Datasheet

25
LT1956/LT1956-5
1956f
current in L2 and C4 flows via the catch diode D3, charging
the negative output capacitor C6. If the negative output is
not loaded enough, it can go severely unregulated (be-
come more negative). Figure 14b shows the maximum
allowable –5V output load current (vs load current on the
5V output) that will maintain the –5V output within 3%
tolerance. Figure 14c shows the –5V output voltage regu-
lation vs its own load current when plotted for three
separate load currents on the 5V output. The efficiency of
the dual output converter circuit shown in Figure 14a is
given in Figure 14d.
POSITIVE-TO-NEGATIVE CONVERTER
The circuit in Figure 15 is a positive-to-negative topology
using a grounded inductor. It differs from the standard
approach in the way the IC chip derives its feedback signal
because the LT1956 accepts only positive feedback sig-
nals. The ground pin must be tied to the regulated negative
output. A resistor divider to the FB pin, then provides the
proper feedback voltage for the chip.
The following equation can be used to calculate maximum
load current for the positive-to-negative converter:
APPLICATIO S I FOR ATIO
WUUU
I
I
VV
VVfL
VV
VV VV
MAX
P
IN OUT
OUT IN
OUT IN
OUT IN OUT F
=
+






++
–
()( )
()()()
()(–.)
(–.)()
2
03
03
I
P
= maximum rated switch current
V
IN
= minimum input voltage
V
OUT
= output voltage
V
F
= catch diode forward voltage
0.3 = switch voltage drop at 1.5A
Example: with V
IN(MIN)
= 5.5V, V
OUT
= 12V, L = 15µH,
V
F
= 0.63V, I
P
= 1.5A: I
MAX
= 0.36A.
INDUCTOR VALUE
The criteria for choosing the inductor is typically based on
ensuring that peak switch current rating is not exceeded.
This gives the lowest value of inductance that can be used,
but in some cases (lower output load currents) it may give
a value that creates unnecessarily high output ripple
voltage.
The difficulty in calculating the minimum inductor size
needed is that you must first decide whether the switcher
will be in continuous or discontinuous mode at the critical
point where switch current reaches 1.5A. The first step is
to use the following formula to calculate the load current
above which the switcher must use continuous mode. If
your load current is less than this, use the discontinuous
mode formula to calculate minimum inductor needed. If
load current is higher, use the continuous mode formula.
Output current where continuous mode is needed:
I
VI
VV VV V
CONT
IN P
IN OUT IN OUT F
>
+++
()()
()( )
22
4
Minimum inductor discontinuous mode:
L
VI
fI
MIN
OUT OUT
P
=
2
2
()()
()( )
OUTPUT**
–12V, 0.25A
V
IN
12V
1956 F15
C2
0.1µF
C
C
R
C
D1
10MQO60N
R1
36.5k
C1
100µF
20V TANT
C3
2.2µF
25V
D2
MMSD914TI
L1*
7µH
C
F
BOOST
LT1956
V
IN
SW
FB
GND
V
C
R2
4.12k
* INCREASE L1 TO 10µH OR 18µH FOR HIGHER CURRENT APPLICATIONS.
SEE APPLICATIONS INFORMATION
** MAXIMUM LOAD CURRENT DEPENDS ON MINIMUM INPUT VOLTAGE
AND INDUCTOR SIZE. SEE APPLICATIONS INFORMATION
+
Figure 15. Positive-to-Negative Converter