Cirrus Logic AN372 - Manual (Page 24)

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.

C min/max  
V
OUT min  
V
BST max  
--------------------------- =
24V 1.2V –  
405V 40.5V +  
----------------------------------------- 5.1% = = [Eq. 31]

C max/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  
2 I
AV


C
------------------
2 29mA 
0.069
-------------------------- = = 840mA = [Eq. 34]
P
IN
V
BST min  
--------------------------
10.5W
405V 40.5V –  
----------------------------------------- = 29mA = [Eq. 35]
T1 
C min/max  
TT  0.051 8 s 408ns =  = = [Eq. 36]
T2 TT T1 – 8 s 408ns – 7.592 s = = = [Eq. 37]