LT6110 [Linear]
Cable/Wire Drop Compensator; 电缆/电线掉落补偿
| 型号： | LT6110 |
| 厂家： | 
Linear |
| 描述： | Cable/Wire Drop Compensator |
| 文件： | 总38页 (文件大小：523K) |
| 中文： | 中文翻译 | 下载： | 下载PDF数据表文档文件 |
LT6110
Cable/Wire Drop
Compensator
FeaTures
DescripTion
The LT®6110 is a precision high side current sense with a
current mode output, designed for controlling the output
voltageofanadjustablepowersupplyorvoltageregulator.
This can be used to compensate for drops in voltage at
a remote load due to resistance in a wire, trace or cable.
n
Improve Voltage Regulation to a Remote Load by 10×
n
Ideal for Resistor-Adjustable Voltage Regulators
n
Gain Configurable with a Single Resistor
High Side Current Sensing:
n
Integrated 20mΩ Sense Resistor for Up to 3A
Ability to Use an External Sense Resistor
The LT6110 monitors load current via a series-connected
internal or external sense resistor. Two current mode out-
puts, one sinking and one sourcing, are provided that are
proportional to the load current. This allows the LT6110 to
adjust the output voltage of a wide variety of regulators.
Either output may be used to monitor the load current.
n
300µV Maximum Input Offset Voltage
n
Output Current Accuracy of 1% Maximum
n
30µA Maximum Supply Current
n
2V to 50V Supply Range
n
Fully Specified from –40°C to 125°C
n
Available in Low Profile (1mm) ThinSOT™ and
Low DC offset allows for the use of a small sense resis-
tor, as well as precise control of small variations in wire
voltage drop.
L, LT, LTC, LTM, Linear Technology, the Linear logo and µModule are registered trademarks
and ThinSOT is a trademark of Linear Technology Corporation. All other trademarks are the
property of their respective owners.
(2mm × 2mm) DFN Packages
applicaTions
n
Automotive and Industrial Power Distribution
n
USB Power
n
DC/DC Converters
n
Plug-In DC Adapters
n
Power over Ethernet
Typical applicaTion
R
WIRE
I
V
LOAD
V
LOAD
REG
V
IN
OUT
IN
REGULATOR
R
IN
REMOTE
LOAD
FB
+
+IN
V
RS –IN
5V
5V
GND
V
V
UNCOMPENSATED
COMPENSATED
LOAD
+
–
LT6110
IOUT
6110 TA01
LOAD
2A
1A
I
LOAD
–
V
IMON
200µs/DIV
6110f
1
For more information www.linear.com/LT6110
LT6110
absoluTe MaxiMuM raTings (Note 1)
+
–
Total Supply Voltage (V to V ).................................55V
Specified Temperature Range (Note 3)
–
+
+IN, –IN, IOUT, IMON to V Voltage ............................ V
+IN, -IN, IOUT, IMON Current.................................10mA
IOUT to IMON Voltage....................................36V, –0.6V
LT6110I................................................–40°C to 85°C
LT6110H............................................. –40°C to 125°C
Junction Temperature .......................................... 150°C
Storage Temperature Range .................. –65°C to 150°C
Lead Temperature (Soldering, 10 sec)
+
V , +IN to IOUT Voltage.............................................36V
+
Differential Input Voltage ............................................ V
R
Current (Note 2)
TS8...................................................................300°C
SENSE
Continuous .............................................................3A
Transient (<0.1 Second)..........................................5A
pin conFiguraTion
TOP VIEW
TOP VIEW
1
2
3
4
8
7
6
5
NC*
+IN
NC* 1
IOUT 2
8 +IN
7 V
6 RS
5 –IN
+
V
IOUT
IMON
9
+
–
V
RS
IMON 3
–
–
–IN
V
V
4
TS8 PACKAGE
8-LEAD PLASTIC TSOT-23
DC PACKAGE
8-LEAD (2mm × 2mm) PLASTIC DFN
T
= 150°C, θ = 195°C/W
JA
JMAX
T
= 150°C, θ = 80.6°C/W
JMAX
JA
*NC PIN NOT INTERNALLY CONNECTED
–
EXPOSED PAD (PIN 9) IS V , MUST BE SOLDERED TO PCB
*NC PIN NOT INTERNALLY CONNECTED
orDer inForMaTion
Lead Free Finish
TAPE AND REEL (MINI)
LT6110ITS8#TRMPBF
LT6110HTS8#TRMPBF
LT6110IDC#TRMPBF
LT6110HDC#TRMPBF
TAPE AND REEL
PART MARKING*
PACKAGE DESCRIPTION
SPECIFIED TEMPERATURE RANGE
LT6110ITS8#TRPBF
LTGCQ
8-Lead Plastic TSOT-23
–40°C to 85°C
–40°C to 125°C
–40°C to 85°C
–40°C to 125°C
LT6110HTS8#TRPBF LTGCQ
8-Lead Plastic TSOT-23
LT6110IDC#TRPBF
LT6110HDC#TRPBF
LGCP
LGCP
8-Lead (2mm × 2mm) Plastic DFN
8-Lead (2mm × 2mm) Plastic DFN
TRM = 500 pieces. *Temperature grades are identified by a label on the shipping container.
Consult LTC Marketing for parts specified with wider operating temperature ranges. *The temperature grade is identified by a label on the shipping container.
Consult LTC Marketing for information on lead based finish parts.
For more information on lead free part marking, go to: http://www.linear.com/leadfree/
For more information on tape and reel specifications, go to: http://www.linear.com/tapeandreel/
6110f
2
For more information www.linear.com/LT6110
LT6110
elecTrical characTerisTics The l denotes the specifications which apply over the full specified
temperature range, otherwise specifications are at TA = 25°C. V+ = 5V, V– = VIMON = 0V, I+IN = 100µA, VIOUT – VIMON = 1.2V, unless
otherwise noted.
SYMBOL
PARAMETER
CONDITIONS
MIN
TYP
MAX
UNITS
+
l
V
Supply Range
2.0
50
V
V
Amplifier Input Offset Voltage
100
300
400
500
550
µV
µV
µV
µV
OS
l
l
l
0°C ≤ T ≤ 85°C (Note 5)
A
85°C ≤ T ≤ 125°C (Note 5)
A
–40°C ≤ T ≤ 0°C (Note 5)
A
Amplifier Input Offset Voltage Change
I
= 10µA to 1mA
0.15
0.3
0.5
1.5
mV/mA
mV/mA
mV/mA
∆V /∆I
OS +IN
+IN
l
l
with I
0°C ≤ T ≤ 85°C (Note 6)
A
+IN
l
l
l
l
Amplifier Input Offset Voltage Change
with IOUT Voltage
V
V
= 0.4V to 5V
= 0V to 1V
0.005
0.3
0.02
mV/V
mV/V
µV/°C
∆V /∆V
IOUT
OS
IOUT
Amplifier Input Offset Voltage Change
with IMON Voltage
1
∆V /∆V
IMON
OS
IMON
Amplifier Input Offset Voltage Drift
Amplifer Input Bias Current (–IN)
1
∆V /∆T
OS
+
I
V = 5V
35
70
200
nA
nA
B
+
I
Amplifier Input Offset Current
Power Supply Rejection Ratio
V = 5V
1
nA
OS
+
l
l
PSRR
V = 2.0V to 36V
96
90
110
100
dB
dB
+
V = 36V to 50V
IOUT Current Error (Note 4)
I
= 10µA
0.6
0.5
0.75
1.5
1.5
1.7
1.6
2
%
%
%
+IN
l
l
(Referred to I
)
0°C ≤ T ≤ 85°C, (Note 6)
A
+IN
2.5
I
= 100µA
1
1.5
2.3
%
%
%
+IN
l
l
0°C ≤ T ≤ 85°C, (Note 6)
A
I
= 1mA
2.5
3
4
%
%
%
+IN
l
l
0°C ≤ T ≤ 85°C, (Note 6)
A
IMON Current Error (Note 4)
(Referred to I
I
= 10µA
3
3.5
5
%
%
%
+IN
l
l
)
+IN
0°C ≤ T ≤ 85°C, (Note 6)
A
I
= 100µA
3
3.5
5
%
%
%
+IN
l
l
0°C ≤ T ≤ 85°C, (Note 6)
A
I
= 1mA
4
5
6
%
%
%
+IN
l
l
0°C ≤ T ≤ 85°C, (Note 6)
A
l
l
∆I
∆I
/V
IOUT Current Error Change with
IOUT Voltage (Note 4)
V
V
= 0.4V to 3.5V
= 0.4V to 5V
0.2
0.4
%/V
%/V
IOUT IOUT
IOUT
IOUT
l
l
l
l
/V
IMON Current Error Change with
IMON Voltage (Note 4)
V
= 0V to 3.1V, V = 5V
IOUT
0.2
%/V
IMON IMON
IMON
+IN Current Range
Supply Current
0.01
1
mA
+
I
V = 5V, I = 0µA
16
30
30
50
µA
µA
S
+IN
+
V = 50V, I = 0µA, V
= 25V
50
100
µA
µA
+IN
IOUT
R
R
Resistance
(Note 2)
0.0165
0.02
180
2
0.0225
Ω
kHz
µs
SENSE
SENSE
BW
Signal Bandwidth (–3dB)
Rise Time
I
= 100µA, R
= 1k
+IN
IOUT
t
r
6110f
3
For more information www.linear.com/LT6110
LT6110
elecTrical characTerisTics
Note 1: Stresses beyond those listed under Absolute Maximum Ratings
may cause permanent damage to the device. Exposure to any Absolute
Maximum Rating condition for extended periods may affect device
reliability and lifetime. In addition to the Absolute Maximum Ratings, the
output current and supply current must be limited to insure that the power
dissipation in the LT6110 does not allow the die temperature to exceed
150°C. See the Applications Information section Power Dissipation for
further information.
Note 4: Specified error is for the LT6110 output current mirror and does
not include errors due to V or resistor tolerances. Since most systems
OS
will not have 100% correction, the total system error can be compensated
to less than the specified error with proper design. See the Applications
Information section for details.
Note 5: Measurement errors limit automatic testing accuracy. These
measurements are guaranteed by design correlation, characterization and
testing to wider limits.
Note 2: R
resistance and maximum R
currents are guaranteed
SENSE
SENSE
Note 6: The 0°C ≤ T ≤ 85°C temperature range is guaranteed by
A
by characterization and process controls.
characterization and correlation to testing at–40°C, 25°C and 85°C.
Note 3: The LT6110I is guaranteed to meet specified performance from
–40°C to 85°C. The LT6110H is guaranteed to meet specified performance
from –40°C to 125°C.
