LT1767/LT1767-1.8/
LT1767-2.5/LT1767-3.3/LT1767-5
10
1767fb
For more information www.linear.com/LT1767
applicaTions inForMaTion
especially with smaller inductors and lighter loads, so
don’t omit this step. Powdered iron cores are forgiving
because they saturate softly, whereas ferrite cores
saturate abruptly. Other core materials fall somewhere
in between.
I
PEAK
=I
OUT
+
V
OUT
V
IN
− V
OUT
2 L
( )
f
( )
V
IN
( )
V
IN
= Maximum input voltage
f = Switching frequency, 1.25MHz
3. Decide if the design can tolerate an “open” core ge
-
ometry like a rod or barrel, which have high magnetic
field radiation, or whether it needs a closed core like a
toroid to prevent EMI problems. This is a tough decision
because the rods or barrels are temptingly cheap and
small and there are no helpful guidelines to calculate
when the magnetic field radiation will be a problem.
4. After making an initial choice, consider the secondary
things like output voltage ripple, second sourcing, etc.
Use the experts in the Linear Technology’s applications
department if you feel uncertain about the final choice.
They have experience with a wide range of inductor
types and can tell you about the latest developments
in low profile, surface mounting, etc.
I
OUT MAX
( )
=
I
P
−
V
OUT
V
IN
− V
OUT
2 L
( )
f
( )
V
IN
( )
Discontinuous operation occurs when
I
OUT(DIS)
=
OUT
2(L)(f)
For V
IN
= 8V, V
OUT
= 5V and L = 3.3µH,
I
OUT MAX
( )
=1.5−
5
8
5
2 3.3 • 10
− 6
( )
1.25 • 10
6
( )
8
( )
Note that the worst case (minimum output current avail-
able) condition
is at the maximum input voltage. For the
same circuit at 15V, maximum output current would be
only 1.1A.
When choosing an inductor, consider maximum load cur
-
rent, core and copper losses, allowable component height,
output
voltage ripple, EMI, fault current in the inductor,
saturation, and of course, cost. The following procedure
is suggested as a way of handling these somewhat com
-
plicated and conflicting requirements.
1.
Choose a value in microhenries from the graphs of
maximum load current. Choosing a small inductor
with lighter loads may result in discontinuous mode
of operation, but the LT1767 is designed to work well
in either mode.
Assume that the average inductor current is equal to
load current and decide whether or not the inductor
must withstand continuous fault conditions. If maxi
-
mum load current is 0.5A, for instance, a 0.5A inductor
may not survive a continuous 2A overload condition.
Also, the instantaneous application of input or release
from shutdown, at high input voltages, may cause
saturation of the inductor. In these applications, the
soft-start circuit shown in Figure 10 should be used.
2. Calculate peak inductor current at full load current
to
ensure that the inductor will not saturate. Peak cur-
rent can
be significantly higher than output current,
Table 1
PART NUMBER VALUE (uH) I
SAT
(Amps) DCR () HEIGHT (mm)
Coiltronics
TP1-2R2 2.2 1.3 0.188 1.8
TP2-2R2 2.2 1.5 0.111 2.2
TP3-4R7 4.7 1.5 0.181 2.2
TP4- 100 10 1.5 0.146 3.0
Murata
LQH1C1R0M04 1.0 0.51 0.28 1.8
LQH3C1R0M24 1.0 1.0 0.06 2.0
LQH3C2R2M24 2.2 0.79 0.1 2.0
LQH4C1R5M04 1.5 1.0 0.09 2.6
Sumida
CD73- 100 10 1.44 0.080 3.5
CDRH4D18-2R2 2.2 1.32 0.058 1.8
CDRH5D18-6R2 6.2 1.4 0.071 1.8
CDRH5D28-100 10 1.3 0.048 2.8
