Lecture 24
Buck Converters
Copyright © 2016 by Mark Horowitz
M. Horowitz, J. Plummer
E40M Lecture 24
1
Reading
•  Course Reader Chapter 9
•  A&L: Chapter 9 – 9.1 to 9.3 (Inductors)
•
Chapter 13 pages 732 – 741 (RL Circuits)
•  Buck converter
–  http://en.wikipedia.org/wiki/Buck_converter
•  Boost converter
–  http://en.wikipedia.org/wiki/Boost_converter
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Roadmap
The last lecture introduced you to our final device in E40M, an
inductor. This component is the complement of a capacitor. It
too stores energy, but in an inductor the energy is proportional
to the current flowing through the device, so it wants to keep its
current constant – the voltage across an inductor sets the rate of
change in the current.
We also introduced a simple “buck” voltage converter, which
was a power inverter driving an LC network. Using impedance,
it seems like none of the high frequency components will get
through, so the only real output is the DC (zero frequency)
component of the waveform. In this lecture we will dig in, and
look at the currents and voltages in this circuit.
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Key Ideas From the Last Lecture - Inductors
•  An inductor is a new type of two terminal device
–  It is linear – double V and you will double i
–  Like a capacitor, it stores energy
•  Ideal inductors don’t dissipate energy
•  Defining equation: V = L di/dt
–  (see next page)
L is inductance (in Henrys)
•  If a sinewave current flows, i = io sin 2πFt
(
)
(
so V = io 2πFLcos 2πFt
•  Thus the impedance of an inductor is
(
)
V 2πFLio cos 2πF
Z= =
= 2πFL if we negelct phase
i
io sin 2πF
(
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)
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)
Key Ideas From the Last Lecture - Inductors
•  The current flowing through an inductor sets its stored energy
–  The only way to change this energy
•  Is to pull power out of the device
•  Is to push power into the device
–  Power is iV
•  So a rapid change in current (a.k.a energy)
–  Requires large power,
–  But i is set by the inductor
•  So it will generate a very large voltage V = Ldi/dt
•  Inductors try to keep their current from changing rapidly
–  And look like current sources for short time periods
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Key Ideas From The Last Lecture - Basic Buck
Converter
12V
12V
0V
Time
•  The CMOS inverter will invert the digital signal, so we can reduce this
circuit to the following:
12V
0V
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Time
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Key Ideas From The Last Lecture - Basic Buck
Converter
12V
0V
Time
•  Maybe this will actually do what we want to do!
•  ZL = 2πFL which = 0 at DC, so DC voltages will go directly to VOUT
•  ZL = 2πFL which at high F → ∞ so high frequencies will be blocked
•  Sounds like a low pass filter!. Today we’ll analyze this more carefully
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Learning Objectives For Today
•  Understand the basic operations of a buck converter
–  Key point to remember is that the output is pretty stable
–  Be able to derive the current and voltage across the inductor
–  Find the output voltage as a function of input duty cycle
•  Vin * Thigh/Tcycle, (the average value of the input waveform)
–  Output voltage doesn’t depend on output current
•  Average input current does depend on output current
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BUCK CONVERTER ENERGY
ANALYSIS i.e. WHAT’S REALLY
GOING ON?
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Buck Converter Circuit – Impedance Analysis
Z2 =
Vout
Vin
1
1
+ 2πFC
R
=
R
1+ 2πFRC
Z1 = 2πFL
R
R
Z2
1
1+ 2πFRC
1+ 2πFRC
=
=
=
=
R
R + 2πFL(1+ 2πFRC)
L
Z1 + Z 2
+ 2πFL
1+ 2πF + LC 2πF
1+ 2πFRC
1+ 2πFRC
R
(
•  So as we expected, DC is passed through to the output and high
frequencies are attenuated as 1/F or 1/F2. So it really is a low pass
filter but it attenuates high frequencies as 1/F2.
