LT1939
12
1939f
APPLICATIONS INFORMATION
Note that the LT1939 will regulate if the input voltage is
taken above the calculated maximum voltage as long as
maximum ratings of the V
IN
and BST pins are not violated.
However operation in this region of input voltage will exhibit
pulse skipping behavior.
Example:
V
OUT1
= 3.3V, I
OUT1
= 1A, Frequency = 1MHz,
Temperature = 25°C,
V
SW
= 0.3V, V
D
= 0.4V, t
ON(MIN)
= 150ns,
t
OFF(MIN)
= 110ns
DC
MAX
= 1 (110ns)1MHz = 89%
V
IN(MIN)
=
3.3 + 0.4
0.89
0.4 + 0.3 = 4.06V
DC
MIN
= t
ON(MIN)
•Frequency = 15%
V
IN(MAX)
=
3.3 + 0.4
0.15
0.4 + 0.3 = 24.57V
Inductor Selection and Maximum Output Current
A good fi rst choice for the inductor value is:
L =
(V
IN
V
OUT1
)•V
OUT1
V
IN
• f
where f is frequency in MHz and L is in µH.
With this value the maximum load current will be ~2A,
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 effi ciency high, the series resistance (DCR) should
be less than 0.05.
For applications with a duty cycle of about 50%, the induc-
tor value should be chosen to obtain an inductor ripple
current less than 40% of peak switch current.
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
1.5A, 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 effi ciency.
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 LT1939 limits its switch
current in order to protect itself and the system from
overload faults. Therefore, the maximum output current
that the LT1939 will deliver depends on the current limit,
the inductor value, switch frequency, and the input and
output voltages. The inductor 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 volt-
age plus the catch diode drop. This gives the peak-to-peak
ripple current in the inductor:
I
L
=
1 DC
()
V
OUT1
+ V
D
()
L•f
where f is the switching frequency of the LT1939 and L
is the value of the inductor. The peak inductor and switch
current is:
I
SW(PK)
= I
LPK
= I
OUT1
+
I
L
2
To maintain output regulation, this peak current must be
less than the LT1939’s switch current limit, I
LIM
. I
LIM
is
guaranteed to be greater than 2.3A over the entire duty
cycle range. The maximum output current is a function
of the chosen inductor value:
I
OUT1(MAX)
= I
LIM
I
L
2
=2.3 –
I
L
2
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 LT1939 will be able to
deliver the required output current. Note again that these
equations assume that the inductor current is continuous.
