LT1375/ LT1376
U
W
U U
APPLICATIONS INFORMATION
POSITIVE-TO-NEGATIVE CONVERTER
Maximum load current:
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
onlypositivefeedbacksignals,thegroundpinmustbetied
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.
V V
( )(
)
IN OUT
IP −
V
V − 0.5
IN
(
OUT)(
)
2 V + V f L
(
IN)( )( )
OUT
IMAX
=
V
+V − 0.5 V + V
OUT
IN
OUT
F
IP = Maximum rated switch current
V = Minimum input voltage
IN
VOUT = Output voltage
V = Catch diode forward voltage
F
D1
1N4148
0.5 = Switch voltage drop at 1.5A
C2
0.1µF
L1*
5µH
INPUT
4.5V TO
20V
BOOST
Example: with VIN(MIN) = 4.7V, VOUT = 5V, L = 10µH, V =
F
V
V
IN
SW
0.5V, IP = 1.5A: IMAX = 0.52A. Note that this equation does
not take into account that maximum rated switch current
(IP) 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 IP value
from the Electrical Characteristics table.
LT1376-5
C3
10µF TO
50µF
+
SENSE
GND
V
C
+
C1
100µF
10V TANT
D2
1N5818
C
C
R
C
OUTPUT**
–5V, 0.5A
* INCREASE L1 TO 10µH OR 20µH FOR HIGHER CURRENT APPLICATIONS.
SEE APPLICATIONS INFORMATION
Operating duty cycle:
** MAXIMUM LOAD CURRENT DEPENDS ON MINIMUM INPUT VOLTAGE
AND INDUCTOR SIZE. SEE APPLICATIONS INFORMATION
V
OUT + V
1375/76 F18
F
DC =
V − 0.3 + VOUT + V
IN
F
Figure 18. Positive-to-Negative Converter
(This formula uses an average value for switch loss, so it
may be several percent in error.)
Inverting regulators differ from buck regulators in the
basicswitchingnetwork. Currentis deliveredtotheoutput
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
maximumswitchcurrent, evenwithlargeinductorvalues.
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
withlargeinductors.Outputripplevoltageforthepositive-
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
calculateoperatingconditions forthepositive-to-negative
converter.
With the conditions above:
5 + 0.5
DC =
= 56%
4.7 − 0.3 + 5 + 0.5
This duty cycle is close enough to 50% that IP can be
assumed to be 1.5A.
OUTPUT DIVIDER
If the adjustable part is used, the resistor connected to
VOUT (R2) should be set to approximately 5k. R1 is
calculated from:
R2 V − 2.42
(
)
OUT
R1=
2.42
24
Linear [ Linear ]
