A704
Note that the inductance of most chocks drop when the chock current has DC-current components. This
inductance drop may cause the actual Io to be lower than the calculation value of Io (please refer the Rcs section of
this application note for more detail)
5: Choose the FET (Q1) and Diode (DF)
Since these power components (diode & MOSFET) maximum peak current exceeds the regulator maximum load
current, these components current rating must be at least 1.2 times greater than the maximum load current. And the
reverse voltage rating of these components should be at least 1.25 times the maximum input voltage. Therefore, the
peak voltage seen by the FET is equal to the maximum input voltage
VQ1 =1.25×Vmax,ac =1.25×( 2 ×240) = 425V
Hence, the current rating of the FET is
I
= 1.25 × I
= 1.25 ×1.3× 0.35 = 0.57 A
o, pk
Q1
In the example, choose at least a 500V 2A N-channel MOSFET. A 600V/2A N-channel MOSFET can allow
more safety margin.
The peak voltage rating of the diode is as same as the Q1 MOSFET. Hence,
= 1.25 ×V = 1.25 × ( 2 × 240) = 425V
V
DF
max, ac
The average current through the diode is:
= 1.25 × I = 1.25 ×1.3× 0.35 = 0.57 A
I
DF
o, pk
Choose 600V 2A Trr=35nS Fast recovery diode in the example.
6. Choose the Sense Resistor (R4)
Since the output ripple current IPP is design for 0.6 times Io current.
V (1− d)T
o
s
I
=
= 0.6× I
o
pp
L
Therefore the Peak current can be given as:
1
2
I pk = Io + I pp = 1.3× Io
Hence, the sense resistor value is given by:
V
V
I
cs
cs
R
=
=
cs
1.3× I
o
pk
In the design example, use RCS = (0.5V/(1.3*0.35) =1.1Ω. Note that factors like inductor value deviation could
cause the actual Io to be higher (or lower) than the ideal value. So we may need to fine-tune the Rcs value for accurate
Io.
Copyright © 2008 ADDtek Corp.
10
A704_V0.6 -- AUGUST 2008
ADDTEK [ ADDTEK CORP ]
