LT1432CN8 [Linear]
5V High Efficiency Step-Down Switching Regulator Controller; 5V高效率降压型开关稳压器控制器
| 型号： | LT1432CN8 |
| 厂家： | 
Linear |
| 描述： | 5V High Efficiency Step-Down Switching Regulator Controller |
| 文件： | 总28页 (文件大小：609K) |
| 中文： | 中文翻译 | 下载： | 下载PDF数据表文档文件 |
LT1432
5V High Efficiency Step-Down
Switching Regulator Controller
U
DESCRIPTIO
EATURE
S
F
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The LT1432 is a control chip designed to operate with the
LT1170/LT1270 family of switching regulators to make a
veryhighefficiency5Vstep-down(buck)switchingregula-
tor. A minimum of external components is needed.
Accurate Preset +5V Output
Up to 90% Efficiency
Optional Burst Mode for Light Loads
Can be Used with Many LTC Switching ICs
Accurate Ultra-Low-Loss Current Limit
Operates with Inputs from 6V to 30V
Shutdown Mode Draws Only 15µA
Uses Small 50µH Inductor
Includedisanaccuratecurrentlimitwhichusesonly60mV
sense voltage and uses “free” PC board trace material for
the sense resistor. Logic controlled electronic shutdown
mode draws only 15µA battery current. The switching
regulator operates down to 6V input.
The LT1432 has a logic controlled “burst” mode to achieve
highefficiencyatverylightloadcurrents(0to100mA)such
as memory keep-alive. In normal switching mode, the
standby power loss is about 60mW, limiting efficiency at
light loads. In burst mode, standby loss is reduced to
approximately 15mW. Output current in this mode is
typically in the 5mA to 100mA range.
O U
PPLICATI
S
A
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Laptop and Palmtop Computers
Portable Data-Gathering Instruments
DC Bus Distribution Systems
Battery-Powered Digital Widgets
The LT1432 is available in 8-pin surface mount and DIP
packages. The LT1170/LT1270 family will also be available
in a surface mount version of the 5-pin TO-220 package.
For 3.3V versions contact Linear Technology Corporation.
U
O
TYPICAL APPLICATI
V
IN
Efficiency
10µH
V
V
IN
SW
3A
+
LT1170
LT1271
C1
330µF
35V
+
OPTIONAL
FB
100
90
100µF
16V
OUTPUT
FILTER
V
GND
D2
C
1N4148
C6
0.02µF
NORMAL MODE
(USE AMPS SCALE)
C3
+
R1
C5
4.7µF
TANT
680Ω
0.03µF
L1
50µH
R2*
0.013Ω
C4
0.1µF
V
OUT
80
×
5V
+
3A**
BURST MODE
(USE mA SCALE)
D1
MBR330p
C2
+
390µF
16V
V
V
C
DIODE
70
V
V
V
IN
LIM
LT1432
GND
LT1271, L = 50µH
MODE
OUT
MODE LOGIC
220pF
60
* R2 IS MADE FROM PC BOARD
COPPER TRACES.
0
0
3A
1A
2A
20mA
40mA
60mA
<0.3V = NORMAL MODE
>2.5V = SHUTDOWN
OPEN = BURST MODE
** MAXIMUM CURRENT IS DETERMINED
BY THE CHOICE OF LT1070 FAMILY.
SEE APPLICATION SECTION.
LT1432 TA02
LT1432 TA01
Figure 1. High Efficiency 5V Buck Converter
1
LT1432
W W W
U
/O
ABSOLUTE AXI U RATI GS
PACKAGE RDER I FOR ATIO
VIN Pin .................................................................... 30V
V+ Pin ..................................................................... 40V
VC ........................................................................... 35V
VLIM and VOUT Pins................................................... 7V
Diode Pin Voltage ................................................... 30V
Mode Pin Current (Note 2) ..................................... 1mA
Operating Temperature Range .................... 0°C to 70°C
Storage Temperature Range ................ –65°C to 150°C
Lead Temperature (Soldering, 10 sec.)................ 300°C
TOP VIEW
ORDER PART
NUMBER
1
2
3
4
8
7
6
5
MODE
GND
V
LIM
V
OUT
LT1432CN8
LT1432CS8
V
C
V
IN
+
DIODE
V
N8 PACKAGE
8-LEAD PLASTIC DIP
S8 PACKAGE
8-LEAD PLASTIC SO
ELECTRICAL CHARACTERISTICS
VC = 6V, VIN = 12V, V+ = 10V, VDIODE = Open, VLIM = VOUT, VMODE = 0V, TJ = 25°C
Device is in standard test loop unless otherwise noted.
PARAMETER
CONDITIONS
V Current = 220µA
MIN
TYP
MAX
UNITS
Regulated Output Voltage
Output Voltage Line Regulation
Input Supply Current (Note 1)
Quiescent Output Load Current
Mode Pin Current
●
●
●
4.9
5.0
5
5.1
20
V
mV
mA
mA
C
V
IN
V
IN
= 6V to 30V
+
= 6V to 30V, V = V + 5V, V = V + 1V
0.3
0.9
0.5
1.2
IN
C
IN
V
MODE
V
MODE
= 0V (current is out of pin)
= 5V (shutdown)
●
●
30
15
50
30
µA
µA
Mode Pin Threshold Voltage
(Normal to Burst)
I
= 10µA (out of pin)
●
0.6
0.9
1.5
V
MODE
V Pin Saturation Voltage
V
V
V
= 5.5V (forced)
= 5.5V (forced)
= 4.5V (forced)
●
●
●
0.25
0.8
60
0.45
1.5
100
64
V
mA
µA
C
OUT
OUT
OUT
V Pin Maximum Sink Current
C
0.45
40
V Pin Source Current
C
Current Limit Sense Voltage (Note 3)
Device in Current Limit Loop
56
60
mV
µA
V
LIM
Pin Current
Device in Current Limit Loop
(current is out of pin)
●
●
30
45
70
+
Supply Current in Shutdown
Burst Mode Output Ripple
Burst Mode Average Output Voltage
Clamp Diode Forward Voltage
Startup Drive Current
V
MODE
> 3V, V < 30V, V and V = 0V
15
100
5
60
µA
IN
C
Device in Burst Test Circuit
Device in Burst Test Circuit
mV
p-p
●
●
●
4.8
30
5.2
V
I = 1mA, All Other Pins Open
F
0.5
45
0.65
V
+
V
V
= 2.5V (forced), V = 5V to 25V,
mA
OUT
IN
+
= 6V to 26V, V = V – 1V, V = V – 1.5V
IN
C
IN
Restart Time Delay
(Note 4)
I = 150µA to 250µA
1
1.8
10
ms
Transconductance, Output to V Pin
●
1500
2000
2800
µmho
C
C
The
range.
●
denotes specifications which apply over the operating temperature
Note 3: Current limit sense voltage temperature coefficient is +0.33%/°C
to match TC of copper trace material.
Note 1: Does not include current drawn by the LT1070 IC. See operating
parameters in standard circuit.
Note 4: V
pin switched from 5.5Vto 4.5V.
OUT
Note 2: Breakdown voltage on the mode pin is 7V. External current must
be limited to value shown.
2
LT1432
ELECTRICAL CHARACTERISTICS
Operating parameters in standard circuit configuration.
IN = +12V, IOUT = 0, unless otherwise noted. These parameters guaranteed where indicated, but not tested.
V
PARAMETER
CONDITIONS
MIN
TYP
MAX
UNITS
Burst Mode Quiescent Input Supply Current
Burst Mode Output Ripple Voltage
1.3
1.8
mA
I
I
= 0
= 50mA
100
130
mV
mV
OUT
OUT
p-p
p-p
Normal Mode Equivalent Input Supply Current
Normal Mode Minimum Operating Input Voltage
Burst Mode Minimum Operating Input Voltage
Efficiency
Extrapolated from I
= 20mA
6
6
mA
V
OUT
100mA < I
< 1.5A
OUT
5mA < I
< 50mA
6.2
V
OUT
Normal Mode I
= 0.5A
= 25mA
91
77
%
%
OUT
OUT
Burst Mode
I
Load Regulation
Normal Mode 50mA < I
< 2A
10
50
25
mV
mV
OUT
Burst Mode 0 < I
< 50mA
OUT
U
W
EQUIVALE T SCHE ATIC
V
V
V
IN
SW
IN
LT1271
FB
GND
V
C
+5V
V
OUT
V
LIM
1
2
60mV
+
–
+
V
V
C
DIODE
V
IN
3
4
6
5
S1**
+
–
S3*
*
S3 IS CLOSED ONLY DURING STARTUP.
** S1 AND S2 ARE SHOWN IN NORMAL
MODE. REVERSE FOR BURST MODE.
