Datasheet

V
OUT
x (V
IN
- V
OUT
)
I
LR
x f
SW
x V
IN
L1 =
8.3 x 10
-11
V
IN(MAX)
x 100 ns
R1 t
V
OUT
8.3 x 10
-11
x f
SW
R1 =
10 k:
= 2.22 k:
- 1
V
OUT
0.6
R4 =
=
V
OUT
0.6
R3
R4
- 1
Design Procedure www.ti.com
3 Design Procedure
The LM3103 is easy to use compared with other devices available on the market because it integrates all
key components, including both the main and synchronous power MOSFETs, in a single package and
requires no loop compensation owing to the use of the Constant On-Time (COT) hysteretic control
scheme. The design of the demonstration board in this application note is detailed below.
Design Parameters:
V
IN
= 8V to 42V, typical 18V
V
OUT
= 3.3V
I
OUT
= 0.75A
Step 1: Calculate the feedback resistors
The ratio of the feedback resistor can be calculated from the following equation:
(1)
As a general practice, R3 and R4 should be chosen from standard 1% resistor values in the range of 1.0
kΩ to 10 kΩ satisfying the above ratio. Now, select R3 = 10 kΩ, with V
OUT
= 3.3V,
(2)
Step 2: Calculate the on-time setting resistor
The switching frequency f
SW
of the demonstration board is affected by the on-time t
on
of the LM3103, which
is determined by R1. If f
SW
and V
OUT
are determined, R1 can be calculated as follows:
(3)
For this demonstration board design, V
OUT
= 3.3V and f
SW
= 500 kHz are chosen. As a result, R1 = 78.52
kΩ. To ensure that the on-time is larger than the minimum limit, which is 100 ns, the value of R1 must
satisfy the following equation:
(4)
Now the maximum V
IN
is 42V, the calculated R1 satisfies the above equation.
Step 3: Determine the inductance
The main parameter affected by the inductor is the amplitude of the inductor current ripple I
LR
. Once I
LR
is
selected, L1 can be determined by:
(5)
For this demonstration board design, I
LR
= 0.3A is selected. Now V
IN
= 18V, V
OUT
= 3.3V, and f
SW
= 500
kHz. As a result, L = 17.97 µH.
4
AN-1678 LM3103 Demonstration Board Reference Design SNVA268A–October 2007–Revised April 2013
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