Noty AN140fa Linear Technology

Application Note 140
October 2013
Basic Concepts of Linear Regulator and Switching Mode Power Supplies
Henry J. Zhang
ABSTRACT
This article explains the basic concepts of linear regulators and switching mode power supplies (SMPS). It is aimed at system engineers who may not be ver
y familiar with power supply designs and selection. The basic operating principles of linear regulators and SMPS are explained and the advantages and disadvantages of each solution are discussed. The buck step-down converter is used as an example to further explain the design considerations of a switching regulator.
INTRODUCTION
Today’s designs require an increasing number of power rails and supply solutions in electronics systems, with loads ranging from a few mA for standby supplies to over 100A for ASIC voltage regulators. It is important to choose the appropriate solution for the targeted application and to meet specified performance requirements, such as high efficiency, tight printed circuit board (PCB) space, accurate output regulation, fast transient response, low solution cost, etc. Power management design is becoming a more frequent and challenging task for system designers, many of whom may not have strong power backgrounds.
A power converter generates output voltage and current for the load from a given input power source. It needs to meet the load voltage or current regulation requirement during steady-state and transient conditions. It also must protect the load and system in case of a component failure. Depending on the specific application, a designer can choose either a linear regulator (LR) or a switching mode power supply (SMPS) solution. To make the best choice of a solution, it is essential for designers to be familiar with the merits, drawbacks and design concerns of each approach.
This article focuses on nonisolated power supply applica­tions and provides an introduction to their operation and design basics.
LINEAR REGULATORS
orks
Let’s start with a simple example. In an embedded system, a 12V bus rail is available from the front-end power supply. On the system board, a 3.3V voltage is needed to power an operational amplifier (op amp). The simplest approach to generate the 3.3V is to use a resistor divider from the 12V bus, as shown in Figure 1. Does it work well? The answer is usually no. The op amp’s V
pin current may vary under
CC
different operating conditions. If a fixed resistor divider is used, the IC V
voltage varies with load. Besides, the
CC
12V bus input may not be well regulated. There may be many other loads in the same system sharing the 12V rail. Because of the bus impedance, the 12V bus voltage var
-
ies with the bus loading conditions. As a result, a resistor
12VDC BUS
R1
3.3V
R2
V
X
Figure 1. Resistor Divider Generates 3.3VDC from 12V Bus Input
L, LT, LTC, LTM, Linear Technology, LTspice, µModule, PolyPhase and the Linear logo are registered trademarks and LTpowerCAD is a trademark of Linear Technology Corporation. All other trademarks are the property of their respective owners.
V
CC
LOAD
+
AN140 F01
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AN140-1
Application Note 140
divider cannot provide a regulated 3.3V to the op amp to ensure its proper operation. Therefore, a dedicated volt­age regulation loop is needed. As shown in Figure 2, the feedback loop needs to adjust the top resistor R1 value to dynamically regulate the 3.3V on V
12V
BUS
R1
3.3V
FEEDBACK
REGULATOR
Figure 2. Feedback Loop Adjusts Series Resistor R1 Value to Regulate 3.3V
+
V
X
.
CC
V
CC
LOAD
+
AN140 F02
This kind of variable resistor can be implemented with a linear regulator, as shown in Figure 3. A linear regulator operates a bipolar or field effect power transistor (FET) in its linear mode. So the transistor works as a variable resistor in series with the output load. To establish the feedback loop, conceptually, an error amplifier senses the DC output voltage via a sampling resistor network R and R reference voltage V
, then compares the feedback voltage VFB with a
B
. The error amplifier output voltage
REF
A
drives the base of the series power transistor via a current amplifier. When either the input V or the load current increases, the V down. The feedback voltage V
FB
voltage decreases
BUS
output voltage goes
CC
decreases as well. As a result, the feedback error amplifier and current amplifier generate more current into the base of the transistor Q1. This reduces the voltage drop V back the V the other hand, if the V
output voltage, so that V
CC
output voltage goes up, in a
CC
similar way, the negative feedback circuit increases V
and hence brings
CE
equals V
FB
REF
. On
CE
VIN = 12V
LINEAR REGULATOR
CURRENT
AMPLIFIER
ERROR
AMPLIFIER
B
+
V
REF
BUS
Q1
C
+
V
CE
C
E
R
A
V
FB
R
B
O
VO = 3.3V
+
V
CC
LOAD
+
V
X
AN140 F03
Figure 3. A Linear Regulator Implements a Variable Resistor to Regulate Output Voltage
to ensure the accurate regulation of the 3.3V output. In summary, any variation of V regulator transistor’s V
is always constant and well regulated.
V
CC
CE
is absorbed by the linear
O
voltage. So the output voltage
Why Use Linear Regulators?
