Datasheet

LT6110
29
6110fa
For more information www.linear.com/LT6110
applicaTions inForMaTion
1. I
LOADCOMP
= 1A.
2. Calculate R
IN1
and R
IN2
: R
IN1
= 576Ω and R
IN2
= 115k.
At I
LOAD
= 1A V
LOAD
= 4.75V and at I
LOAD
= 3.5A V
LOAD
= 5V.
Wire Drop Compensation Using a µModule Regulator
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
the external feedback resistor (R
F
) required for wire drop
compensation with an LT6110 (the R
INT
value is listed in
the µModule regulator data sheet).
Design Procedure:
1. Choose the compensation current I
IOUT
(100µA typically).
2. Calculate R
F
, R
G
and R
IN
.
R
F
=
R
INT
•
I
LOAD
I
IOUT
• R
SENSE
+R
WIRE
( )
R
INT
–
I
LOAD
I
IOUT
• R
SENSE
+R
WIRE
( )
R
G
=
R
F
•R
INT
R
F
+R
INT
( )
•
V
FB
V
REG
– V
FB
( )
R
IN
=
I
LOAD
•R
SENSE
I
IOUT
Example: Use 24ft, 18AWG wire to regulate a 3V, 10A load,
using an LTM4600 µModule regulator.
R
INT
of LTM4600 is 100k and the feedback voltage V
FB
= 0.6V.
The R
WIRE
of 24ft, 18AWG is 0.15Ω.
The R
SENSE
resistor is a 6mΩ PCB trace.
I
LOAD
= 10A and set I
IOUT
= 100µA.
Calculate R
F
, R
G
and R
IN
.
For I
IOUT
= 100µA, R
IN
= (10 • 0.006)/0.0001 = 600Ω and
to the nearest 1% resistor R
IN
= 604Ω.
If R
IN
= 604Ω then I
IOUT
= 99.34µA [I
IOUT
= (I
LOAD
•
R
SENSE
/R
IN
)].
R
F
=
10
5
•
10
99.34 • 10
–6
• 0.006 + 0.15
( )
10
5
–
10
99.34 • 10
–6
• 0.006 + 0.15
(
)
R
F
= 18.7k (to the nearest 1% value).
R
G
=
18.7 • 10
3
( )
• 10
5
• 0.6
18.7 • 10
3
+10
5
( )
• 3 – 0.6
(
)
R
G
= 3.92k (to the nearest 1% value).
Regulator Loop Stability
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.
Figure 22 shows a small-signal model for a current mode
buck regulator with an LT6110 in the control loop. The open
loop transfer function from the error amplifier output (V
C
),
to the modulator output (V
REG
), to the feedback divider
output (V
FB
) is: (V
REG
/V
C
) (V
FB
/V
REG
) (V
C
/V
FB
).
The loop’s DC gain is equal to the product of the modulator
gain (g
m
• R
LOAD
), the error amplifier gain (g
e
• R
e
) and
the feedback ratio (V
REF
/V
REG
).
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
Figure 21. An LT6110 with a µModule Regulator
V
IN
µModule
REGULATOR
R
INT
IOUT
+IN –IN
R
IN
LT6110
V
+
V
–
RS
R
F
R
G
IMON
+
–
V
REG
V
FB
I
LOAD
I
OUT
1/2R
WIRE
6110 F21
R
SENSE
V
LOAD
+
–
LOAD
1/2R
WIRE