Transcript
R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
6.1. Circuit Manipulations L
1
+ Vg
+ –
2
C
R
V –
Begin with buck converter: derived in Chapter 1 from first principles • Switch changes dc component, low-pass filter removes switching harmonics • Conversion ratio is M = D
Fundamentals of Power Electronics
2
Chapter 6: Converter circuits
6.1.1. Inversion of source and load Interchange power input and output ports of a converter Buck converter example
V2 = DV1
Port 1
L
1
+ + –
Port 2
+ 2
V1
V2
–
–
Power flow Fundamentals of Power Electronics
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Chapter 6: Converter circuits
Inversion of source and load
Interchange power source and load: Port 1
L
1
+
Port 2
+ 2
V1
V2
–
–
+ –
Power flow
V1 = 1 V2 D
V2 = DV1 Fundamentals of Power Electronics
4
Chapter 6: Converter circuits
Realization of switches as in Chapter 4
Port 1
• Reversal of power flow requires new realization of switches • Transistor conducts when switch is in position 2 • Interchange of D and D’
V1 = 1 V2 D'
Fundamentals of Power Electronics
L
Port 2
+
+
V1
V2
–
–
+ –
Power flow
Inversion of buck converter yields boost converter
5
Chapter 6: Converter circuits
6.1.2. Cascade connection of converters Converter 1 Vg
+ –
V1 = M 1(D) Vg
Converter 2
+ V1 –
V = M (D) 2 V1
+ V –
D
V1 = M 1 (D)Vg
V = M 2 (D)V1
Fundamentals of Power Electronics
V = M(D) = M (D)M (D) 1 2 Vg
6
Chapter 6: Converter circuits
Example: buck cascaded by boost L1
1
L2
2
+
+ Vg
1
2
+ –
V1
C1
C2
R
V –
{ {
–
Buck converter
Boost converter
V1 =D Vg
V = D Vg 1 – D
V = 1 V1 1 – D Fundamentals of Power Electronics
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Chapter 6: Converter circuits
Buck cascaded by boost: simplification of internal filter Remove capacitor C1 L1
1
L2
2
+ Vg
1
2
+ –
C2
R
V –
Combine inductors L1 and L2 L
1
iL
2
+ Vg
+ –
2
1
V
Noninverting buck-boost converter
– Fundamentals of Power Electronics
8
Chapter 6: Converter circuits
Noninverting buck-boost converter L
1
iL
2
+ Vg
1
2
+ –
V –
subinterval 1 iL Vg
+ –
subinterval 2 +
+
V
Vg
V iL
– Fundamentals of Power Electronics
+ –
9
– Chapter 6: Converter circuits
Reversal of output voltage polarity subinterval 1
subinterval 2
noninverting buck-boost
Vg
+ –
V
Vg
+ –
Vg
+
+ –
V
V iL
–
iL
inverting buck-boost
+
+
iL
iL Vg
+ –
10
+ V –
–
Fundamentals of Power Electronics
–
Chapter 6: Converter circuits
Reduction of number of switches: inverting buck-boost Subinterval 1 +
iL Vg
Subinterval 2
+ –
+
iL + –
Vg
V
V –
–
One side of inductor always connected to ground — hence, only one SPDT switch needed: 1
Vg
+ –
+
2
iL
V
V =– D Vg 1–D
– Fundamentals of Power Electronics
11
Chapter 6: Converter circuits
Discussion: cascade connections
• Properties of buck-boost converter follow from its derivation as buck cascaded by boost Equivalent circuit model: buck 1:D transformer cascaded by boost D’:1 transformer Pulsating input current of buck converter Pulsating output current of boost converter
• Other cascade connections are possible Cuk converter: boost cascaded by buck
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Chapter 6: Converter circuits
6.1.4. Differential connection of load to obtain bipolar output voltage dc source
load
Converter 1
+ V1
V1 = M(D) Vg
+
–
V Vg
+ –
–
D
Converter 2
V = V1 – V2 The outputs V1 and V2 may both be positive, but the differential output voltage V can be positive or negative.