Typical perForMance characTerisTics
VOS Distribution
VOS vs Supply Voltage
VOS vs Supply Voltage
200
175
150
125
100
75
400
300
200
100
0
400
300
200
100
0
+
V
V
V
= 5V
800 UNITS
I
V
V
= 100µA
= 0.4V
IMON
I
V
V
= 100µA
= 25V
IOUT
+IN
IOUT
+IN
= 1.2V
IOUT
IMON
= 0V
= 0V
= 0V
IMON
I
= 100µA
+IN
T
= 125°C
A
T
= 85°C
A
T
= 85°C
A
T
= 125°C
A
T
= 0°C
A
T
= 0°C
A
T
= –40°C, –55°C
A
T
= 25°C
A
50
T
= –40°C, –55°C
A
T
= 25°C
A
–100
–200
–100
–200
25
0
–350 –250 –150 –50 50 150 250 350
0
5
10 15 20 25 30 35 40
35
40
45
50
INPUT OFFSET VOLTAGE (µV)
SUPPLY VOLTAGE (V)
SUPPLY VOLTAGE (V)
6110 G01
6110 G02
6110 G03
VOS Temperature Coefficient
VOS vs IOUT Voltage
VOS vs IOUT Voltage
12
400
300
200
100
0
10
9
8
7
6
5
4
3
2
1
0
+
+
+
V
V
V
= 5V
40 UNITS
–40°C TO 125°C
V
V
I
= 36V
= 0V
V
V
I
= 50V
= 25V
IMON
= 1.2V
= 0V
IOUT
IMON
+IN
IMON
+IN
10
8
= 100µA
= 100µA
+IN
I
= 100µA
T
= 25°C
A
T
= 85°C
A
T
= 125°C
= –55°C
A
T
= 125°C
A
6
T
= 85°C
A
T
A
= 25°C
4
T
A
T
= 0°C
T
= –55°C
A
A
T
= –40°C
10
2
–100
A
T
= 0°C
35
T = –40°C
A
A
0
–200
–3.0 –2.0 –1.0
0
1.0
2.0
3.0
0.1
1
40
30
40
45
50
INPUT OFFSET VOLTAGE TEMPERATURE COEFFICIENT (µV/°C)
IOUT VOLTAGE (V)
IOUT VOLTAGE (V)
6110 G04
6110 G05
6110 G06
6110f
4
For more information www.linear.com/LT6110
LT6110
Typical perForMance characTerisTics
VOS vs VSENSE Voltage
VOS vs IMON Voltage
VOS vs +IN Current
400
300
200
100
0
10
9
8
7
6
5
4
3
2
1
0
1000
800
600
400
0
+
+
+
V
= 5V
V
+IN
= V = 36V
V
V
= 5V
= 1.2V
IOUT
I
= 100µA
OUT
T
= 25°C
A
T
= 85°C
T
= 125°C
A
A
T
= 85°C
A
T = 125°C
A
T
= 85°C
T
= 125°C
A
A
T
= 0°C
A
T
= 25°C
A
T
= –40°C, –55°C
A
T
= 25°C
A
–100
–200
200
–200
T
= –40°C
T
= –55°C
A
A
T
= 0°C
A
T
= –40°C, –55°C
10
A
0
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
(V)
0.1
1
40
0.001
0.01
0.1
1
2
V
IMON VOLTAGE (V)
OUTPUT CURRENT (mA)
SENSE
6110 G07
6110 G08
6110 G09
IOUT Current Error vs Supply
Voltage
IOUT Current Error vs Supply
Voltage
IOUT Current Error Distribution
300
250
200
150
100
50
3
2
3
2
+
V
V
V
= 5V
800 UNITS
V
I
= 0.4V
V
+IN
= 25V
IOUT
IOUT
+IN
= 1.2V
= 100µA
I
= 100µA
IOUT
IMON
+IN
= 0V
= 100µA
T
= –40°C
A
I
T = –40°C
A
T
= 0°C
A
T
= –55°C
A
T
= –55°C
A
T
= 0°C
A
1
1
0
0
T
= 85°C, 125°C
A
T
= 85°C, 125°C
A
T
= 25°C
T
= 25°C
A
A
–1
–2
–3
–1
–2
–3
0
–1.2 –0.8 –0.4
0
0.4
0.8
1.2
0
5
10 15 20 25 30 35 40
35
40
45
50
IOUT CURRENT ERROR (%)
SUPPLY VOLTAGE (V)
SUPPLY VOLTAGE (V)
6110 G10
6110 G11
6110 G12
IOUT Current Error vs Output
Voltage
IOUT Current Error vs Output
Voltage
IOUT Current Error vs +IN Current
3
2
3
2
3
2
+
+
+
V
V
I
= 5V
= 0V
V
V
I
= 36V
= 0V
V
V
= 5V
= 1.2V
IMON
+IN
IMON
+IN
OUT
T
= 0°C
T
= –40°C
A
A
= 100µA
= 100µA
T
= –40°C
T
= –40°C
A
A
T
= –55°C
A
T = –55°C
A
1
T
= –55°C
A
1
1
0
T
= 85°C, 125°C
A
0
0
T
= 85°C, 125°C
T
= 85°C, 125°C
A
A
–1
–2
–3
–4
T
= 25°C
T
A
= 25°C
A
T
= 25°C
A
T
= 0°C
A
T
= 0°C
A
–1
–2
–3
–1
–2
–3
0.001
0.01
0.1
1
2
0
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
0
10
20
30
40
+IN CURRENT (mA)
OUTPUT VOLTAGE (V)
OUTPUT VOLTAGE (V)
6110 G13
6110 G14
6110 G15
6110f
5
For more information www.linear.com/LT6110
LT6110
Typical perForMance characTerisTics
IMON Current Error vs Supply
IMON Current Error vs Supply
Voltage
IMON Current Error Distribution
Voltage
300
250
200
150
100
50
7
6
7
6
+
V
V
V
= 5V
800 UNITS
V
I
= 0.4V
V
= 25V
IOUT
IOUT
= 1.2V
= 100µA
I
= 100µA
IOUT
IMON
+IN
+IN
= 0V
= 100µA
I
+IN
5
5
4
4
T
= –55°C
T = –55°C
A
A
3
3
T
= –40°C
T
= 0°C
T
= –40°C
A
A
A
T
= 0°C
2
2
A
T
= 25°C
A
T = 25°C
A
1
1
T
= 85°C, 125°C
A
T
= 85°C, 125°C
A
0
0
0
–1
–1
0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4
0
5
10 15 20 25 30 35 40
35
40
45
50
IMON CURRENT ERROR (%)
SUPPLY VOLTAGE (V)
SUPPLY VOLTAGE (V)
6110 G16
6110 G17
6110 G18
Minimum IOUT to IMON Voltage
vs Temperature
IMON Current Error vs Output
Voltage
IMON Current Error vs Output
Voltage
1.0
0.8
0.6
0.4
0.2
0
7
6
7
6
+
+
+
V
V
= 5V
= 0V
V
= 5V
= 100µA
V
V
= 36V
= 0V
I
IMON
+IN
IMON
= 100µA
+IN
∆IOUT ERROR < 1%
I
5
5
4
4
T
T
= –55°C
= –40°C
A
T
= –55°C
= –40°C
A
I
= 1mA
+IN
3
3
A
T
= 0°C
T
A
T
A
T = 0°C
A
2
2
I
I
= 100µA
= 10µA
+IN
1
1
T
= 125°C
+IN
A
T = 125°C
A
0
0
T
= 85°C
= 25°C
T
= 25°C
T = 85°C
A
A
A
A
–1
–1
–55 –35 –15
5
25 45 65 85 105 125
0
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
0
5
10 15 20 25 30 35 40
TEMPERATURE (°C)
OUTPUT VOLTAGE (V)
OUTPUT VOLTAGE (V)
6110 G19
6110 G20
6110 G21
IMON Current Error vs +IN Current
Supply Current vs Supply Voltage
Supply Current vs +IN Current
5
4
60
50
40
30
20
10
0
10
1.0
+
+
I
= 0µA
V
V
= 5V
= 1.2V
V
V
= 5V
= 1.2V
+IN
OUT
OUT
T
= –55°C
A
3
T
= –40°C
A
T = 85°C
A
T
= 0°C
A
2
T
= 125°C
A
1
T = 25°C
A
T
= 125°C
A
T
= 125°C
A
0
T
= –40°C
A
0.1
T
= 25°C
T
= 85°C
A
A
T
= 85°C
A
–1
–2
–3
T
= –55°C
A
T
= –40°C, –55°C
A
T
= 25°C
0.01
A
0.01
0.001
0.01
0.1
1
2
0
5
10 15 20 25 30 35 40 45 50
0.001
0.1
1
2
+IN CURRENT (mA)
SUPPLY VOLTAGE (V)
+IN CURRENT (mA)
6110 G22
6110 G23
6110 G24
6110f
6
For more information www.linear.com/LT6110
LT6110
Typical perForMance characTerisTics
Input Bias Current vs Supply
Voltage
Output Short-Circuit Current vs
Temperature
RSENSE vs Temperature
180
160
140
120
100
80
50
40
30
20
10
0
35
30
25
20
15
10
5
+
V
V
= V
IOUT
IMON
= 0V
SHORT DURATION = 1ms
+
V
= 36V
= 5V
T
= 125°C
A
T
= –40°C
A
T
= –55°C
A
60
+
V
40
T
= 85°C
A
T
= 25°C
A
20
0
0
5
10 15 20 25 30 35 40 45 50
–55 –35 –15
5
25 45 65 85 105 125
–55 –35 –15
5
25 45 65 85 105 125
SUPPLY VOLTAGE (V)
TEMPERATURE (°C)
TEMPERATURE (°C)
6110 G25
6110 G26
6110 G27
Frequency Response
PSRR vs Frequency
10
+
100
90
80
70
60
50
40
30
20
10
0
+
V
V
V
= 5V
V
I
= 5V
= 100µA
= 1.2V
IOUT
+IN
5
0
= 0V
I
= 10µA
R
= R
= 1k
IMON
+IN
IN
IOUT
I
= 1mA
+IN
–5
I
= 100µA
+IN
–10
–15
–20
–25
–30
I
I
I
= 10µA, R = R
= 10k
IOUT
IOUT
+IN
+IN
+IN
IN
= 100µA, R = R
= 1k
IN
= 1mA, R = R
= 100Ω
10k
IN
IOUT
10
100
1k
100k
1M
10
100
1k
10k
100k
1M
FREQUENCY (Hz)
FREQUENCY (Hz)
6110 G28
6110 G29
0µA to 10µA
IOUT Current Step Response
0µA to 100µA
IOUT Current Step Response
0µA to 1mA
IOUT Current Step Response
V
V
V
SENSE
SENSE
SENSE
50mV/DIV
50mV/DIV
50mV/DIV
V
V
V
IOUT
IOUT
IOUT
6110 G30
6110 G31
6110 G32
20µs/DIV
+
20µs/DIV
+
20µs/DIV
+
V
R
R
R
= 5V
V
R
R
R
= 5V
V = 5V
= 10k
= 0Ω
= 10k TO 1.2V
= 1k
= 0Ω
= 1k TO 1.2V
R
R
R
= 100Ω
+IN
–IN
+IN
–IN
+IN
–IN
= 0Ω
= 100Ω TO 1.2V
IOUT
IOUT
IOUT
6110f
7
For more information www.linear.com/LT6110
LT6110
Typical perForMance characTerisTics
0µA to 30µA
IMON Current Step Response
0µA to 300µA
IMON Current Step Response
0µA to 3mA
IMON Current Step Response
V
V
V
SENSE
SENSE
SENSE
50mV/DIV
50mV/DIV
50mV/DIV
V
V
V
IOUT
IOUT
IOUT
6110 G34
6110 G33
6110 G35
20µs/DIV
+
20µs/DIV
+
20µs/DIV
+
V
R
R
R
= 5V
V
R
R
R
= 5V
V = 5V
= 1k
= 10k
= 0Ω
= 3.4k TO GND
R
R
R
= 100Ω
+IN
–IN
+IN
–IN
+IN
–IN
= 0Ω
= 340Ω TO GND
= 0Ω
= 34Ω TO GND
IMON
IMON
IMON
VSENSE = 5mV Step Response
VSENSE = 50mV Step Response
VSENSE = 500mV Step Response
V
V
SENSE
SENSE
20mV/DIV
200mV/DIV
V
SENSE
20mV/DIV
V
V
V
IOUT
IOUT
IOUT
50mV/DIV
50mV/DIV
50mV/DIV
6110 G36
6110 G37
6110 G38
100µs/DIV
+
20µs/DIV
20µs/DIV
+
+
V
R
R
R
= 5V
V = 5V
V
= 5V
= 49.9Ω
= 0Ω
= 1k TO 1.2V
R
R
R
= 499Ω
R
R
R
= 4.99k
+IN
–IN
+IN
–IN
IOUT
+IN
–IN
IOUT
= 0Ω
= 1k TO 1.2V
= 0Ω
= 1k TO 1.2V
IOUT
VSENSE = 1V Step Response
Unbalanced Inputs
VSENSE = 1V Step Response
Balanced Inputs
V
V
SENSE
500mV/DIV
SENSE
500mV/DIV
V
V
IOUT
50mV/DIV
IOUT
50mV/DIV
6110 G39
6110 G40
20µs/DIV
+
20µs/DIV
+
V
R
R
R
= 5V
V
R
R
R
= 5V
= 10k
= 0Ω
= 1k TO 1.2V
= 10k
= 10k
+IN
–IN
+IN
–IN
= 1k TO 1.2V
IOUT
IOUT
6110f
8
For more information www.linear.com/LT6110
LT6110
pin FuncTions (TSOT-23/DFN)
+
NC (Pin 1/Pin 8): Not Internally Connected.
V (Pin 7/Pin 2): Positive Power Supply. Connect to the
more positive side of the sense resistor. A minimum ca-
pacitance of 0.1µF is required from V to V .
IOUT (Pin 2/Pin 7): Sinking Current Output. IOUT will sink
+
–
a current that is equal to V
/R
V
is the voltage
SENSE IN. SENSE
developed across the sense resisor.
+IN (Pin 8/Pin 1): Positive Input to the Internal Sense
Amplifier. The internal sense amplifier will drive +IN to the
IMON (Pin 3/Pin 6): Sourcing Current Output. IMON will
+
same potential as –IN. A resistor, R , tied from V to +IN
+IN
source a current that is equal to 3 • V
/R .
SENSE IN
sets the IOUT and IMON output currents as defined in the
–
V (Pin 4/Pin 5): Negative Power Supply. Normally con-
the IOUT and IMON pin functions description.
nected to ground.
–
Exposed Pad (Pin 9, DFN Only): V . Must be soldered
to the PCB.
–IN (Pin 5/Pin 4): Negative Input to the Internal Sense
Amplifier. Must be tied to system load side of the sense
resistor, either directly or through a resistor.
RS (Pin 6/Pin 3): Internal Sense Resistor. Connect to the
load to use. Leave open when using an external sense
resistor.