S2**
MODE
CONTROL
7
GND
MODE
8
LT1432 F02
Figure 2
3
LT1432
TYPICAL PERFOR A CE CHARACTERISTICS
U W
Minimum Input Voltage – Normal
Mode (1270/1271)
Efficiency vs Input Voltage
Efficiency vs Load Current
100
90
7.5
7.0
6.5
6.0
5.5
5.0
100
90
LT1271
LT1270/1271
J
LT1270
L = 50µH
I
= 0.5A
LOAD
T = 25°C
L = 50µH
I
= 1A
LOAD
I
= 2A
LT1170
L = 25µH
LOAD
LT1271
LT1270
80
80
70
70
T
J
= 25°C
LT1271, L = 50µH
T
J
= 25°C
60
60
0
5
10
15
20
25
30
0
0.5
1.0
1.5
2.0
2.5
3.0
0
1
2
3
4
5
INPUT VOLTAGE (V)
LOAD CURRENT (A)
OUTPUT CURRENT (A)
LT1432 G01
LT1432 G03
LT1432 G02
Minimum Input Voltage – Normal
Mode (1070 Family)
Minimum Input Voltage – Normal
Mode (1170 Family)
Burst Mode Minimum Input
Voltage
7.5
7.0
6.5
7.5
7.0
6.5
7.0
6.5
6.0
5.5
5.0
LT1070 FAMILY(40kHz)
J
LT1170 FAMILY(100kHz)
J
T = 25°C
J
T = 25°C
T = 25°C
LT1170
LT1171
LT1172
LT1072
LT1170
LT1071
LT1070
6.0
5.5
5.0
6.0
5.5
5.0
LT1070
0
1
2
3
4
5
0
1
2
3
4
5
0
10
20
30
40
50
OUTPUT CURRENT (A)
OUTPUT CURRENT (A)
LOAD CURRENT (mA)
LT1432 G04
LT1432 G05
LT1432 G06
Shutdown Current vs Input
Voltage
Current Limit Sense Voltage*
Battery Current in Shutdown*
50
40
30
20
40
30
20
10
0
80
70
60
50
40
T = 25°C
J
V
V
= 30V
= 6V
IN
IN
10
0
0
10
15
20
25
30
5
50
TEMPERATURE (°C)
75
50
JUNCTION TEMPERATURE (°C)
0
100
0
75
100
25
25
INPUT VOLTAGE (V)
LT1432 G11
LT1432 G07
*DOES NOT INCLUDE LT1271 SWITCH LEAKAGE.LT1432 G08
* TEMPERATURE COEFFICIENT OF SENSE VOLTAGE IS
DESIGNED TO TRACK COPPER RESISTANCE.
4
LT1432
U W
TYPICAL PERFOR A CE CHARACTERISTICS
No Load Battery Current in Burst
Mode
Transconductance – VOUT to VC
Current
Incremental Battery Current * in
Burst Mode
4000
3000
2000
1000
40
2.0
1.5
1.0
0.5
0
5
4
3
2
1
0
T = 25°C
J
T
= 25°C
∆I(V PIN)
J
C
Gm =
∆V
OUT
50
75
0
100
0
5
10
15
20
25
0
5
10
15
20
25
25
JUNCTION TEMPERATURE (°C)
BATTERY VOLTAGE (V)
BATTERY VOLTAGE (V)
LT1432 G10
LT1432 G12
LT1432 G09
* TO CALCULATE TOTAL BATTERY CURRENT IN BURST
MODE, MULTIPLY LOAD CURRENT BY INCREMENTAL
FACTOR AND ADD NO-LOAD CURRENT.
Burst Mode Load Regulation
Mode Pin Current
Line Regulation
40
25
0
60
40
T
= 25°C
T = 25°C
J
T = 25°C
J
J
BURST MODE
20
0
20
–25
–50
–75
0
NORMAL MODE
–20
–40
–20
–40
MODE DRIVE MUST
SINK ≈ 30µA AT 0V
10
INPUT VOLTAGE (V)
15
0
20
5
0
20
60
LOAD CURRENT (mA)
80
40
100
0
2
4
6
8
10
MODE PIN VOLTAGE (V)
LT1432 G14
LT1432 G13
LT1432 G15
Restart Load Current
Restart Time Delay
Startup Switch Characteristics
40
30
20
10
0
4
3
2
1
0
5
0
T = 25°C
J
V
= 4.5V
OUT
NOTE VERTICAL &
HORIZONTAL SCALE
CHANGES AT 0,0
–20
–40
–60
–80
50
75
0
100
25
50
JUNCTION TEMPERATURE (°C)
0
75
100
25
–2
–1
0
10
20
30
+
JUNCTION TEMPERATURE (°C)
V
TO V VOLTAGE
IN
LT1432 G16
LT1432 G16
LT1432 G18
5
LT1432
PPLICATI
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Basic Circuit Description
The circuit in Figure 1 is a basic 5V positive buck converter
which can operate with input voltage from 6V to 30V. The
power switch is located between the VSW pin and GND pin
on the LT1271. Its current and duty cycle are controlled by
the voltage on the VC pin with respect to the GND pin. This
voltage ranges from 1V to 2V as switch current increases
from zero to full scale. Correct output voltage is main-
tained by the LT1432 which has an internal reference and
erroramplifier(see EquivalentSchematicinFigure 2). The
amplifier output is level shifted with an internal open
collector NPN to drive the VC pin of the switcher. The
normal resistor divider feedback to the switcher feedback
pin cannot be used because the feedback pin is referenced
to the GND pin, which is switching up and down. The
feedback pin (FB) is simply bypassed with a capacitor.
This forces the switcher VC pin to swing high with about
200µA sourcing capability. The LT1432 VC pin then sinks
this current to control the loop. Transconductance from
the regulator output to the VC pin current is controlled to
approximately 2000µmhos by local feedback around the
LT1432erroramplifier(S2closedinFigure2).Thisisdone
to simplify frequency compensation of the overall loop. A
word of caution about the FB pin bypass capacitor (C6):
this capacitor value is very non-critical, but the capacitor
must be connected directly to the GND pin or tab of the
switcher to avoid differential spikes created by fast switch
currents flowing in the external PCB traces. This is also
true for the frequency compensation capacitors C4 and
C5. C4 forms the dominant loop pole with a loop zero
added by R1. C5 forms a higher frequency loop pole to
control switching ripple at the VC pin.
The LT1432 is a dedicated 5V buck converter driver chip
intended to be used with an IC switcher from the LT1070
family. This family of current mode switchers includes
current ratings from 1.25A to 10A, and switching frequen-
cies from 40kHz to 100kHz as shown in the table below.
SWITCH
CURRENT
OUTPUT CURRENT IN
BUCK CONVERTER
DEVICE
FREQUENCY
LT1270A
LT1270
LT1170
LT1070
LT1271
LT1171
LT1071
LT1172
LT1072
10A
8A
5A
5A
4A
2.5A
2.5A
1.25A
1.25A
60kHz
60kHz
100kHz
40kHz
60kHz
100kHz
40kHz
100kHz
40kHz
7.5A
6A
3.75A
3.75A
3A
1.8A
1.8A
0.9A
0.9A
The maximum load current which can be delivered by
these chips in a buck converter is approximately 75% of
theirswitchcurrentrating.Thisispartlyduetothefactthat
buck converters must operate at very high duty cycles
when input voltage is low. The “current mode” nature of
the LT1070 family requires an internal reduction of peak
currentlimitathighdutycycles, sothesedevicesarerated
at only 80% of their full current rating when duty cycle is
80%. A second factor is inductor ripple current, half of
which subtracts from maximum available load current.
See Inductor Selection for details. The LT1070 family was
originally intended for topologies which have the negative
side of the switch grounded, such as boost converters. It
has an extremely efficient quasi-saturating NPN switch
which mimics the linear resistive nature of a MOSFET but
consumes much less die area. Driver losses are kept to a
minimum with a patented adaptive antisat drive that main-
tains a forced beta of 40 over a wide range of switch
currents. This family is attractive for high efficiency buck
converters because of the low switch loss, but to operate
as a positive buck converter, the ground pin of the IC must
be floated to act as the switch output node. This requires
a floating power supply for the chip and some means for
level shifting the feedback signal. The LT1432 performs
these functions as well as adding current limiting, mi-
cropower shutdown, and dual mode operation for high
conversionefficiencywithbothheavyandverylightloads.
Afloating5Vpowersupplyfortheswitcherisgeneratedby
D2 and C3 which peak detect the output voltage during
switch “off” time. The diode used for D2 is a low capaci-
tance type to avoid spikes at the output. Do not substitute
a Schottky diode for D2 (they are high capacitance). This
is a very efficient way of powering the switcher because
power drain does not increase with regulator input volt-
age. However, the circuit is not self-starting, so some
means must be used to start the regulator. This is per-
formed by the internal current path of the LT1432 which
allows current to flow from the input supply to the V+ pin
during startup.
6
LT1432
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D1, L1 and C2 act as the conventional catch diode and
output filter of the buck converter. These components
should be selected carefully to maintain high efficiency
and acceptable output ripple. See other sections of this
data sheet for detailed discussions of these parts.
2
I
V
(
)
OUT OUT
P =
40V
IN
Diode loss;
V V
– V I
OUT OUT
(
)(
)
F
IN
Current limiting is performed by R2. Sense voltage is only
60mV to maintain high efficiency. This also reduces the
value of the sense resistor enough to utilize a printed
circuitboardtraceasthesenseresistor. Thesensevoltage
has a positive temperature coefficient of 0.33%/°C to
match the temperature coefficient of copper. See Current
Limiting section for details.
P =
V
IN
(Use VF vs IF graph on diode data sheet, assuming IF =
IOUT
)
RS = Inductor series resistance
RSW = Switch resistance of LT1271, etc.
IF = Diode current
The basic regulator has three different operating modes,
defined by the mode pin drive. Normal operation occurs
when the mode pin is grounded. A low quiescent current
“burst” mode can be initiated by floating the mode pin.
Input supply current is typically 1.3mA in this mode, and
output ripple voltage is 100mVp-p. Pulling the mode pin
above 2.5V forces the entire regulator into micropower
shutdown where it typically draws less than 20µA. See
Mode Pin Drive for details.