+ V2
V2 = M(D') Vg
Differential load voltage is
–
D' Fundamentals of Power Electronics
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Chapter 6: Converter circuits
Differential connection using two buck converters
}
Buck converter 1 1
Converter #1 transistor driven with duty cycle D
+ 2
V1 +
–
V Vg
–
+ –
Converter #2 transistor driven with duty cycle complement D’ Differential load voltage is
2 1
V = DVg – D'V g
+ V2
Simplify:
–
{
V = (2D – 1)Vg
Buck converter 2
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Chapter 6: Converter circuits
Conversion ratio M(D), differentially-connected buck converters V = (2D – 1)Vg
M(D) 1
0
0.5
1
D
–1 Fundamentals of Power Electronics
17
Chapter 6: Converter circuits
Simplification of filter circuit, differentially-connected buck converters Original circuit
Bypass load directly with capacitor
}
Buck converter 1 1
1
+ 2
–
Vg
2
V1 +
+
V
V
–
+ –
Vg
2
–
+ – 2
1
+
1
V2
{
–
Buck converter 2
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18
Chapter 6: Converter circuits
Simplification of filter circuit, differentially-connected buck converters Combine series-connected inductors
Re-draw for clarity C
1
1
Vg
2
+ –
2
2
L iL
+
V – 1
R + Vg
V
+ –
– 2 1
Fundamentals of Power Electronics
H-bridge, or bridge inverter Commonly used in single-phase inverter applications and in servo amplifier applications
19
Chapter 6: Converter circuits
Differential connection to obtain 3ø inverter dc source
3øac load
Converter 1
+
V1 = M(D 1) Vg
V1
Vn = 1 V1 + V2 + V3 3 n
D1
– v a
+ –
+
– Vg
+
V2 = M(D 2) Vg
V2
+
V3 = M(D 3) Vg
V3
Control converters such that their output voltages contain the same dc biases. This dc bias will appear at the neutral point Vn. It then cancels out, so phase voltages contain no dc bias.
– D3
Fundamentals of Power Electronics
Phase voltages are Van = V1 – Vn Vbn = V2 – Vn Vcn = V3 – Vn
+
– D2
Converter 3
Vn
+ vbn – n – vc
Converter 2
With balanced 3ø load, neutral voltage is
20
Chapter 6: Converter circuits
3ø differential connection of three buck converters 3 ac load dc source
+
+
V1 – v a
Vg
n
– + – +
+
–
n – vc
V2
Vn
+ vbn –
+ V3 –
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21
Chapter 6: Converter circuits
3ø differential connection of three buck converters Re-draw for clarity: dc source
– v a
n
+
3 ac load
Vg
+ –
Vn
+ vbn – n – vc
+
“Voltage-source inverter” or buck-derived three-phase inverter
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Chapter 6: Converter circuits
The 3ø current-source inverter dc source
– v a
n
+
3 ac load
Vg
+ –
Vn
+ vbn – n – vc
+
• Exhibits a boost-type conversion characteristic
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23
Chapter 6: Converter circuits
Single-input single-output converters containing one inductor • Use switches to connect inductor between source and load, in one manner during first subinterval and in another during second subinterval • There are a limited number of ways to do this, so all possible combinations can be found • After elimination of degenerate and redundant cases, eight converters are found: dc-dc converters buck
boost
buck-boost
noninverting buck-boost
dc-ac converters bridge
Watkins-Johnson
ac-dc converters current-fed bridge Fundamentals of Power Electronics
inverse of Watkins-Johnson 25
Chapter 6: Converter circuits
Converters producing a unipolar output voltage M(D) = D
1. Buck
M(D)
1
1
+ Vg
+ –
2
M(D) =
2. Boost
V
0.5
–
0
1 1–D
D
0
0.5
1
D
3
1
2
V
1 0
–
Fundamentals of Power Electronics
