Cirrus Logic AN379 - Manual (Page 21)

AN379
AN379REV2 21
4 Design Example
The required operating parameters for the analytical process are outlined in the table below.
Notes: 1. It is desirable to use a power FET that does not exceed 50V breakdown.
2. System efficiency estimate is 77%
3. Rectifying Schottky diode voltage V
F
is equal to 0.5V
4.1 Buck Design Steps
The example design is a 12VAC application. A maximum buck switching frequency of 150kHz is selected,
which corresponds to a switching period TT
BUCK
of 6.67µs.
Step 1) Select a Value for Boost Output Voltage
Full-scale boost output voltage V
BST(full)
is determined by sense resistor R
BST
which is set to 604k . Use
Equation 7 on page 10 to calculate the full scale boost output voltage V
BST(full)
. Solve Equation 40 for boost
voltage V
BST(full)
:
For a 12VAC system, full-scale voltage V
BST(full)
is approximately 40V. The nominal boost output voltage V
BST
is 28V and droops to its lowest value towards the end of each half line-cycle until the boosting process starts
again in the next half line-cycle. The CS1680 limits the boost output voltage to ±15%. Therefore, minimum
boost output voltage V
BST(min)
and maximum boost output voltage V
BST(max)
are calculated using Equation 41
and 42, respectively:
Step 2) Select an Appropriate FET
Buck converters have an optimal operating range above 50% duty cycle, although any duty cycle above 30%
is close to optimal. When FET Q
BUCK
is turned off, the maximum drain voltage V
Drain(max)
occurs and is
calculated from Equation 43:
Use Equation 9 on page 11 to determine FET breakdown voltage V
Breakdown
. Solve Equation 44 for
voltage V
Breakdown
:
Parameters Symbol Value
Output Power P
OUT
5.15W
AC Input Voltage V
IN
12 VAC

10%
Output Voltage V
OUT
12V

10%
Load Current I
OUT
430 mA
Maximum Buck Switching Frequency * F
BUCKSW(max)
150 kHz
* Increasing the switching frequency may reduce the size of the magnetics but increase switching losses in the FET
and inductor.
V
BST full  
I
ref
R
BST
20k  +    64 A 604k  20k +    39.9V = = = [Eq. 40]
V
BST min  
28V = 0.85  24V = [Eq. 41]
V
BST max  
28V = 1.15 32V =  [Eq. 42]
V
Drain max  
V
BST BOP  
V
Diode
+ 40V 0.9375    0.5V + 38V = = = [Eq. 43]
V
Breakdown
V
BST BOP  
V
Diode
V
M in arg
+ + 37.5V 0.5V V
M in arg
+ + = = [Eq. 44]