VF = Diode forward voltage at IF = IOUT
Inductorcorelossdependsonpeak-to-peakripplecurrent
in the inductor, which is independent of load current for
any load current large enough to establish continuous
current in the inductor. Believe it or not, core loss is also
independent of the physical size of the core. It depends
only on core material, inductance value, and switching
frequency for fixed regulator operating conditions. In-
creasing inductance or switching frequency will reduce
core loss, because of the resultant decrease in ripple
current. Forhighefficiency, lowlosscoressuchasferrites
or Magnetics Inc. molypermalloy or KoolMµ are recom-
mended. The lower cost Type 52 powdered iron from
Phillips is acceptable only if larger inductance is used and
the increased size and slight loss in efficiency is accept-
able. In a typical buck converter using the LT1271 (60kHz)
with a 12V input, and a 50µH inductor, core loss with a
Type 52 powdered iron core is 203mW. A molypermalloy
core reduces this figure to 28mW. With a 1A output, this
translates to 4% and 0.56% core loss respectively – a big
difference in a high efficiency converter. For details on
inductor design and losses, see Application Note 44.
Efficiency
Efficiency in normal mode is maximum at about 500mA
load current, where it exceeds 90%. At lower currents, the
operating supply current of the switching IC dominates
losses. The power loss due to this term is approximately
8mA × 5V, or 40mW. This is 4% of output power at a load
current of 200mA. At higher load currents, losses in the
switch, diode, and inductor series resistance begin to
increase as the square of current and quickly become the
dominant loss terms.
Loss in inductor series resistance;
2
P = RS (IOUT
)
What are the benefits of using an active (synchronous)
switch to replace the catch diode? This is the trendy thing
to do, but calculations and actual breadboards show that
theimprovementinefficiencyisonlyafewpercentatbest.
This can be shown with the following simplified formulas:
Loss in switch on resistance;
2
V
R
I
OUT
OUT SW
(
)
(
)
P =
V
IN
V V
– V I
OUT OUT
(
)(
)
F
IN
Loss in switch driver current;
Diode Loss =
V
IN
7
LT1432
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PPLICATI
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in a lap-top computer. In this mode, hysteresis is added to
the error amplifier to make it switch on and off, rather than
maintain a constant amplifier output. This forces the
switching IC to either provide a rapidly increasing current
or to go into full micropower shutdown. Current is deliv-
ered to the output capacitor in pulses of higher amplitude
and low duty cycle rather than a continuous stream of low
amplitude pulses. This maximizes efficiency at light load
byeliminatingquiescentcurrentintheswitchingICduring
the period between bursts.
2
V
– V
R I
SW OUT
(
)(
)(
)
IN
OUT
FET Switch Loss =
V
IN
(Ignoring gate drive power)
The change in efficiency is:
2
Diode Loss – FET Loss Efficiency
(
)(
)
V
V
(
)(
)
IN OUT
This is equal to:
The result of pulsating currents into the output capacitor
is that output ripple amplitude increases, and ripple fre-
quency becomes a function of load current. The typical
output ripple in burst mode is 150mVp-p, and ripple
frequencycanvaryfrom50Hzto2kHz.Thisisnotnormally
a problem for the logic circuits which are kept “alive”
during sleep mode.
2
V – V
V – R
× I
OUT
E
)( )
(
)(
IN
OUT
F
FET
V
V
(
)(
)
IN OUT
IfVF (diodeforwardvoltage)=0.45V,VIN =10V,VOUT =5V,
RFET =0.1Ω, IOUT =1A, andefficiency=90%, theimprove-
ment in efficiency is only:
Some thought must be given to proper sequencing be-
tween normal mode and burst mode. A heavy (>100mA)
load in burst mode can cause excessive output ripple, and
an abnormally light load (10mA to 30mA, see curves) in
normal mode can cause the regulator to revert to a quasi-
burst mode that also has higher output ripple. The worst
condition is a sudden, large increase in load current
(>100mA) during this quasi-burst mode or just after a
switch from burst mode to normal mode. This can cause
the output to sag badly while the regulator is establishing
normalmodeoperation(≈100µs). Toavoidproblems, itis
suggested that the power-down sequence consist of re-
ducing load current to below 100mA, but greater than the
minimumfornormalmode, thenswitchingtoburstmode,
followed by a reduction of load current to the final sleep
value. Power-up would consist of increasing the load
current to the minimum for normal mode, then switching
to normal mode, pausing for 1ms, followed by return to
full load.
2
10V – 5V 0.45V – 0.1Ω ×1A 0.9
(
)(
)(
)
= 2.8%
10V 5V
( )(
)
This does not take FET gate drive losses into account,
which can easily reduce this figure to less than 2%. The
added cost, size, and complexity of a synchronous switch
configuration would be warranted only in the most ex-
treme circumstances.
Burst mode efficiency is limited by quiescent current drain
in the LT1432 and the switching IC. The typical burst mode
zero-load input power is 27mW. This gives about one
month battery life for a 12V, 1.2AHr battery pack. Increas-
ing load power reduces discharge time proportionately.
Full shutdown current is only about 15µA, which is consid-
erably less than the self-discharge rate of typical batteries.
Burst Mode Operation
If this sequence is not possible, an alternative is to
minimize normal mode settling time by adding a 47kΩ
resistor between V+ and VC pins. The output capacitor
should be increased to >680µF and the compensation
capacitors should also be as small as possible, consistent
with adequate phase margin. These modifications will
Burst mode is initiated by allowing the mode pin to float,
where it will assume a DC voltage of approximately 1V. If
AC pickup from surrounding logic lines is likely, the mode
pin should be bypassed with a 200pF capacitor. Burst
mode is used to reduce quiescent operating current when
theregulatoroutputcurrentisverylow,asin“sleep”mode
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often allow the power-down sequence to consist of simul-
taneousturn-offofload currentandswitchto burst mode.
Power-up is accomplished by switching to normal mode
and simultaneously increasing load current to the lowest
possible value (30mA to 500mA), followed by a short
pause and return to full load current.
5V/DIV
0
Full Shutdown
When the mode pin is driven high, full shutdown of the
regulatoroccurs. Regulatorinputcurrentwillthenconsist
of the LT1432 shutdown current (≈15µA) plus the switch
leakage of the switching IC (≈1µA to 25µA). Mode input
current (≈15µA at 5V) must also be considered. Startup
from shutdown can be in either normal or burst mode, but
one should always check startup overshoot, especially if
the output capacitor or frequency compensation compo-
nents have been changed.
1A/DIV
0
5µs/DIV
Figure 3
Switching Waveforms in Normal Mode
5V/DIV
0
The waveforms in Figures 3 through 10 were taken with
an input voltage of 12V. Figure 3 shows the classic buck
converterwaveformsofswitchoutputvoltage(5V/DIV)at
the top and switch current (1A/DIV) underneath, at an
output current of 2A. The regulator is operating in “con-
tinuous” mode as evidenced by the fact that switch
current does not start at zero at switch turn-on. Instead,
itjumpstoaninitialvalue, thencontinuestoslopeupward
during the duration of switch on time. The slope of the
current waveform is determined by the difference be-
tween input and output voltage, and the value of inductor
used.
1A/DIV
0
5µs/DIV
Figure 4
V – V
(
)
dl
dt
IN
OUT
=
5V/DIV
0
L
According to theory, the average switch current during
switch on time should be equal to the 2A output current
and this is confirmed in the photograph. The peak switch
current, however, is about 2.4A.This peak current must
be considered when calculating maximum available load
current because both the LT1432 and the LT1070 family
current limit on instantaneous switch current.
0.5A/DIV
0
5µs/DIV
Figure 5
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Note that the switch output voltage is nearly identical to
the 12V input during switch on time, a necessary require-
ment for high efficiency, and indicative of an efficient
switch topology. Also note the fast, clean edges on the
switching waveforms, an additional requirement for high
efficiency. The “overlap time” of switch current and volt-
age, which leads to AC switching losses, is only 10ns.
1A/DIV
Figure 4 shows the same waveforms when load current
has been reduced to 0.25A, and Figure 5 is at 25mA (note
the scale change for current in Figure 5). The regulator is
now into discontinuous mode as shown by the fact that
switch current has no initial jump, but starts its upward
slope from zero. This implies that the inductor current has
dropped to zero during switch off time, and that is shown
by the “ringing” waveform on the rising edge of switch
voltage. The switch has not yet been turned on, but the
voltage at its output rises and rings as the “input” end of
the inductor tries to settle to the same voltage as its
“output” end (5V).
5µs/DIV
Figure 6. Input Capacitor Current
This ringing is not an oscillation. It is the result of stored
energy in the catch diode capacitance. This energy is
transferred to the inductor as the inductor voltage at-
tempts to rise to 5V. The inductor and diode capacitance
tank circuit continues to ring until the stored energy is
dissipated by losses in the core and parasitic resistances.
The relatively undamped nature in this case is good
because it shows low losses and that translates to high
efficiency. EMI is not increased by operating in this mode.
0.5A/DIV
5µs/DIV
Figure 7. Output Capacitor Ripple Current
Figure 6 shows input capacitor current (1A/DIV) with IOUT
= 2A. The theoretical peak-to-peak value (ignoring sloping
waveforms) is equal to output current, and this is indeed
what the top waveform shows. The RMS value is approxi-
mately equal to one half output current. This is a major
consideration because the physical size of a capacitor with
1Aripplecurrentratingmaymakeitthelargestcomponent
in the regulator (see output capacitor section). Clever
desigers may hit on the idea of utilizing battery impedance
or remote input capacitors to divert some of the current
away from the actual local capacitor to reduce its size. This
is not too practical as shown by the middle waveform in
Figure 6, which shows input capacitor current when an
additional large capacitor is added about 6" away from the
local capacitor. The wiring inductance and parasitic resis-
tance limit the shunting effect and local capacitor current
isreducedonlyslightly.thebottomwaveformshowsinput
capacitor current with output current reduced to 0.25A.
Figure 7 shows output capacitor ripple current at loads of
2A, 0.25A, and 25mA respectively starting from the top.