1
4
+ + –
0.5
M(D)
2
Vg
0
26
Chapter 6: Converter circuits
Converters producing a unipolar output voltage
3. Buck-boost
M(D) = – 1
D 1–D
0.5
1
D
0
0.5
1
D
–1
+
2
0 0
–2
Vg
+ –
V
–3 –4
–
4. Noninverting buck-boost
M(D) =
M(D)
D 1–D
M(D)
2
1
4
+ Vg
+ –
2
3
1
2
V
1 0
–
Fundamentals of Power Electronics
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Chapter 6: Converter circuits
Converters producing a bipolar output voltage suitable as dc-ac inverters M(D) = 2D – 1
5. Bridge
M(D) 1
1
Vg
+ –
2
+ V – 2
0
1
0.5
1
D
0.5
1
D
–1
M(D) = 2D – 1 D
6. Watkins-Johnson 2
Vg
1
+ –
+ V
1
2
M(D) 1
Vg
+ –
0 –1
2
V –
–
Fundamentals of Power Electronics
1
+
or
28
–2 –3
Chapter 6: Converter circuits
Converters producing a bipolar output voltage suitable as ac-dc rectifiers M(D) =
7. Current-fed bridge
M(D)
1 2D – 1
2 1
1
Vg
+ –
1
Vg
2
+ – 2
1
M(D) =
D 2D – 1 +
+ –
V
2
D
1 0 –1
–
Fundamentals of Power Electronics
1
2
or Vg
0.5
M(D)
1
V
D
–2
1
+
1
–1
+ V – 2
8. Inverse of Watkins-Johnson
0.5 0
2
–
29
–2
Chapter 6: Converter circuits
Several members of the class of two-inductor converters
´ 1. Cuk
M(D) = –
D 1–D
0
0.5
1
D
0
0.5
1
D
0 –1
+ Vg
1
+ –
2
–2 –3
V
–4
M(D)
–
M(D) =
2. SEPIC
2
Vg
+ –
1
D 1–D
M(D) 4
+
3 2
V
1
–
Fundamentals of Power Electronics
0
30
Chapter 6: Converter circuits
Several members of the class of two-inductor converters
M(D) =
3. Inverse of SEPIC 1
D 1–D
M(D) 4
+ Vg
+ –
3 2
2
V
1 0
–
M(D) = D 2
4. Buck 2
0
0.5
1
D
0
0.5
1
D
M(D)
1
Vg
+ –
2
1
V
0.5
2
1
Fundamentals of Power Electronics
+
–
31
0
Chapter 6: Converter circuits
6.3. Transformer isolation Objectives: • Isolation of input and output ground connections, to meet safety requirements • Reduction of transformer size by incorporating high frequency isolation transformer inside converter • Minimization of current and voltage stresses when a large step-up or step-down conversion ratio is needed —use transformer turns ratio • Obtain multiple output voltages via multiple transformer secondary windings and multiple converter secondary circuits
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32
Chapter 6: Converter circuits
A simple transformer model Multiple winding transformer i1(t)
n1 : n 2
Equivalent circuit model i1(t)
i2(t)
+
+
+
v1(t)
v2(t)
v1(t)
–
–
i1'(t)
n1 : n2
i2(t) +
iM(t) LM
v2(t)
–
– i3(t)
i3(t) + v3(t)
v3(t) –
–
: n3
: n3
Fundamentals of Power Electronics
+
v1(t) v2(t) v3(t) n 1 = n 2 = n 3 = ... 0 = n 1i 1' (t) + n 2i 2(t) + n 3i 3(t) + ...
Ideal transformer
33
Chapter 6: Converter circuits
The magnetizing inductance LM Transformer core B-H characteristic
• Models magnetization of transformer core material
B(t)
• Appears effectively in parallel with windings • If all secondary windings are disconnected, then primary winding behaves as an inductor, equal to the magnetizing inductance
v1(t) dt
saturation
slope
LM H(t)
i M (t)
• At dc: magnetizing inductance tends to short-circuit. Transformers cannot pass dc voltages • Transformer saturates when magnetizing current iM is too large Fundamentals of Power Electronics
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Chapter 6: Converter circuits
Volt-second balance in LM The magnetizing inductance is a real inductor, obeying i1(t)
di (t) v1(t) = L M M dt
+ v1(t)
integrate: i M (t) – i M (0) = 1 LM
t 0
Ts 0
i2(t) +
iM(t) LM
v2(t) – i3(t) + v3(t) – : n3 Ideal transformer
v1(t)dt
Fundamentals of Power Electronics
n1 : n2
–
v1( )d
