Datasheet

LT6110
17
6110fa
For more information www.linear.com/LT6110
applicaTions inForMaTion
To program the regulator output voltage and compensate
for V
DROP
at I
LOAD(MAX)
the following procedure can be
used:
Make I
SET
>> I
ADJ
, if I
SET
= 33.3 • I
ADJ
then I
SET
= 2.5mA
R1=
V
REF
I
SET
=
1.25V
2.5mA
= 499Ω
For 12V output with no-load current:
R2+R
G
( )
=
V
LOAD
– V
REF
I
SET
+I
ADJ
=
10.75V
2.575mA
= 4.175
k
Resistor R
G
is used to develop the maximum load current
compensation voltage. A smaller value for R
G
minimizes
the voltage programming error at no load but requires
more current from the LT6110 IMON pin to compensate
for cable drop loss. The IMON pin current is most accurate
over a range from 30µA to 3mA.
R
G
=
V
DROP
I
IMON
For 1.375V of compensation, using a convenient value 1k
resistor for R
G
will require 1.375mA from the IMON pin
which is near the mid range of accurate current levels.
With this selection for R
G
then:
R2 = 4.175k – 1k = 3.175k
use a 3.16k standard 1% tolerance value to set the no-load
output voltage to 12V.
To program the LT6110 compensation current requires
a selection for R
IN
:
R
IN
=
V
SENSE
I
+IN
=
V
SENSE
I
IMON
3
V
SENSE
= 5A • 25mΩ = 125mV and
I
IMON
3
=
1.375mA
3
= 460µA so
R
IN
=
125mV
460µA
= 271Ω
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
I
LOAD
is always greater than 10mA) is added to the output.
The voltage that appears at the IMON pin can impact the
accuracy of the compensation circuit and should be noted.
In this example the voltage will be a maximum at full load
current and voltage compensation. This voltage is:
V
IMON(MAX)
= (I
SET
+ I
ADJ
+ I
IMON
) • R
G
= (2.5mA + 75µA
+ 1.375mA) • 1k = 3.95V.
IOUT
+IN –IN
R
IN
LT6110
V
+
V
–
RS
IMON
I
IMON
I
IMON
= 3 I
+IN
+
–
R
WIRE
0.25Ω
V
LOAD
12V
5A
6110 F06
20mΩ
R2
3.16k
R1
499Ω
R
G
1k
V
REG
V
IN
I
ADJ
10µF
1.62k
R
SENSE
0.025Ω
I
SET
10µF
OUTIN
LT1083
ADJ
R1=
1.25V
I
SET
R2 =
R1• V
LOAD
– V
REF
( )
I
ADJ
•R1+ V
REF
– R
G
R
G
=
I
LOAD
• R
SENSE
+R
WIRE
( )
3 • I
+IN
V
IMON(MAX)
= R
G
• I
SET
+I
ADJ
+ 3 •I
+IN
( )
Figure 6. Wire Compensation Using a High Current Adjustable Regulator