User Manual

AN372
24 AN372REV1
4 Design Example
The required operating parameters for the analytical process are outlined in the table below.
4.1 Buck Design Steps
A switching frequency of 125kHz is selected, which corresponds to a switching period TT of 8µs.
Step 1) Choosing a Buck Topology
Normal Buck Circuit Design
A normal buck approach is used to calculate the critical duty cycle

C(min/max)
at minimum output voltage
V
OUT(min)
and maximum boost voltage V
BST(max)
. Solve for 
C(min/max)
using Equation 10:
A normal buck approach is used to calculate the critical duty cycle

C(max/min)
at maximum output voltage
V
OUT(max)
and minimum boost voltage V
BST(min)
. Solve for 
C(max/min)
using Equation 10:
Assuming an efficiency
 = 92% the input power P
IN
required from boost voltage V
BST
is
The current is triangular and flows only during the critical duty cycle

C
which is a fraction of the switching
cycle, the peak current I
PK(FB)
is
where
I
AV
= Average current
Calculate the current at minimum boost voltage V
BST(min)
Calculate time T1 and T2 at maximum boost voltage V
BST(max)
Parameters Symbol Value
Output Power
P
OUT
9.6W
AC Line Input Voltage
V
IN
230V
Output Voltage
V
OUT
24V5%
Load Current
I
OUT
400mA
Maximum Switching Frequency*
F
sw(max)
125kHz
* Increasing F
SW
reduces the size of the magnetics but increases switching losses in the FET.

Cmin/max
V
OUT min
V
BST max
---------------------------
=
24V 1.2V–
405V 40.5V+
-----------------------------------------
5.1%==
[Eq. 31]

Cmax/min
V
OUT max
V
BST min
----------------------------
=
24V 1.2V+
405V 40.5V–
-----------------------------------------
6.9%==
[Eq. 32]
P
IN
P
OUT

-----------------
=
9.6W
0.92
--------------
10.5W==
[Eq. 33]
I
PK FB
2I
AV


C
----------------- -
229mA
0.069
------------------------- -
== 840mA=
[Eq. 34]
P
IN
V
BST min
--------------------------
10.5W
405V 40.5V–
-----------------------------------------
= 29mA=
[Eq. 35]
T1 
Cmin/max
TT 0.051 8s 408ns===
[Eq. 36]
T2 TT T1– 8s 408ns– 7.592s===
[Eq. 37]