6110f
9
For more information www.linear.com/LT6110
LT6110
block DiagraM
I
LOAD
R
WIRE
V
+
V
–
REG
SENSE
+
V
IN
OUT
V
LOAD
IN
R
REGULATOR
ADJ
F
0.1µF
I
+IN
R
IN
+
GND
+IN
V
RS
–IN
R
SENSE
0.020Ω
1k
NC
+
–
IOUT
R
G
–
IMON
V
6110 F01
R
WIRE
–
V
LOAD
Figure 1. Block Diagram and Typical Connection
6110f
10
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LT6110
applicaTions inForMaTion
INTRODUCTION
The LT6110 has two output pins, IOUT and IMON. Either
pinmaybeusedtoprovideacurrentthatisproportionalto
the load current. The IOUT pin provides a sinking current
to compensate regulators with a ground referred voltage-
reference, such as the LT3980. The IMON pin provides a
sourcing current to compensate regulators with an output
referred reference like the LT1083 and current-referenced
regulatorsliketheLT3080.Asanaddedfeature,theoutput
current from either pin can be converted to a voltage via a
simple resistor, creating a voltage that is also proportional
to load current. This voltage may be used to measure
or monitor the load current. Either or both pins may be
used for regulator control, and either or both pins may
be used for monitoring, allowing substantial flexibility in
system design.
The LT6110 provides a simple and effective solution to
a common problem in power distribution. When a load
draws current through a long or thin wire, wire resistance
causes an IR drop that reduces the voltage delivered to
the load. A regulator IC cannot detect this drop without a
Kelvin sense at the load, which requires a multi-conductor
wire that is not supported in some applications.
The LT6110 detects the load current and sets a propor-
tional current at an output that can be used to control the
output voltage of an adjustable regulator to compensate
for the drop in the wire.
TheaccuracyandwideoutputcurrentrangeoftheLT6110
allowittocompensateforeithersmallorlargevoltagedrops
to a high degree of precision. The LT6110 can sense the
load current with its internal sense resistor or an external
senseresistorcanbeusedtoimproveaccuracyandhandle
currents greater than 3A. Resistor-programmable gain
gives substantial flexibility to the compensation circuit. A
signal bandwidth of 180kHz enables fast response time
to load changes and provides good loop characteristics
so that the power supply circuit remains stable.
THEORY OF OPERATION
The outputs of the LT6110 are proportional to a sense
voltage, V
, developed across an internal or external
SENSE
SENSE
sense resistor, R
(see Figure 1).
A sense amplifier loop forces +IN to the same voltage as
+
–IN. Connecting an external resistor, R , between V and
IN
+IN forces a voltage across R equal to V
, creating
/R . This current
IN
SENSE
The LT6110 requires that the resistance of the wire be
known. However, that resistance does not have to be very
accurate for the LT6110 to provide good compensation
since the regulation at the load is the product of the error
due to the wire resistance and the error in the LT6110
compensation circuit.
a current into +IN, I , equal to V
+IN
SENSE IN
is precisely mirrored to IOUT. The emitter currents of
the three transistors in the mirror are combined to form
the IMON output current. Ideally, the IOUT sink current
is equal to I and the IMON source current is equal to
+IN
three times I
.
+IN
For example, a 5V regulator circuit has 10% regulation at
the load due to a wire resistance drop of 0.5V. Even if the
wire resistance doubled, causing an error in the LT6110
compensation circuit of 50%, the regulation at the load
is still reduced to 10% • 50% = 5%.
+
–
V and V
The LT6110 is designed to operate with a supply voltage
+
–
(V to V ) up to 50V. However, when using a supply volt-
age greater than 36V additional care must be taken not
+
to exceed the absolute maximum ratings. The V to IOUT
voltagemustbekeptlessthan36Vtoavoidthebreakdown
of internal transistors.
For systems that are better controlled, the load regula-
tion can be improved to far exceed that possible without
the LT6110. As an example, for a known wire resistance,
and with an external 1% sense resistor, the same 10%
load regulation in the previous example can be reduced
to less than 0.5%.
+
The V pin needs to be bypassed with at least a 0.1µF
capacitor placed close to the pin.
6110f
11
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LT6110
applicaTions inForMaTion
+IN and –IN
DESIGN PROCEDURE
The design of an LT6110 compensation circuit is a simple
3-step process. To start, the following parameters must
be known:
The +IN and –IN inputs can have a maximum differential
voltage equal to the supply voltage. This protects the
LT6110 if the –IN pin (the remote load side) is accidentally
shorted to ground. In this case, the IOUT current must be
limited to less than 2mA (see the Limiting the Regulator
Boost Voltage section).
R
R
, total wire resistance to the load
WIRE
, resistor used to sense the load current
SENSE
R , feedback resistor of the regulator
F
The +IN to IOUT voltage must be kept below 36V to avoid
the breakdown of internal transistors.
I , maximum load current
LOADMAX
The circuit in Figure 2 shows an adjustable voltage regula-
tor with an LT6110 compensation circuit. The regulator
has an internal ground referred voltage reference to set
it’s output voltage. There are two wires to the load, one
IOUT and IMON
The IOUT to IMON outputs can have a maximum differ-
ential voltage of 36V for IOUT above IMON and –0.6V for
IOUT below IMON. A 36V Zener diode can be connected
from IOUT to IMON to prevent damage to the output NPN
transistor in the event of a fault condition. In this case, a
low leakage Zener diode should be used to reduce error
in the IOUT current.
source (R
) and one return (R
). Since it is the
SWIRE
RWIRE
most common configuration it will be used for the follow-
ing design example. Current referenced regulators and
regulators with an output referred reference are covered
in later sections.
Step 1: Determine the drop in voltage at the load due to
the wire resistance and sense resistor at the maximum
load current.
RS (Internal R
)
SENSE
The internal sense resistor can reliably carry a continuous
current up to 3A and transient currents of 5A for up to 0.1
seconds. For currents greater than this, an external sense
resistor should be used. The internal sense resistor has a
temperature coefficient similar to copper.
V
DROP
V
DROP
= (R
+ R
+ R
) • I
SWIRE
RWIRE
SENSE LOADMAX
= (0.125Ω + 0.125Ω + 0.02Ω) • 2A = 0.54V
V
DROP
R
SWIRE
I
LOAD
0.125Ω
V
REG
V
IN
I
+IN
R
R
F
V
REGULATOR
FB
I
= 2A
SENSE
LOADMAX
3.65k
IN
+
V
–
+
+IN
RS –IN
+
LOAD
R
SENSE
R
G
20mΩ
CIRCUIT
OR
BATTERY
–
V
LOAD
+
–
LT6110
IOUT
R
–
RWIRE
IMON
V
0.125Ω
V
6110 F02
DROP
Figure 2. 2-Wire Compensation, One Wire Is Connected to the Load and One Wire Is the Ground Return Wire
6110f
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Step2:Determinetheresistoronthe+INpin,R required
external sense resistor with a tighter tolerance. See the
section on External Current Sense Resistors for more
information.
IN,
to cancel V
.
DROP
The regulator output voltage will increase as current is
pulled from the IOUT pin through the feedback resistor,
R , creating a compensation voltage.
F
In most cases, the internal sense resistor, wire resistance
tolerances and temperature mismatch of the R
WIRE
and
SENSE
R
resistances will contribute the largest portion of the
V
COMP
= I
• R
IOUT F
overall compensation circuit error. See the sections on
Error Sources, Copper Wire Information and Temperature
Errors for a comprehensive discussion.
To cancel the voltage drop at the load, set V
equal
COMP
to V
.
DROP
V
= I
• R = V
F DROP
COMP
IOUT
ADDITIONAL DESIGN CONSIDERATIONS
IOUT Current
Since the IOUT current is equal to the current going into
the +IN pin and the current in the +IN pin is equal to the
sense voltage divided by R , R can be determined by
the following equations:
IN IN
The recommended range of IOUT current is 30µA ≤ I
IOUT
≤ 300µA for the best precision. For performance outside
of this range, see the Typical Performance Curves to
determine typical errors.
VSENSE
RIN
IIOUT =I+IN
=
If the IOUT current is less than 30µA, the feedback resis-
tor may need to be adjusted to reduce the error in the
compensation circuit.
where V
= I
• R
SENSE
SENSE
LOADMAX
Combining the above equations,
RF
VDROP
In the previous example,
R = ILOADMAX •RSENSE
•
(
)
IN
VSENSE 0.04
I
=
=
=148µA
IOUT
3.65k
0.54V
RIN
270
R = 2A •0.02Ω •
= 270Ω
(
)
IN
Since this is within the recommended range no further
adjustment is needed.
Step 3: The final step is to consider the errors in the
compensation circuit to determine if the resulting voltage
error at the load meets the desired performance.
SeethesectiononCompensatingaLowQuiescentCurrent
Design for IOUT current less than 30µA.
Forexample,theinternalR
oftheLT6110hasatypical
SENSE
Load Regulation
toleranceof 7.5%.Iftheothererrorsinthecompensation
circuit such as V , IOUT current error and the resistor
OS
Load regulation is often specified as an error in output
voltage ata given load current, asin the previous example,
but it is also specified as a percentage of the regulator
outputvoltage. Iftheoutputvoltageoftheregulatorcircuit
in Figure 2 is 5V, the resulting compensated load regula-
tion, in percent, would be the following:
tolerances of R and R add an additional 2.5% error,
F
IN
then the total error in the compensation circuit would be
10%resultinginavoltageerrorattheloadofthefollowing:
V
V
= V
• Compensation Error
COMP
LOADERROR
= 0.54V • (±10%) = ±0.054V
LOADERROR
VLOADERROR
A 10× improvement.
LoadRegCOMP % =
•100
( )
VREG
If this is not adequate for the given application, steps can
be taken to reduce the sources of error, such as using an
0.054V
5V
LoadRegCOMP % =
•100= 1.1%
( )
6110f
13
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LT6110
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Without the compensation circuit (no R
regulation in percent would be,
) the load
Kelvin Sense Connection to R
SENSE
SENSE
To reduce R
error due to trace resistance, the –IN pin
SENSE
–0.5V
5V
and R resistor should be connected as close to R
IN
SENSE
LoadRegUNCOMP % =
( )
•100= –10%
as possible, as reflected in Figure 2.
The regulator’s output will also change due to its own
load regulation effects (per the regulator’s specification).
In general, this change in voltage is small compared to
the wire-drop, and can be ignored. If it is considered to
be a significant source of error, it can be included as part
of the wire-drop compensation. To include the regulator’s
load regulation effect, simply add the voltage drop due to
Compensating a Low Quiescent Current Design
Switching regulator circuits are used for high power ef-
ficiency. Many are required to maintain high efficiency at
light or no load conditions. In these cases the quiescent
operating current is minimized by using larger valued
resistors to program the output voltage so very little cur-
rent is wasted in the feedback network.
theregulator’sloadregulationatI
toV
, when
LOADMAX
DROP
A large value for resistor R could require too low of a
calculating the compensation circuit parameters.
F
compensating current (<30µA) from IOUT of the LT6110.
PCB Trace Resistance
In this situation the feedback resistor, R , can be split
F
into two resistor values. A small value resistor to conduct
Printed circuit trace resistance between the output of the
regulator and the load will cause additional voltage drops.
Aswiththeregulator’sloadregulationeffects, thesedrops
I
from the LT6110 and compensate the output voltage
IOUT
when the load current is high, and a second, larger valued
resistor, to keep the no-load quiescent current drain low.
can be compensated for by adding them to V
when
DROP
With this arrangement, as shown in Figure 3, I
can
IOUT
calculatingthecompensationcircuitparameters.Thisalso
allows the use of narrower traces to deliver power to the
load and still retain good load regulation. See the PCB
Copper Resistor section for more information on how to
determine trace resistance.
be designed for 100µA to preserve V
compensation
DROP
accuracy. At no load the quiescent current drawn through
the feedback resistors, I , can be kept very low.
Q
I
R
WIRE
LOAD
V
REG
V
LOAD
V
IN
I
+IN
V
R
FA
REGULATOR
FB
SENSE
R
IN
+
V
–
I
Q
+
<30µA
+IN
RS –IN
R
R
LOAD
FB
20mΩ
G
+
–
LT6110
IOUT
–
IMON
V
6110 F03
Figure 3. Low Quiescent Current Wire Compensation Using Three Regulator Resistors
6110f
14
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In Figure 3 R is split into R and R . V is the no-load
Compensating a Current Referenced
Regulator Power Source
F
FA
FB REG
quiescent output voltage of the regulator. The design of
these two feedback resistors follows:
Figure 5 shows a cable drop compensation circuit using a
currentreferencedregulator,theLT3080.Aprecision10µA
VDROP
RFA =
set current, I , is sourced through two series connected
I
SET
IOUT
resistors to program the output voltage for the remote
load. To compensate for the load connecting cable drop
requires sourcing an additional current into this resistor
pair to increase the output voltage. The LT6110 provides
a sourced current at the IMON pin which is also directly
proportional to the current flowing to the load. This cur-
rent is three times the normal IOUT current. The following
equations are used to design this circuit:
I
can be sized to be 100µA at full load current and
IOUT
onlythisresistorcreatestheV
compensationvoltage.
DROP
VREG – V
FB
RFB =
–RFA
IQ
I is the no-load quiescent current flowing through the
resistor string.
