Fundamentals of power electronics [presentation slides] 2nd ed r. erickson ww

Fundamentals of Power Electronics
Second edition

Robert W. Erickson
Dragan Maksimovic
University of Colorado, Boulder

Fundamentals of Power Electronics 1 Chapter 1: Introduction

Chapter 1: Introduction

1.1. Introduction to power processing
1.2. Some applications of power electronics
1.3. Elements of power electronics
Summary of the course


1.1 Introduction to Power Processing

Power Switching Power
input converter output

Control
input
Dc-dc conversion: Change and control voltage magnitude
Ac-dc rectification: Possibly control dc voltage, ac current
Dc-ac inversion: Produce sinusoid of controllable
magnitude and frequency
Ac-ac cycloconversion: Change and control voltage magnitude
and frequency


Control is invariably required

Power Switching Power
input converter output

Control
input

feedforward feedback
Controller

reference


High efficiency is essential

1
Pout
η= η
Pin
0.8
1
Ploss = Pin – Pout = Pout η – 1

0.6

High efficiency leads to low
power loss within converter
Small size and reliable operation 0.4

is then feasible
Efficiency is a good measure of
converter performance 0.2
0 0.5 1 1.5
Ploss / Pout


A high-efficiency converter

Pin Pout
Converter

A goal of current converter technology is to construct converters of small
size and weight, which process substantial power at high efficiency


Devices available to the circuit designer

+
–
DTs s T
Linear-
mode Switched-mode
Resistors Capacitors Magnetics Semiconductor devices



+
–
DTs s T
Linear-
mode Switched-mode

Signal processing: avoid magnetics



+
–
DTs s T
Linear-
mode Switched-mode

Power processing: avoid lossy elements


Power loss in an ideal switch

Switch closed: v(t) = 0 +
i(t)
Switch open: i(t) = 0
v(t)
In either event: p(t) = v(t) i(t) = 0

Ideal switch consumes zero power
–


A simple dc-dc converter example

I
10A
+

Vg + Dc-dc
converter R V
– 5Ω 50V
100V
–

Input source: 100V
Output load: 50V, 10A, 500W
How can this converter be realized?


Dissipative realization

Resistive voltage divider
I
10A
+
+ 50V –
Vg + Ploss = 500W R V
– 5Ω 50V
100V
–
Pin = 1000W Pout = 500W


Dissipative realization

Series pass regulator: transistor operates in
active region
I
+ 50V – 10A
+

Vg linear amplifier –+ Vref
+ R V
– and base driver
100V 5Ω 50V
Ploss ≈ 500W
–
Pin ≈ 1000W Pout = 500W


Use of a SPDT switch

I
1 10 A
+ +

Vg 2
+ vs(t) R v(t)
– 50 V
100 V
– –

vs(t)
Vg
Vs = DVg

0
DTs (1 – D) Ts t
switch
position: 1 2 1

The switch changes the dc voltage level

vs(t)
Vg
D = switch duty cycle
Vs = DVg 0≤D≤1
0
Ts = switching period
DTs (1 – D) Ts t
switch
position: fs = switching frequency
1 2 1
= 1 / Ts

DC component of vs(t) = average value:
Ts
Vs = 1 vs(t) dt = DVg
Ts 0


Addition of low pass filter

Addition of (ideally lossless) L-C low-pass filter, for
removal of switching harmonics:
1
i(t)
+ +
L
Vg 2
+ vs(t) C R v(t)
–
100 V
– –
Pin ≈ 500 W Pout = 500 W
Ploss small

• Choose filter cutoff frequency f0 much smaller than switching
frequency fs
• This circuit is known as the “buck converter”

Addition of control system
for regulation of output voltage

Power Switching converter Load
input
+
i

vg + v
–
Sensor
– H(s) gain

Transistor Error
gate driver signal
δ Pulse-width vc G (s) ve –+ Hv
δ(t) modulator c