The linear regulator has been widely used by industry for a very long time. It was the basis for the power supply industry until switching mode power supplies became prevalent after the 1960s. Even today, linear regulators are still widely used in a wide range of applications.
In addition to their simplicity of use, linear regulators have other performance advantages. Power management sup
-
pliers have developed many integrated linear regulators.
, V
A typical integrated linear regulator needs only V
IN
OUT
, FB and optional GND pins. Figure 4 shows a typical 3-pin linear regulator, the LT1083, which was developed more than 20 years ago by Linear Technology. It only needs an input capacitor, output capacitor and two feedback resistors to set the output voltage. Almost any electrical engineer can design a supply with these simple linear regulators.
AN140-2
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P
OUTPUT+PLOSS
VO•I
O
V
IN
1.2V TO 36V
F
5V AT 7.5A
V
≥ 6.5V
AN140 F04
IN
+
10µF
*REQUIRED FOR STABILITY
Application Note 140
work with low headroom (VIN – VO) are called low dropout
LT1083
IN
ADJ
OUT
121Ω 1%
365Ω 1%
+
10µF* TANTALUM
regulators (LDOs).
It is also clear that a linear regulator or an LDO can only provide step-down DC/DC conversion. In applications that require V negative V
voltage to be higher than VIN voltage, or need
O
voltage from a positive VIN voltage, linear
O
regulators obviously do not work.
Figure 4. Integrated Linear Regulator Example: 7.5A Linear Regulator with Only Three Pins
One Drawback – A Linear Regulator Can Burn a Lot of Power
A major drawback of using linear regulators can be the excessive power dissipation of its series transistor Q1 operating in a linear mode. As explained previously, a lin
­ear regulator transistor is conceptually a variable resistor. Since all the load current must pass through the series transistor
. In this case, the efficiency of a linear regulator can be
I
O
, its power dissipation is P
Loss
= (V
– VO)
IN
quickly estimated by:
ηLR=
P
OUTPUT
=
VO•IO+(VIN– VO)I
O
O
=
V
(1)
So in the Figure 1 example, when the input is 12V and output is 3.3V, the linear regulator efficiency is just 27.5%. In this case, 72.5% of the input power is just wasted and generates heat in the regulator. This means that the transis
­tor must have the thermal capability to handle its power/ heat dissipation at worst case at maximum V
and full
IN
load. So the size of the linear regulator and its heat sink may be large, especially when V
is much less than VIN.
O
Figure 5 shows that the maximum efficiency of the linear regulator is proportional to the V
O/VIN
ratio.
On the other hand, the linear regulator can be very efficient
is close to VIN. However, the linear regulator (LR) has
if V
O
another limitation, which is the minimum voltage differ­ence between V
and VO. The transistor in the LR must
IN
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
is too close to VIN, the LR may be unable to regulate
V
O
output voltage anymore. The linear regulators that can
100
80
60
40
EFFICIENCY %
20
0
0
0.40.2 VO/V
IN
0.6
0.8 1
AN140 F05
Figure 5. Maximum Linear Regulator Efficiency vs, VO/VIN Ratio
Linear Regulator with Current Sharing for High Power [8]
For applications that require more power, the regulator must be mounted separately on a heat sink to dissipate the heat. In all-surface-mount systems, this is not an option, so the limitation of power dissipation (1W for example) limits the output current. Unfortunately, it is not easy to directly parallel linear regulators to spread the generated heat.
SET
LT3080
R
SET
V
OUT
+ –
= R
SET
• 10µA
OUT
AN140 F06
V
OUT
2.2µ
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V
CONTROL
1µF
IN
V
IN
Figure 6. Single Resistor Setting LDO LT3080 with a Precision Current Source Reference
AN140-3
Application Note 140
4.5V TO 30V
OUT
Replacing the voltage reference shown in Figure 3 with a precision current source, allows the linear regulator to be directly paralleled to spread the current load and thus spread dissipated heat among the ICs. This makes it pos
­sible to use linear regulators in high output current, all­surface-mount applications, where only a limited amount of heat can be dissipated in any single spot on a board.
T3080 is the first adjustable linear regulator that can
The L be used in parallel for higher current. As shown in Figure 6, it has a precision zero TC 10µA internal current source connected to the noninverting input of the operational amplifier R
SET
from 0V to (V
. With an external single voltage setting resistor
, the linear regulator output voltage can be adjusted
IN
– V
DROPOUT
).
