Transcript the buck converter

Introduction to DC-DC
Conversion
EE174 – SJSU
Tan Nguyen
OBJECTIVES
• Introduction of DC-DC Converter
• Types of DC-DC Converters
• Linear regulator (LR)
• Switching mode power supply (SMPS)
• Advantages and Disadvantages
Introduction
• Batteries are often shown on a schematic diagram as the source of DC
voltage but usually the actual DC voltage source is a power supply.
• DC to DC converters are important portable electronic devices used
whenever we want to change DC electrical power efficiently from one
voltage level to another.
• A power converter generates output voltage and current for the load from
a given input power source.
• Depending on the specific application, either a linear regulator (LR) or a
switching mode power supply (SMPS) solution to be chosen.
Typical Application of DC-DC converter
• Car battery 12V must be stepped down to 3-5V DC voltage to run DVD/CD player
• Laptop computers or cellular phone battery voltage must be stepped down to run
several sub-circuits, each with its own voltage level requirement different from
that supplied by the battery.
• Single cell 1.5 V DC must be stepped up to 5V operate an electronic circuitry.
• A 6V or 9V DC must be stepped up to 500V DC or more, to provide an insulation
testing voltage.
• A 12V DC must be stepped up to +/-40V or so, to run a car hifi amplifier circuitry.
• A 12V DC must be stepped up to 650V DC or so, as part of a DC-AC sinewave
inverter.
Linear Regulator
How a Linear Regulator Works
In an embedded system, a 15V bus rail is available from the front-end power supply. On the
system board, a 10V voltage is needed to power an Op-Amp circuit. The simplest approach to
generate the 10V is to use a voltage divider from the 15V bus, as shown below:
Does this circuit work well? The answer is usually no.
The output 10V is unchanged under the following
conditions:
• Vin is stable.
• The resistor values are unchanged under any condition.
• The load is not vary under different operating conditions.
The Basic Linear Regulator
A linear regulator uses a voltage-controlled current source to force a fixed voltage to appear at the
regulator output terminal. The control circuitry senses the output voltage, and adjusts the current
source (as required by the load) to hold the output voltage at the desired value. The design limit of the
current source defines the maximum load current the regulator can source and still maintain
regulation.
Linear Regulator Operation
The pass device (Q1) is made up of an NPN Darlington driven by a PNP transistor. The current flowing out
the emitter of the pass transistor (load current IL) is controlled by Q2 and the voltage error amplifier. The
current through the R1, R2 resistive divider is assumed to be negligible compared to the load current.
Ideally, VX = VREF  Error Amp = 0  VOUT is constant.
VX
Q2
When VOUT decreases, VX < VREF  The error amplifier
will be high which turns on Q2 and Q1  VIN apply to the
circuit that adjusts the output to desired level.
When VOUT increases, VX > VREF  The error amplifier
will be low which turns off Q2 and Q1  VIN disconnect
from the output that adjusts the output to desired level.
Linear Regulator Types
• Standard (NPN Darlington) Regulator
• Low Dropout or LDO Regulator
• Quasi LDO Regulator
Note:
The single most important difference between these three types is the dropout voltage, which is
defined as the minimum voltage drop required across the regulator to maintain output voltage
regulation.
THE STANDARD (NPN) REGULATOR
In order to maintain output regulation, the pass transistor requires a minimum voltage across it given by:
VD(MIN) = 2 VBE + VCE
Allowing for the -55°C to +150°C temperature range, this minimum VD(MIN) is set at 2.5V to 3V by the
manufacturer to guarantee specified performance limits. The actually falls out of regulation will be
between 1.5V and 2.2V
Example:
1) If VIN = 12V, what is the VOUT_max?
2) If VIN = 12V, VOUT = 6V, R1 = 1 kΩ,
R2 = 2kΩ, what is VREF?
R1
VX
R2
Solutions:
1) VOUT_max = 12 – 3 = 9V
2) R1 = 1 kΩ and R2 = 2kΩ  VX = 4V
 VREF = 4V
THE LOW-DROPOUT (LDO) REGULATOR
The minimum voltage drop required across the LDO regulator to maintain regulation is just the voltage
across the PNP transistor:
VD(MIN) = VCE
The maximum specified dropout voltage of an LDO regulator is usually about 0.7V to 0.8V at full current,
with typical values around 0.6V.
R1
VX
R2
Example:
1) If VIN = 10V, what is the VOUT_max?
2) If VIN = 10V, VREF = 2.5V and VOUT = 5V,
what are values of R1, R2?