Notethatripplecurrentisindependentofloadcurrentuntil
the load drops well into the discontinuous region. The
small steps superimposed on the triangular ripple are
caused by loading of the diode which pumps the power
supply capacitor on the LT1271. Amplitude of the ripple
current is about 0.7Ap-p in this case, or approximately
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tures, so be sure to check ESR ratings at the lowest
expected operating temperature. Ripple voltage can be
reduced by increasing the inductor value, but this has
rapidly diminishing returns because of typical size re-
straints.
0.2A RMS. Theoretically the output capacitor size would
be minimized by using one which just met this ripple
current, but in practice, this would yield such high output
ripple voltage that an additional output filter would have to
be added. A better solution in the case of buck converters
is usually just to increase the size of the output capacitor
to meet output ripple voltage requirements.
Figure 9 shows diode current under normal load condi-
tions of 2A, and with the output shorted. Current limit has
been set at 3A. Average diode current at IOUT = 2A is only
about 1A because of duty cycle considerations. Under
short circuit conditions, duty cycle is nearly 100% for the
diode (switch duty cycle is near zero), and diode average
current is nearly 3A. Designs which must tolerate continu-
ous short circuit conditions should be checked carefully
for diode heating. Foldback current limiting can be used if
necessary.
50mV/DIV
Figure10showsinductorcurrent(0.5A/DIV)witha2Aand
100mA load. Average inductor current is always equal to
output current, but it is obvious that with 100mA load,
inductor current drops to zero for part of the switching
cycle, indicating dicontinuous mode. When selecting an
inductor, keep in mind that RMS current determines
copperlosses,peak-to-peakcurrentdeterminescoreloss,
and peak current must be calculated to avoid core satura-
tion. Also, remember that during short circuit conditions,
inductorcurrentwillincreasetothefullcurrentlimitvalue.
Inductor failure is normally caused by overheating of the
winding insulation with resultant turn-to-turn shorts.
Foldback current limiting will be helpful.
1A/DIV
0
5µs/DIV
Figure 8. Output Ripple Current
1A/DIV
0
1A/DIV
0
0.5A/DIV
5µs/DIV
Figure 9. Diode Current
Figure 8 shows output ripple voltage at the top and switch
current below. Peak-to-peak ripple voltage is 80mV. This
implies an output capacitor effective series resistance
(ESR) of 80mV/0.7A = 0.11Ω. Capacitor ESR varies sig-
nificantly with temperature, increasing at low tempera-
0
5µs/DIV
Figure 10. Inductor Current
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Switching Waveforms in Burst Mode
In burst mode, the LT1432 amplifier is converted to a
comparator with hysteresis. This causes its VC pin current
drive to be either zero (output low), or full “on” at about
0.8mA(outputhigh).TheLT1271thereforeiseitherdriven
to full on condition or forced into complete micropower
shutdown. This makes a dramatic reduction in quiescent
currentlossesbecausetheswitchingregulatorchipdraws
supply current only during the relatively short “on” peri-
ods. This burst mode results in a battery drain of only
1.2mA with zero output load, even though the nominal
quiescent current of the switcher chip is 7mA. This low
battery drain is accomplished at the expense of higher
output ripple voltage, but the ripple is still well within the
normal requirements for logic chips.
100mV/DIV
100mV/DIV
5ms/DIV
Figure 11. Burst Mode Output Ripple Voltage
Figure 11 shows burst mode output ripple at load currents
of 0 (top trace), and 50mA (bottom trace). Ripple ampli-
tudeisnominallysetbythe100mVhysteresisbuiltintothe
LT1432, but in most applications, other effects come into
play which can significantly modify this value. The first is
delay in turning off the switcher. This causes the output to
overshoot slightly and therefore increases output ripple.
Delay is caused by the compensation capacitors used to
maintain a stable loop in the normal mode. Another effect,
however, is the ESR of the output capacitor. The surge
currentfromtheswitchercreatesastepacrossthecapaci-
tor ESR which prematurely trips the LT1432 comparator,
reducing ripple amplitude. A second delay occurs in
turning the switcher back on when the output falls below
itslowerlevel. Thisdelayissomewhatlonger, butbecause
theoutputnormallyfallsatamuchslowerratethanitrises,
this delay is not significant until output current exceeds
10mA. Falling rate is set by the output capacitor (including
any secondary filter capacitor), and the actual load cur-
rent, dVOUT/dt = IOUT/COUT. The slope in the top traces
implies a load current of approximately 2mA. This is the
sum of the 1mA output quiescent current of the LT1432
and the 1mA drawn by the VC pin and shunted through the
internal Schottky diode during the switcher “off” period.
100mV/DIV
100mV/DIV
5ms/DIV
Figure 12. Burst Mode Output Ripple Voltage
more than doubled. Figure 12 shows the same conditions
except that a 47kΩ resistor is connected from the LT1271
VIN pin to the VC pin to provide more start-up current.
These additions reduce ripple amplitude at 50mA load
current to a value only slightly higher than the no-load
condition.
Although it is difficult to see in Figures 11 and 12, there is
a narrow spike on the leading edge of the ripple caused by
the burst current and capacitor ESR. Figure 13 shows this
spike in more detail, both with and without an output filter.
The bottom trace at IOUT = 50mA shows increased ripple
caused by turn-on delay. Note that ripple frequency has
increased from 50Hz to about 600Hz and amplitude has
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Current Limiting
The LT1432 has true switching current limit with a sense
voltage of 60mV. This low sense voltage is used to
maintain high efficiency with normal loads and to make it
possible to use the printed circuit board trace material as
the sense resistor. The sense resistor value must take
ripple current into account because the LT1432 limits on
the peak of the inductor ripple current. Errors in the sense
resistor must also be allowed for.
100mV/DIV
100mV/DIV
V
SENSE
R
=
SENSE
I
RIP
2
I
1.2 *+
(
)
MAX
50µs/DIV
Figure 13
RSENSE = Required sense resistor
VSENSE = 60mV
FROM
INDUCTOR
IMAX = Maximum load current, including any surge
longer than 50µs
“L” IS MEASURED FROM
POINT “X” TO POINT “Y”
* 1.2 is a fudge factor for errors in R
and V
.
SENSE
SENSE
“X”
TO V
LIM
PIN
“W”
I
RIP
= 1/2 Peak to Peak Inductor Ripple Current
2
V
V – V
OUT
(
)
OUT IN
=
2V (f)(L)
IN
TO V
OUT
PIN
f = Frequency
“Y”
L = Inductance
Use VIN maximum
KEEP THIS DISTANCE SHORT
FOR BEST LOAD REGULATION.
}
Example: IMAX = 2A, f = 60kHz, maximum VIN = 15V,
L = 50µH;
LT 432 F14
TO LOAD
Figure 14. PC trace Current Limit Sense Resistor
5 15 – 5
with Kelvin Contacts
(
)
I
RIP
2
=
= 0.55A
= 0.02Ω
3
–6
2 15 60E 50E
( )
Time scale has been expanded to 50µs/DIV. The spike
consists of several switching cycles of the LT1271 as
shown in the lower trace. In the upper trace, the output
filterhassmoothedtheswitchingfrequencycontentofthe
spike, but the actual spike amplitude is only modestly
reduced. Increasing the output filter constants from 10µH
and 220µF to 20µH and 330µF would eliminate most of
the spike.
60mV
2A 1.2 + 0.55A
R
=
SENSE
(
)
The formula for RSENSE shows a 1.2 multiplier term in the
denominatorwhichmakestypicalcurrentlimit20%above
full load current. This accounts for small errors in the PCB
trace resistance. Trace resistance errors are kept to a
minimum by using internal traces (on multilayer boards)
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because these traces do not have errors caused by plating
operations. The suggested trace width for 1/2oz foil is
0.03"foreach1Aofcurrentlimittokeeptracetemperature
rise reasonable. 3A current limit would require the width
to be 0.09". 1oz foil can reduce trace width to 0.02" per
amp. Inductance in the trace is not critical so the trace can
be wound serpentine or any other shape that fits available
space. Kelvin connections should be used as shown in
Figure 14 to avoid errors due to termination resistance.
0.06 0.02
(
)
= 1.2 Inches
Length =
0.001
Currentlimitingmaintainstrueswitchingaction,butpower
dissipation in the IC switch and catch diode will shift
depending on output voltage. At output voltages near the
correct regulated value, power will be distributed between
switch and the diode according to the usual calculations.
Under short circuit conditions, switch duty cycle will drop
The length of the sense resistor trace can be calculated to a very low value, and power will concentrate in the
from:
diode, which will be running at near 100% duty cycle. If
continuousshortsmustbetolerated,thecatchdiodemust
be sized to handle the full current limit value, or foldback
current can be used.
W R
(
)
Inches
SENSE
Length =
R
CU
W = width of copper trace (≈0.03" per amp for 1/2oz
copper foil)
Foldback Current Limiting
Foldback current limiting makes the short circuit current
limit somewhat lower than the full load current limit to
reduce component stress under short circuit conditions.
This is shown in Figure 15 with the addition of R3 and R4.
The voltage drop across R3 adds to the 60mVcurrent limit
RCU = resistivity of PCB trace, expressed as Ω per square.
Itisfoundbycalculatingtheresistanceofasectionoftrace
with equal length and width. For typical 1/2oz material,
RCU is approximately 1mΩ per square. In the example
shown above, with width = 2A times 0.03" = 0.06";
V
IN
V
V
SW
IN
10µH
6V – 25V
3A
LT1271
+
C1
200µF
35V
FB
+
OPTIONAL
100µF
16V
V
C
GND
OUTPUT
D2
FILTER
1N4148
C6
R1
680Ω
C5
0.03µF
0.02µF
C3
+
4.7µF
C4
0.1µF
L1
50µH
TANT
R2
0.025Ω
V
OUT
×
5V
+
3A
C2
470µF
16V
D1
MBR330p
R3
100Ω
+
V
C
V
DIODE
V
V
IN
LIM
R4
12.5k
LT1432
GND
MODE
V
OUT
<0.3V = NORMAL MODE
>2.5V = SHUTDOWN
OPEN = BURST MODE
220pF
LT1432 F15
Figure 15. Adding Foldback Current Limiting
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voltage. This extra sense voltage is set by output voltage
and R4 under normal loads, but drops to near zero when
the output is shorted.
