.2c indicates that the apparent power of the primary side is equal to the secondary apparent power (7-12), so that: S prim = 1.05 · Pgi . The apparent power of the transformer is: 1
S = __ 2 · ( Sprim + Ssec ) = 1.05 · Pgi
S = 1.05 · Pgi
(7-13)
In German literature the apparent power is called the “Bauleistung”. If we connect the bridge directly to a three phase supply obviously the value of the rectified voltage is directly related to the value of the supply voltage.
!
Remark In studying rectifiers we assume ideal behaviour, this also includes ignoring the magnetising current of the transformer.
5.3 Twelve-pulse rectifier When we connect two three-phase rectifiers which are 30° out of phase with each other in series we produce a 12-pulse rectifier (fig. 7-19). By connecting one secondary of the supply transformer in star and the other in delta we achieve the 30° phase displacement. The fifth and seventh harmonics disappear on the AC-side (see p. 8.56: 12-pulse circuit). A configuration as shown in fig. 7-19 can be used to supply a “three level” inverter. Such an inverter (NPC-inverter = neutral point clamped inverter) can be used to supply large AC-motors (> 1MW) .
+V
L1 L2
N
L3 S1
S2
S1 / S2: transformer secondaries
Fig. 7-19: Twelve-pulse rectifier used to supply an NPC-inverter
-V
7.21
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
6. TABLE 7-3 SINGLE-PHASE RECTIFIERS HALF WAVE
v
i0
Zb
v0
v
t
Fig. 7-20
v0
0
D2
i0
v D3
D2
v0
v^
v0
Zb
D1
v
t
0
FULL WAVE (bridge)
i0
v
v^
0
Pdi = Vdi · Id
D1
D
Vdi = average DC output voltage, at no load Id = DC current, through load
Ideal diode and transformer. Mains frequency f
FULL WAVE (center tap)
v0
v^ t
Fig. 7-21
Zb
v0
D4
v^ t
0
Fig. 7-22
LOAD IRMS
VRMS
Ripple frequency Ripple factor r (%) TRANSFORMER
load R R R R
1.57 · Id
1.11 · Id
1.11 · Id
1.57 · Vdi
1.11 · Vdi
1.11 · Vdi
f 121
2 · f 47.2
2 · f 47.2
VRMS secondary
load: R/L
2.22 · Vdi
2.22 · Vdi (total)
1.11 · Vdi
IRMS secondary
sine wave square wave sine wave
1.57 · Id −−−− 3.49 · Pdi
0.785 · Id 0.71 · Id 1.23 · Pdi
1.11 · Id Id 1.23 · Pdi
square wave
−−−−
1.57 · Pdi
1.11 · Pdi
sine wave
3.49 · Pdi
1.23 · Pdi
1.23 · Pdi
square wave
−−−−
1.11 ·Pdi
1.11 · Pdi
Id
0.5 · Id
0.5 · I d
VA secondary VA primary
PER DIODE (minimal specifications) load Average current IFAV
Effective current IF(RMS) Peak current IFPM Maximum reverse V RRM IRMS Form factor a = ____ IAV
R/L R
1.57 · Id
0786 · Id
0.786 · Id
R
3.14 · Id
1.57 · Id
1.57 · Id
L
−−−−
R/L
3.14 · Vdi
Id
3.14 · Vdi
Id
1.57 · Vdi
R
1.57
1.57
1.57
7.22
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
7. TABLE 7-4 THREE-PHASE RECTIFIERS HALF WAVE Vdi = average DC output voltage, at no load Id = DC current, through load
FULL WAVE (center tap)
SECONDARY
SECONDARY
L 1 PRIMARY
L
2
v
SECONDARY
L 3
D 1
v
D D
Ideal diodes and transformer Mains frequency f
Z
i0
b
v0
T
D D 2 3
D 1
v
Z b D 4
D 5
D
i0
v0
3
v0
Fig. 7-23
^ 0
v^
t
0
v0
v^
D D D D D D D D 4 1 6 3 2 5 4 1 D 1
D
i0
Zb v0
D D D 4 6 2
T
iD
D D D 1 3 5
v0
6
2
0
Pdi = Vdi · Id
FULL WAVE (bridge)
t
T 0
iD
5
t
Fig. 7-24
v^
D6 D1 D 2 D3 D 4 D 5 D 6 1 2 3 4 5 6 1
D1
t
D5 t
0
Fig. 7-25
LOAD load IRMS
R
Ripple frequency
R R
VRMS
R
Ripple factor r (%)
1.02 · Id
1.02 · Vdi
Id
Vdi
Id
Vdi
3 · f 18.3
6 · f 4.3
6 ·f 4.3
TRANSFORMER VRMS secondary phase
sine/square wave
0.855 · Vdi
0.74 · Vdi
0.428 · Vdi
IRMS secondary
sine/square wave
0.58 · Id
0.408 · Id
0.816 · Id
VA secondary
sine/square wave
1.48 · Pdi
1.81· Pdi
1.05 · Pdi
VA primary
sine/square wave
1.21 · Pdi
1.28 · Pdi
1.05 · Pdi
PER DIODE (minimal specifications) load Average current IFAV
R/L
0.333 · Id
0.167 · Id
0.333 · Id
Effective current IF(RMS)
R/L