Compensator
Reference
dTs Ts t input vref


The boost converter

2
+
L
1
Vg + C R V
–

–

5Vg
4Vg
3Vg
V
2Vg
Vg
0
0 0.2 0.4 0.6 0.8 1
D

A single-phase inverter

vs(t)
1 + – 2
Vg +
– + v(t) –
2 1
load

vs(t) “H-bridge”
Modulate switch
duty cycles to
obtain sinusoidal
t low-frequency
component


1.2 Several applications of power electronics

Power levels encountered in high-efficiency converters
• less than 1 W in battery-operated portable equipment
• tens, hundreds, or thousands of watts in power supplies for
computers or office equipment
• kW to MW in variable-speed motor drives
• 1000 MW in rectifiers and inverters for utility dc transmission
lines


A laptop computer power supply system

Inverter Display
backlighting

iac(t) Charger
Buck Microprocessor
vac(t) PWM converter
Rectifier Power
management

ac line input Boost Disk
85–265 Vrms Lithium
battery converter drive


Power system of an earth-orbiting spacecraft

Dissipative
shunt regulator

+
Solar
array vbus

–

Battery Dc-dc Dc-dc
charge/discharge converter converter
controllers

Batteries
Payload Payload


An electric vehicle power and drive system

ac machine ac machine

Inverter Inverter control bus

battery
µP
+ system
controller
3øac line Battery
charger DC-DC
vb converter
50/60 Hz
Vehicle
– electronics
Low-voltage
dc bus

Inverter Inverter

Variable-frequency
Variable-voltage ac
ac machine ac machine


1.3 Elements of power electronics

Power electronics incorporates concepts from the fields of
analog circuits
electronic devices
control systems
power systems
magnetics
electric machines
numerical simulation


Part I. Converters in equilibrium

Inductor waveforms Averaged equivalent circuit
D' VD
vL(t) RL D Ron D' RD D' : 1
Vg – V

+
–
+
DTs D'Ts
t Vg + V R
–V – I
switch
position: 1 2 1 –

iL(t)
iL(DTs)
∆iL
Predicted efficiency
I
iL(0) Vg – V –V
100%
0.002
L L 90%
0.01
80%
0 DTs Ts t
70% 0.02

60% 0.05

η 50% RL/R = 0.1

40%

Discontinuous conduction mode 30%

20%

Transformer isolation 10%

0%
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

D


Switch realization: semiconductor devices

iA(t)
The IGBT collector
Switching loss
transistor
waveforms Qr
Vg
gate iL
vA(t)

0 0
emitter t

Emitter iB(t)
diode
waveforms iL
vB(t)
Gate 0 0
t
area
–Qr –Vg
n n n n
p p

minority carrier
n- injection tr

p pA(t)
= vA iA
area
~QrVg
Collector area
~iLVgtr
t0 t1 t2 t


Part I. Converters in equilibrium

2. Principles of steady state converter analysis

3. Steady-state equivalent circuit modeling, losses, and efficiency

4. Switch realization

5. The discontinuous conduction mode

6. Converter circuits


Part II. Converter dynamics and control

Closed-loop converter system Averaging the waveforms
Power Switching converter Load gate
input drive
+

vg(t) + v(t) R
–
feedback
connection t
–

transistor actual waveform v(t)
gate driver compensator including ripple
δ(t) pulse-width vc G (s) –+ v
modulator c
averaged waveform <v(t)>Ts
with ripple neglected
δ(t) vc(t) voltage
reference vref t

dTs Ts t t

Controller

Vg – V d(t)
L
1:D D' : 1
+
–
+
Small-signal
+
averaged vg(t)
– I d(t) I d(t) C v(t) R

equivalent circuit –


Part II. Converter dynamics and control

7. Ac modeling

8. Converter transfer functions

9. Controller design
10. Input filter design

11. Ac and dc equivalent circuit modeling of the discontinuous
conduction mode

12. Current-programmed control


Part III. Magnetics

n1 : n2

transformer i1(t) iM(t) i2(t)
the layer 3i
design LM
proximity
3
–2i
2Φ
R1 R2 effect 2i
layer
2
–i
ik(t)
Φ
i
layer d
1
: nk Rk

