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LT1766EGN-5#PBF

P1-P3P4-P6P7-P9P10-P12P13-P15P16-P18P19-P21P22-P24P25-P27P28-P30
LT1766/LT1766-5
25
1766fc
RiseTime
RC V
V
SS OUT
BE
=
()( )( )
4
Using the values shown in Figure 10,
Rise Time ms=
()( )
()
=
47 10 15 10 5
07
5
39
••
.
–
The ramp is linear and rise times in the order of 100ms are
possible. Since the circuit is voltage controlled, the ramp
rate is unaffected by load characteristics and maximum
output current is unchanged. Variants of this circuit can
be used for sequencing multiple regulator outputs.
DUAL OUTPUT SEPIC CONVERTER
The circuit in Figure 14 generates both positive and nega-
tive 5V outputs with a single piece of magnetics. The two
inductors shown are actually just two windings on a stan-
dard Coiltronics inductor. The topology for the 5V output
is a standard buck converter. The – 5V topology would be
a simple fl yback winding coupled to the buck converter
if C4 were not present. C4 creates a SEPIC (single-ended
primary inductance converter) topology which improves
regulation and reduces ripple current in L1. Without C4,
the voltage swing on L1B compared to L1A would vary
due to relative loading and coupling losses. C4 provides a
low impedance path to maintain an equal voltage swing in
L1B, improving regulation. In a fl yback converter, during
switch on-time, all the converter’s energy is stored in L1A
only, since no current fl ows in L1B. At switch off, energy
V
OUT1
5V
(SEE DN100
FOR MAX I
OUT
)
V
OUT2
–5V
†
* L1 IS A SINGLE CORE WITH TWO WINDINGS
COILTRONICS #CTX50-3A
†
IF LOAD CAN GO TO ZERO, AN OPTIONAL
PRELOAD OF 1k TO 5k MAY BE USED TO
IMPROVE LOAD REGULATION
D1, D3: 10MQ060N
V
IN
7.5V
TO 60V
GND
1766 F14
C2
0.33μF
C
F
220pF
D1
C1
100μF
10V
TANT
C5
100μF
10V
TANT
C3
2.2μF
100V
CER
C4
100μF
10V
TANT
D2
1N4148W
D3
L1A*
50μH
L1B*
R1
15.4k
R2
4.99k
++
+
R
C
2.2k
C
C
0.022μF
BOOST
V
IN
LT1766
SHDN
SYNC
SW
FB
V
C
GND
Figure 14. Dual Output SEPIC Converter
OUTPUT
5V
1A
INPUT
40V
1766 F13
C2
0.33μF
C1
100μF
C
SS
15nF
C
F
220pF
D1
C3
2.2μF
50V
CER
D2
1N4148W
L1
47μH
R1
15.4k
R3
2k
C
C
0.022μF
R2
4.99k
R4
47k
Q1
BOOST BIAS
V
IN
LT1766
SHDN
SYNC
SW
FB
V
C
GND
+
R
C
2.2k
Figure 13. Buck Converter with Adjustable Soft-Start
is transferred by magnetic coupling into L1B, powering
the –5V rail. C4 pulls L1B positive during switch on-time,
causing current to fl ow, and energy to build in L1B and
C4. At switch off, the energy stored in both L1B and C4
supply the –5V rail. This reduces the current in L1A and
changes L1B current waveform from square to triangular.
For details on this circuit, including maximum output cur-
rents, see Design Note 100.
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 LT1766 accepts only positive feedback signals.
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:
I
I
VV
VVfL
VV
VV VV
MAX
P
IN OUT
OUT IN
OUT IN
OUT IN OUT F
=
+
⎡
⎣
⎢
⎤
⎦
⎥
++
–
()( )
()()()
()(–.)
(–.)()
2
03
03
APPLICATIONS INFORMATION
LT1766/LT1766-5
26
1766fc
APPLICATIONS INFORMATION
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 = 18μH,
V
F
= 0.63V, I
P
= 1.5A: I
MAX
= 0.280A.
OUTPUT DIVIDER
Refer to Applications Information Feedback Pin Functions
to calculate R1 and R2 for the (negative) output voltage
(from Table 1).
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
()()
()( )
Minimum inductor continuous mode:
L
VV
fV V I I
VV
V
MIN
IN OUT
IN OUT P OUT
OUT F
IN
=
++
+
⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎤
⎦
⎥
()( )
()( ) –
()
21
For a 40V to –12V converter using the LT1766 with peak
switch current of 1.5A and a catch diode of 0.63V:
IA
CONT
>
+++
=
()(.)
()( .)
.
40 1 5
440124012063
0 573
22
For a load current of 0.25A, this says that discontinuous
mode can be used and the minimum inductor needed is
found from:
LH
MIN
==μ
212 025
200 10 1 5
13 3
32
()(.)
(•)(.)
.
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 formula
will give an
approximate
value for RMS ripple current.
