LT3992EFE View Datasheet(PDF) - Linear Technology
Part Name
Description
MFG CO.
'LT3992EFE' PDF : 36 Pages View PDF
LT3992
Applications Information
The maximum input voltage is determined by the absolute
maximum ratings of the VIN and BST pins and by the
frequency and minimum duty cycle. The minimum duty
cycle is defined as:
DCMIN = tON(MIN) • Frequency
Maximum input voltage as:
VIN(MAX)
=
VOUT + VD
DCMIN
–
VD
+
VSW
Note that the LT3992 will regulate if the input voltage is
taken above the calculated maximum voltage as long as
maximum ratings of the VIN and BST pins are not violated.
However operation in this region of input voltage will
exhibit pulse skipping behavior.
Example:
VOUT = 3.3V, IOUT = 1A, frequency = 1MHz, temperature
= 25°C, VSW = 0.1V, B = 50 (from boost characteristics
specification), VD = 0.4V, tON(MIN) = 180ns:
DCMAX
=
1
1+ 1
= 98%
50
VIN(MIN)
=
3.3+ 0.4
0.98
–
0.4
+
0.1=
3.48V
DCMIN = tON(MIN) •Frequency = 0.18
VIN(MAX)
=
3.3+ 0.4
0.18
–
0.4
+
0.1=
20.2V
SW1
1/(2 • tP)
SW2
CLKOUT
tDCLKOSW1
2 • tP
tP
tP/2
tP
tDCLKOSW2
3992 F05
In cases where multiple input voltages are present, or the
VIN/VOUT 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:
VOUT = 3.3V, IOUT = 1A, frequency = 1MHz, temperature
= 25°C, VSW = 0.1V, B = 50 (from boost characteristics
specification), VD = 0.4V, tON(MIN) = 180ns. RDIV = 102kΩ.
DCMIN1 = tON(MIN1) • Frequency/4 = 0.045
VIN1(MAX)
=
3.3+ 0.4
0.045
–
0.4
+
0.1=
>60V
Inductor Selection and Maximum Output Current
A good first choice for the LT3992 inductor value is:
L = VOUT
f
where f is frequency in MHz and L is in µH.
With this value 3A of load current will be available over
the entire input voltage range. 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:
ISW(PK)
=
IL(PK)
=
IOUT
+
ΔIL
2
Figure 5. Timing Diagram RT/SYNC = 28.0k,
tP = 1µs, VDIV = 0.75V
3992fa
14
For more information www.linear.com/LT3992
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