5V 100Ω
(
)
R4 =
100mV – 60mV +100Ω 40µA – 0.042 0.55
(
)
(
)
The 40µA bias current flowing out of the VLIM pin must be
accountedforwhencalculatingavalueforR4.Thiscurrent
flows through R3, causing a 4mV decrease in sense
voltage for R3 = 100Ω. The following formulas define
current limit conditions:
= 7.45kΩ
Current limit at VOUT = 5V
100
7.45k
0.042Ω
60mV – 40µA 100Ω + 5V
0.042 0.55
(
)
(
)(
)
=
Current limit at VOUT = 5V
= 2.38A
R3
R4
I
RIP
2
60mV – I R3 + V
– R
( ) (
)
(
)
B
OUT
SENSE
Current limit (output shorted)
=
R
SENSE
60mV – 100Ω 40µA
(
)
= 1.33A
=
60mV – I (R3)
B
Short Circuit Current =
0.042Ω
R
SENSE
Minimum Input Voltage
V
LIM
(1.2)
R
=
SENSE
Minimum input voltage for a buck converter using the
LT1432 is actually limited by the IC switcher used with it.
There are three factors which contribute to the minimum
voltage. At very light loads, the charge pump technique
used to provide the floating power for the switcher chip is
unable to provide sufficient current. See Figure 16 for the
minimum load required as a function of input voltage
when operating in the normal mode.
I
MAX
V
R3
( )
OUT
R4 =
I
RIP
2
V – 60mV +I R3 + R
( ) (
)
S
B
SENSE
VS = Desired full load sense voltage.
IMAX = Peak load current (for any time greater than
50µs)
IB = VLIM pin bias current (≈40mA)
At moderate to heavy loads, switch on-resistance and
maximum duty cycle will limit minimum input voltage.
GraphsintheTypicalPerformanceCharacteristicssection
showminimuminputvoltageasafunctionofloadcurrent.
At moderate loads, maximum switch duty cycle is the
limiting factor. The LT1070 family, operating at 40kHz has
a maximum duty cycle of about 94%. The LT1170 family
runsat100kHzandhasamaximumdutycycleof90%.The
LT1270 and LT1271 operate at 60kHz with a maximum
duty cycle of 92%. The curves were generated using the
expected worst case duty cycle for these devices over the
commercial operating temperature range (0°C to 100°C
junction temperature). Note that the lower frequency
devices will operate at lower input voltage because of their
higher duty cycle. These devices will require larger induc-
tors, however. (Yet another example of the universal “no
free lunch” syndrome).
To maintain high efficiency and avoid any startup prob-
lems with loads that have non-linear V/I characteristics, a
100mV (average) sense voltage is suggested for foldback
current limiting. The suggested value for R3 is 100Ω. This
is a compromise value to keep errors due to VLIM bias
current low, and to minimize current drain on the output
created by the R3/R4 path. From the previous design
example, with IMAX = 2A and IRIP/2 = 0.55A, and assuming
R3 = 100Ω, VLIM = 100mV:
100mV
R
=
= 0.042Ω
SENSE
2A 1.2
(
)
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At heavy loads, switch on-resistance increases minimum
input voltage. With an LT1071 for instance, minimum
input is 6.1V at 1A load, but increases to 6.3V at 2A load.
If absolute minimum input voltage is needed, use lower
frequency devices with higher current rating than is actu-
ally needed. The LT1070, for instance, operates down to
6.15V at 2A. Current limit is defined by the LT1432, so
higher current switchers used in lower current applica-
tions do not degrade performance or reliability.
selected. As inductance increases, core loss goes down.
Unfortunately, increased inductance requires more turns
of wire and therefore copper loss will increase. The trick is
to find the smallest inductor whose inductance is high
enough to limit core loss, and whose series resistance is
low enough to limit copper loss. Historically, inductor
manufacturers have a tendency to be ultra conservative
whendesigninginductors,andunlessyouareveryspecific
about your constraints and requirements, they will more
oftenthannotcomeupwithaunitwhichis50%largerthan
the optimum. Part of this is due to manufacturing consid-
erations. The trade-off of core loss and copper loss is
optimized by “filling the winding window” with wire, but
especially for toroids this can require more expensive
winding techniques than the widely used “single layer”
design. The lesson here is to spend time with the manufac-
turer exploring the cost trade-offs of different inductor
designs. The following guidelines may be helpful in this
regard.
Minimum Load Current in Normal Mode
There is a minimum load current requirement in normal
mode. This is caused by the necessity to “pump” the IC
switcher floating power supply capacitor during switch
“off” time. This pumping current comes from inductor
current, so load current must not be allowed to drop too
low, or the floating bias supply for the switcher will
collapse. Minimum load current is a function of input
voltage as shown in Figure 16.
1. For most buck converter applications using the
LT1070, LT1170, or LT1270 families of parts at 40kHz to
100kHz, inductor value will be in the range of 25µH to
200µH. The lower values would be used for higher output
currents and/or higher frequencies, with higher values
used for low output current, low frequency applications.
Lower inductance obviously means smaller size, but at
some point the core loss will begin to hurt, or the large
peak-to-peak inductor currents will cause high output
ripple voltage or limit available output current. The follow-
ing formula is a rough guide for picking an initial inductor
value:
80
70
60
50
40
30
20
10
0
5
10
INPUT VOLTAGE (V)
20
0
25
15
LT1432 F15
8
Figure 16. Minimum Normal Mode Load Current
L =
I
(
f
)( )
MAX
IMAX = maximum load current, including surges
f = switching frequency
Inductor Selection
Inductorselectionwouldbeeasyifmoneyandspacedidn’t
count. Unfortunately, these two factors usually count the
most, and compromises must be made. High efficiency
converters generally cannot afford the core loss found in
low cost powdered iron cores, forcing the use of more
expensivecoressuchasferrite,molypermalloy,orKoolMµ.
Actual core loss is independent of core size for a fixed
inductor value, but it is very dependent on inductance
This formula assumes that a switcher IC is selected which
has a maximum switch current of 1.5 to 2.5 times maxi-
mum load current. For a 2.5A design using the LT1271 at
60kHz, L would calculate to 53µH. This formula is very
arbitrary, so do not hesitate to modify the calculated value
by as much as 2:1 if the need arises. Keep in mind that all
the IC switchers have a peak current rating which is a
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function of duty cycle. Care must be taken to ensure that
the sum of output current plus 1/2 inductor p-p ripple
current does not exceed the switch current limit at the
highest duty cycle (lowest input voltage).
Cost may also be higher. Ferrite core material saturates
“hard,” which means that inductance collapses abruptly
when peak design current is exceeded. This may be a
problem in current limit or if peak load requirements are
not well characterized.
V
V
+ Vf
OUT
Duty Cycle (maximum)
=
3. Molypermalloy (from Magnetics, Inc.) is a very
good,lowlosscorematerialfortoroids,butitis(naturally)
ratherexpensive.AreasonablesubstituteisKoolMµ(same
manufacturer). Toroidsare very space efficient, especially
when you can convince the manufacturer to use several
layers of wire. Because they generally lack a bobbin,
mounting is more difficult. Newer designs for surface
mount are available (Coiltronics), which are nested in a
ring that does not increase the height significantly.
IN MIN
(
)
Vf = Diode forward voltage
V
V – V
(
)(
)
OUT IN
OUT
=
1/2 p-p Ripple Current
(Use minimum VIN +2V)
2 V f L
(
)( )( )
IN
A 2.5A design using an LT1271 at 60kHz, with a minimum
input voltage of 7V and a 50µH inductor, would have a
maximum duty cycle of (5 + 0.5)/7 = 79%. 1/2 p-p ripple
current would be:
Catch Diode
The catch diode carries load current only during switch
“off” time. Its average current is therefore dependent on
switch duty cycle. At high input voltages, the diode con-
ducts most of the time, and as VIN approaches VOUT, it
conducts only a small fraction of the time. The current
rating of the diode should be higher than maximum load
current for two reasons. First, conservative diode current
improves efficiency because the diode forward voltage is
lower, and second, short circuit conditions result in near
100% diode duty cycle at currents higher than full load
unless some form of foldback current limiting is used.
Schottky diodes are a must for their low forward drop and
fast switching times.
5 7 + 2 – 5
( )(
)
= 0.37A
3
–6
2 7 + 2 60E 50E
(
)
Output current plus 1/2 ripple current = 2.5 + 0.37 = 2.9A.
The switch current rating for the LT1271 is shown on the
data sheet as 4A for duty cycle below 50% and 2.67 (2–
DC) for duty cycles greater than 50%. With DC = 79%,
switch current rating would be 2.67 (2 – 0.79) = 3.23A, so
this meets the guidelines. It should be noted that if normal
running load current conditions result in switch currents
that are close to the maximum switch ratings, efficiency
will drop. Switch voltage loss at maximum switch current
rating is typically 0.7V, and this represents a significant
loss, especially at low input voltages. In most laptop
computer designs, surge currents from hard or floppy
disks require an oversized switcher, so normal running
currents are typically less than one half rated switch
current and efficiency is high except during the short
surge periods.
Maximum diode reverse voltage is equal to maximum
input voltage. However, do not over-specify the diode for
breakdown voltage. Schottky diodes are made with lighter
silicon doping as breakdown ratings increase. This gives
higher forward voltage and degrades regulator efficiency.