Magnetizing current is determined by integral of the applied winding voltage. The magnetizing current and the winding currents are independent quantities. Volt-second balance applies: in steady-state, iM(Ts) = iM(0), and hence
0= 1 Ts
i1'(t)
35
Chapter 6: Converter circuits
Transformer reset • “Transformer reset” is the mechanism by which magnetizing inductance volt-second balance is obtained • The need to reset the transformer volt-seconds to zero by the end of each switching period adds considerable complexity to converters • To understand operation of transformer-isolated converters: • replace transformer by equivalent circuit model containing magnetizing inductance • analyze converter as usual, treating magnetizing inductance as any other inductor • apply volt-second balance to all converter inductors, including magnetizing inductance
Fundamentals of Power Electronics
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Chapter 6: Converter circuits
6.3.1. Full-bridge and half-bridge isolated buck converters Full-bridge isolated buck converter
Q1
D1
Q3
D3
i1(t)
1 : n
D5
+ –
i(t) +
+
+ Vg
L
iD5(t)
vT(t)
vs(t)
C
R
v
– – Q2
D2
Fundamentals of Power Electronics
Q4
: n
D4
37
–
D6
Chapter 6: Converter circuits
Full-bridge, with transformer equivalent circuit
Q1
Vg
D1
Q3
+ –
D3
i1(t)
i1'(t)
+
iM(t)
vT(t)
LM
1 : n
D5
L
iD5(t) i(t) + vs(t)
+ C
R
v
– – Q2
D2
Q4
: n
D4
Ideal
D6
–
iD6(t)
Transformer model
Fundamentals of Power Electronics
38
Chapter 6: Converter circuits
Full-bridge: waveforms iM(t)
Vg LM
vT(t)
• During first switching period: transistors Q1 and Q4 conduct for time DTs , applying voltseconds Vg DTs to primary winding
– Vg LM
Vg 0
0 –Vg
• During next switching period: transistors Q2 and Q3 conduct for time DTs , applying voltseconds –Vg DTs to primary winding
i(t) ∆i
I
vs(t)
nVg
nVg
iD5(t)
0
0
0.5 i
0.5 i
i 0 0
conducting devices:
DTs Q1 Q4 D5
Ts D5 D6
Fundamentals of Power Electronics
t Ts+DTs
Q2 Q3 D6
2Ts
• Transformer volt-second balance is obtained over two switching periods • Effect of nonidealities?
D5 D6
39
Chapter 6: Converter circuits
Effect of nonidealities on transformer volt-second balance Volt-seconds applied to primary winding during first switching period: (Vg – (Q1 and Q4 forward voltage drops))( Q1 and Q4 conduction time) Volt-seconds applied to primary winding during next switching period: – (Vg – (Q2 and Q3 forward voltage drops))( Q2 and Q3 conduction time) These volt-seconds never add to exactly zero. Net volt-seconds are applied to primary winding Magnetizing current slowly increases in magnitude Saturation can be prevented by placing a capacitor in series with primary, or by use of current programmed mode (Chapter 12) Fundamentals of Power Electronics
40
Chapter 6: Converter circuits
Operation of secondary-side diodes D5
: n
L
iD5(t)
i(t)
vs(t)
C
R
– : n
vs(t)
• Output filter inductor current divides approximately equally between diodes
D6 nVg 0
0
0.5 i
0.5 i
• Secondary amp-turns add to approximately zero
i 0 0
conducting devices:
v –
nVg
iD5(t)
• During second (D′) subinterval, both secondary-side diodes conduct
+
+
DTs Q1 Q4 D5
Ts D5 D6
Fundamentals of Power Electronics
t Ts+DTs
Q2 Q3 D6
2Ts D5 D6
41
• Essentially no net magnetization of transformer core by secondary winding currents Chapter 6: Converter circuits
Volt-second balance on output filter inductor : n
D5
L
iD5(t)
i(t)
i(t)
+
+ vs(t)
C
– : n
R
v
∆i
I
vs(t)
nVg
– iD5(t)
D6
nVg 0
0
0.5 i
0.5 i
i 0 0
V = vs
conducting devices:
V = nDVg M(D) = nD
Fundamentals of Power Electronics
DTs Q1 Q4 D5
Ts D5 D6
t Ts+DTs
