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3
AN-H48
In this example,
L1 = 2.9mH
.
The peak current rating of the inductor will be:
I
P
= 0.35 • 1.15 = 0.4A
(9)
The rms current through the inductor will be the same as
the average current for the chosen 30% ripple.Right inductor
for this application is an off-the-shelf 2.7mH, 0.54A (peak),
0.33A (rms) inductor.
Step 5: Choose the FET (Q1) and Diode (D2)
The peak voltage seen by the FET is equal to the maximum
input voltage. Using a 50% safety rating,
V
FET
= 1.5 • (√2 • 135) = 286V
(10)
The maximum rms current through the FET depends on the
maximum duty cycle, which is 50% by design. Hence, the
current rating of the FET is:
I
FET
≈ I
O,MAX
• √0.5 = 0.247A
(11)
Typically a FET with about 3 times the current is chosen to
minimize the resistive losses in the switch.
For this application chose a 300V, <1A MOSFET, such as a
BSP130 from Phillips. Actual MOSFET type should be de
-
termined by the transistor permitted power dissipation on
printed board. For example, a BSP130 SOT-223 package
limits the dissipation to less than a Watt at 50+ Celsius, even
if the MOSFET peak current capability is 1.5A. A good rule
of thumb is to limit overall MOSFET power dissipation to
not more than 3-5% of total output power, by making a right
transistor choice. In choosing MOSFET transistors for such
LED drivers, going bigger does not mean getting better, just
the opposite. Using TO-220 transistor 500/4A/2W instead
of SOT-223 transistor 300V/0.5A/6W does more harm than
good, reducing overall efficiency by several percent.
The peak voltage rating of the diode is the same as the FET.
Hence,
V
DIODE
= V
FET
= 286V
(12)
The average current through the diode is:
I
DIODE
= 0.5 • I
O,MAX
= 0.175A
(13)
Choose a 300V, 1A ultra-fast diode.
Step 6: Choose the Sense Resistor (R
2
)
The sense resistor value is given by:
R
2
=
0.25
(14)
1.15 • I
O,MAX
if the internal voltage threshold is being used. Otherwise,
substitute the voltage at the LD pin instead of the 0.25V in
equation (14).
For this design, R
2
= 0.55Ω. Also calculate the resistor power
dissipation:
P
R2
= (I
O,MAX
)
2
• R
2
= 0.067W
(15)
A 0.1W resistor is good for this application.
Note:
Capacitor C
3
is a bypass capacitor. A typical value of 1.0 to
2.2µF, 16V is recommended.
Design for DC/DC Applications
The same procedure can be used for DC/DC applications
(like the HV9910DB3). The only modifications are that the
input diode bridge and input hold-up capacitor are not re-
quired. A small input capacitance to absorb high frequency
ripple current is all that is required. This capacitance can be
computed using equation (7).
Appendix
The more accurate equations for computing the required ca
-
pacitance values are:
(16)
(17)
For the example in this application note, the actual minimum
capacitance required from the above equations is 19µF (as
compared to 26µF from equation (5)).
1
1
t
1
=
2 x π x freq
sin
-1
V
MIN,DC
√2 x V
MIN,AC
2 x V
O,MAX
x I
O,MAX
x t
1
+ 4 x freq
2 x V
2
MIN,AC
- V
2
MIN,DC
x η
C1 ≥
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022509