Datasheet
24
LT1375/LT1376
13756fd
APPLICATIONS INFORMATION
WUU
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POSITIVE-TO-NEGATIVE CONVERTER
The circuit in Figure 18 is a classic positive-to-negative
topology using a grounded inductor. It differs from the
standard approach in the way the IC chip derives its
feedback signal, however, because the LT1376 accepts
only positive feedback signals, the ground pin must be tied
to the regulated negative output. A resistor divider to
ground or, in this case, the sense pin, then provides the
proper feedback voltage for the chip.
OUTPUT**
–5V, 0.5A
INPUT
4.5V TO
20V
1375/76 F18
C2
0.1µF
C
C
R
C
D2
1N5818
C1
100µF
10V TANT
C3
10µF TO
50µF
D1
1N4148
L1*
5µH
BOOST
LT1376-5
V
IN
V
SW
SENSE
GND
V
C
* INCREASE L1 TO 10µH OR 20µH FOR HIGHER CURRENT APPLICATIONS.
SEE APPLICATIONS INFORMATION
** MAXIMUM LOAD CURRENT DEPENDS ON MINIMUM INPUT VOLTAGE
AND INDUCTOR SIZE. SEE APPLICATIONS INFORMATION
+
+
Figure 18. Positive-to-Negative Converter
Inverting regulators differ from buck regulators in the
basic switching network. Current is delivered to the output
as
square waves with a peak-to-peak amplitude much
greater than load current
. This means that
maximum load
current will be significantly less than the LT1376’s 1.5A
maximum switch current, even with large inductor values
.
The buck converter in comparison, delivers current to the
output as a triangular wave superimposed on a DC level
equal to load current, and load current can approach 1.5A
with large inductors. Output ripple voltage for the positive-
to-negative converter will be much higher than a buck
converter. Ripple current in the output capacitor will also
be much higher. The following equations can be used to
calculate operating conditions for the positive-to-negative
converter.
Maximum load current:
I
I
VV
VVfL
VV
VV VV
MAX
P
IN OUT
OUT IN
OUT IN
OUT IN OUT F
=
−
()( )
+
()()()
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
()
−
()
+ −
()
+
()
2
05
05
.
.
I
P
= Maximum rated switch current
V
IN
= Minimum input voltage
V
OUT
= Output voltage
V
F
= Catch diode forward voltage
0.5 = Switch voltage drop at 1.5A
Example: with V
IN(MIN)
= 4.7V, V
OUT
= 5V, L = 10µH, V
F
=
0.5V, I
P
= 1.5A: I
MAX
= 0.52A. Note that this equation does
not take into account that maximum rated switch current
(I
P
) on the LT1376 is reduced slightly for duty cycles
above 50%. If duty cycle is expected to exceed 50% (input
voltage less than output voltage), use the actual I
P
value
from the Electrical Characteristics table.
Operating duty cycle:
DC
VV
VVV
OUT F
IN OUT F
=
+
− ++03.
(This formula uses an average value for switch loss, so it
may be several percent in error.)
With the conditions above:
DC =
+
− ++
=
505
47 03 5 05
56
.
.. .
%
This duty cycle is close enough to 50% that I
P
can be
assumed to be 1.5A.
OUTPUT DIVIDER
If the adjustable part is used, the resistor connected to
V
OUT
(R2) should be set to approximately 5k. R1 is
calculated from:
R
RV
OUT
1
2242
242
=
−
()
.
.