current
density
J
4226

transformer 3622 0.1

size vs. 0.08
Pot core size

2616 2616
2213 2213

Bmax (T)
switching 1811 1811
0.06

0.04
frequency 0.02

0
25kHz 50kHz 100kHz 200kHz 250kHz 400kHz 500kHz 1000kHz
Switching frequency


Part III. Magnetics

13. Basic magnetics theory

14. Inductor design

15. Transformer design


Part IV. Modern rectifiers,
and power system harmonics

Pollution of power system by A low-harmonic rectifier system
rectifier current harmonics ig(t)
boost converter
i(t)
iac(t) + +
L D1

vac(t) vg(t) Q1 C v(t) R

– –
vcontrol(t) vg(t) ig(t)
PWM
Rs
multiplier X va(t)
v (t)
+– err Gc(s)
vref(t)
= kx vg(t) vcontrol(t) compensator
controller
100%
100%
91%
percent of fundamental
Harmonic amplitude,

THD = 136%
80% 73% Distortion factor = 59% iac(t) Ideal rectifier (LFR) i(t)
60% 52% + 2
p(t) = vac / Re +
40% 32%
Model of
vac(t) Re(vcontrol) v(t)
20% 19% 15% 15%
13% 9% the ideal
0%
1 3 5 7 9 11 13 15 17 19
rectifier – –

Harmonic number ac dc
input output
vcontrol


Part IV. Modern rectifiers,
and power system harmonics

16. Power and harmonics in nonsinusoidal systems

17. Line-commutated rectifiers

18. Pulse-width modulated rectifiers


Part V. Resonant converters

The series resonant converter

Q1 Q3 L C
D1 D3 1:n
+

Vg +
– R V

Q2 Q4
–
D2 D4
Zero voltage
switching
1 Q = 0.2 vds1(t) Vg
0.9
Q = 0.2
0.8
0.35
0.7
0.5
0.6
0.35 Q1 X D2 t
conducting
M = V / Vg

0.75 devices: Q4 D3
0.5
0.5 1
0.4 turn off commutation
0.75
1 1.5 Q 1, Q 4 interval
0.3
1.5
2
2
0.2
Dc 0.1
3.5
5
10
3.5
5
10
characteristics 0
Q = 20

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Q = 20

F = fs / f0


Part V. Resonant converters

19. Resonant conversion
20. Soft switching


Appendices

A. RMS values of commonly-observed converter waveforms
B. Simulation of converters
C. Middlebrook’s extra element theorem
L iLOAD
D. Magnetics design tables 1 2
50 µH
3
+

1

2 CCM-DCM1
Vg
+ C R1
– 500 µF R v
11 kΩ
28 V

5
20 dB 4
|| Gvg || Open loop, d(t) = constant –
0 dB R2

4

3
85 kΩ
Xswitch C2
–20 dB R=3Ω R3 C3
L = 50 µΗ 2.7 nF 1.1 nF
fs = 100 kΗz
–40 dB 120 kΩ
+12 V
8 7 6 – 5
–60 dB Closed loop R = 25 Ω +
vx –vy LM324
VM = 4 V vz
–80 dB vref R4
5 Hz 50 Hz 500 Hz 5 kHz 50 kHz Epwm + 47 kΩ
f value = {LIMIT(0.25 vx, 0.1, 0.9)} –
5V
.nodeset v(3)=15 v(5)=5 v(6)=4.144 v(8)=0.536


Chapter 2
Principles of Steady-State Converter Analysis

2.1. Introduction
2.2. Inductor volt-second balance, capacitor charge
balance, and the small ripple approximation
2.3. Boost converter example
2.4. Cuk converter example
2.5. Estimating the ripple in converters containing two-
pole low-pass filters
2.6. Summary of key points

Fundamentals of Power Electronics 1 Chapter 2: Principles of steady-state converter analysis

