lt1766efe-trpbf Linear Technology Corporation, lt1766efe-trpbf Datasheet - Page 25

• Manufacturer Part Number: lt1766efe-trpbf
• Description: High Voltage 1.5a, 200khz Step-down Switching Regulators
• Manufacturer: Linear Technology Corporation

APPLICATIO S I FOR ATIO
I
V
V
V
0.3 = Switch voltage drop at 1.5A
Example: with V
V
OUTPUT DIVIDER
Refer to Applications Information Feedback Pin Functions
to calculate R1 and R2 for the (negative) output voltage
(from Table 1).
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.
P
IN
OUT
F
F
** MAXIMUM LOAD CURRENT DEPENDS ON MINIMUM INPUT VOLTAGE
5.5V TO
INPUT
* INCREASE L1 TO 30µH OR 60µH FOR HIGHER CURRENT APPLICATIONS.
= Maximum rated switch current
†
= Catch diode forward voltage
= 0.63V, I
SEE APPLICATIONS INFORMATION
FOR V
PATH OF D2 IS REQUIRED TO ENSURE BOOST PIN MAXIMUM RATING IS
NOT EXCEEDED. SEE APPLICATIONS INFORMATION (BOOST PIN VOLTAGE)
= Minimum input voltage
AND INDUCTOR SIZE. SEE APPLICATIONS INFORMATION
48V
2.2µF
100V
= Output voltage
CER
†
C3
IN
> 44V AND V
Figure 15. Positive-to-Negative Converter
P
V
= 1.5A: I
IN
GND
IN(MIN)
C
OUT
LT1766
BOOST
F
= –12V, ADDITIONAL VOLTAGE DROP IN THE
U
V
V
C
MAX
SW
FB
R
= 5.5V, V
C
C
U
C
= 0.280A.
D1
10MQO60N
C2
0.33µF
1N4148W
OUT
W
D2
18µH
†
L1*
= 12V, L = 18µH,
R2
4.99k
44.2k
R1
+
C1
100µF
25V
TANT
U
OUTPUT**
–12V, 0.25A
1766 F15
Output current where continuous mode is needed:
Minimum inductor discontinuous mode:
Minimum inductor continuous mode:
For a 40V to –12V converter using the LT1766 with peak
switch current of 1.5A and a catch diode of 0.63V:
For a load current of 0.25A, this says that discontinuous
mode can be used and the minimum inductor needed is
found from:
In practice, the inductor should be increased by about
30% over the calculated minimum to handle losses and
variations in value. This suggests a minimum inductor of
18µH for this application.
Ripple Current in the Input and Output Capacitors
Positive-to-negative converters have high ripple current in
the input capacitor. For long capacitor lifetime, the RMS
value of this current must be less than the high frequency
ripple current rating of the capacitor. The following for-
mula will give an approximate value for RMS ripple cur-
rent. This formula assumes continuous mode and large
inductor value . Small inductors will give somewhat higher
ripple current, especially in discontinuous mode. The
exact formulas are very complex and appear in Application
Note 44, pages 29 and 30. For our purposes here a fudge
factor (ff) is used. The value for ff is about 1.2 for higher
L
I
I
L
L
CONT
CONT
MIN
MIN
MIN
=
=
=
>
>
2
(
2
200 10
( )(
(
f V
V
4 40 12 40 12 0 63
4
2 12 0 25
OUT
( )( )
(
(
( )( . )
f I
V
•
IN
IN
P
)(
+
+
+
I
3
2
OUT
( ) ( . )
V
V
)( . )
40 1 5
OUT
OUT
)(
1 5
(
)
V
2
IN
)(
)
LT1766/LT1766-5
(
⎡
⎢
⎣
2
) ( )
I
+
V
V
2
P
IN
=
IN
I
–
P
)(
2
13 3
+
I
OUT
+
2
V
V
.
OUT
OUT
µ
. )
⎛
⎜
⎝
H
1
)
+
+
(
=
V
V
F
0 573
OUT
)
.
V
IN
+
25
A
V
F
1766fb
)
⎞
⎟
⎠
⎤
⎥
⎦