20
LTC3720
3720f
APPLICATIO S I FOR ATIO
WUUU
By positioning the output voltage 60mV above the regula-
tion point at no load, it will only drop 65mV below the
regulation point after the load step, well within the ±100mV
tolerance.
Implementing active voltage positioning requires setting a
precise gain between the sensed current and the output
voltage. Because of the variability of MOSFET on-resis-
tance, it is prudent to use a sense resistor with active
voltage positioning. In order to minimize power lost in this
resistor, a low value is chosen of 0.003Ω. The nominal
sense voltage will now be:
V
SNS(NOM)
= (0.003Ω)(15A) = 45mV
To maintain a reasonable current limit, the voltage on the
V
RNG
pin is reduced to its minimum value of 0.5V, corre-
sponding to a 50mV nominal sense voltage.
Next, the gain of the LTC3720 error amplifier must be
determined. The change in I
TH
voltage for a corresponding
change in the output current is:
∆=
∆
=
()
Ω
()()
=
I
V
V
RI
AV
TH
RNG
SENSE OUT
12
24 0 003 15 1 08..
The corresponding change in the output voltage is deter-
mined by the gain of the error amplifier and feedback
divider. The LTC3720 error amplifier has a transconduc-
tance g
m
that is constant over both temperature and a wide
± 40mV input range. Thus, by connecting a load resis-
tance R
VP
to the I
TH
pin, the error amplifier gain can be
precisely set for accurate active voltage positioning.
∆=
∆IgR
V
V
V
TH m VP
OUT
OUT
08.
Solving for this resistance value:
R
VI
Vg V
VV
VmS mV
k
VP
OUT TH
m OUT
=
∆
∆
==
(. )
(. )(. )
(. )(. )( )
.
08
15 108
08 17 125
953
The gain setting resistance R
VP
is implemented with two
resistors, R
VP1
connected from I
TH
to ground and R
VP2
connected from I
TH
to INTV
CC
. The parallel combination of
these resistors must equal R
VP
and their ratio determines
nominal value of the I
TH
pin voltage when the error
amplifier input is zero. To center the load line around the
regulation point, the I
TH
pin voltage must be set to corre-
spond to half the output current. The relation between I
TH
voltage and the output current is:
I
V
V
RI I V
V
V
AAV
V
TH NOM
RNG
SENSE OUT L()
–.
.
..–..
.
=
∆
+
=
Ω
()
+
=
12 1
2
08
12
05
0 003 7 5
1
2
47 08
117
Solving for the required values of the resistors:
R
V
VI
R
V
VV
k
k
R
V
I
R
V
V
kk
VP
TH NOM
VP
VP
TH NOM
VP
1
2
5
5
5
5117
953
12 44
55
117
953 4073
==
=
===
––.
.
.
.
..
()
()