Solutions:
1) VOUT_max = 10.0 – 0.8 = 9.2V
2) Want VX = VREF = 2.5V so for VOUT = 5V
 R1 = R2 = 1 kΩ
THE QUASI LOW-DROPOUT REGULATOR
The minimum voltage drop required across the Quasi-LDO regulator to maintain regulation is given by:
VD(MIN) = VBE + VCE
The dropout voltage for a quasi-LDO is usually specified at about 1.5V(max). The actual dropout
voltage is temperature and load current dependent, but could never be expected to go lower than
about 0.9V (25°C) at even the lightest load.
R1
VX
R2
Example:
1) If VIN = 5V, what is the VOUT_max?
2) If VIN = 9V, VREF = 2V and VOUT = 5V,
R1= 3 kΩ, what is value of R2?
Solutions:
1) VOUT_max = 5.0 – 1.5 = 3.5V
2) Want VX = 2V for VOUT = 5V and
R1 = 3 kΩ  R2 = 2 kΩ
Comparison of the Linear Regulators
≈ 0.8 V (WC)
≈ 1.5 V (WC)
≈ 3 V (WC)
Example:
Given VIN = 5V, require output VOUT = 3.3V and 2 Amax, what is the best choice for the design?
Solution: Quasi-LDO
LINEAR REGULATORS
• The linear regulator is a DC-DC converter to provide a constant
voltage output without using switching components.
• The linear regulator is very popular in many applications for its
low cost, low noise and simple to use.
• It was the basis for the power supply industry until switching
mode power supplies became prevalent after the 1960s.
• Power management suppliers have developed many integrated
linear regulators.
• The linear regulator has limited efficiency and can not boost
voltage to make Vout > Vin.
ADJUSTABLE LINEAR REGULATORS
A typical integrated linear regulator needs only VIN, VOUT, FB and optional GND pins. Figure below shows a
typical 3-pin linear regulator, it only needs an input capacitor, output capacitor and two feedback resistors to
set the output voltage.
LINEAR REGULATORS DRAWBACK
• A major drawback of using linear regulators can be the excessive power dissipation of its series transistor Q1
operating in a linear mode.
• Since all the load current must pass through the series transistor, its power dissipation is PLoss = (VIN – VO) •IO.
• The efficiency of a linear regulator can be estimated by:
LINEAR REGULATORS DRAWBACK
• The linear regulator can be very efficient only if VO is close to VIN.
• The linear regulator (LR) has another limitation, which is the minimum voltage
difference between VIN and VO. The transistor in the LR must be operated in its linear
mode. So it requires a certain minimum voltage drop across the collector to emitter of
a bipolar transistor or drain to source of a FET. When VO is too close to VIN, the LR may
be unable to regulate output voltage anymore.
• The linear regulators that can work with low headroom (VIN – VO) are called low
dropout regulators (LDOs).
• The linear regulator or an LDO can only provide step-down DC/DC conversion.
LINEAR REGULATORS APPLICATIONS
There are many applications in which linear regulators provide superior solutions to switching supplies:
1. Simple/low cost solutions. Linear regulator or LDO solutions are simple and easy to use, especially for
low power applications with low output current where thermal stress is not critical. No external power
inductor is required.
2. Low noise/low ripple applications. For noise-sensitive applications, such as communication and radio
devices, minimizing the supply noise is very critical.
3. Fast transient applications. The linear regulator feedback
loop is usually internal, so no external compensation
is required.
4. Low dropout applications. For applications where output voltage is close to the input voltage, LDOs
may be more efficient than an SMPS.
We see that price sensitive applications prefer linear regulators over their sampled-time counterparts.
The design decision is especially clear cut for makers of:
• communications equipment
• small devices
• battery operated systems
• low current devices
• high performance microprocessors with sleep mode (fast transient recovery required)
LINEAR REGULATORS VS SWITCHING REGULATORS
Regulators Linear regulators are less energy efficient than switching regulators. Why do we
continue using them?
Depending upon the application, linear regulators have several redeeming features:
• lower output noise is important for radios and other communications equipment
• faster response to input and output transients
• easier to use because they require only filter capacitors for operation
• generally smaller in size (no magnetics required)
• less expensive (simpler internal circuitry and no magnetics required)
Furthermore, in applications using low input-to-output voltage differentials, the efficiency is not
all that bad! For example, in a 5V to 3.3V microprocessor application, linear regulator efficiency
approaches 66%. And applications with low current subcircuits may not care that regulator
efficiency is less than optimum as the power lost may be negligible overall.
Introduction to DC-DC
Converter – Cont.
EE174 – SJSU
Tan Nguyen
SWITCHING MODE POWER SUPPLY (SMPS)
• The switching-mode power supply is a power supply that provides the power supply function through low loss
components such as capacitors, inductors, and transformers -- and the use of switches that are in one of two states,
on or off.
• It offers high power conversion efficiency and design flexibility.