2.1 Introduction
Buck converter

1
SPDT switch changes dc + +
component 2
Vg + vs(t) R v(t)
–

– –

vs(t)
Switch output voltage Vg
waveform D'Ts
DTs

Duty cycle D: 0
0≤D≤1 0 DTs Ts t
Switch
complement D′: position: 1 2 1
D′ = 1 - D


Dc component of switch output voltage

vs(t)
Vg
〈vs〉 = DVg
area =
DTsVg
0
0 DTs Ts t

Fourier analysis: Dc component = average value

Ts
vs = 1 vs(t) dt
Ts 0

vs = 1 (DTsVg) = DVg
Ts


Insertion of low-pass filter to remove switching
harmonics and pass only dc component

L
1
+ +

2
Vg + vs(t) C R v(t)
–

– –

V

Vg
v ≈ vs = DVg

0
0 1 D


Three basic dc-dc converters
(a)
1
L M(D) = D
1 0.8
iL (t) +
0.6
Buck

M(D)
2
Vg + C R v 0.4
–
0.2
– 0
0 0.2 0.4 0.6 0.8 1
D

(b) 5
L 2 1
M(D) = 1 – D
+ 4
iL (t)
Boost 3

M(D)
1
Vg + C R v 2
–
1
–
0
0 0.2 0.4 0.6 0.8 1
D

D
(c) 0 0.2 0.4 0.6 0.8 1
0

Buck-boost 1 2 + –1

iL (t) –2
M(D)

Vg + C R v
– L –3

– –4 M(D) = 1 –D
–
D
–5


Objectives of this chapter

G Develop techniques for easily determining output
voltage of an arbitrary converter circuit
G Derive the principles of inductor volt-second balance
and capacitor charge (amp-second) balance
G Introduce the key small ripple approximation
G Develop simple methods for selecting filter element
values
G Illustrate via examples


2.2. Inductor volt-second balance, capacitor charge
balance, and the small ripple approximation

Actual output voltage waveform, buck converter
iL(t) L
1
Buck converter + vL(t) – +
iC(t)
containing practical 2
low-pass filter Vg +
–
C R v(t)

–

Actual output voltage v(t) Actual waveform
waveform v(t) = V + vripple(t)
V
v(t) = V + vripple(t)
dc component V

0
t


The small ripple approximation

v(t) Actual waveform
v(t) = V + vripple(t)
v(t) = V + vripple(t) V

dc component V

0
t

In a well-designed converter, the output voltage ripple is small. Hence,
the waveforms can be easily determined by ignoring the ripple:

vripple < V

v(t) ≈ V


Buck converter analysis:
inductor current waveform
iL(t) L
1
+ vL(t) – +
iC(t)
original Vg +
2
C R v(t)
converter –

–

switch in position 1 switch in position 2

iL(t) L L

+ vL(t) – + + vL(t) – +
iC(t) iC(t)

Vg + C R v(t) Vg + iL(t) C R v(t)
– –

– –


Inductor voltage and current
Subinterval 1: switch in position 1

iL(t) L
Inductor voltage
+ vL(t) – +
iC(t)
vL = Vg – v(t)
Vg + C R v(t)
–
Small ripple approximation:
vL ≈ Vg – V –

Knowing the inductor voltage, we can now find the inductor current via
diL(t)
vL(t) = L
dt

Solve for the slope:
diL(t) vL(t) Vg – V ⇒ The inductor current changes with an
= ≈
dt L L essentially constant slope


Inductor voltage and current

L
Inductor voltage
+ vL(t) – +
iC(t)
vL(t) = – v(t)
Vg + iL(t) C R v(t)
–
Small ripple approximation:
–
vL(t) ≈ – V

Knowing the inductor voltage, we can again find the inductor current via
diL(t)
vL(t) = L
dt

Solve for the slope:
diL(t) ⇒ The inductor current changes with an
≈– V
dt L essentially constant slope


Inductor voltage and current waveforms

vL(t)
Vg – V
DTs D'Ts
t
–V
Switch
position: 1 2 1 diL(t)
vL(t) = L
dt
iL(t)
iL(DTs)
I ∆iL
iL(0) Vg – V –V
L L