Q
V
V
= I • (R
+ R
)
REG
SET
SET1
SET2
Figure 4 is a circuit using the LT6110 and a three resistor
voltage setting technique to compensate the voltage loss
due to a 2A load connected through 6 feet of stranded
copper wire (300mΩ of wire resistance). The LT3980 is a
2A buck switching regulator programmed for 5V out with
= I
• R
SENSE
SENSE
LOAD
VSENSE
RIN
I+IN
=
only 10µA of current, I , through the feedback resistor
Q
I
= 3 • I
IMON
+IN
string when there is no load current. At the full 2A load the
LT6110 uses the internal 20mΩ sense resistor to produce
100µA at IOUT to compensate for the 640mV drop.
V
V
BD
IN
IN
RUN/SS BOOST
LT3980
0.47µF
10µH
100k
V
REG
PGOOD
SW
DA
FB
47µF
10pF
6.49k
DFLS240L
402Ω
97.6k
NC
+IN
R
T
0.1µF
LT6110
+
IOUT
V
1.5nF
422k
15k
V
FB
= 0.79V
20mΩ
V
C
R
WIRE
IMON
GND
GND
RS
0.3Ω
80.6k
100pF
5V
2A
–IN
100µF
6110 F04
Figure 4. LT3980 Buck Regulator with LT6110 Cable Drop Compensation Circuit
6110f
15
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R
WIRE
0.5Ω
V
3V
1A
LOAD
R
IN
LT3080
IN
200Ω
V
IN
+
+IN
V
RS –IN
I
SET
LOAD
20mΩ
+
–
+
–
LT6110
V
IOUT
REG
SET
R
SET1
301k
–
IMON
V
6110 F05
I
MON
MON
R
SET2
+
1.69k
I
= 3 I
IN
Figure 5. Wire Loss Compensation Using a Current Referenced LDO
To compensate for V
at I
set:
As shown in Figure 6 an LT6110 can add cable drop com-
pensation by using the current sourced from the IMON
pin. In this type of circuit the voltages appearing at the
IOUT and IMON pins can be higher so care not to exceed
their voltage ratings is important. To preserve accuracy
DROP
LOAD(MAX)
VDROP
IIMON
RSET2
=
and
RSET1
–
the voltage at IMON should be kept within 5V of V , or
VREG
ISET
=
–RSET2
ground in this example. By using two resistors for the bot-
tom resistor in the voltage regulator programming string,
the cable drop compensation voltage can be added to a
voltage near ground appearing at the IMON pin.
As an example, to compensate this 3V regulator for a
500mVcabledropwitha1AloadcurrentsetI for100µA
+IN
for best accuracy. Then:
The following equations are used to design this circuit
using an LT1083, 7A adjustable voltage regulator:
R
SET1
= 301k and R
= 1.69k using nearest 1%
SET2
tolerance standard resistor values.
V
REF
=1.25VbetweenOUTandADJpins,I =75µAtyp
ADJ
1A •20mΩ
100µA
VREF
R1
RIN =
= 200Ω
ISET
=
IADJ
V
(I
= 0) = (I + I ) • (R2 + R ) + V
SET ADJ G REF
LOAD LOAD
Compensating an Output Referred
Adjustable Voltage Regulator
V
= I
• R
SENSE
SENSE
LOAD
VSENSE
RIN
Many adjustable voltage regulators are biased from a
floating voltage reference that sets a voltage between the
output pin and an adjust pin. Three terminal fixed voltage
regulators can also be made adjustable by biasing up
the ground terminal. A feedback resistor string is used
to program the output voltage. The amount of current
through these resistors is scaled to a level to minimize
error caused by any bias current at the adjust pin.
I+IN
=
I
= 3 • I
+IN
IMON
Asanexample,Figure6isa12Vregulatorfora5Aremotely
connected load with a wire resistance of 250mΩ. For the
higher load current an external 25mΩ sense resistor is
6110f
16
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R
WIRE
0.25Ω
R
SENSE
0.025Ω
V
12V
5A
LOAD
R
IN
+
+IN
V
RS –IN
1.25V
ISET
R1=
20mΩ
R1• VLOAD – VREF
IADJ •R1+ VREF
(
)
–RG
R2=
+
–
LT6110
LT1083
V
IOUT
REG
V
IN
10µF
IN
OUT
ILOAD • RSENSE +RWIRE
(
)
RG
=
R1
499Ω
ADJ
I
1.62k
10µF
3•I+IN
V
IMON(MAX) =RG • ISET +IADJ +3•I+IN
ADJ
(
)
I
SET
–
IMON
V
6110 F06
R2
3.16k
I
IMON
IMON
I
= 3 I
+IN
R
1k
G
Figure 6. Wire Compensation Using a High Current Adjustable Regulator
used. The cable drop voltage for such a high current ap-
plication is significant:
For 1.375V of compensation, using a convenient value 1k
resistor for R will require 1.375mA from the IMON pin
G
which is near the mid range of accurate current levels.
V
DROP
= I
• (R
+ R ) = 5A • 275mΩ
WIRE
LOAD(MAX)
SENSE
= 1.375V
To program the regulator output voltage and compensate
for V at I the following procedure can be
With this selection for R then:
G
R2 = 4.175k – 1k = 3.175k
DROP
LOAD(MAX)
usea3.16kstandard1%tolerancevaluetosettheno-load
output voltage to 12V.
used:
Make I >> I , if I = 33.3 • I then I = 2.5mA
SET
ADJ
SET
ADJ
SET
To program the LT6110 compensation current requires
VREF 1.25V
=
a selection for R :
IN
R1=
= 499Ω
ISET 2.5mA
VSENSE VSENSE
RIN =
=
I
I+IN
IMON
For 12V output with no-load current:
3
VLOAD – VREF 10.75V
V
SENSE
= 5A • 25mΩ = 125mV and
R2+R =
=
= 4.175k
(
)
G
ISET +IADJ
2.575mA
I
1.375mA
IMON
3
=
= 460µA so
Resistor R is used to develop the maximum load current
3
G
compensation voltage. A smaller value for R minimizes
G
125mV
460µA
RIN =
= 271Ω
the voltage programming error at no load but requires
more current from the LT6110 IMON pin to compensate
forcabledroploss. TheIMONpincurrentismostaccurate
over a range from 30µA to 3mA.
use a 274Ω standard value.
The IOUT pin can be connected to the 12V regulator out-
put. The LT1083 requires a minimum output load current
of 10mA so an additional 1.62k resistor (not required if
VDROP
RG =
IIMON
I
is always greater than 10mA) is added to the output.
LOAD
6110f
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The voltage that appears at the IMON pin can impact the
accuracyofthecompensationcircuitandshouldbenoted.
In this example the voltage will be a maximum at full load
current and voltage compensation. This voltage is:
example circuit that indicates the various error terms to be
considered. For this example a 5V regulator is to provide
2A maximum to a remote load connected through 6 feet
(~2 meters) of 28AWG (19/40) stranded hook-up wire. At
2A of current, 28 gauge is the thinnest, low cost wire suit-
able. From this an estimate of the wire resistance can be
made. From wire tables the DC resistance of 28AWG can
V
= (I + I + I
) • R = (2.5mA + 75µA
IMON(MAX)
+ 1.375mA) • 1k = 3.95V.
SET ADJ IMON G
be determined making R
= 6ft • 58.7mΩ/ft = 352mΩ.
WIRE
ERROR SOURCES
At 2A full load current this will create a V
of 704mV.
DROP
Without the LT6110 compensator the regulation of the 5V
supply at the load would be 14%.
The LT6110 output current allows for reliable compensa-
tion for small or large connection wiring voltage drops.
The voltage regulation at the remote load can be improved
dramatically using the LT6110. With properly designed
cable drop compensation the load voltage variation will
be reduced to only the error in the compensation voltage
created. This error voltage is a combination of several
circuit characteristics.
To emphasize error terms this example design will use
the internal 20mΩ sense resistor of the LT6110 and will
assume that the feedback resistor network in the voltage
regulator cannot be modified or optimized for compensa-
tion. The R used to develop the compensation voltage
F
is fixed at 10k and the reference voltage at the feedback
node where the compensator connects is 1.2V. From
these parameters the basic compensation circuit can be
The first step in determining the error is to determine the
amount of compensation voltage required. Figure 7 is an
+ V
–
DROP
I
+ V
–
WIRE
+ V
–
SENSE
LOAD
V
REG
V
IN
IN
OUT
V
LOAD
+
R
≥0.1µF
I
WIRE
R
F
V
REGULATOR
COMP
6 FT, AWG 28
19/40 STRANDED
WIRE
10k
–
+IN
R
IN
ADJ
GND
+
LT6110
V
+IN
V
RS
–IN
LOAD
R
SENSE
* +
OS
0.020Ω
–
1k
NC
I
+IN
+
–
V
IOUT
IOUT
BIAS
I
IOUT
IOUT
ERROR
–
IMON
V
6110 F07
I
IMON
V
IMON
IMON
ERROR
∆VOS ∆VOS
∆VOS
*
VOS = VOS
+
+
+
+TCVOS •∆T
∆I+IN ∆V
∆V
IOUT
IMON
Figure 7. Cable Drop Compensation Error Sources
6110f
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easily designed:
is ohms and is accounted for as a small resistance in
serieswithR .Thevoltageacrossthissmallresistance
IN
V
SENSE
at full load is 20mΩ • 2A or 40mV
is included in the total offset voltage term. The change
The compensation voltage, V , required is:
COMP
in I current is relative to 100µA where the LT6110
+IN
is trimmed for accuracy.
V
+ V
, 704mV + 40mV, or 744mV
SENSE
WIRE
•
•
∆V /∆V
is a change in the offset voltage caused
OS
IOUT
To create this compensation voltage will require a current
throughfeedbackresistorR ofV /R , 744mV/10kfor
by a change in the voltage applied to the IOUT pin
F
COMP
F
specified in mV/V. The change in V is relative to
1.2V DC where the LT6110 is trimmed for accuracy.
IOUT
an I
of 74.4µA. This is well within the most accurate
IOUT
rangeofcurrent(30µAto300µA)flowingintotheIOUTpin.
∆V /∆V
is a change in the offset voltage caused
OS
IMON
To create this current at full load requires an R value of
IN
by a change in the voltage applied to the IMON pin
V
/I
,40mV/74.4µA,or537.6Ω.Usingthenearest
SENSE IOUT
specified in mV/V.
standard 1% tolerance value of 536Ω will be sufficient.
Without considering any error terms other than this slight
• IOUTcurrenterroristheaccuracyoftheinternalcurrent
mirror. This is a percent deviation from I
change in value for R results in nearly perfect cable drop
.
IN
+IN
compensation. The theoretical load regulation would be
• IMON current error is the accuracy of the total internal
improved from 14% to less than 0.01%.
mirror current sourced to the IMON output. This is a
percent deviation from 3 • I
The single largest source of compensation error comes
from any change in the connecting wire resistance from
thedesignassumptions.Thiscouldbecausedbytempera-
ture, aging and possibly corrosion. In the compensator
circuit itself component tolerances and errors terms will
combinetodeviatefromthenearperfectdesignedamount
of compensation. Figure 7 shows this simple example
design and indicates the various error sources within the
LT6110. All of the error terms can be determined from
the Electrical Characteristics Table. The error terms for
any compensator design include:
.
+IN
• Temperature Related Errors (see Temperature Errors
section)
Table 1 is an example of the stack-up of all error terms in
the design of Figure 7. This table uses typical variances to
be seen at 25°C. It is not a rigorous worst case analysis
over all possible operating conditions, but instead serves
toillustratewhattoexpectforloadregulationimprovement
under nominal conditions.
Inthisexample,includingalltypicalerrorterms,theLT6110
stillprovidesafactorof10improvementinvoltageregula-
tion at the remote load. To obtain the same level of load
voltage stability without using the LT6110 would require
reducing the amount of cable drop loss. The easiest way
to do so would be to increase the wire gauge used to
connect to the load. For a 70mV change in load voltage
at 2A full load current would require a wire resistance of
only 33mΩ and for a 6 foot length 16AWG gauge wire is
required. A larger wire gauge can be significantly more
costly and is less flexible in routing to the load. These are
two significant design compromises to be considered.
• R
tolerance
SENSE
• R tolerance
IN
• R tolerance
F
• V , the offset voltage in µV of the internal current
OS
sense amplifier
•
∆V /∆I is an error term caused by the finite gain
OS +IN
of the current sense amplifier.
This is the change in the offset voltage as the sense
voltage and resulting input current varies from 0 to the
maximum value. It is a factor specified in mV/mA which
6110f
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Table 1. Compensation Error Using Typical Variances Expected at 25°C.