• It can step down or step up output voltage.
• The term switchmode was widely used for this type of power supply until Motorola, Inc., who used the trademark
SWITCHMODE TM for products aimed at the switching-mode power supply market, started to enforce their
trademark. Switching-mode power supply or switching power supply are used to avoid infringing on the trademark.
• Typical switching frequencies lie in the range 1 kHz to 1 MHz, depending on the speed of the semiconductor
devices.
• Types of SMPS:
• Buck converter: Voltage to voltage converter, step down.
• Boost Converter: Voltage to voltage converter, step up.
• Buck-Boost or FlyBack Converter: Voltage-Voltage, step up and down (negative voltages)
• Cuk Converter: Current-Current converter, step up and down
These converters typically have a full wave rectifier front-end to produce a high DC voltages
SIMPLE SWITCHING MODE POWER SUPPLY
42%
58%
PULSE WIDTH MODULATION (PWM)
• The switch control signal, which controls the on and
off states of the switch, is generated by comparing a
signal level control voltage vcontrol with a repetitive
waveform.
• The switching frequency is the frequency of the
sawtooth waveform with a constant peak.
• The duty ratio D can be expressed as
t on v control
D
 ^
Ts
V st
THE BUCK CONVERTER
• The buck converter is known as voltage step-down converter, current step-up converter, chopper, direct
converter.
• The buck converter simplest and most popular switching regulator.
The Buck Converter Circuit Diagram
DC-DC Buck Converter Module 4.5-14V to 0.8-9.5V 6A
Adjustable Set-Down Regulator
Size:30mm(L)*18mm(W)*14(H) mm
THE BUCK CONVERTER
During switch is close Vi = Vo, when switch is open
at T1, Vo starts to discharge, with higher capacitor
C value  slower discharge rate (improve ripple).
If switch VSW is close and open as shown below.
Observe output Vo and current spike at ISW every
time the switch is close.
Use high capacitor
C value to improve
the output voltage
Vo ripple, but still
have issue with
huge current spike
ISW when switch is
close can easily
burn out the
switch.
THE BUCK CONVERTER
The current spikes ISW can be controlled by adding
an inductor (L) between the switch (SW) and the
capacitor (C ). Since Inductor acts as a storage
energy. When the switch is close, the inductor will
absorb the energy and when the switch is open,
the inductor will supply that energy to the
capacitor result in smooth out ISW.
However, there is another issue with inductor current
path. The inductor current must have continuous to
flow through the inductor but during the switch open,
there is no current path for the inductor current. To
fix this issue, adding a free-wheeling diode to the
circuit below it, this will keep the inductor current
continues to flow.
Freewheeling
Diode
THE BUCK CONVERTER SUMMARY
Two Mode of Operations:
1. Continuous Conduction Mode: Inductor current IL does not
reach zero, when output current IO is very large.
2. Discontinuous Conduction Mode: Inductor current IL will
reach zero, when output current IO is very small.
•
•
•
•
•
•
LC low-pass filter: to pass the DC component while
attenuating the switching components.
diode is reversed biased during ON period, input
provides energy to the load and to the inductor
energy is transferred to the load from the inductor
during switch OFF period
Interchange of energy between inductor and capacitor
is referred as flywheel effect.
in the steady-state, average inductor voltage is zero
in the steady-state, average capacitor current is zero
THE BUCK CONVERTER CONTINUOUS MODE
1. Continuous Conduction Mode: Inductor current IL does not reach zero, when output current IO is very large.
THE BUCK CONVERTER DISCONTINUOUS MODE
2. Discontinuous Conduction Mode: Inductor current IL will reach zero, when output current IO is very small.
When the switch is ON (short), Diode reversed bias (open):
VL = Vi – VO = constant > 0
When the switch is OFF (open), Diode forwarded bias:
VL = – VD – VO ≈ – VO = constant < 0
Calculate IL,max and relationship of Vo and Vi:
THE BUCK CONVERTER EXAMPLE
Given a buck converter design with fsw = 200 kHz (TS = 5 μsec), L = 33 μH, C = 10 μF, I0 = 1 A and D = 50%
duty cycle. Find:
a) VO if Vi = 10 V in continuous mode
b) Output current and voltage ripples
c) Current IL,max
d) VO if Vi = 10 V in discontinuous mode
Solutions:
a) VO = D Vi = 0.5 x 10 = 5V
b)
c)
d) VO = 3.32 V
= (33 μH)-1 (10 – 5) x 0.5 x 5 μsec = 0.38 A
Buck Converter Design Example
For a buck converter, R=1 ohm, Vd=40 V, V0=5 V, fs=4 kHz. Find the duty ratio
and “on” time of the switch.