An MBR350 (3A, 50V) has almost 100mV higher forward
voltage than the MBR330 (3A, 30V).
2. Ferrite designs have very low core loss, so design
goals can concentrate on copper loss and preventing
saturation. The downside is that the finished unit will
almost surely be larger than a molypermalloy toroid de-
sign because of the basic topological limitations of the
ferrite/bobbinarrangement.Newerlow-profileferritecores
are even less space efficient than older configurations.
Diode current ratings are predicated on proper thermal
mounting techniques. Check the manufacturers assump-
tions carefully before assuming that a 3A diode is actually
capable of carrying 3A continuously. Pad size may have to
be larger than normal to meet the mounting requirements
for full current capability.
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in mind when adding an output filter is that if the filter
capacitor is small, it may allow large output perturbations
if large load transients occur. This effect should be care-
fully checked before finalizing any filter design. For more
details on output filters, consult Application Notes 19
and 44.
Input Supply Bypass Capacitor
The input capacitor on a step-down (buck) switching
regulator must handle switching currents with a peak-to-
peak amplitude at least equal to the output current. The
RMS value of capacitor current is approximately equal to:
1/2
]
I
V
V – V
(
)
OUT
[
OUT IN OUT
Output Capacitor
I
=
RMS
V
IN
To avoid overheating, the output capacitor must be large
enough to handle the ripple current generated by the main
inductor. It must also have low enough effective series
resistance (ESR) to meet output ripple voltage require-
ments. RMS ripple current in the output capacitor is given
by:
This formula has a maximum at VIN = 2VOUT, where IRMS
is equal to IOUT/2. This simple worst case condition is
commonly used for design because even significant de-
viations from VIN/2 do not offer much relief. A 2A output
(transient loads can be ignored if they last less than 30
seconds) therefore requires an input capacitor with a 1A
ripple current rating. Don’t cheat, and read the output
capacitor section for details on ripple current! The input
capacitormaywellbethelargestcomponentintheswitch-
ing regulator. Spend time playing with aspect ratios of
various capacitor families and don’t hesitate to parallel
several units to achieve a low profile.
V
V – V
OUT
(
)
OUT IN
I
=
RIPPLE(RMS)
3.5V (f)(L)
IN
(use maximum VIN)
For VIN = 15V, f = 60kHz, L = 50µH,
5(15 – 5)
I
=
RIPPLE(RMS)
3
–6
Output Voltage Ripple
3.5 15 60E 50E
( )
Output voltage ripple is determined by the main inductor
value, switching frequency, input voltage, and the ESR
(effective series resistance) of the output capacitor. The
following formula assumes a load current high enough to
establish continuous current in the inductor.
= 0.32A
RMS
Ripplecurrentratingsarespecifiedoncapacitorsintended
for switching applications, but the number is subject to
muchmanipulation. Thehighfrequencynumberisgreater
than the low frequency value, and theoretically one can
multiply the ripple number by significant amounts at
temperaturesbelowthetypical85°Cor105°Cratingpoint.
The problem is that the ripple ratings are already unreal-
istically high at the rated temperature because they are
typically based on a 2000 hour life. I assume this is an
unacceptablelifetimenumber, sotherippleratingmustbe
reduced to extend life. The net result of all this fiddling
with the numbers is generally a headache, but it is prob-
ably conservative to use the stated high frequency rating
at temperatures below 60°C for a 105°C capacitor, and
assume that the unit will last at least 50,000 hours.
Remember to factor in actual operating time at elevated
temperatures. Laptop computers, for instance, might be
expected to operate no more than four hours a day on
Output Ripple Voltage = Vp-p
V
V – V
ESR
)(
(
)
V
p-p
OUT IN
OUT
=
V (f)(L)
IN
With VIN = 12V, ESR = 0.05Ω, f = 60kHz, and L = 50µH
5(12 – 5)(0.05)
V
=
= 48.6mV
p-p
p-p
3
–6
12 60E 50E
( )
If low output ripple voltage is a requirement, larger output
capacitors and/or inductors may not be the answer. An
output filter can be added at modest cost which will
attenuate ripple much more space-effectively than an
oversized output capacitor or inductor. The thing to keep
18
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board space may not increase prohibitively. See the dis-
cussionofwaveformsforloadtransientresponseimplica-
tions when adding a filter.
average, so a ten year life is only 15,000 hours. The
manufacturer shouldbe consulted for afinalblessing. See
Application Note 46 for specific formulas for calculating
the life time or allowed ripple current in capacitors.
If modest reductions in output ripple are required, one can
increase the size of the main inductor and/or the output
capacitor. Buck converters are easier than other types
because the main inductor acts as a filter element. The
square wave voltage is converted to a triangular current
before being fed to the output capacitor. Actually, at
switching frequencies, the output capacitor is resistive
and output ripple voltage is determined not by the capaci-
tor value in µF, but rather by the capacitor effective series
resistance(ESR). Thisparameterisdeterminedbycapaci-
tor volume within any given family, so to get ESR down,
onemuststillusea“bigger”capacitor. Theproblemisthat
often the main inductor/capacitor becomes physically too
large if low output ripple is needed. Inverters, such as the
positive to negative converter, tend to have much higher
output ripple voltage because the main inductor is not a
filter element – it simply acts as an energy storage device
for shuttling essentially square wave currents from input
tooutput. Unlikethebuckconverter, thesecurrentscanbe
much higher in amplitude than the output current.
The reason for all this attention to ripple rating is that
everyone is in a size squeeze, and the temptation is to use
the smallest possible components. Do not cheat here
folks, or you may be faced with costly field failures.
ESR on the output capacitor determines output voltage
ripple, so this is also of much concern. Mother Nature has
decreed that for a given capacitor technology, ESR is a
direct function of the volume of the capacitor. In other
words,ifyouwantlowESRyoumustconsumespace.This
is quickly confirmed by scanning the ESR numbers for a
wide range of capacitor values and voltage ratings within
a given family of capacitors. It is immediately obvious that
can size determines ESR, not capacitance, or voltage
rating. Theonlywayto cheaton this limitation is tofindthe
bestfamilyofcapacitors.ManufacturerssuchasNichicon,
Chemicon, and Sprague should be checked. Sanyo makes
a very low ESR capacitor type know as OSCON, utilizing a
semiconductordielectric. Itsmajordisadvantageissome-
what higher price, and a tendency to make regulator
feedbackloopsunstablebecauseofitsextremelylowESR.
Most switching regulator loops depend to some extent on
the output capacitor ESR for a phase lead!
An output filter of very modest size can reduce normal
mode output ripple voltage by a factor of ten or more. The
formulaforfilterattenuationinbuckconvertersandinvert-
ers is shown below.
Output Filters
ESR
8 L f
Attenuation =
(BUCK CONVERTER)
Outputripplevoltageattheswitchingfrequencyisafactof
life with switching regulators. Everyone knows that this
ripple must be held below some level to guarantee that it
doesnotaffectsystemperformance. Thequestionis, what
is that level? For sensitive analog systems with wide
bandwidths, supply ripple may have to be a 1mV or less.
Digital systems can often tolerate 400mVp-p ripple with no
effect on performance. In most of these digital applica-
tions of the LT1432 as a buck converter, an output filter is
not needed because output ripple is normally in the 25mV
to 100mVp-p range without a filter. Note that burst mode
ripple is at low frequencies where small output filters are
not effective. The decision to add an output filter does
allow the main filter capacitor to get smaller, so the overall
ESR
(INVERTER)
(The factor “4” is an
approximation
assuming worst case
duty cycle of 50%)
(
)
Attenuation =
4 L f
A 10µH, 100µF (ESR = 0.4Ω) filter on a buck converter
using a 60kHz LT1271 will give an attenuation of:
0.4
= 0.083
–6
3
8 10E
60E
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100mVoutputrippleonthemaincapacitorwillbereduced
to (0.083)(100) = 8.3mV at the output of the filter.
boards,theirmaximumratedcurrentmustalsobeconsid-
ered. For currents greater than 1A, multiple vias may have
to be used.
Layout Considerations
4. The catch diode has large square wave currents
flowing in it. Connect the anode directly to the ground
plane and the cathode directly to the IC ground pin.
Although buck converters are fairly tolerant with regard to
layout issues, there are still several important things to
keep in mind. Most of these revolve around spikes created
by switching high currents at high speeds. If 3A of current
is switched in 30ns, the rate of change of current is 10E8
A/S. Voltage generated across wires will be equal to this
rate multiplied by the approximate 20nH per inch of wire.
This calculates to 2V per inch of wire or trace!! Needless
to say, connections should be kept short if the circuitry
connected to these lines is sensitive to narrow spikes.
5. The ground pin of the LT1432 is the reference point
for output voltage. It should be routed separately to power
ground as near to the load as is reasonable.
Transient Response
Load transient response may be important in portable
applications where parts of the system are switched on
and off to save power. There are two types of problems
that differ by time scale. The first occurs very rapidly and
is caused by the surge current created in charging the
supply bypass capacitors on the switched load. This can
be a very serious problem if large (>0.1µF) capacitors
mustbecharged. Noregulatorcanrespondfastenoughto
handlethesurgeiftheloadswitchon-resistanceislowand
itisdrivenquickly.Thesolutionhereistolimittherisetime
of the switch drive so that the load rise time is limited to
approximately 25 × CLOAD. A 1µF load capacitor would
require a 25µs load rise time, etc. This limits surge to
about 200mA. This time frame is still too quick for a
switching regulator to adjust to, but the surge is limited to
a low enough value that the output capacitor will attenuate
the surge voltage to an acceptable level.
1. The input bypass capacitor must be kept as close to
the switcher IC as possible, and its ground return must go
directly to the ground plane with no other component
grounds tied to it. The output capacitor should also
connect directly to the ground plane.