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)
2
Bode Plot
Log F
0
Vout
Vin
Vout
=
Vin
dB
1
L
1+ 2πF + LC 2πF
R
(
Let L = 10 µH
C = 600 µF
R = 0.1 ohms
•  DC is passed to the output, high frequencies are attenuated
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)
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Some Reasonable Values
•  L = 10µH
C = 600 µF
•  If the input is 0 to 12 V with a basic
frequency of 200 KHz
–  LC(2πF)2 ≈ 9500
–  So the output at 200 KHz is ≈ 1.25 mV
•  But if the input is at 12 V 1/12 of the time
and 0 V the rest of the time,
–  The DC output voltage = 1 V
•  So we have a 12 V to 1 V converter! And
we could actually produce other output
voltages by varying the input duty cycle
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E40M Lecture 25
Vout
=
Vin
1
L
1+ 2πF + LC 2πF
R
(
12
)
2
EveryCircuit Simulation – Buck Converter
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Buck Converter As A Filter
VIN
VOUT
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Setting The Output Voltage
12V
VOUT (DC)
0V
Time
12V
VOUT (DC)
0V
•  We’ll derive an expression for VOUT later.
•  Now let’s talk about how this works in terms of energy flow,
current flow etc.
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Is This Energy Efficient?
•  Ideal C, L don’t dissipate power
•  Can make the R of the transistors small
12V
•  So it doesn’t seem to dissipate much energy
•  If it is energy efficient
–  Power @ 12V supply ≈ Power @1V
•  But this means i@1V = 12 i@12V
–  Is this possible?
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How Can This Work?
t1
t2
12V
12V
0V
12V
T
0V
•  Suppose VOUT is 1V and the load R draws 10A (i.e. R = 0.1 ohm).
Remember that the current i through an inductor changes slowly so
iINDUCTOR ≈ 1A.
•  This current can come from the power supply only when the PMOS
device is ON, which is a small fraction of the time.
(
)(
)( ) ( )(
)( )
ESupplied = 12V 10A t1 = 1V 10A T
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Current Flow In Buck Converter
12V
12V
t1
t2
12V
0V
Time
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BREAKING BREAK
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Why Wall Wart Got Smaller
http://hackedgadgets.com/2007/08/22/diy-wall-wartusb-power-supply/
http://www.dxing.info/equipment/wall_warts_bryant.dx
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HOW IT ALL WORKS
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Inductor Current
12V
V =L
diL
dt
VOUT
diL
so
=−
dt
L
iL = C
dVOUT VOUT
+
dt
R
•  We could solve the circuit using KCL and KVL
•  But that will give
–  A 2nd order differential equation
•  Would prefer a simpler method if possible
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Make Two Approximations
(both are good approximations)
•  The circuit will settle into some repeating cycle
VOUT and iL must be the same at the beginning and end of each cycle
•  The output voltage is approximately constant
–  Want stable output voltage
This means the voltage across the inductor is roughly constant
in each phase of the converter. Thus the inductor current will
increase or decrease linearly during each phase
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Operation with Our Assumptions
Time
= VOUT/R
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Inductor Current
12V
12V
V =L
t1
di
dt
t2
VOUT
di
=−
dt
L
di 12 − VOUT
=
dt
L
12 − VOUT
Δicharge = t1 ⋅
L
M. Horowitz, J. Plummer
Δidischarge
E40M Lecture 25
VOUT
= −t 2 ⋅
L
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Balancing the Equations
Since the current at the start and end of the cycle are the same,
Δicharge + Δidischarge = 0
So we substitute the equations in:
t1 ⋅
12 − VOUT
L
t1
= t2 ⋅
VOUT
L
VOUT
t1
= 12 ⋅
t1 + t 2
12V
t2
12V
0V
T
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Time
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Output Voltage Only Depends on Duty Cycle
•  The output voltage doesn’t depend on load current (R)!
•  The output is the “average” voltage created by inverter
–  It is Vdd times percentage of time the output is high
–  The LC circuit is “filtering” the square wave voltage
•  Smoothing out the signal.
•  The “ripple” in the current is α 1 / L
Δidischarge = −t 2 ⋅
VOUT
L
•  Our model did not include resistance of transistor, inductor
–  If it did, the output voltage would depend on the current
•  But this would be like having an ideal “average” voltage source
plus a series resistor
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Buck Converter With Diode
12V
12V
0V
Time
12V
•  Note that we actually could use a diode
rather than the NMOS switch in the Buck
converter.
•  This is often how it is actually
implemented since a diode is usually
cheaper than a transistor.
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Learning Objectives
•  Understand the basic operations of a buck converter
–  Key point to remember is that the output is pretty stable
–  Be able to derive the current and voltage across the inductor
–  Find the output voltage as a function of input duty cycle
•  Vin * Thigh/Tcycle, (the average value of the input waveform)
–  Output voltage doesn’t depend on output current
•  Average input current does depend on output current
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