Datasheet

L1 =
0.30A x 463 kHz x 75V
10V x (75V - 10V)
= 63 PH
L1 =
I
OR
x F
S(min)
x V
IN(max)
V
OUT1
x (V
IN(max)
- V
OUT1
)
L1 Current
0 mA
I
OR
1/Fs
I
PK+
I
O
I
PK-
R
ON
=
10V
1.18 x 10
-10
x 625 kHz
= 136 k:
LM5010
SNVS307F –SEPTEMBER 2004–REVISED FEBRUARY 2013 www.ti.com
APPLICATIONS INFORMATION
EXTERNAL COMPONENTS
The procedure for calculating the external components is illustrated with a design example. The circuit in Typical
Application Circuit and Block Diagram is to be configured for the following specifications:
• V
OUT
= 10V
• V
IN
= 15V to 75V
• F
S
= 625 kHz
• Minimum load current = 150 mA
• Maximum load current = 1.0A
• Softstart time = 5 ms
R1 and R2: The ratio of these resistors is calculated from:
R1/R2 = (V
OUT
/2.5V) - 1 (7)
R1/R2 calculates to 3.0. The resistors should be chosen from standard value resistors in the range of 1.0 kΩ - 10
kΩ. Values of 3.0 kΩ for R1, and 1.0 kΩ for R2 will be used.
R
ON
, F
S
: R
ON
sets the on-time, and can be chosen using Equation 2 to set a nominal frequency, or from
Equation 5 if the on-time at a particular V
IN
is important. A higher frequency generally means a smaller inductor
and capacitors (value, size and cost), but higher switching losses. A lower frequency means a higher efficiency,
but with larger components. If PC board space is tight, a higher frequency is better. The resulting on-time and
frequency have a ±25% tolerance. Re-arranging Equation 2,
(8)
The next larger standard value (137 kΩ) is chosen for R
ON
, yielding a nominal frequency of 618 kHz.
L1: The inductor value is determined based on the load current, ripple current, and the minimum and maximum
input voltage (V
IN(min)
, V
IN(max)
). Refer to Figure 13.
Figure 13. Inductor Current
To keep the circuit in continuous conduction mode, the maximum allowed ripple current is twice the minimum
load current, or 300 mAp-p. Using this value of ripple current, the inductor (L1) is calculated using the following:
(9)
where F
S(min)
is the minimum frequency (F
S
- 25%).
(10)
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