FIGURE 7 DESIGN EXAMPLE. TOTAL VDROP TO COMPENSATE = 744mV,
I+IN = 74.6µA
FOR MAXIMUM V
FOR MINIMUM V
COMP
COMP
TERM
DESIGN VALUE/SPEC UNITS COMMENT/CALCULATION
TYPICAL ERROR VALUE TYPICAL ERROR VALUE
R
R
20
536
0
mΩ
Ω
Internal Sense Resistor
7.50%
–0.5%
–100
21.5
533
–7.50%
0.5%
100
18.5
539
SENSE
IN
V
OS
µV
–100
0.135
0.0045
0.27
100
0.15
0.005
0.3
mV/mA Relative to I = 100µA
–10%
–10%
–10%
10%
0.165
0.0055
0.33
∆V /∆I
+IN
OS +IN
mV/V Relative to V
mV/V Relative to V
= 0.8V
= 0V
10%
∆V /∆V
IOUT
OS
IOUT
10%
∆V /∆V
IMON
OS
IMON
Total Offset
Total V
V
+ ∆V /∆I • (100µA – 74.6µA) + ∆V /∆V
• (1.2V – 0.8V) + ∆V /∆V
• 0V
OS
OS +IN
OS
IOUT
OS
IMON
µV
%
%
–94.5
0.004
0.015
106.4
–0.004
–0.015
OS
I
I
Error
Error
0
0
% IOUT Current Error Relative to I
0.4%
–0.4%
IOUT
+IN
% IMON Current Error Relative to 3 • I
1.50%
–1.50%
IMON
+IN
Summary of Terms
V
40
mV
µA
µA
µA
kΩ
mV
%
I
• R
SENSE
43
37
68.5
SENSE
LOAD(MAX)
I
I
I
74.6
74.4
223.8
10
(V
– Total V )/R
IN
80.8
+IN
SENSE
OS
I
+IN
• (1 + I Error)
IOUT
81.1
68.2
IOUT
3 • I • (1 + I Error)
IMON
246.0
10.05
815
202.4
9.95
IMON
+IN
R
Fixed Resistor Value in Power Source
• R
0.5%
–0.5%
F
V
COMP
V
COMP
746
0
I
679
IOUT
F
Error
9.26%
–9.04%
With Compensation
V
2
mV
%
V
COMP
– V
DROP
71
–65
LOAD_ERROR
Load Regulation
0.04
1.42%
–1.31%
FREQUENCY RESPONSE AND TRANSIENTS
inFigure9. Anyringingwhilesettlingoutcanbesmoothed
by additional filtering components in the control loop. A
small feedback capacitor across the regulator feedback
The LT6110 has a –3dB bandwidth of 180kHz. This
smooth frequency response is shown in Figure 8. This
defines the response time from the sensed input voltage
to the compensation output currents. Power sources will
typically have a large output capacitance making their
loop response bandwidth much slower than the LT6110.
The cable drop compensation loop is much faster than
the power source so there should be little impact on loop
stability in driving a remote load.
resistor,R ,canprovideeffectivesmoothingoftransients.
F
Specific values to use depend on the particular application
component values.
One important consideration for transients is a sudden
open or removal of the load current from a high current
condition. There is a risk of overvoltage at the load before
the LT6110 can reduce the compensation voltage. A good
solution to this potential issue is to bypass the remote
load with a capacitance greater than the capacitance at the
output of the regulator or power source. Figure 10 shows
a load removal transient using a 100µF load. Fortunately
the amount of compensation in most applications should
not be so large as to cause a serious overvoltage risk but
should always be considered.
For fast or step change variations in load current some
transients will be observed at the power source output
and at the remote load due to the finite reaction time of
the compensation loop. The amount of voltage transient
seen will depend mostly on the size and quality of the
supply bypass capacitors used at each end of the load
connecting wire. An example of these transients is shown
6110f
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0dB
R
–IN
R
+IN
C1
1k
–3dB
0
+
–30
–60
–90
–120
+IN
V
RS –IN
20mΩ
+
–
1
I
10
100
1000
6110 F08
LT6110
IOUT
= 100µA
FREQUENCY (kHz)
IOUT
Figure 8. LT6110 Frequency Response
–
IMON
V
6110 F11
V
REG
500mV/DIV
Figure 11. LT6110 Frequency Compensation
V
LOAD
500mV/DIV
Loop compensation with an LT6110 RC filter is not
required if the regulator’s loop is compensated with a
zero in the feedback divider (refer to the Regulator Loop
Stability section).
2A
1A
6110 F09
100µs/DIV
Figure 9. VLOAD Compensated
EXTERNAL CURRENT SENSE RESISTORS
The LT6110 internal current sense resistor, R
, is
SENSE
V
REG
provided for convenient use in many applications with a
maximum load current less than 3A. For higher current
or greater precision wire loss compensation an external
500mV/DIV
V
LOAD
500mV/DIV
sense resistor can be used. The external R
resistor
SENSE
2A
0A
can be a low valued current sense or shunt resistor, the
DC resistance (DCR) of an inductor, or the resistance of
a printed circuit board trace. Figure 12 shows an LT6110
circuit configuration using an external sense resistor. The
internal resistor at the RS pin is left open circuited.
6110 F10
C
= 100µF
10ms/DIV
LOAD
Figure 10. Removing Load
In addition to using a regulator capacitortoadjust the loop
response, an RC pole in the LT6110 circuit can provide
frequencycompensation.Figure11showsanLT6110with
an input RC filter. Using the input RC filter introduces a
second pole to the LT6110 one pole response (Figure 9).
The LT6110 poles become a zero in the regulator’s open-
loop response that includes the LT6110 in it’s feedback
EXTERNAL
SENSE
TO LOAD
TO V
REG
R
R
IN
+
+IN
V
RS –IN
20mΩ
path (providing the same function as the regulator’s R
+
–
F
LT6110
IOUT
with a shunt capacitor). Typically the RC pole frequency
should be about one-tenth of the regulator’s switching
frequency.
–
IMON
V
6110 F12
1
fCLK
fPOLE
=
=
2πR–INC1 10
Figure 12. Using an External RSENSE
(Resistor, Inductor or PCB Trace)
6110f
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ThevalueofR
externaldeterminestheV
voltage.
Precision Current Shunt Resistor
A precision, very low V error, compensation circuit
SENSE
is 100µA then a V
SENSE
If I
of 50mV is large enough
IOUT
SENSE
LOAD
to minimize the compensating IOUT current error due to
can be implemented with an LT6110 and a precision ex-
V
to less than 1% (see Figure 13).
OS
ternal R
. A 1% to 5% tolerance or better R
SENSE
SENSE
compensation current
resistor significantly reduces I
100
IOUT
+
0.4V ≤ V
≤ V – 1.5V
IOUT
error due to part to part variations. In addition, the low
temperature coefficient (TCR of typically 100ppm/°C) of
anexternalsenseresistorgreatlyreducesthecontribution
–
V
V
= V = 0V
IMON
OS(MAX)
10
1
of R
to the total voltage drop loss at higher operating
SENSE
temperatures. Figure 14 shows a 5V, 3.5A buck regulator
with an LT6110 using an external R
of typical current sense resistors.
. Table 2 is a list
SENSE
I
I
I
= 300µA
= 100µA
= 30µA
IOUT
IOUT
IOUT
0.1
0
10 20 30 40 50 60 70 80 90 100
V
(mV)
SENSE
6110 F13
Figure 13. VSENSE
V
IN
V
BD
IN
8V TO 36V
22µF
4.7µF
RUN/SS
BOOST
0.047µF
6.8µH
SW
FB
V
C
47µF
9.53k
47pF
15k
1nF
866Ω
LT3972
GND
NC
+IN
MBRA340
+
IOUT
V
523k
100k
EXTERNAL
0.1µF
LT6110
100pF
R
RT
SENSE
25mΩ
5ꢀ
R
0.79V
WIRE
IMON
RS
63.4k
0.25Ω
SYNC
GND
–IN
V
LOAD
10 FT
24AWG
5V
LOAD
I = 600kHz
3.5A
6110 F14
Figure 14. LT6110 with an External RSENSE and LT3972 Buck Regulator
Table 2. Surface Mount RSENSE Resistors
PART NUMBER THICK FILM
IRC LRC-LRF-2512
VALUE RANGE
TOLERANCE
1% to 5%
1% to 5%
1% to 5%
1% to 5%
TCR
POWER
2W
SIZE
2512
2512
2512
2512
2mΩ to 1Ω
10mΩ to 1Ω
33mΩ to 51Ω
1mΩ to 20mΩ
100ppm
200ppm
200ppm
100ppm
Stackpole Electronics CSR2512
Vishay RCWE2512
2W
2W
Panasonic ERJM1W
2W
Susumu PRL1632
Susumu PRL3264
10mΩ to 100mΩ
10mΩ to 100mΩ
1% to 2%
1% to 2%
100ppm (20mΩ to 51mΩ)
100ppm (20mΩ to 51mΩ)
1W
2W
1206
2512
6110f
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24
22
20
18
16
14
12
10
8
Copper Resistor Made from an R Inductor
F
An inductor made of copper wire will have a small DC
resistance, DCR or R , with a temperature coefficient
COIL
2oz COPPER
thatmatchesthatofthecopperwireconnectingtheremote
load. Copper wire resistance has a positive temperature
coefficient of approximately +3900ppm/°C. If the current
sense resistor and the remote load are in the same operat-
ingenvironmentandsubjecttoanincreaseintemperature,
1oz COPPER
6
4
theresistanceincreaseinR
willincreasebothV
2
SENSE
SENSE
0
andtheLT6110compensationcurrenttodirectlytrackand
cancel the increase in wire voltage drop to the load(refer
to the Temperature Errors section). Table 3 shows a list
of small air core inductors suitable for use as external
0
50 100 150 200 250 300 350 400 450 500
PCB TRACE WIDTH (MILS)
6110 F15
Figure 15. PCB Trace Current vs Trace Width
(<10°C Temperature Rise)
R
resistors.
SENSE
Example: Design a 2oz copper PCB trace sense resistor to
Table 3. Coilcraft Air Core Inductors for External RSENSE
compensateforwirevoltagedropforanI
of10A.
COILCRAFT PART
NUMBER
INDUCTANCE DCR NOMINAL (mΩ)
I
RMS
LOAD(MAX)
(nH)*
( 6ꢀ TYPICAL)
(A)
A V
of 60mV is large enough to minimize the com-
SENSE
0908SQ-27N
2222SQ-221
27
8.5
9.8
4.4
5
pensatingIOUTcurrenterrorduetotheinputoffsetvoltage
221
of the LT6110.
1010 US-141
146
3.1
14
VSENSE
ILOAD(MAX) 10A
60mV
*Inductance is not relevant for current sense.
RPCB
=
=
= 6mΩ
PCB Copper Resistor
Using Figure 15, the 2oz copper minimum trace width for
10A is 150mils. This sets the current handling capability
of the trace.
In a high load current application without a high preci-
sion load regulation specification, the cost of an external
SENSE
R
resistor can be eliminated using the resistance of
a printed circuit board, PCB, trace as a sense resistor. The
The resistance of the trace resistor is set by the length of
the trace. Each 150mil wide square of 2oz copper will have
a resistance of 0.25mΩ. A total resistance of 6mΩ will
require 24 squares (6mΩ/0.25mΩ/square). The length of
the PCB trace will then be 24sq × 150mils or 3.6 inches.
resistance,R ,isafunctionofcopperresistivity(ρ),PCB
PCB
copper thickness (T), trace width (W) and trace length (L),
R
= ρ (L/(T • W)). The typical manufacturing of PCB
PCB
fabrication limits the trace resistance tolerance to 15%.
A simplified R calculation sets the length equal to the
PCB
A serpentine layout can be used to reduce the footprint of
width (L/W = 1) and approximates 0.5mΩ and 0.25mΩ
per square trace area for 1oz and 2oz copper respectively.
The maximum current of a PCB trace depends on the
trace cross sectional area, trace width (W) times cop-
per thickness (T) and the amount of heating of the trace
permitted. Figure 15 plots PCB trace current vs PCB trace
width for 1oz (T = 1.4mils) and 2oz (T = 2.8mils) copper
for less than 10˚C temperature rise (this graph provides
a conservative maximum trace current estimate based on
the ANSI IPC2221 standard).
R
. Figure 16 shows a serpentine layout for a 6mΩ PCB
PCB
sense resistor and the V
connections to the LT6110.
SENSE
The corners of the serpentine resistor count as 3/4 of a
square. In Figure 16, R consists of six 3.5 square rect-
PCB
angular traces (two whole squares and two 3/4 squares).
The R six rectangular traces equal 21 0.15in × 0.15in
PCB
squares. Using a 2oz copper trace the resistance of the
21 squares is 5.25mΩ at 25°C (21 • 0.25mΩ per square).
An additional very small trace resistance is due to the
0.015in × 0.15in trace that connects the rectangular
6110f
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5.4mΩ 15ꢀ AT 25°C PCB RESISTOR
21 2oz COPPER SQUARES
ONE SQUARE
0.15 INCH × 0.15 INCH
TO
TO
LOAD
3/4 CORNER SQUARES
0.15 INCH × 0.15 INCH
REGULATOR
A
B
A 3.5-SQUARE COLUMN
— 3/4 SQUARE
R
IN
21 SQUARES (6 COLUMNS)
— ONE SQUARE
— ONE SQUARE
— 3/4 SQUARE
6110 F16
A
B
Figure 16. LT6110 and PCB Trace Resistor Layout
tracesatthetopandbottomcornersquares. Therearefive
connecting traces and their total resistance is 0.125mΩ
([0.015 inch/0.15 inch] • 0.25mΩ • 5).
