LT1940 Linear Technology, LT1940 Datasheet - Page 8

• Manufacturer Part Number: LT1940
• Description: Dual Monolithic 1.4A/ 1.1MHz Step-Down Switching Regulator
• Manufacturer: Linear Technology

LT1940
APPLICATIO S I FOR ATIO
The maximum input voltage is determined by the absolute
maximum ratings of the V
minimum duty cycle DC
This limits the maximum input voltage to ~14V with
V
a restriction on the operating input voltage; the circuit will
tolerate transient inputs up to the absolute maximum
rating.
Inductor Selection and Maximum Output Current
A good first choice for the inductor value is:
where V
L is in H. With this value the maximum load current will
be ~1.4A, independent of input voltage. The inductor’s
RMS current rating must be greater than your maximum
load current and its saturation current should be about
30% higher. To keep efficiency high, the series resistance
(DCR) should be less than 0.1 . Table 1 lists several
vendors and types that are suitable.
Of course, such a simple design guide will not always
result in the optimum inductor for your application. A
larger value provides a slightly higher maximum load
current, and will reduce the output voltage ripple. If your
load is lower than 1.4A, then you can decrease the value of
the inductor and operate with higher ripple current. This
allows you to use a physically smaller inductor, or one with
a lower DCR resulting in higher efficiency. Be aware that if
the inductance differs from the simple rule above, then the
maximum load current will depend on input voltage. There
are several graphs in the Typical Performance Character-
istics section of this data sheet that show the maximum
load current as a function of input voltage and inductor
value for several popular output voltages. Also, low
inductance may result in discontinuous mode operation,
which is okay, but further reduces maximum load current.
For details of maximum output current and discontinuous
mode operation, see Linear Technology Application
Note 44. Finally, for duty cycles greater than 50%
8
OUT
V
L = (V
INMAX
= 1.8V and ~19V with V
D
OUT
is the voltage drop of the catch diode (~0.4V) and
= (V
+ V
OUT
D
)/1.2
+ V
U
D
)/DC
MIN
U
IN
MIN
= 0.15:
and BOOST pins and by the
OUT
– V
= 2.5V. Note that this is
D
W
+ V
SW
.
U
(V
to avoid subharmonic oscillations. See AN19. The discus-
sion below assumes continuous inductor current.
The current in the inductor is a triangle wave with an
average value equal to the load current. The peak switch
current is equal to the output current plus half the peak-to-
peak inductor ripple current. The LT1940 limits its switch
current in order to protect itself and the system from
overload faults. Therefore, the maximum output current
that the LT1940 will deliver depends on the current limit,
the inductor value, and the input and output voltages. L is
chosen based on output current requirements, output
voltage ripple requirements, size restrictions and effi-
ciency goals.
When the switch is off, the inductor sees the output
voltage plus the catch diode drop. This gives the peak-to-
peak ripple current in the inductor:
where f is the switching frequency of the LT1940 and L is
the value of the inductor. The peak inductor and switch
current is
To maintain output regulation, this peak current must be
less than the LT1940’s switch current limit I
least 1.8A at low duty cycle and decreases linearly to 1.5A
at DC = 0.8. The maximum output current is a function of
the chosen inductor value:
If the inductor value is chosen so that the ripple current is
small, then the available output current will be near the
switch current limit.
One approach to choosing the inductor is to start with the
simple rule given above, look at the available inductors,
and choose one to meet cost or space goals. Then use
these equations to check that the LT1940 will be able to
deliver the required output current. Note again that these
equations assume that the inductor current is continuous.
Discontinuous operation occurs when I
I
OUT
L
I
I
SWPK
OUTMAX
/2 as calculated above.
I
L
/V
= (1 – DC)(V
IN
= I
< 0.5), there is a minimum inductance required
= I
LPK
LIM
= I
– I
OUT
OUT
L
+ I
/2 = 1.8A • (1 – 0.21 • DC) – I
+ V
L
D
/2.
)/(L • f)
OUT
LIM
is less than
. I
LIM
is at
1940i
L
/2