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
OUTPUT**
–12V, 0.25A
INPUT
†
5.5V TO
48V
1766 F15
C2
0.33μF
C
C
R
C
D1
10MQO60N
R1
44.2k
C1
100μF
25V
TANT
C3
2.2μF
100V
CER
D2
†
1N4148W
L1*
18μH
C
F
BOOST
LT1766
V
IN
V
SW
FB
GND
V
C
R2
4.99k
* INCREASE L1 TO 30μH OR 60μH FOR HIGHER CURRENT APPLICATIONS.
SEE APPLICATIONS INFORMATION
** MAXIMUM LOAD CURRENT DEPENDS ON MINIMUM INPUT VOLTAGE
AND INDUCTOR SIZE. SEE APPLICATIONS INFORMATION
†
FOR V
IN
> 44V AND V
OUT
= –12V, ADDITIONAL VOLTAGE DROP IN THE
PATH OF D2 IS REQUIRED TO ENSURE BOOST PIN MAXIMUM RATING IS
NOT EXCEEDED. SEE APPLICATIONS INFORMATION (BOOST PIN VOLTAGE)
+
Figure 15. Positive-to-Negative Converter
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 diffi culty in calculating the minimum inductor size
needed is that you must fi rst decide whether the switcher
will be in continuous or discontinuous mode at the critical
point where switch current reaches 1.5A. The fi rst 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
LT1766/LT1766-5
27
1766fc
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
load currents and L ≥15μH. It increases to about 2.0 for
smaller inductors at lower load currents.
Input Capacitor I ff I
V
V
RMS OUT
OUT
IN
= ()( )
ff = 1.2 to 2.0
The output capacitor ripple current for the positive-to-
negative converter is similar to that for a typical buck
regulator—it is a triangular waveform with peak-to-peak
value equal to the peak-to-peak triangular waveform of the
inductor. The low output ripple design in Figure 15 places
the input capacitor between V
IN
and the regulated negative
output. This placement of the input capacitor signifi cantly
reduces the size required for the output capacitor (versus
placing the input capacitor between V
IN
and ground).
The peak-to-peak ripple current in both the inductor and
output capacitor (assuming continuous mode) is:
I
P-P
P-P
=
==
+
++
=
DC V
fL
DC Duty Cycle
VV
VVV
I RMS
I
IN
OUT F
OUT IN F
COUT
•
•
()
12
The output ripple voltage for this confi guration is as low
as the typical buck regulator based predominantly on the
inductor’s triangular peak-to-peak ripple current and the
ESR of the chosen capacitor (see Output Ripple Voltage
in Applications Information).
Diode Current
Average
diode current is equal to load current.
Peak
diode
current will be considerably higher.
Peak diode current:
Continuous Mode
I
VV
V
VV
LfV V
Discontinuous Mode
IV
Lf
OUT
IN OUT
IN
IN OUT
IN OUT
OUT OUT
=
+
+
+
=
()()()
()()( )
()( )
()()
2
2
Keep in mind that during start-up and output overloads,
average diode current may be much higher than with nor-
mal loads. Care should be used if diodes rated less than
1A are used, especially if continuous overload conditions
must be tolerated.
BOOST Pin Voltage
To ensure that the BOOST pin voltage does not exceed its
absolute maximum rating of 68V with respect to device
GND pin voltage, care should be taken in the generation of
boost voltage. For the conventional method of generating
boost voltage, shown in Figure 1, the voltage at the BOOST
pin during switch on time is approximately given by:
V
BOOST
(GND pin) = (V
IN
– V
GNDPIN
) + V
C2
where:
V
C2
= (D2+) – V
D2
– (D1+) + V
D1
= voltage across the boost capacitor
For the positive-to-negative converter shown in Figure 15,
the conventional Buck output node is grounded (D2+) = 0V
and the catch diode (D1+) is connected to the negative
output = V
OUT
= –12V. Absolute maximum ratings should
also be observed with the GND pin now at –12V. It can be
seen that for V
D1
= V
D2
:
V
C2
= (D2+) – (D1+) = |V
OUT
| = 12V
The maximum V
IN
voltage allowed for the device (GND
pin at –12V) is 48V.
The maximum V
IN
voltage allowed without exceeding the
BOOST pin voltage absolute maximum rating is given by:
V
IN(MAX)
= Boost (Max) + (V
GNDPIN
) – V
C2
V
IN(MAX)
= 68 + (–12) – 12 = 44V
To increase usable V
IN
voltage, V
C2
must be reduced. This
can be achieved by placing a zener diode V
Z1
(anode at
C2+) in series with D2.
Note: A maximum limit on V
Z1
must be observed to
ensure a minimum V
C2
is maintained on the boost
capacitor; referred to as V
BOOST(MIN)
in the Electrical
Characteristics.
APPLICATIONS INFORMATION
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LT1766EGN-5#PBF

Mfr. #:
Buy LT1766EGN-5#PBF
Manufacturer:
Analog Devices / Linear Technology
Description:
Switching Voltage Regulators 1.5A 200kHz High Voltage Step-down Regulator
Lifecycle:
New from this manufacturer.
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