Datasheet

LT3692A
15
3692afc
For more information www.linear.com/3692A
applicaTions inForMaTion
The maximum input voltage is determined by the absolute
maximum ratings of the V
IN
and BST pins and by the
frequency and minimum duty cycle. The minimum duty
cycle is defined as:
DC
MIN
= t
ON(MIN)
• Frequency
Maximum input voltage as:
V
IN(MAX)
=
V
OUT
+ V
D
DC
MIN
– V
D
+ V
SW
Note that the LT3692A 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
OUT
= 3.3V, I
OUT
= 1A, frequency = 1MHz, temperature
= 25°C, V
SW
= 0.1V, B = 50 (from boost characteristics
specification), V
D
= 0.4V, t
ON(MIN)
= 140ns:
DC
MAX
=
1
1+
1
50
= 98%
V
IN(MIN)
=
3.3 + 0.4
0.98
– 0.4 + 0.1= 3.48V
DC
MIN
= t
ON(MIN)
• Frequency = 0.14
V
IN(MAX)
=
3.3 + 0.4
0.14
– 0.4 + 0.1= 26.1V
In cases where multiple input voltages are present, or the
V
IN
/V
OUT
ratio for channel 1 is significantly different than
channel 2, channel 1’s frequency can be divided by a factor
of 2, 4 or 8 from the programmed value by setting the DIV
pin resistor to the appropriate value. Dividing channel 1’s
frequency will increase the maximum input voltage by the
same ratio. Channel 1’s external components will have to
be chosen according to the resulting frequency.
Example:
V
OUT
= 3.3V, I
OUT
= 1A, frequency = 1MHz, temperature
= 25°C, V
SW
= 0.1V, B = 50 (from boost characteristics
specification), V
D
= 0.4V, t
ON(MIN)
= 140ns. V
DIV
= 0.75V.
DC
MIN1
= t
ON(MIN1)
• Frequency/2 = 0.07
V
IN1(MAX)
=
3.3 + 0.4
0.07
– 0.4 + 0.1=>39V
Inductor Selection and Maximum Output Current
A good first choice for the LT3692A inductor value is:
L =
V
OUT
f
where f is frequency in MHz and L is in µH.
With this value the maximum load current will be ~3.5A,
independent of input voltage. The inductor’s RMS cur
-
rent rating must be greater than your maximum load
current and its saturation current should be higher than
the maximum peak switch current, and will reduce the
output voltage ripple.
If the maximum load for a single channel is lower than
2.5A, then you can decrease the value of the inductor and
operate with higher ripple current, or you can adjust the
maximum switch current for the channel via the ILIM pin.
This allows you to use a physically smaller inductor, or one
with a lower DCR resulting in higher efficiency.
The peak inductor and switch current is:
I
SW(PK)
= I
L(PK)
= I
OUT
+
∆I
L
2
To maintain output regulation, this peak current must be
less than the LT3692A’s switch current limit, ILIM. ILIM
Figure 5. Timing Diagram RT/SYNC = 28.0k,
t
P
= 1µs, V
DIV
= 0.75V
2 • t
P
t
P
t
P
1/(2 • t
P
)
t
P
/2
t
DCLKOSW2
t
DCLKOSW1
SW1
SW2
CLKOUT
3692a F05