The error sources due to temperature of an LT6110
circuit are:
The IOUT current error vs temperature coefficient is
–50ppm/°C
Temperature Errors
The V temperature coefficient is 1µV/°C
OS
In addition to the initial errors at 25°C the errors due to
a temperature variation must be included. The ambient
temperature variation of the LT6110 and the wire can
have the following cases: The LT6110 and wire are at
the same temperature, the LT6110 and wire are at much
different temperatures or the temperature of the LT6110
circuit is known and the wire temperature can only be ap-
proximated. The design procedure targets a load voltage
The R and R resistors temperature coefficient is
IN
F
100ppm/°C
The internal R
+3400ppm/°C
resistor temperature coefficient is
SENSE
An additional temperature error is due to R
. The
WIRE
copper wire temperature coefficient is +3900ppm/°C
equal to V
DROP
at maximum load current and cancels
REG(NOM)
by setting I
The IOUT current, V , R and R errors are small com-
OS IN
F
V
• R = V
. If, over the specified
IOUT
F
DROP
pared to the errors of the internal R
and R
OS IN
. For
SENSE
WIRE
temperature range, {I
• R – V
} is not zero volts,
IOUT
F
DROP
a 50°C temperature rise the IOUT current, V , R and R
F
then there will be an error to the expected load voltage
at maximum load current (for example, if V = 5V at
resistor error is 0.25%, 50µV and 0.5% respectively and
LOAD
} is 5mV then the
the internal R
respectively.
and R
error is 17% and 19.5%
SENSE
WIRE
25°C and at 75°C {I
LOAD
• R – V
IOUT
F
DROP
V
error is 100 • (5mV/5V) = 0.1%).
Using the example of V
= 5V, I
= 2A, I
=
IOUT
LOAD
LOAD
WIRE
Since I
= V
/R , the temperature errors must
SENSE IN
IOUT
71.2µA, R = 10k, R = 562Ω and R
= 0.336Ω the
F
IN
include the errors due to R , R
and V .
IN SENSE
OS
V
error due to the following three example cases is
LOAD
calculated:
6110f
24
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Case 1: LT6110 and the wire are at 75°C and the V
Copper Wire Information
LOAD
error is –0.36%. If the R
temperature coefficient
SENSE
The wire used in the power distribution of electronic sys-
tems is annealed (heated and cooled) copper wire and is
specified for it’s resistance per unit length, weight per unit
mass and current capacity. In the American Wire Gauge
standard, AWG is the gauge number and corresponds to
thediameterofasolidwire(asthegaugenumberincreases
thewirediameterdecreases, thewireresistanceincreases
and the current capacity decreases). Stranded copper
wire is an insulated bundle of packed and twisted bare
solid strands and its resistance, weight or cost depends
on the type of coating (tin, silver or nickel) and stranding
options (how the strands are grouped and twisted). The
stranded wire’s flexibility is useful for building and rout-
ing wire harness. The current capacity of copper wire is
inversely proportional to its gauge number, number of
wire conductors and operating temperature (increasing
gauge, conductors and temperature, decreases current
capacity). In addition the wire insulation temperature rat-
ing determines the maximum operating current (typical
insulation ratings range from 80°C to 200°C).
matchesthewire’stemperaturecoefficientof3900ppm/°C
then the V error is reduced. Using the copper wire
LOAD
resistance of an inductor as an R
external the V
SENSE
LOAD
error is reduced to –0.025%.
Case 2: The LT6110 is at 75°C, the wire is at 25°C and the
error is 2.3%. The 2.3% error is mostly due to the
V
LOAD
internal R
temperature coefficient. Using an external
SENSE
100ppm/°C R
reduces the V
error to 0.05%.
SENSE
LOAD
In addition, using a thermistor across R to increase the
IN
IOUT current as the temperature increases can reduce the
temperature induced V
error.
LOAD
Case 3: The LT6110 is at 25°C, the wire is at 75°C and the
error is –2.6%. The error is due only to the copper
V
LOAD
wire resistance increase vs temperature. The Case 3 error
can be reduced by designing for the maximum R at
WIRE
a specified temperature. Copper wire specifications from
a reliable manufacturer are required.
The maximum current per wire is a function of the wire
temperatureriseduetocurrent,themaximumwireinsula-
tion temperature and the number of cable wires (refer to
the Copper Wire Information section).
Copper wire resistance increases directly with operating
temperature. The temperature coefficient of copper α is
equal to 0.0039/°C at 20°C (a useful linear approxima-
Table 4 is a random list of AWG wire resistance versus
current based on lab measurements.
tion from 0°C to 100°C). If R
is the resistance at a
LOW
T
temperature and R
is the resistance at a T
LOW
HIGH HIGH
Table 4. A Random List of Wire Resistance vs Current at 20°C
AWG 18
AWG 20
AWG 22
AWG 24
AWG 26
AWG 28
AWG 30
STRANDS/GAUGE STRANDS/GAUGE STRANDS/GAUGE STRANDS/GAUGE STRANDS/GAUGE STRANDS/GAUGE STRANDS/GAUGE
16/30
7/28
7/30
19/36
19/38
7/36
7/38
Current
(AMPS)
R
R
R
R
R
R
R
WIRE
WIRE
WIRE
WIRE
WIRE
WIRE
WIRE
(mΩ/ft)
6.53
6.54
6.56
6.59
6.62
6.65
6.71
6.79
6.83
6.91
(mΩ/ft)
(mΩ/ft)
(mΩ/ft)
22.47
22.66
22.99
23.38
23.78
(mΩ/ft)
37.97
38.41
39.08
40.21
(mΩ/ft)
(mΩ/ft)
1
2
9.61
9.63
9.68
9.73
9.82
9.90
10.02
10.15
15.42
15.51
15.66
15.84
15.99
16.32
62.31
63.32
65.23
102.36
109.14
3
4
5
6
7
8
9
10
6110f
25
For more information www.linear.com/LT6110
LT6110
applicaTions inForMaTion
temperature then the wire’s resistance vs temperature is:
Example: Find the weight of one hundred thousand feet of
18AWG wireandcompareittotheweightofa24AWG wire:
R
HIGH
= R
• (1 + α • (T
– T
)).
LOW
LOW
HIGH
Table 4 shows 6.5mΩ/ft for 18AWG and 22.43mΩ/ft for
24AWG.
An approximation to the temperature rise in a wire due
to current can be derived from the wire’s resistance vs
temperature equation using the wire’s resistance increase
The weight of the 18AWG wire is:
vs safe operating current. If R
is the wire resistance at
–6
2
–3
LOW
(31.39 • 10 ) • [(100000) /(6.5 • 10 • 100000)] =
a low current and R
is the wire resistance at a higher
HIGH
483 pounds.
current and T
is equal to T
– T
then the tem-
RISE
HIGH
LOW
The weight of the 24AWG wire is:
perature rise in a wire is:
–6
2
–3
(31.39 • 10 ) • [(100000) / (22.43 • 10 • 100000)]
T
(°C) = 256.4 • (R
/R
– 1).
RISE
HIGH LOW
= 141 pounds.
Table 4 is a list of measured copper wire resistance ver-
sus current at 20°C for an arbitrary group of 18AWG to
30AWG wires.
Theweightofthe18AWG is3.4× theweightofthe24AWG.
Using an LT6110 simplifies wire drop compensation and
provides the option to specify the smallest size and lowest
cost of copper wire.
Example: Find the wire temperature rise for 3A flowing in
a 28AWG wire. The 28AWG wire on Table 4 has 62.31mΩ/
ft R
resistance at 1A and 65.23mΩ/ft R
resistance
LOW
at 3A.
HIGH
The US Department of Commerce, National Bureau of
Standards Handbook 100 is a comprehensive source of
copper wire information.
T
RISE
for 3A is equal to 256.4 • (65.23/62.31 – 1) = 12°C.
An LT6110 wire drop compensation design requires reli-
able information of wire resistance and current capacity.
Published copper wire tables are a convenient quick-start
guidetocopperwireinformation.Howeveraccuratecopper
wire data is obtained by actual measurements of samples
of copper wire to be used from a reputable manufacturer.
A statistically small sample of copper wire is sufficient for
measurements (the average measured mass resistivity
deviationofalargesampleofcopperwireisonly 0.26%).
Power Dissipation
The LT6110 power dissipation is at a minimum for I
100µAorless.IftheI currentisatitsspecifiedmaximum
of1mAorgreaterthenthemaximumpowerdissipationand
operating temperature must be considered. The LT6110
power dissipation is the sum of three components:
+IN
+IN
V
V
• I
,
IOUT IOUT
• (I + I ) and
SUPPLY
REG
+IN
The International Annealed Copper Standard of mass
resistivity is:
2
I
• R
(if the internal R
is used)
LOAD
SENSE
SENSE
Example of an extreme power dissipation case:
–6
2
153.28 • 10 (Ω-kg)/m in Metric and
V
V
= 50V, I = 1mA.
+IN
–6
2
REG
31.39 • 10 (Ω-lb)/ft in English units.
= 36V, I
= 1mA,
IOUT
IOUT
Mass resistivity is the product of Resistance/Length and
Mass/Length and is useful for estimating the weight of
copper wire required and its cost (the cost of copper wire
depends on its weight and the price fluctuation of copper
in the commodities market).
I
=2.7mA(I
isafunctionofI .SeetheI
SUPPLY
SUPPLY +IN SUPPLY
vs I plot under Typical Performance Characteristics).
+IN
I
= 2A and R
= 20mΩ
LOAD
SENSE
Calculate LT6110 power dissipation:
The weight of copper wire is:
2
Power = 36 • 0.001 + 50 • (0.001 + 0.0027) + 2 • 0.02
–6
–6
2
153.28•10 (Lengthinm )/(ResistanceinΩ)inkilograms
2
Power = 0.301 Watts
or31.39•10 (Lengthinft )/(ResistanceinΩ)inpounds.
6110f
26
For more information www.linear.com/LT6110
LT6110
applicaTions inForMaTion
I
I
LOAD
R
LOAD
WIRE
DROP
V
V
REG
REG
V
LOAD
V
V
IN
IN
I
I
SWITCHING
REGULATOR
SWITCHING
REGULATOR
+IN
SUPPLY
V
R
FA
LOAD
R
IN
R
IN
I
OUT
FB
FB
I
+
+
IOUT
+IN
V
RS –IN
+IN
V
RS –IN
R
R
20mΩ
20mΩ
FB
R
IOUT
+
G
+
–
–
V
IOUT
IOUT
IOUT
–
–
LT6110
6110 F17
LT6110
IMON
V
IMON
V
6110 F18
Figure 17. LT6110 Power Dissipation
Figure 18. Limiting Regulator Voltage Boost (VREGMAX)
The maximum operating ambient temperature T
is
2. Calculate R
:
AMAX
IOUT
equal to T
– θ • Power.
JMAX
JA
R
FA
V –
LOAD
• V
•R
FB
FA
T
is 150°C and θ is 195°C/W for a TSOT-23 pack-
JMAX
age and
JA
R
G
R
=
IOUT
V
– V
LOAD
REGMAX
T
T
is 150°C and θ is 80.6°C/W for a DFN package.
JMAX
JA
Example:Limittheoutputofa5Vregulatortolessthan6V.
= 5V, I = 2A and I = 100µA.
= 150°C – 0.301W • 195°C/W = 91°C for the TSOT-
23 package and
AMAX
V
LOAD
LOADMAX
IOUT
T
= 150°C – 0.301W • 80.6°C/W = 126°C for the
R = 6.49k, R = 422k and R = 80.6k, R = 402Ω, V
AMAX
DFN package.
FA
FB
G
IN
FB
= 0.79V (Figure 4).
Calculate R
:
IOUT
Limiting the Regulator Boost Voltage (V
)
REGMAX
6490
80600
6–5
= 32k and 5.649V ≤ V
In some wire drop compensation applications it may be
necessary to limit the maximum voltage at the regulator
output to ensure the safe operation of all load circuitry.
5–
•0.79 •6490
RIOUT
=
Addingaresistor,R
,inserieswiththeoutputpinlimits
IOUT
R
≤ 6V.
REGMAX
IOUT
the maximum compensation current. This in turn limits
the maximum voltage boosting at the regulator output,
Limiting V
IOUT
V . The increasing I
REGMAX
current through R
IOUT IOUT
The absolute maximum voltage at the IOUT pin (V
) is
IOUT
drops the voltage at the IOUT pin to a minimum level and
limits the maximum IOUT current (refer to the Minimum
IOUTtoIMONVoltagevsTemperaturegraphunderTypical
Performance Characteristics). If the limited IOUT current
is greater than 1mA, a 0.1µF capacitor should be placed
from the IOUT pin to ground to ensure stable operation.
36V. If V
is greater than 36V then a Zener diode from
IOUT
the IOUT pin to the regulator resistors and a resistor from
–
the IOUT pin to V can limit the V
voltage to ≤36V.