0 DTs Ts t


Determination of inductor current ripple magnitude

iL(t)
iL(DTs)
I ∆iL
iL(0) Vg – V –V
L L

0 DTs Ts t

(change in iL) = (slope)(length of subinterval)
Vg – V
2∆iL = DTs
L

Vg – V Vg – V
⇒ ∆iL = DTs L= DTs
2L 2∆iL


Inductor current waveform
during turn-on transient

iL(t)

Vg – v(t)
L
iL(nTs) iL((n + 1)Ts)
– v(t)
iL(Ts) L
iL(0) = 0
0 DTs Ts 2Ts nTs (n + 1)Ts t

When the converter operates in equilibrium:
i L((n + 1)Ts) = i L(nTs)


The principle of inductor volt-second balance:
Derivation

Inductor defining relation:
di (t)
vL(t) = L L
dt
Integrate over one complete switching period:
Ts
iL(Ts) – iL(0) = 1 vL(t) dt
L 0

In periodic steady state, the net change in inductor current is zero:
Ts
0= vL(t) dt
0

Hence, the total area (or volt-seconds) under the inductor voltage
waveform is zero whenever the converter operates in steady state.
An equivalent form:
T
0= 1 s v (t) dt = v
Ts 0 L L

The average inductor voltage is zero in steady state.

Inductor volt-second balance:
Buck converter example

vL(t)
Vg – V Total area λ
Inductor voltage waveform,
previously derived:
DTs t

–V
Integral of voltage waveform is area of rectangles:
Ts
λ= vL(t) dt = (Vg – V)(DTs) + ( – V)(D'Ts)
0

Average voltage is
vL = λ = D(Vg – V) + D'( – V)
Ts
Equate to zero and solve for V:
0 = DVg – (D + D')V = DVg – V ⇒ V = DVg


The principle of capacitor charge balance:
Derivation

Capacitor defining relation:
dv (t)
iC(t) = C C
dt
Integrate over one complete switching period:
Ts
vC(Ts) – vC(0) = 1 iC(t) dt
C 0

In periodic steady state, the net change in capacitor voltage is zero:
Ts
0= 1 iC(t) dt = iC
Ts 0

Hence, the total area (or charge) under the capacitor current
waveform is zero whenever the converter operates in steady state.
The average capacitor current is then zero.


2.3 Boost converter example

L 2

iL(t) + vL(t) – +
iC(t)
Boost converter 1
with ideal switch Vg + C R v
–

–

L D1

Realization using iL(t) + vL(t) – +
iC(t)
power MOSFET Q1
and diode Vg + C R v
– +
DTs Ts
–
–


Boost converter analysis

L 2

iL(t) + vL(t) – +
iC(t)
1
original Vg + C R v
converter –

–

switch in position 1 switch in position 2
L L

iL(t) + vL(t) – + iL(t) + vL(t) – +
iC(t) iC(t)

Vg + C R v Vg + C R v
– –

– –



Inductor voltage and capacitor current
vL = Vg
L
iC = – v / R
iL(t) + vL(t) – +
iC(t)

Vg + C R v
Small ripple approximation: –

vL = Vg –
iC = – V / R



Inductor voltage and capacitor current

vL = Vg – v L
iC = i L – v / R iL(t) + vL(t) – +
iC(t)

Vg + C R v
Small ripple approximation: –

–
vL = Vg – V
iC = I – V / R


Inductor voltage and capacitor current waveforms

vL(t)
Vg

DTs D'Ts
t

Vg – V

iC(t) I – V/R

DTs D'Ts
t
– V/R


Inductor volt-second balance

vL(t)
Net volt-seconds applied to inductor Vg
over one switching period: DTs D'Ts
Ts t
vL(t) dt = (Vg) DTs + (Vg – V) D'Ts
0
Vg – V

