LT1375 Linear Technology, LT1375 Datasheet - Page 24

• Manufacturer Part Number: LT1375
• Description: 1.5A/ 500kHz Step-Down Switching Regulators
• Manufacturer: Linear Technology

4.5V TO
LT1375/LT1376
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.
APPLICATIONS
24
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.
INPUT
10µF TO
20V
** MAXIMUM LOAD CURRENT DEPENDS ON MINIMUM INPUT VOLTAGE
50µF
* INCREASE L1 TO 10µH OR 20µH FOR HIGHER CURRENT APPLICATIONS.
C3
SEE APPLICATIONS INFORMATION
AND INDUCTOR SIZE. SEE APPLICATIONS INFORMATION
+
Figure 18. Positive-to-Negative Converter
V
IN
GND
LT1376-5
BOOST
U
V
C
SENSE
R
V
C
C
SW
INFORMATION
C
U
D2
1N5818
C2
0.1µF
W
1N4148
D1
5µH
L1*
+
U
C1
100µF
10V TANT
OUTPUT**
–5V, 0.5A
1375/76 F18
Maximum load current:
I
V
V
V
0.5 = Switch voltage drop at 1.5A
Example: with V
0.5V, I
not take into account that maximum rated switch current
(I
above 50%. If duty cycle is expected to exceed 50% (input
voltage less than output voltage), use the actual I
from the Electrical Characteristics table.
Operating duty cycle:
(This formula uses an average value for switch loss, so it
may be several percent in error.)
With the conditions above:
This duty cycle is close enough to 50% that I
assumed to be 1.5A.
OUTPUT DIVIDER
If the adjustable part is used, the resistor connected to
V
calculated from:
P
IN
OUT
F
OUT
P
= Maximum rated switch current
) on the LT1376 is reduced slightly for duty cycles
I
= Catch diode forward voltage
DC
DC =
R
MAX
= Minimum input voltage
1
= Output voltage
=
(R2) should be set to approximately 5k. R1 is
P
=
R V
= 1.5A: I
=
V
4 7 0 3 5 0 5
2
IN
.
⎡
⎢
⎢
⎣
(
I
P
−
−
OUT
−
2 42
0 3 .
V
5 0 5
.
2
OUT
IN(MIN)
MAX
.
+
(
(
−
V
V
+
+ +
OUT
OUT
2 42
( )( )
.
+
V
V
= 0.52A. Note that this equation does
.
IN
OUT
V
= 4.7V, V
+
F
+
)
V
V
V
+
.
OUT
IN
IN
V
)( )( )
=
−
F
f L
56
0 5
.
OUT
%
)
⎤
⎥
⎥
⎦
(
V
( )
= 5V, L = 10µH, V
V
OUT
OUT
+
(
V
V
F
IN
)
−
P
0 5
P
can be
.
value
13756fd
)
F
=