Figure 7 shows how easy it is to parallel LT3080s for current sharing. Simply tie the SET pins of the LT3080s together, the two regulators share the same reference voltage. Because the operational amplifiers are precisely trimmed, the offset voltage between the adjustment pin and the output is less than 2mV. In this case, only 10mΩ ballast resistance, which can be the sum of a small external resistor and PCB trace resistance, is needed to balance the load current with better than 80% equalized sharing. Need even more power? Even paralleling 5 to 10 devices is reasonable.
V
IN
LT3080
Applications Where Linear Regulators Are Preferable
There are many applications in which linear regulators or LDOs provide superior solutions to switching supplies, including:
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. Linear regulators have very low output voltage ripple because there are no elements switching on and off frequently and linear regulators can have very high bandwidth. So there is little EMI problem. Some special LDOs, such as Linear Technology’s LT1761 LDO family, have as low as 20μV
noise voltage on the output. It is almost
RMS
impossible for an SMPS to achieve this low noise level. An SMPS usually has mV of output ripple even with very low ESR capacitors.
3. Fast transient applications. The linear regulator feed
­back loop is usually internal, so no external compensa­tion is required. Typically, linear regulators have wider control loop bandwidth and faster transient response than that of SMPS.
V
CONTROL
+ –
SET
V
V
CONTROL
IN
V
IN
1µF
Figure 7. Paralleling of Two LT3080 Linear Regulators for Higher Output Current
LT3080
+ –
SET
165k
OUT
OUT
10mΩ
10mΩ
AN140 F07
AN140-4
V
3.3V 2A
100µF
Low dropout applications. For applications where
4. output voltage is close to the input voltage, LDOs may be more efficient than an SMPS. There are very low dropout LDOs (VLDO) such as Linear’s LTC1844, LT3020 and LTC3025 with from 20mV to 90mV dropout voltage and up to 150mA current. The minimum input voltage can be as low as 0.9V. Because there is no AC switch
­ing loss in an LR, the light load efficiency of an LR or an LDO is similar to its full load efficiency. An SMPS usually has lower light load efficiency because of its AC switching losses. In batter
y powered applications in which light load efficiency is also critical, an LDO can provide a better solution than an SMPS.
In summary, designers use linear regulators or LDOs because they are simple, low noise, low cost, easy to use and provide fast transient response. If V
is close to VIN,
O
an LDO may be more efficient than an SMPS.
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Application Note 140
T
S
SWITCHING MODE POWER SUPPLY BASICS
Why Use a Switching Mode Supply?
A quick answer is high efficiency. In an SMPS, the tran
­sistors are operated in switching mode instead of linear mode. This means that when the transistor is on and conducting current, the voltage drop across its power path is minimal. When the transistor is off and blocking high voltage, there is almost no current through its power path. So the semiconductor transistor is like an ideal switch. The power loss in the transistor is therefore minimized. High efficiency, low power dissipation and high power density (small size) are the main reasons for designers to use SMPS instead of linear regulators or LDOs, especially in high current applications. For example, nowadays a 12V
3.3V
switching mode synchronous buck step-down
OUT
IN
,
supply can usually achieve >90% efficiency vs less than
27.5% from a linear regulator. This means a power loss or size reduction of at least eight times.
The Most Popular Switching Supply—the Buck Converter
Figure 8 shows the simplest and most popular switching regulator, the buck DC/DC converter. It has two operating modes, depending on if the transistor Q1 is turned on or off. To simplify the discussion, all the power devices are assumed to be ideal. When switch (transistor) Q1 is turned on, the switching node voltage V L current is being charged up by (V
SW
IN
= V
and inductor
IN
– VO). Figure 8(a) shows the equivalent circuit in this inductor charging mode. When switch Q1 is turned off, inductor current goes through the freewheeling diode D1, as shown in Figure 8(b). The switching node voltage V L current is discharged by the V
= 0V and inductor
SW
load. Since the ideal
O
inductor cannot have DC voltage in the steady state, the average output voltage V
V
= AVG[VSW]=
O(DC)
can be given as:
O
ON
V
IN
T
(2)
Q1
I
C(IN)
DUTY
CYCLE
Q1
+
V
GS1
+
+
V
IN
+
I
C(IN)
+
V
IN
A. INDUCTOR CHARGING MODE
+
+
V
IN
B. INDUCTOR DISCHARGING MODE
SW
SW
SW
+
V
V
L
O
L
I
L
D1
+
V
L
I
L
I
L
+
V
L
I
L
D1
I
L
I
CO
+
C
V
O
L
I
CO
+
C
V
O
L
I
CO
+
C
V
GS1
LOAD
O
V
SW
V
L
LOAD
O
O
LOAD
I
C(IN)
I
I
L
CO
V
O
D • T
V
IN
S
T
S
VIN – V
O
V
O
AN140 F08
Figure 8. Buck Converter Operating Modes and Typical Waveforms
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