2. The frequency compensation components shown in
Figure 1 (R1 + C4, and C5) and the feedback pin bypass
capacitor(C6)areshownconnectedtothefloatingground
pin of the IC switcher. This ground pin is also the high
current path for the switch. To avoid differential spikes
being coupled into the VC and FB pins, these components
must tie together and then be connected through a direct
trace to the IC switcher ground pin. No other components
should be connected anywhere on this trace and the trace
area should be minimized. A separate wide trace must be
used to connect the IC ground pin to the catch diode and
inductor. Smaller traces can be used to connect the
floating supply capacitor (C3) and the diode pin of the
LT1432tothewidetracereasonablyclosetotheICground
pin.
A second problem is the change in DC load current.
Switching regulators take many switching cycles to re-
spond to sudden output load changes. During this time,
the output shifts by an amount equal to ∆load (ESR + t/C),
where ESR is the series resistance of the output capacitor,
t is the time for the regulator to shift output current, and C
is the output capacitor value. For example, if the load
change is 0.5A, ESR is 0.1Ω, t is 30µs, and C = 390µF, the
shift in output voltage would be:
3. Traces which carry high current must be sized
correctly. To limit temperature rise to 20°C, using 1oz
copper, the trace width must be 20 mils for each ampere
of current. 1/2oz copper requires 30 mils/A. These high
current paths include the IC switcher ground pin and
switch pin, the inductor, the catch diode, the current limit
sense resistor, and the input bypass capacitor. If vias are
used to connect these components on multiple layer
30µs
390µF
∆V
= 0.5A 0.1Ω +
= 0.088V
OUT
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Figure17showstheeffectofa500mAtransientload(0.3A
to 0.8A) on the LT1432, both with and without an output
filter. The top trace with no filter shows about a 60mV
deviation with a settling time of 300µs. Astute switching
regulatordesignersmaynoticethelackofswitchingripple
in this trace. To make a clean display the actual trace was
fed through a one pole filter with 16µs time constant to
eliminate most of the switching ripple. This had very little
effectontheshapeoramplitudeoftheresponsewaveform
(you’ll have to trust me on this one). In the middle trace,
an output filter of 10µH and 200µF was added to the
regulator to achieve very low output ripple. The load
transient response is obviously degraded because the
second filter capacitor, following normal design practice,
is somewhat smaller than the main output capacitor, and
therefore also has higher ESR. Note the slight ringing
caused by the “Q” of the output filter. Calculated ringing
Mode Pin Drive
ThemodepindefinesoperatingconditionsfortheLT1432.
A low state programs the IC to operate in “normal” mode
as a constant frequency, current mode, buck converter.
Floating the pin converts the internal error amplifier to a
comparator which puts the LT1432 into a low-power
“burst” mode. In this mode, the pin assumes an open
circuit voltage of approximately 1V. To ensure stable
operation, current into or out of the pin must be limited to
2µA. Ifthepinisroutednearanyswitchingorlogicsignals
it should be bypassed with a 200pF capacitor to avoid
pickup.
Driving the mode pin high causes the LT1432 to go into
complete shutdown. An internal resistor limits mode pin
current to about 15µA at 5V. A 7V zener diode is also in
parallel with the pin, so input voltages higher than 6.5V
must be externally limited with a resistor. The current/
voltage characteristics of the mode pin are shown in
Typical Performance Characteristics. Note that the drive
signal must sink about 30µA when pulling the mode pin to
its worst case low threshold of 0.6V. This should not be a
problemforanystandardopendrainorthree-stateoutput.
frequency is 1/(2π√LC) = 3.4kHz. Also note the small step
in DC level between the two load conditions on the filtered
output. To maintain good loop stability, the added filter is
left “outside” the feedback loop. Therefore, the DC resis-
tance of the 10µH inductor will add to load regulation. The
10mV step implies a resistance of 10mV/0.5A = 0.02Ω.
The message in all this is to be careful when adding output
filters if transient load response or load regulation is
critical. The second filter capacitor may have to be as large
as the main filter capacitor.
If all three states are desired and a three-state drive is not
available, the circuit shown in Figure 18 can be used. Two
separate logic inputs are used. Both low will allow the
mode pin to float for burst mode. “A” high, “B” low will
generate shutdown, and “B” high, “A” low forces normal
mode operation. Both high will also force normal mode
operation, but this is not an intended state and R1 is
included to limit overload of “A” if this occurs. C1 is
suggested if the mode pin line can pick up capacitively
coupled stray switching or logic signals.
100mV/DIV
100mV/DIV
D1
R1
1N914
10k
TO
A
MODE PIN
VN2222L
C1
200pF
B
0.5A/DIV
LT1432 F18
0.5ms/DIV
Figure 18. Two Input Mode Drive
Figure 17
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Internal Restart Sequence
a regulated output voltage of minus 5V, the auxiliary
winding output would have to be about minus 7V. Maxi-
mum output current from the 7V output would be 1.25W/
7V = 178mA. Note that the power restriction is the total for
all auxiliary outputs.
At very light load currents (>10mA), coupled with low
inputvoltages(<8.5V), itispossibleforthebasicarchitec-
ture used by the LT1432 to assume a stable output state
of less than 5V. To avoid this possibility, the LT1432 has
an internal timer which applies a temporary 20mA load to
the output if the output is below its regulated value for
more than 1.8ms. This action is normally transparent to
the user.
The formula to calculate turns ratio for the auxiliary
windings versus main winding is simple:
N
V
+ V = 2V + V
(
)
MAIN
[
AUX DO DA
]
N
=
AUX
5V + V
D
Auxiliary Outputs – “Free” Extra Voltages
NMAIN = Number of turns on main inductor winding
NAUX = Number of turns on auxiliary winding
VDA = Auxiliary diode forward voltage
Semi-regulated secondary outputs may be added to buck
converters by adding additional windings to the main
inductor. These outputs will have a typical regulation of 5
to 10%, but have one very important limitation. The total
output power of the auxiliary windings is limited by the
output power of the main output. If this limit is exceeded,
the auxiliary winding voltages will begin to collapse,
although the main 5V output is unaffected by collapse of
the secondary. The auxiliary power available is also a
function of input voltage. At higher input voltages signifi-
cantly more power is available.
VD = Main 5V catch diode forward voltage
VDO = Allowance for regulation of auxiliary winding and
dropout voltage of low-dropout linear regulator used on
auxiliary winding. Set equal to zero if no regulator is used.
2.0
1.5
1.0
0.5
0
Figure 19 shows the ratio of maximum auxiliary power to
main output power, versus input voltage. The auxiliary
output was loaded until its output voltage dropped 10%.
For applications which push the limit of theoretically
available current, care should be used in winding the
inductor. The effects of leakage inductance and series
resistance are magnified at low input voltage where aux-
iliary winding currents are many times DC load current.
Also, be aware that output voltage ripple on the 5V main
outputcanincreasesignificantlywhentheauxiliaryoutput
is heavily loaded. The inductor is acting partially like a
transformer, so the AC current delivered to the 5V output
capacitorincreasesinamplitudeandshiftsfromatri-wave
to a trapezoid with much faster edges.
10
15
0
20
5
INPUT VOLTAGE (V)
LT1432 F19
Figure 19. Auxiliary Power vs 5V Power
It is not necessary to use a linear regulator on the auxiliary
winding if 5 to 10% regulation is adequate. Line regulation
will be fairly good, but variations in auxiliary voltage will
occur with load changes on either the auxiliary winding or
the 5V output. For relatively constant loads, regulation will
be significantly better.
A typical example would be a +5V buck converter with a
minimum load of 500mA. Output power is 5V × 0.5A =
2.5W. Maximum power from the auxiliary windings would
be1.25Wforinputvoltagesof9Vandabove. Ifweassume
a low dropout linear regulator on the auxiliary output, with
22
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+
a load to the 5V output to bootstrap itself. Figure 21 shows
maximum current out of a 14V auxiliary (used to power a
12V linear regulator) connected in this fashion. The aux-
iliarywindingvoltageisactually9V.Notethatforlighter5V
loads, there is an inflection point in the curves at about
11V. Thatisbecausetheoreticallythebootstrappingeffect
should allow one to draw unlimited power from the
auxiliary winding when duty cycle exceeds 50%. The
actual available current above 50% duty cycle is limited by
parasitic losses. At high 5V loads, the inflection disap-
pears for the same reason. The curves asymptotically
approach 1 amp at high input voltage because the criteria
used to generate the curves was a drop in auxiliary output
voltage to 13.5V, and again parasitic resistance limits
output current.
AUXILIARY
WINDING
+
AUXILIARY
OUTPUT
–
MAIN
WINDING
+5V OUTPUT
L1
+
D1
POSITIVE
POSITIVE
REGULATOR
REGULATED
+
OUTPUT
+
+
POSITIVE
REGULATOR
+
NEGATIVE
REGULATED
OUTPUT
Auxiliary windings deliver current in triangular or quasi-
square waves only during switch off time. Therefore the
amplitude of these pulses will be somewhat higher than
the DC auxiliary load current, especially at low input
voltage. This means that in the “stacked” connection,
ripple voltage on the 5V output will increase with auxiliary
load current.
LT1432 F20
Figure 20
Figure 20 shows how to connect the auxiliary windings.
Dots indicate winding polarity. Pay attention here -- his-
tory shows that with a 50% chance of connecting up the
auxiliary correctly when you ignore the dots, in actual
practice you will be wrong 90% of the time.
1.0
14V LOAD INCREASED
UNTIL V = 13.5V
0.8
Thefloatingoutputcanhaveeitherendgrounded,depend-
ing on the need for a positive or negative output. Also
shown are the connections for both positive and negative
outputs using a linear regulator. Note that the two circuits
are identical! The floating auxiliary winding allows the use
of a positive low-dropout regulator for negative outputs.