Equate to zero and collect terms:
Vg (D + D') – V D' = 0

Solve for V:
Vg
V =
D'
The voltage conversion ratio is therefore

M(D) = V = 1 = 1
Vg D' 1 – D


Conversion ratio M(D) of the boost converter

5
M(D) = 1 = 1
4 D' 1 – D

3
M(D)

2
1
0
0 0.2 0.4 0.6 0.8 1
D


Determination of inductor current dc component

iC(t) I – V/R

Capacitor charge balance: DTs D'Ts
t
Ts
iC(t) dt = ( – V ) DTs + (I – V ) D'Ts
– V/R
0 R R

Collect terms and equate to zero: I
Vg/R
– V (D + D') + I D' = 0 8
R
Solve for I: 6
4
I= V
D' R 2

Eliminate V to express in terms of Vg: 0
0 0.2 0.4 0.6 0.8 1
Vg D
I= 2
D' R


Determination of inductor current ripple

Inductor current slope during iL(t)
subinterval 1: ∆iL
I
diL(t) vL(t) Vg Vg Vg – V
= =
dt L L L L
Inductor current slope during
subinterval 2: 0 DTs Ts t
diL(t) vL(t) Vg – V
= =
dt L L
Change in inductor current during subinterval 1 is (slope) (length of subinterval):
Vg
2∆iL = DTs
L
Solve for peak ripple:
Vg • Choose L such that desired ripple magnitude
∆iL = DTs
2L is obtained


Determination of capacitor voltage ripple

Capacitor voltage slope during v(t)
subinterval 1:
dvC(t) iC(t) – V V ∆v
= = –V I – V
dt C RC
RC C RC
Capacitor voltage slope during
subinterval 2: 0 DTs Ts t
dvC(t) iC(t) I
= = – V
dt C C RC
Change in capacitor voltage during subinterval 1 is (slope) (length of subinterval):

– 2∆v = – V DTs
RC
Solve for peak ripple: • Choose C such that desired voltage ripple
magnitude is obtained
∆v = V DTs • In practice, capacitor equivalent series
2RC
resistance (esr) leads to increased voltage ripple

2.4 Cuk converter example

L1 C1 L2
Cuk converter, i2 +
i1 + v1 –
with ideal switch
1 2
Vg + C2 v2 R
–

–

L1 C1 L2
Cuk converter:
practical realization i1 i2 +
+ v1 –
using MOSFET and
diode Vg + Q1 D1 C2 v2 R
–

–


Cuk converter circuit
with switch in positions 1 and 2

Switch in position 1: L1 L2 i2
MOSFET conducts i1 + vL1 – iC1 + vL2 –
+
– iC2
Capacitor C1 releases Vg + v1 C1 C2 v2 R
–
energy to output
+ –

i1 L1 L2 i2
iC1
Switch in position 2: + vL1 – + vL2 – +
+ iC2
diode conducts
Vg + C1 v1 C2 v2 R
Capacitor C1 is –

charged from input – –


Waveforms during subinterval 1
MOSFET conduction interval

Inductor voltages and L1 L2 i2
capacitor currents: +
i1 + vL1 – – iC1 + vL2 – iC2
vL1 = Vg +
Vg v1 C1 C2 v2 R
–
vL2 = – v1 – v2
+ –
i C1 = i 2
v
i C2 = i 2 – 2
R
Small ripple approximation for subinterval 1:
vL1 = Vg
vL2 = – V1 – V2
i C1 = I 2
V
i C2 = I 2 – 2
R


Waveforms during subinterval 2
Diode conduction interval

Inductor voltages and L2
i1 L1 i2
capacitor currents: iC1
+
+ vL1 – + vL2 –
+ iC2
vL1 = Vg – v1 +
Vg C1 v1 C2 v2 R
–
vL2 = – v2
– –
i C1 = i 1
v
i C2 = i 2 – 2
R

Small ripple approximation for subinterval 2:
vL1 = Vg – V1
vL2 = – V2
i C1 = I 1
V
i C2 = I 2 – 2
R