IOUT
The Zener diode voltage, V
, is typically specified as
ZENER
a nominal voltage with a minimum and a maximum. For
limiting V , use the minimum Zener voltage rating,
IOUT
. V
The R
resistor limits the regulator’s voltage to an
IOUT
V
is typically specified at a current
ZENERMIN ZENERMIN
of 2mA to 5mA and at the low LT6110 I
arbitrary value higher than V
+ R • I
.
LOAD
FA IOUT
currents
IOUT
Design Procedure:
(≤1mA), the actual V
can be up to 2V less than
ZENERMIN
theminimumvoltagelistedinadiodedatasheet.Therefore
1. Select a V
voltage > V
+ R • I
.
REGMAX
LOAD
FA IOUT
select a Zener diode with a minimum voltage at least 2V
6110f
27
For more information www.linear.com/LT6110
LT6110
applicaTions inForMaTion
I
I
LOAD
R
WIRE
R
LOAD
WIRE
V
V
REG
REG
V
V
LOAD
LOAD
V
V
IN
IN
SWITCHING
REGULATOR
SWITCHING
REGULATOR
V
V
DROP
DROP
R
FA
R
FA
LOAD
LOAD
R
R
IN1
IN
I
OUT
FB
FB
+
+IN
V
RS –IN
R
IN2
R
R
R
R
20mΩ
FB
FB
V
ZENER
+
+IN
V
RS –IN
G
+
G
–
20mΩ
V
IOUT
IOUT
+
–
R
Z
V
IOUT
IOUT
10M
–
LT6110
IMON
V
6110 F19
–
LT6110
IMON
V
Figure 19. Limiting the Voltage at the IOUT Pin (VOUT ≤ 36V)
greater than the calculated V
voltage.
ZENERMIN
V
REG
V
+I
•R
+
FA
V
IOUT IOUT
REGMAX
V
≥ V
–
R
I
ZENERMIN
LOAD
REGMAX
FA
• V
I
I
LOADMAX
LOADCOMP
FB
R
G
WIRE DROP
COMPENSATION
THRESHOLD
V
= V
+ I
• (R
+ R ).
WIRE
REGMAX
LOAD
LOADMAX
SENSE
6110 F20
Example: Limit V
to 20V.
IOUT
Figure 20. Setting the Wire Drop Compensation Threshold
V
= 48V and I
= 2A, R
= 1Ω.
LOAD
LOADMAX
WIRE
threshold, I
, for the start of wire drop compen-
LOADCOMP
R
= 20mΩ, R = 20.5k, R = 453k, R = 12.4k, R
SENSE
FA
FB
G
IN
sation. When the load current is equal to I
the
LOADCOMP
= 402Ω, V = 1.223V, I
= 100µA.
FB
IOUT
maximum error in voltage at the load occurs. For I
greater than I
decreases to zero at I
LOAD
Calculate V
Calculate V
= 48 + 2(0.02 + 1) = 50.04V.
REGMAX
:
ZENERMIN
the error in voltage at the load
LOADMAX
LOADCOMP
.
Design Procedure:
–6
20+ 100•10
•
(
)
1. Choose a threshold current.
2. Calculate R and R
3
:
IN2
IN1
VZENERMIN ≥50.04– 20.5•10 +
(
)
VLOAD +ILOADMAX •RWIRE
20.5•103
ILOADMAX •RSENSE
I
•1.223
IOUT
3
RIN1
=
=
–
12.4•10
VLOAD
ILOADCOMP •RSENSE
I
IOUT
–1
V
= 26V.
ZENERMIN
VLOAD
LOADCOMP •RSENSE
The minimum Zener diode voltage must be ≥28V.
RIN2
–1 •RIN1
I
Setting the Wire Compensation Threshold
Example:Designthestartofwiredropcompensationat1A.
= 5V, I = 3.5A, R = 0.25Ω, R
With light load currents, wire drop compensation may not
be desirable. An additional resistor, R , from the +IN
V
=
SENSE
IN2
LOAD
25mΩ and I
LOADMAX
WIRE
pin to ground provides the option to set a load current
= 100µA.
IOUT
6110f
28
For more information www.linear.com/LT6110
LT6110
applicaTions inForMaTion
I
LOAD
V
1/2R
The R
resistor is a 6mΩ PCB trace.
REG
WIRE
SENSE
µModule
REGULATOR
R
V
SENSE
IN
+
I
= 10A and set I
= 100µA.
R
INT
R
LOAD
IOUT
F
LOAD
V
R
LOAD
IN
I
V
OUT
FB
–
1/2R
+
Calculate R , R and R .
+IN
V
RS –IN
F
G
IN
WIRE
R
G
For I
= 100µA, R = (10 • 0.006)/0.0001 = 600Ω and
IN
IOUT
to the nearest 1% resistor R = 604Ω.
+
–
IN
IOUT
If R = 604Ω then I
SENSE IN
= 99.34µA [I
= (I
•
IN
IOUT
IOUT
LOAD
R
/R )].
–
LT6110
IMON
V
6110 F21
10
105 •
• 0.006+0.15
(
)
99.34•10–6
RF =
10
Figure 21. An LT6110 with a µModule Regulator
105 –
• 0.006+0.15
(
)
99.34•10–6
1. I
= 1A.
LOADCOMP
R = 18.7k (to the nearest 1% value).
F
2. Calculate R and R : R = 576Ω and R = 115k.
IN1
IN2 IN1
IN2
18.7•103 •105 •0.6
AtI
=1AV
=4.75VandatI
=3.5AV
=5V.
(
)
LOAD
LOAD
LOAD
LOAD
RG =
18.7•103 +105 • 3–0.6
Typically a µModule® regulator contains a resistor (R
)
INT
(
)
)
(
from the regulator’s output to the error amplifier’s input.
The µModule resistor is inaccessible and is in parallel to
R = 3.92k (to the nearest 1% value).
G
the external feedback resistor (R ) required for wire drop
F
compensation with an LT6110 (the R value is listed in
Regulator Loop Stability
INT
the µModule regulator data sheet).
A regulator’s control loop response is optimized for a
variety of load, input voltage and temperature conditions.
Adding an LT6110 to a regulator circuit does not disturb
control loop stability. However an LT6110 adds a pole
that reduces the loop’s phase margin. The effect of the
LT6110 pole in the loop is easily compensated by a zero
in the feedback divider.
Design Procedure:
1.ChoosethecompensationcurrentI
(100µAtypically).
IOUT
2. Calculate R , R and R .
F
G
IN
ILOAD
RINT
RINT
•
• R
+RWIRE
+RWIRE
(
)
)
SENSE
I
IOUT
RF =
Figure 22 shows a small-signal model for a current mode
buckregulatorwithanLT6110inthecontrolloop.Theopen
ILOAD
–
• R
(
SENSE
I
IOUT
looptransferfunctionfromtheerroramplifieroutput(V ),
C
RF •RINT
R +R
V
FB
to the modulator output (V ), to the feedback divider
RG =
RIN =
•
REG
V
– V
FB
(
)
(
)
F
INT
REG
output (V ) is: (V /V ) (V /V ) (V /V ).
FB
REG
C
FB REG
C
FB
ILOAD •RSENSE
Theloop’sDCgainisequaltotheproductofthemodulator
gain (g • R ), the error amplifier gain (g • R ) and
I
m
LOAD
e
e
IOUT
the feedback ratio (V /V ).
REF REG
Example:Use24ft, 18AWG wiretoregulatea3V, 10Aload,
using an LTM4600 µModule regulator.
The overall regulator control loop frequency response is
determined by a combination of several poles and zeros.
Loop compensation is provided by the R1 and C1 zero
at the error amplifier’s output. This zero adds a positive-
going phase near the loop’s crossover frequency and is
R
of LTM4600 is 100k and the feedback voltage V
FB
INT
= 0.6V.
The R
of 24ft, 18AWG is 0.15Ω.
WIRE
6110f
29
For more information www.linear.com/LT6110
LT6110
applicaTions inForMaTion
BUCK REGULATOR MODEL
V
IN
–
+
MODULATOR
g
IS THE
m
MODULATOR
TRANSCONDUCTANCE
SLOPE COMP
OSCILLATOR
R
Q
S
LT6110 MODEL
L
V
REG
R
C
O
R
IN
C
FA
COMP
I
R
WIRE
IOUT
R
SENSE
V
LOAD
R
ESR
V
C
+
–
V
REF
R
R
FB
R
V
LOAD
FB
R1
C1
R
e
C2
6110 F22
G
ERROR
LT6110
AMPLIFIER
g IS THE ERROR
e
AMPLIFIER
TRANSCONDUCTANCE
Figure 22. A Small-Signal Model: Current Mode Buck Regulator with an LT6110
adjusted for an optimum phase margin. Regulator loop
compensation,transientresponseandstabilityarecovered
in depth in AN76.
C
zero is best adjusted during a load transient test.
COMP
Start with a C
160kHz (the LT6110 pole), then increase C
transient that settles with minimal overshoot or ringing.
value for a zero equal to or less than
COMP
for a load
COMP
An LT6110 in the control loop introduces a pole near
160kHz (from the Typical Performance Curves) and this
pole reduces the loop’s optimized phase margin resulting
in load transient overshoot and possibly ringing. Adding a
Figure23showsanLT3980buckregulatorwithanLT6110
circuit used for transient response testing and with the
added zero to restore the loop’s phase margin. During
the circuit’s load transient testing, a C
producesaloadtransientthatsettleswithoutovershootor
capacitor, C
in parallel with the regulator’s feedback
COMP
value of 1nF
COMP
resistance, R introduces a zero to compensate the ef-
FA
fects of the LT6110 pole. The frequency of the R and
FA
V
V
BD
IN
IN
RUN/SS BOOST
LT3980
0.47µF
10µH
100k
V
REG
C
PGOOD
SW
DA
FB
R
IN
R
COMP
FA
47µF
402Ω
1nF
97.6k
6.49k
NC
+IN
R
T
0.1µF
LT6110
+
R
IOUT
V
1.5nF
FB
15k
412k
20mΩ
IMON
V
C
R
WIRE
GND
RS
R
G
0.3Ω
100pF
V
LOAD
80.6k
GND
–IN
5Ω
8Ω
10µF
1.6A
2k
1A
180pF
6110 F23
Figure 23. Load Transient Response Test Circuit Using an LT3980 Buck Regulator with an LT6110
6110f
30
For more information www.linear.com/LT6110
LT6110
applicaTions inForMaTion
ringing (a 10% C
tolerance is adequate). An optional
Figure 24b shows a load transient response of the LT3980
buck regulator with LT6110 line drop compensated load
voltage. The load transient has an overshoot due to the
LT6110 decreasing the phase margin.
COMP
connection for C
is in parallel with R and R (from
COMP
FA FB
value for the smallest
V
to V ) to reduce the C
REG
FB
COMP
capacitor size.
Figures 24a through 24c illustrate a typical loop opti-
mization procedure when an LT6110 is included in the
regulator’s loop.
Figure 24c shows a load transient response of the LT3980
buckregulatorwithanLT6110andwithaC
capacitor
COMP
added to compensate for the LT6110 in the loop. The load
transient settles without overshoot as the phase margin
is restored.
Figure 24a shows a load transient response of the LT3980
buck regulator with an optimum phase margin without
linedropcompensation. Theloadtransientsettleswithout
overshoot.
V
V
V
V
REG
REG
200mV/DIV
200mV/DIV
200mV/DIV
200mV/DIV
LOAD
LOAD
1.6A
1A
1.6A
1A
I
I
LOAD
LOAD
6110 F24a
6110 F24b
200µs/DIV
200µs/DIV
Figure 24a. Transient Response of Buck Regulator without
LT6110 Line Drop Compensation
Figure 24b. Transient Response Buck Regulator with an LT6110
in the Loop
V
V
REG
200mV/DIV
200mV/DIV
LOAD
1.6A
1A
I
LOAD
6110 F24c
200µs/DIV
Figure 24c. Capacitor CCOMP Compensates for the LT6110 in the
Regulator’s Loop
6110f
31
For more information www.linear.com/LT6110
LT6110
Typical applicaTions
LT6110 with External RSENSE and LT3690 Buck Regulator at 3.3V
V
IN
V
IN
6.5V TO 25V
10µF
EN
BST
SW
0.68µF
4.7µH
UVLO
SS
1000pF
R
IN
10.2k
47pF
BIAS
PG
340Ω
LT3690
GND
1
2
8
7
V
C
NC
+IN
100µF
22k
680pF
+
IOUT
V
301k
100k
0.8V
0.1µF
LT6110
R
*
FB
RT
SENSE
3
4
6
5
R
WIRE
V
CCINT
IMON
RS
8.5mΩ
0.25Ω
0.47µF
SYNC
GND
–IN
32.4k
600kHz
V
3.3V
4A
LOAD
LOAD
*THE CURRENT SENSE RESISTOR
IS THE DCR OF A LOW COST INDUCTOR.
COILCRAFT 0908SQ-27N (27nH)
6110 TA02
WIRE DROP COMPENSATION: V
= 3.3V, I
= 4A, USING 10ft, 24AWG WIRE.