Q2 Q3 D6
2Ts D5 D6
buck converter with turns ratio
42
Chapter 6: Converter circuits
Half-bridge isolated buck converter Q1
D1
Ca
i1(t)
1 : n
D3
+ –
i(t) +
+
+ Vg
L
iD3(t)
vT(t)
vs(t)
C
R
v
– – Q2
D2
Cb
: n
–
D4
• Replace transistors Q3 and Q4 with large capacitors • Voltage at capacitor centerpoint is 0.5Vg • vs(t) is reduced by a factor of two • M = 0.5 nD Fundamentals of Power Electronics
43
Chapter 6: Converter circuits
6.3.2. Forward converter n1 : n 2 : n 3
D2
L + D3
Vg
+ –
C
R
V –
Q1
D1
• Buck-derived transformer-isolated converter • Single-transistor and two-transistor versions • Maximum duty cycle is limited • Transformer is reset while transistor is off Fundamentals of Power Electronics
44
Chapter 6: Converter circuits
Forward converter with transformer equivalent circuit D2
n1 : n 2 : n 3
Vg
iM
+
LM + –
i1'
–
+
v1
v2
v3
–
+
–
Q1
i1
+
vQ1
+
+
D3
vD3 –
i3
i2
L
C
R
V –
D1
–
Fundamentals of Power Electronics
45
Chapter 6: Converter circuits
Forward converter: waveforms v1
Vg
• Magnetizing current, in conjunction with diode D1, operates in discontinuous conduction mode
0
n – n 1 Vg 2
iM
Vg LM
vD3
Conducting devices:
n Vg – n1 2 LM
• Output filter inductor, in conjunction with diode D3, may operate in either CCM or DCM
0
n3 n 1 Vg
0
0
DTs
D2Ts Ts
D3Ts
Q1 D2
D1 D3
D3
Fundamentals of Power Electronics
t
46
Chapter 6: Converter circuits
Subinterval 1: transistor conducts n1 : n 2 : n 3
Vg
+ –
iM
+
LM
i1'
–
+
v1
v2
v3
–
+
–
i1 Q1 on
Fundamentals of Power Electronics
i2
L
D2 on
+
+ vD3 –
i3
C
R
V –
D1 off
47
Chapter 6: Converter circuits
Subinterval 2: transformer reset L
n1 : n 2 : n 3
Vg
+ –
iM
+
LM
i1'
–
+
v1
v2
v3
–
+
–
i1 Q1 off
+
+
D3 on vD3 –
i3
C
R
V –
i2 = iM n1 /n2 D1 on
Fundamentals of Power Electronics
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Chapter 6: Converter circuits
Subinterval 3 L
n1 : n2 : n3
iM =0 LM Vg
+ –
i1'
+
v1
v2
v3
–
+
–
i1
i2
+
+ –
+
D3 on vD3 –
i3
C
R
V –
Q1 off D1 off
Fundamentals of Power Electronics
49
Chapter 6: Converter circuits
Magnetizing inductance volt-second balance v1
Vg 0
n – n 1 Vg 2
iM
Vg LM
n Vg – n1 2 LM
0
v1 = D Vg + D2 – Vg n 1 /n 2 + D3 0 = 0 Fundamentals of Power Electronics
50
Chapter 6: Converter circuits
Transformer reset From magnetizing current volt-second balance: v1 = D Vg + D2 – Vg n 1 /n 2 + D3 0 = 0
Solve for D2:
n D2 = n 2 D 1
D3 cannot be negative. But D3 = 1 – D – D2. Hence D3 = 1 – D – D2 ≥ 0
D3 = 1 – D 1 + Solve for D
D≤
1 n 1+ 2 n1
Fundamentals of Power Electronics
n2 ≥0 n1 for n1 = n2:
51
D≤ 1 2
Chapter 6: Converter circuits
What happens when D > 0.5 magnetizing current waveforms, for n1 = n2
iM(t)
D < 0.5
DTs D2Ts D3Ts
iM(t)
D > 0.5
DTs Fundamentals of Power Electronics
t
D2Ts 52
2Ts
t
Chapter 6: Converter circuits
Conversion ratio M(D) : n3
D2
L + D3
C
R
V –
vD3
Conducting devices:
n3 n 1 Vg
0
0
DTs
D 2T s Ts
D 3T s
Q1 D2
D1 D3
D3
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53
n vD3 = V = n 3 DVg 1 t
Chapter 6: Converter circuits
Maximum duty cycle vs. transistor voltage stress Maximum duty cycle limited to
D≤
1 n 1+ 2 n1
which can be increased by increasing the turns ratio n2 / n1. But this increases the peak transistor voltage:
n max vQ1 = Vg 1 + n 1 2 For n1 = n2 D≤ 1 2
Fundamentals of Power Electronics
and
max(vQ1) = 2Vg
54
Chapter 6: Converter circuits
The two-transistor forward converter Q1
D3
D1
L +
1:n
Vg
+ –
D4
C
R
V –
D2
Q2
V = nDVg
Fundamentals of Power Electronics
max(vQ1) = max(vQ2) = Vg
D≤ 1 2
55
Chapter 6: Converter circuits