Thesepositiveregulatorsaremorereadilyavailable, espe-
cially at lower current levels.
I(+5) = 1A
I(+5) = 400mA
0.6
0.4
I(+5) = 200mA
I(+5) = 200mA
0.2
I(+5) = 50mA
0
8
12 14
16
18
20
10
INPUT VOLTAGE (V)
There is a way to “cheat” somewhat on auxiliary power for
positive outputs higher than the 5V main output. The
auxiliary winding return can be connected to the 5V
output. This reduces the winding voltage so that more
current is available, and at the same time it actually adds
LT1432 F21
Figure 21
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ThecircuitinFigure22willconvertavariablepositiveinput
voltage to a regulated –5V output. By selecting different
members of the LT1070 family, this basic design can
provide up to 6A output current at high input voltages, and
up to 3A with a five volt input supply. As shown using an
LT1271, maximumloadcurrenthasbeenreducedto1Aby
utilizing the current limit circuit in the LT1432. Unlike a
positive buck converter, it is not possible to sense output
currentdirectly.Instead,switch/inductorcurrentissensed.
This would normally result in a DC output current limit
valuethatchangesconsiderablywithinputvoltage,butthe
additionofR2andR3alterspeakcurrentlimitasafunction
of input voltage to correct for this effect. Maximum load
current and short circuit current are shown as a function
of input voltage in Figure 23. A 0.02Ω sense resistor was
used, so other values of current limit can be scaled from
this value.
This circuit uses the same basic connections between the
LT1432 and the LT1271 as the buck converter. The differ-
enceisinthewaypowerflowsinthecatchdiode, inductor,
and switch. In a buck converter, current flows simulta-
neously in the switch, inductor, and output. This makes
maximum output current approximately equal to maxi-
mum switch current. In inverting designs, current deliv-
ered to the output is zero during switch on-time. The
switchallowscurrenttoflowdirectlyfromtheinputsupply
through the inductor to ground. At switch turn-off, induc-
INPUT
4.5V – 25V
V
V
IN
SW
+
C1
330µF
35V
LT1271
FB
GND
V
C
C6
0.02µF
C5
0.03µF
D2
1N4148
+
C3
22µF
16V
R1
680Ω
C4
0.047µF
L1
50µH
R4
0.02Ω
R2
100Ω
+
DIODE
V
V
C
+
C2
V
V
LIM
IN
1000µF
16V
LT1432
D1
MBR330p
V
OUT
MODE
GND
R3
100k
–5V
OUTPUT
×
×
10µH
3A
OPTIONAL
OUTPUT
FILTER
100µF
16V
+
LT1432 F22
Figure 22. Positive-to-Negative Converter
24
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3.0
T
= 25°C
LIM
J
R
= 0.02Ω
2.5
2.0
1.5
V
OUT
= 0 (SHORT CIRCUIT)
ISWITCH
1A/DIV
∆V
OUT
= 1%
0
1.0
2
IDIODE
1A/DIV
1
0
5
10
15
20
25
INPUT VOLTAGE (V)
0
LT1432 F23
5µs/DIV
Figure 23. Positive-to-Negative Converter
Output Current
Figure 24. Positive-to-Negative Converter
Switch and Diode Current
tor current is diverted through the catch diode to the
output. Figure 24 shows switch current (1A/DIV) with the
upper waveform, and catch diode current (which is deliv-
eredtotheoutput)inthelowerwaveform, witha+5Vinput
and1Aload.Notethatswitch,inductor,anddiodecurrents
aremuchhigherthanoutputcurrentasrequiredbythefact
that current is delivered to the output during only part of
a switch cycle. An approximate formula for peak switch
current required in an inverting design is:
5 + 0.4
I
= 1 1+
SW PEAK
(
)
4.7 + 5
(
)
4.7 – 1 0.25
(
)
4.7
4.75 5
( )
+
–6
3
2 50E
60E 4.75 + 5
(
)
= 2.29 + 0.4 = 2.69A
The first term (2.29A) represents the minimum switch
current required if the inductor were infinitely large. A
finite inductor value requires additional switch current.
The 0.4A represents one-half the peak-to-peak inductor
ripple current. The end result is that peak switch current is
almost three times output load current. This multiplier
drops rapidly at higher input voltages, so worst case is
calculated at lower input voltage.
V
+ V
F
OUT
I
= I
1+
SW PEAK
OUT
(
)
V + V
(
)
IN
OUT
V – I
R
(
)
IN OUT SW
V
IN
V V
(
)
IN OUT
+
2 L f V + V
( )( )(
)
IN
OUT
VF = Forward voltage of catch diode
RSW = Switch on-resistance
L = Inductor value
Figure 25 shows the efficiency of this converter. At higher
input voltages and modest output currents efficiency
hovers around 85%, quite good for a 5V output inverter.
Lowinputvoltagereducesefficiencybecauseofincreased
currents in the switch, catch diode, and inductor. High
input voltage and low output current also show lower
efficiency due to quiescent currents in the ICs. Note that
the efficiency is actually significantly improved in this
regard over a more conventional design because the
f = Switching frequency
If VIN is 4.7V (minimum),
VF = 0.4V, RSW = 0.25Ω,
L = 50µH, f = 60kHz, and IOUT = 1A;
25
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LT1271operatesfromaconstant5Vsupplyvoltagerather
than the high input voltage.
design. It will then select the LT1070 family of ICs which
normally are not used in positive to negative converters.
Efficiency calculations will be somewhat in error at higher
input voltages because the program assumes full input
voltage across the IC. Later versions of SwitcherCAD will
have a special section for this particular design.
Output voltage ripple in an inverter can be much higher
than a buck converter because current is delivered to the
output capacitor in high amplitude square waves rather
than a DC level with superimposed tri-wave. C2 is there-
fore somewhat larger than in a buck design. Also C2 must
be rated to handle the large RMS current pulses fed into it.
This RMS current is approximately equal to:
100
90
V
= 20V
IN
V
V
OUT
V
IN
= 10V
80
70
60
50
I
OUT
V
= 5V
IN
IN
For 1A output current, with 5V input, this computes to
1ARMS in the output capacitor. A small additional output
filter would reduce output ripple voltage, but it does not
change the current rating requirement for the main output
capacitor. The reader is referred to a switching regulator
CAD program (SwitcherCAD) supplied by LTC for further
insight into converters. It is suggested that the reader fool
theprogrambyaskingforanegativeinput, positiveoutput
0
0.2
0.4
0.6
0.8
1.0
OUTPUT CURRENT (A)
LT1432 F25
Figure 25. Positive-to-Negative Converter Efficiency
26
LT1432
W
W
SCHE ATIC DIAGRA
V
IN
1k
1k
10k
D1
Q2
100k*
Q40
Q1
+
V
V
C
Q3
Q4
Q39
Q38
D3
Q5
Q6
1.5k
Q37
V
LIM
DIODE
Q36
D5
Q35
D2
V
OUT
Q34
Q8
9k
Q9
Q33
Q7
Q20
2k
TO Q31
11k
7k
Q31 Q32
Q19
Q18
MODE
D4
1k
Q10
100k
Q23
Q28
Q22
Q30
Q16
Q14
C2
Q25
Q21
C
A
20pF
4k
Q12
Q29
50pF
200k*
30k*
Q11
3k
1k
Q27
TO
Q9
Q17
40k*
150Ω
Q15
Q13
Q26
Q24
150k*
600Ω
10k
2k
720Ω
10k
600Ω
600Ω
30k*
C
B
GND
*
LT1432 F26
INDICATES PINCH RESISTOR
Information furnished by Linear Technology Corporation is believed to be accurate and reliable.
However, no responsibility is assumed for its use. Linear Technology Corporation makes no represen-
tation that the interconnection of circuits as described herein will not infringe on existing patent rights.
27
LT1432
U
PACKAGE DESCRIPTIO
N8 Package
8-Lead Plastic DIP
0.400
(10.160)
MAX
0.300 – 0.320
(7.620 – 8.128)
0.130 ± 0.005
(3.302 ± 0.127)
0.045 – 0.065
(1.143 – 1.651)
8
1
7
6
5
4
0.065
(1.651)
TYP
0.250 ± 0.010
(6.350 ± 0.254)
0.009 - 0.015
(0.229 - 0.381)
0.125
0.020
(3.175)
MIN
+0.025
–0.015
(0.508)
MIN
0.045 ± 0.015
(1.143 ± 0.381)
3
2
0.325
+0.635
8.255
(
)
–0.381
0.100 ± 0.010
(2.540 ± 0.254)
0.018 ± 0.003
(0.457 ± 0.076)
N8 1291
TJMAX
θJA
100°C 150°C/W
S8 Package
8-Lead Small Outline
0.189 – 0.197
(4.801 – 5.004)
0.010 – 0.020
(0.254 – 0.508)
7
5
8
6
× 45°
0.053 – 0.069
(1.346 – 1.753)
0.004 – 0.010
(0.102 – 0.254)
0.008 – 0.010
(0.203 – 0.254)
0.228 – 0.244
(5.791 – 6.198)
0.150 – 0.157
(3.810 – 3.988)
0.016 – 0.050
0.406 – 1.270
0.050
(1.270)
BSC
0.014 – 0.019
(0.356 – 0.483)
0°– 8° TYP
1
3
4
2
S8 1291
TJMAX
θJA
100°C 170°C/W
LT/GP 0392 10K REV 0
Linear Technology Corporation
1630 McCarthy Blvd., Milpitas, CA 95035-7487
28
LINEAR TECHNOLOGY CORPORATION 1992
(408) 432-1900 FAX: (408) 434-0507 TELEX: 499-3977
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