0.577 · Id
0.408 · Id
0.577 · Id
R
1.21 · Id
1.05 · Id
1.05 · Id
Peak current IFPM
L
Id
Id
Id
Maximum reverse V RRM
R/L
2.09 · Vdi
2.09 · Vdi
1.05 · Vdi
IRMS ____ Form factor a = I
R/L
1.73
2.44
1.73
AV
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD
7.23
8. EVALUATION 7.1 Consider a single phase rectifier loaded with a motor in series with an extra filter coil (Lm and R m ). Sketch the switching matrix (p. 1.3) for this configuration. 7.2 In the configuration shown in fig. 7-18 the transformer has a winding ratio 1:1. a) Sketch ip , ip , iL , as a function of time. 1 2 2 b) Calculate the effective value of iL . 2 c) Determine the primary apparent power and prove these is equal to the secondary apparent power. 7.3 Consider an SKN400 (Semikron) with an IFAV = 200A. The diode is used in a three phase bridge rectifier. Calculate with expression (2-4) the dissipation per diode. The form factor a = 1.73. 7.4 Is the number of (diodes) switches in the configuration of fig. 7-14 correct according to the theory on p. 1.3 ? 7.5 Why may we express diode power P max = IFM · V RRM in kVA? 7.6 A three-phase bridge rectifier is resistively and inductively loaded. The average output voltage V di should be 100 V. For a load of 5Ω ( = Rb ) , calculate: diode specifications: IFAV ; IFRMS , V RRM transformer specifications: V RMS (secondary phase voltage) ; IRMS (secondary phase current) ; apparent power 7.7 How much energy is stored in the coil shown in fig. 7-6? 7.8 In German literature what is meant by “Bauleistung” of a transformer? 7.9 What is meant by “reference line voltage” of a three-phase bridge rectifier? 7.10 Why does the exponentially decaying term of expression (7-1) practically disappear after t = 5 . τ ? 7.11 How large is v D in fig. 7-6 if the diode is conducting the nominal current? 7.12 What is meant by ”the equal surface area criterion”? 7.13 What is meant by “conduction angle” of a rectifier?
..
7.14 We sometimes refer to natural commutation points. What does this mean? 7.15 Consider the configuration shown in fig. 7-1. Supply voltage 230 V - 50 Hz. 3 · T ____ Load: 10 Ω - 100mH. The switch is closed at t1 = 4 . Determine the current flow. 7.16 Considered of configuration shown in fig. 7-1. Supply voltage 230 V - 50 Hz. Load: 0 Ω -100mH. Determine the current flow in the following situations: 1. t1 = 0 ; 2. t1 = T/4 3.t1 = T/2 . 7.17 Apparently the diodes in fig. 7-14 are numbered in an odd manner. Write down the conduction order of the diode pairs and discover for yourself the logic.
7.24
UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD 7.18 A three-phase bridge circuit with diodes is consecutively loaded as shown in fig. 7-26. Impedance (b) is highly inductive, while in configuration (c) a small inductance and a large capacitance are used. Sketch the theoretical current flow of the three load conditions.
i
Fig. 7-26:
i
R
(a) 7 .19 7.20
i R
L
L
R
C
(c)
(b)
Try to sketch, starting from fig. 7-10, the configuration of a half wave three-phase rectifier resistively and inductively loaded. In addition sketch the waveform of the output voltage and current. Fig. 7-27 shows a full wave three-phase rectifier (the so-called double star circuit). Sketch the waveform of the output voltage and current through diode D1 and diode D4 .
L1
L2
L3
Rb D1
D2
D3
D4
D5
D6
v0
i0
Fig. 7-27: Double star circuit (three-phase full wave rectifier)
7.21 What is the pulse number of the configuration of fig. 7-27? 7.22 What is the pulse number of the configuration in fig. 7-10? 7.23 Which ideal components would we need so that approximations 1, 2, 3, 4 (no. 3 p. 7.11) would not be approximations but exact properties?