Equate average values to zero

The principles of inductor volt-second and capacitor charge balance
state that the average values of the periodic inductor voltage and
capacitor current waveforms are zero, when the converter operates in
steady state. Hence, to determine the steady-state conditions in the
converter, let us sketch the inductor voltage and capacitor current
waveforms, and equate their average values to zero.
Waveforms:

Inductor voltage vL1(t)
Volt-second balance on L1:
vL1(t)
Vg

DTs D'Ts vL1 = DVg + D'(Vg – V1) = 0
t

Vg – V1



Inductor L2 voltage
vL2(t) – V2

DTs D'Ts
– V1 – V2 t Average the waveforms:

vL2 = D( – V1 – V2) + D'( – V2) = 0
Capacitor C1 current
i C1 = DI 2 + D'I 1 = 0
iC1(t)
I1

DTs D'Ts
I2 t



Capacitor current iC2(t) waveform

iC2(t)

I2 – V2 / R (= 0) V2
i C2 = I 2 – =0
DTs D'Ts t R

Note: during both subintervals, the
capacitor current iC2 is equal to the
difference between the inductor current
i2 and the load current V2/R. When
ripple is neglected, iC2 is constant and
equal to zero.


Cuk converter conversion ratio M = V/Vg

D
0 0.2 0.4 0.6 0.8 1
0
-1
-2
M(D)

V2
-3 M(D) = =– D
Vg 1–D
-4
-5


Inductor current waveforms

Interval 1 slopes, using small i1(t)
ripple approximation:
∆i1
I1
di 1(t) vL1(t) Vg Vg Vg – V1
= =
dt L1 L1 L1 L1
di 2(t) vL2(t) – V1 – V2
= = DTs Ts t
dt L2 L2

DTs Ts t
Interval 2 slopes:
– V1 – V2 – V2
di 1(t) vL1(t) Vg – V1 L2 L2
= = I2
dt L1 L1 ∆i2
di 2(t) vL2(t) – V2
= = i2(t)
dt L2 L2


Capacitor C1 waveform

Subinterval 1:
v1(t)
dv1(t) i C1(t) I 2 ∆v1
= =
dt C1 C1 V1
I2 I1
C1 C1
Subinterval 2:
DTs Ts t
dv1(t) i C1(t) I 1
= =
dt C1 C1


Ripple magnitudes

Analysis results Use dc converter solution to simplify:

VgDTs VgDTs
∆i 1 = ∆i 1 =
2L 1 2L 1
V + V2 VgDTs
∆i 2 = 1 DTs ∆i 2 =
2L 2 2L 2
– I DT VgD 2Ts
∆v1 = 2 s
2C 1 ∆v1 =
2D'RC 1

Q: How large is the output voltage ripple?


2.5 Estimating ripple in converters
containing two-pole low-pass filters

Buck converter example: Determine output voltage ripple
L
1
iL(t) +
iC(t) iR(t)
2
Vg + C vC(t) R
–

–

iL(t)
Inductor current iL(DTs)
I ∆iL
waveform.
iL(0) Vg – V –V
What is the L L
capacitor current?
0 DTs Ts t


Capacitor current and voltage, buck example

iC(t)
Total charge
Must not q
neglect ∆iL t
inductor Ts /2
current ripple!
DTs D'Ts

If the capacitor
voltage ripple is
vC(t)
small, then
essentially all of
∆v
the ac component V
∆v
of inductor current
flows through the
t
capacitor.


Estimating capacitor voltage ripple ∆v

iC(t) Current iC(t) is positive for half
Total charge of the switching period. This
q
positive current causes the
∆iL t
capacitor voltage vC(t) to
Ts /2
increase between its minimum
DTs D'Ts and maximum extrema.
During this time, the total
charge q is deposited on the
vC(t) capacitor plates, where
∆v q = C (2∆v)
V
∆v
(change in charge) =
t C (change in voltage)


Fundamentals of power electronics [presentation slides] 2nd ed r. erickson ww

Fundamentals of power electronics [presentation slides] 2nd ed r. erickson ww

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