LOAD
LOADMAX
MEASURED V
REGULATION FOR 0 ≤ I
≤ 4A AT 25°C:
LOAD
LOAD
WITHOUT COMPENSATION: ∆V
= 1000mV (250mV/A)
LOAD
WITH COMPENSATION: ∆V
= 16mV (4mV/A)
LOAD
6110f
32
For more information www.linear.com/LT6110
LT6110
Typical applicaTions
LT6110 with External RSENSE and LT3690 Buck Regulator at 5V
V
IN
V
IN
8.5V TO 36V
10µF
EN
BST
SW
0.68µF
10µH
UVLO
SS
1000pF
R
IN
20.5k
47pF
BIAS
PG
340Ω
LT3690
GND
1
2
8
7
V
C
NC
+IN
47µF
22k
680pF
+
IOUT
V
316k
100k
0.8V
0.1µF
LT6110
R
*
FB
RT
SENSE
3
4
6
5
R
WIRE
V
CCINT
IMON
RS
8.5mΩ
0.5Ω
0.47µF
SYNC
GND
–IN
32.4k
600kHz
V
5V
4A
LOAD
LOAD
*THE CURRENT SENSE RESISTOR
IS THE DCR OF A LOW COST INDUCTOR.
COILCRAFT 0908SQ-27N (27nH)
6110 TA03
WIRE DROP COMPENSATION: V
= 5V, I
= 4A, USING 20ft, 24AWG WIRE.
LOAD
LOADMAX
MEASURED V
REGULATION FOR 0 ≤ I
≤ 4A AT 25°C:
LOAD
LOAD
WITHOUT COMPENSATION: ∆V
= 2000mV (500mV/A)
LOAD
WITH COMPENSATION: ∆V
= 24mV (6mV/A)
LOAD
6110f
33
For more information www.linear.com/LT6110
LT6110
Typical applicaTions
LT6110 with External PCB RSENSE and LTM4600 µRegulator at 3.3V
V
IN
+
150µF 22µF
22µF
6V TO 24V
V
REG
V
IN
V
REG
22µF 22µF + 470µF
LTM4600HV
V
1M
REG
R
IN
523Ω
RF
1
2
8
7
RUN/SS
100k
NC
+IN
18.7k
0.6V
0.1µF
+
IOUT
V
PCB
R
6mΩ
20ꢀ
RG
3.92k
0.1µF
FCB SGND PGND VOSET
LT6110
SENSE
3
4
6
5
R
WIRE
IMON
RS
0.075Ω
GND
–IN
+
LOAD
–
V
LOAD
3V
R
10A
WIRE
WIRE DROP COMPENSATION: V
= 3.3V, I
= 10A, USING 24ft, 18AWG WIRE WITH GROUND RETURN.
LOAD
LOADMAX
0.075Ω
MEASURED V
REGULATION FOR 0 ≤ I
≤ 10A AT 25°C:
LOAD
LOAD
WITHOUT COMPENSATION: ∆V
= 1500mV (150mV/A)
LOAD
WITH COMPENSATION: ∆V
= 50mV (5mV/A)
LOAD
6110 TA04
LT6110 with External PCB RSENSE and LTM4600 µRegulator at 1.8V
V
IN
+
150µF 22µF
22µF
1M
5V TO 24V
V
REG
V
IN
V
REG
22µF 22µF + 470µF
LTM4600HV
V
REG
R
IN
523Ω
RF
1
2
8
7
RUN/SS
100k
NC
+IN
18.7k
0.6V
0.1µF
+
IOUT
V
PCB
R
6mΩ
20ꢀ
RG
7.87k
0.1µF
FCB SGND PGND VOSET
LT6110
SENSE
3
4
6
5
R
WIRE
IMON
RS
0.075Ω
GND
–IN
+
LOAD
–
V
1.8V
10A
LOAD
R
WIRE
WIRE DROP COMPENSATION: V
= 1.8V, I
= 10A, USING 24ft, 18AWG WIRE WITH GROUND RETURN.
LOAD
LOADMAX
0.075Ω
MEASURED V
REGULATION FOR 0 ≤ I
≤ 10A AT 25°C:
LOAD
LOAD
WITHOUT COMPENSATION: ∆V
= 1500mV (150mV/A)
LOAD
WITH COMPENSATION: ∆V
= 50mV (5mV/A)
LOAD
6110 TA05
6110f
34
For more information www.linear.com/LT6110
LT6110
Typical applicaTions
LT6110 with External RSENSE and LTC3789 Buck-Boost Regulator at 12V
INTV
V
CC
IN
5V TO 36V
+
C
270µF
100k
A
390pF
0.22µF
5.6Ω
10Ω
1
2
27
PGOOD
SW1
V
FB
26
25
24
23
22
Q
A
TG1
BOOST1
PGND
SS
Si7848BDP
0.01µF
1nF
3
4
D
+
–
A
SENSE
SENSE
L
DFLS160
Q
B
4.7µH
BG1
0.01µF
Si7848BDP
14.7k
1µF
5
V
IN
I
TH
LTC3789
10µF
6
SGND
D1
21
20
10mΩ
7
B240A
INTV
EXTV
MODE/PLLIN
FREQ
CC
121k
8
CC
D
B
DFLS160
9
RUN
400kHz
19
18
10
11
12
Q
C
V
BG2
V
IN
INSNS
SiR496DP
V
REG
BOOST2
V
REG
V
OUTSNS
C
B
I
V
REG
LIM
100pF
0.22µF
R
IN
619Ω
BZT52C6V2S
13
14
17
16
15
Q
D
SiR496DP
+
5.62k
TG2
SW2
TRIM
I
OSENSE
1
2
8
7
D2
B240A
NC
+IN
2.2µF
–
I
OSENSE
+
IOUT
V
133k
10k
EXTERNAL
LT6110
R
SENSE
3
4
6
5
0.1µF
R
WIRE
0.1Ω
100Ω
IMON
RS
25mΩ
5ꢀ
10mΩ
2.2µF
100Ω
1k
GND
–IN
+
LOAD
–
+
1k
330µF
0.8V
= 5A, USING 20ft, 20AWG WIRE WITH GROUND RETURN.
12V
5A
R
WIRE
0.1Ω
WIRE DROP COMPENSATION: V
= 12V, I
LOADMAX
6110 TA06
LOAD
MEASURED V
REGULATION FOR 0 ≤ I
≤ 5A AT 25°C:
= 1000mV (250mV/A)
LOAD
WITHOUT COMPENSATION: ∆V
LOAD
LOAD
= 25mV (5mV/A)
WITH COMPENSATION: ∆V
LOAD
6110f
35
For more information www.linear.com/LT6110
LT6110
package DescripTion
Please refer to http://www.linear.com/designtools/packaging/ for the most recent package drawings.
TS8 Package
8-Lead Plastic TSOT-23
(Reference LTC DWG # 05-08-1637 Rev A)
2.90 BSC
(NOTE 4)
0.40
MAX
0.65
REF
1.22 REF
1.4 MIN
1.50 – 1.75
(NOTE 4)
2.80 BSC
3.85 MAX 2.62 REF
PIN ONE ID
RECOMMENDED SOLDER PAD LAYOUT
PER IPC CALCULATOR
0.22 – 0.36
8 PLCS (NOTE 3)
0.65 BSC
0.80 – 0.90
0.20 BSC
DATUM ‘A’
0.01 – 0.10
1.00 MAX
0.30 – 0.50 REF
1.95 BSC
TS8 TSOT-23 0710 REV A
0.09 – 0.20
(NOTE 3)
NOTE:
1. DIMENSIONS ARE IN MILLIMETERS
2. DRAWING NOT TO SCALE
3. DIMENSIONS ARE INCLUSIVE OF PLATING
4. DIMENSIONS ARE EXCLUSIVE OF MOLD FLASH AND METAL BURR
5. MOLD FLASH SHALL NOT EXCEED 0.254mm
6. JEDEC PACKAGE REFERENCE IS MO-193
6110f
36
For more information www.linear.com/LT6110
LT6110
package DescripTion
Please refer to http://www.linear.com/designtools/packaging/ for the most recent package drawings.
DC8 Package
8-Lead Plastic DFN (2mm × 2mm)
(Reference LTC DWG # 05-08-1719 Rev A)
0.70 ±0.05
2.55 ±0.05
0.64 ±0.05
1.15 ±0.05
(2 SIDES)
PACKAGE
OUTLINE
0.25 ±0.05
0.45 BSC
1.37 ±0.05
(2 SIDES)
RECOMMENDED SOLDER PAD PITCH AND DIMENSIONS
APPLY SOLDER MASK TO AREAS THAT ARE NOT SOLDERED
R = 0.115
TYP
5
8
R = 0.05
TYP
0.40 ±0.10
PIN 1 NOTCH
2.00 ±0.10 0.64 ±0.10
(4 SIDES)
(2 SIDES)
R = 0.20 OR
0.25 × 45°
CHAMFER
PIN 1 BAR
TOP MARK
(SEE NOTE 6)
(DC8) DFN 0409 REVA
4
1
0.23 ±0.05
0.45 BSC
0.75 ±0.05
0.200 REF
1.37 ±0.10
(2 SIDES)
BOTTOM VIEW—EXPOSED PAD
0.00 – 0.05
NOTE:
1. DRAWING IS NOT A JEDEC PACKAGE OUTLINE
2. DRAWING NOT TO SCALE
3. ALL DIMENSIONS ARE IN MILLIMETERS
4. DIMENSIONS OF EXPOSED PAD ON BOTTOM OF PACKAGE DO NOT INCLUDE
MOLD FLASH. MOLD FLASH, IF PRESENT, SHALL NOT EXCEED 0.15mm ON ANY SIDE
5. EXPOSED PAD SHALL BE SOLDER PLATED
6. SHADED AREA IS ONLY A REFERENCE FOR PIN 1 LOCATION ON THE
TOP AND BOTTOM OF PACKAGE
6110f
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 representa-
37
tion that the interconnection of its circuits as described herein will not infringe on existing patent rights.
LT6110
Typical applicaTion
LT6110 with Internal RSENSE and LT3975 Buck Regulator at 3.3V
V
IN
B00ST
V
IN
7V TO 42V
0.47µF
4.7µH
100µF
10µF
EN
SW
PDS360
100µF
V
LT3975
REG
OUT
FB
R
IN
1µF
16.5k
499Ω
1
2
8
7
NC
+IN
+
SS
RT
78.7k
f = 600kHz
IOUT
V
10pF
1M
10nF
1.197V
0.1µF
LT6110
3
4
6
5
R
SYNC GND
WIRE
IMON
RS
590k
0.32Ω
GND
–IN
+
LOAD
–
V
3.3V
2.5A
LOAD
R
WIRE
WIRE DROP COMPENSATION: V
= 3.3V, I
= 2.5A, USING 6ft, 30AWG WIRE WITH GROUND RETURN.
LOAD
LOADMAX
0.32Ω
MEASURED V
REGULATION FOR 0 ≤ I
≤ 2.5A AT 25°C:
LOAD
LOAD
WITHOUT COMPENSATION: ∆V
= 1600mV (640mV/A)
LOAD
WITH COMPENSATION: ∆V
= 25mV (10mV/A)
LOAD
6110 TA07
relaTeD parTs
PART NUMBER DESCRIPTION
COMMENTS
LT1787
LTC4150
LT6100
LTC6101
LTC6102
LTC6103
LTC6104
LT6105
LT6106
LT6107
LT6700
LT4180
Bidirectional High Side Current Sense Amplifier
Coulomb Counter/Battery Gas Gauge
2.7V to 60V, 75µV Offset, 60µA Quiescent, 8V/V Gain
Indicates Charge Quantity and Polarity
Gain-Selectable High Side Current Sense Amplifier
High Voltage High Side Current Sense Amplifier
Zero Drift High Side Current Sense Amplifier
Dual High Side Current Sense Amplifier
4.1V to 48V, Gain Settings: 10, 12.5, 20, 25, 40, 50V/V
Up to 100V, Resistor Set Gain, 300µV Offset, SOT-23
Up to 100V, Resistor Set Gain, 10µV Offset, MSOP8/DFN
4V TO 60V, Resistor Set Gain, 2 Independent Amps, MSOP8
4V TO 60V, Separate Gain Control for Each Direction, MSOP8
–0.3V to 44V Input Range, 300µV Offset, 1% Gain Error
2.7V to 36V, 250µV Offset, Resistor Set Gain, SOT23
2.7V to 36V, –55°C 150°C, Fully Tested: –55°C, 25°C, 150°C
1.4V to 18V, 6.5µA Supply Current
Bidirectional High Side Current Sense Amplifier
Precision Rail-to-Rail Input Current Sense Amplifier
Low Cost High Side Current Sense Amplifier
High Temperature High Side Current Sense Amplifier
Dual Comparator with 400mV Reference
Virtual Remote Sense Controller
Automatically Detects Line Impedance to Improve Load Regulation
6110f
LT 0413 • PRINTED IN USA
LinearTechnology Corporation
1630 McCarthy Blvd., Milpitas, CA 95035-7417
38
●
●
LINEAR TECHNOLOGY CORPORATION 2013
(408)432-1900 FAX: (408) 434-0507 www.linear.com/LT6110
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