D = V0 /Vd = 5/40 = 0.125 = 12.5%
Ts = 1/fs = 1/4000 = 0.25 ms = 250 μs
Ton = DTs = 31.25 μs
Toff = Ts – ton = 218.75 μs
When the switch is “on”: VL = Vd - V0 = 35 V
When the switch is “off”: VL = -V0 = - 5 V
I0 = IL = V0 / R = 5 A
Id = D I0 = 0.625 A
Power Losses in a Buck Converter
There are two types of losses in an SMPS:
• DC conduction losses.
• AC switching losses.
DC conduction losses in Buck converter
• The conduction losses of a buck converter primarily result from
voltage drops across transistor Q1, diode D1 and inductor L when
they conduct current.
• A MOSFET is used as the power transistor.
The
conduction loss of the MOSFET = IO2 x RDS(ON) x D,
where RDS(ON) is the on-resistance of MOSFET Q1.
• The conduction power loss of the diode = IO • VD • (1 – D),
where VD is the forward voltage drop of the diode D1.
• The conduction loss of the inductor = IO2 x RDCR,
where RDCR is the copper resistance of the inductor
winding.
Power Losses in a Buck Converter
Therefore, the conduction loss of the buck converter is approximately:
PCON_LOSS = (IO2 x RDS(ON) x D) + (IO • VD • [1 – D]) + (IO2 x RDCR)
Considering only conduction loss, the converter efficiency is:
Example:
For 12V input buck supply  3.3V/10AMAX output buck supply.
• Use 27.5% duty cycle provides a 3.3V output voltage.
Vout = Vin x D = 12 x 0.275 = 3.3 V
• MOSFET RDS(ON) = 10 mΩ
• Diode forward voltage VD = 0.5V (freewheeling diode)
• Inductor RDCR = 2 mΩ
Conduction loss at full load:
PCON_LOSS = (IO2 x RDS(ON) x D) + (IO x VD x [1 – D]) + (IO2 x RDCR)
= (102 x 0.01 x 0.275) + (10 x 0.5 x [1 – 0.275]) + (102 x 0.002)
= 0.275W + 3.62W + 0.2W = 4.095W
Buck converter efficiency:
AC Switching Losses in Buck Converter
1. MOSFET switching losses. A real transistor requires time to be
turned on or off. So there are voltage and current overlaps
during the turn-on and turn-off transients, which generate AC
switching losses.
2. Inductor core loss. A real inductor also has AC loss that is a
function of switching frequency. Inductor AC loss is primarily
from the magnetic core loss.
3. Other AC related losses. Other AC related losses include the
gate driver loss and the dead time (when both top FET Q1 and
bottom FET Q2 are off) body diode conduction loss.
Basic Nonisolated DC/DC SMPS Topologies
BUCK COVERTER
Basic dc-dc converters and their dc conversion ratios M(D) = V/Vg.
Mobile Device Using Linear versus Switch-Mode Regulator
DC-DC Converter Technology Comparison
Parameter
Linear Regulator
Switching Regulator
Efficiency
Low
High
EMI Noise
Low
High
Output Current
Low to Medium
Low to High
Boost (Step up)
No
Yes
Buck (Step down)
Yes
Yes
Size
Small
Large
Cost
Inexpensive
High cost
Sample of Linear and Switch-Mode Regulator Output
References:
http://en.wikipedia.org/wiki/DC-to-DC_converter
https://www.jaycar.com/images_uploaded/dcdcconv.pdf
Linear Technology - Application Note 140
buck converter tutorial abuhajara
http://www.smpstech.com/tutorial/t03top.htm#SWITCHINGMODE
Notes from Fang Z. Peng Dept. of Electrical and Computer Engineering MSU
https://www.google.com/webhp?sourceid=chromeinstant&rlz=1C1OPRB_enUS587US587&ion=1&espv=2&ie=UTF8#q=picture+of+noise+on+buck+output
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=0CCQQFjABa
hUKEwj329J4YvIAhVLy4AKHZiyADY&url=http%3A%2F%2Fusers.ece.utexas.edu%2F~kwasinski%2F_6_EE
462L_DC_DC_Buck_PPT.ppt&usg=AFQjCNH1PIzP73b3t11mgGhnUBBg-sVNXg&cad=rja
http://ecee.colorado.edu/ecen4517/materials/Encyc.pdf
https://www.valuetronics.com/Manuals/Lambda_%20linear_versus_switching.pdf
References:
http://en.wikipedia.org/wiki/DC-to-DC_converter
https://www.jaycar.com/images_uploaded/dcdcconv.pdf
Linear Tecnology - Application Note 140
http://www.ti.com/lit/an/snva558/snva558.pdf
http://www.micrel.com/_PDF/other/LDOBk.pdf