LT3510FE(RevC) データシートの表示（PDF） - Linear Technology
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LT3510FE Datasheet PDF : 28 Pages
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LT3510
APPLICATIONS INFORMATION
maximum duty cycle. The duty cycle is the fraction of time
that the internal switch is on during a clock cycle. Unlike
most fixed frequency regulators, the LT3510 will not switch
off at the end of each clock cycle if there is sufficient volt-
age across the boost capacitor (C3 in Figure 1) to fully
saturate the output switch. Forced switch off for a minimum
time will only occur at the end of a clock cycle when the
boost capacitor needs to be recharged. This operation
has the same effect as lowering the clock frequency for a
fixed off time, resulting in a higher duty cycle and lower
minimum input voltage. The resultant duty cycle depends
on the charging times of the boost capacitor and can be
approximated by the following equation:
DCMAX
=
1
1+ 1
B
where B is 2A divided by the typical boost current from
the Electrical Characteristics.
This leads to a minimum input voltage of:
VIN(MIN)
=
VOUT + VD
DCMAX
–
VD
+
VSW
where VSW is the voltage drop of the internal switch.
Figure 3 shows a typical graph of minimum input voltage
vs load current for the 3.3V and 1.8V application on the
first page of this data sheet. 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 LT3510 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.
6.0
VOUT = 3.3V
5.5
5.0
START-UP
4.5
4.0
3.5
3.0
1
RUNNING
10
100
1000
CURRENT (mA)
10000
3510 F03
Figure 3. Minimum Input Voltage vs Load Current
Example:
VOUT = 3.3V, IOUT = 1A, Frequency = 1MHz, Temperature
= 25°C
VSW = 0.1V, B = 40 (from boost characteristics specifica-
tion), VD = 0.4V, tON(MIN) = 200ns
DCMAX
=
1
1+ 1
= 98%
40
VIN(MIN)
=
3.3 + 0.4
0.98
–
0.4
+
0.1=
3.48V
DCMIN = tMIN(ON) • f = 0.200
VIN(MAX)
=
3.3 + 0.4
0.200
–
0.4
+
0.1=
18.2V
Inductor Selection and Maximum Output Current
A good first choice for the inductor value is:
L = (VIN – VOUT ) • VOUT
VIN • 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
3510fc
12
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