LT1956IGN#TRPBF Linear Technology, LT1956IGN#TRPBF Datasheet - Page 25

• Manufacturer Part Number: LT1956IGN#TRPBF
• Description: IC SW REG STP-DWN HI VOLT 16SSOP
• Manufacturer: Linear Technology
• Type: Step-Down (Buck)r

Specifications of LT1956IGN#TRPBF

• Internal Switch(s): Yes
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 1.2 ~ 45 V
• Current - Output: 1.5A
• Frequency - Switching: 500kHz
• Voltage - Input: 5.5 ~ 60 V
• Operating Temperature: -40°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 16-SSOP
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Power - Output: -

APPLICATIO S I FOR ATIO
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:
** MAXIMUM LOAD CURRENT DEPENDS ON MINIMUM INPUT VOLTAGE
* INCREASE L1 TO 10 H OR 18 H FOR HIGHER CURRENT APPLICATIONS.
SEE APPLICATIONS INFORMATION
AND INDUCTOR SIZE. SEE APPLICATIONS INFORMATION
12V
V
2.2 F
IN
25V
C3
Figure 15. Positive-to-Negative Converter
V
IN
GND
LT1956
BOOST
U
C
F
V
C
SW
FB
R
C
C
U
C
D1
10MQO60N
C2
0.1 F
MMSD914TI
W
D2
7 H
L1*
R2
4.12k
36.5k
R1
+
U
C1
100 F
20V TANT
OUTPUT**
–12V, 0.25A
1956 F15
I
V
V
V
0.3 = switch voltage drop at 1.5A
Example: with V
V
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:
Minimum inductor discontinuous mode:
P
IN
OUT
F
F
= maximum rated switch current
I
I
= catch diode forward voltage
= 0.63V, I
L
MAX
CONT
= minimum input voltage
MIN
= output voltage
2
I
(
P
P
V
4
OUT
( )( )
–
(
= 1.5A: I
f I
V
2
IN
IN(MIN)
(
(
V
P
)(
V
OUT
OUT
I
(
2
OUT
V
V
IN
OUT
MAX
(
)(
= 5.5V, V
)
V
V
V
IN
V
)(
LT1956/LT1956-5
OUT
IN
IN
= 0.36A.
) ( )
V
2
)( )( )
IN
– . )(
I
)
P
f L
0 3
2
V
OUT
OUT
V
(
V
OUT
OUT
= 12V, L = 15 H,
V
)(
F
V
)
V
IN
F
)
– . )
25
0 3
1956f