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| LM3409MY-NOPB Datasheet, PDF (30/43 Pages) Texas Instruments – PFET Buck Controller for High Power LED Drivers | |||
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LM3409, LM3409HV, LM3409-Q1
SNVS602J â MARCH 2009 â REVISED MAY 2013
6. PFET
Determine minimum Q1 voltage rating and current rating:
VT-MAX = VIN-MAX = 42V
IT
= D x ILED
=
VO x ILED
VIN x K
=
14V x1.02A
24V x 0.90
= 660 mA
A 70V, 5.7A PFET is chosen with RDS-ON = 190m⦠and Qg = 20nC. Determine IT-RMS and PT:
IT-RMS = ILED x
D
x
¨¨¨©§1+112
x
¨¨©§
'iL-PP
ILED
¸¸¹·2
¸·
¸¸¹
IT-RMS = 1.02A x
14V
24V x 0.90
x
¨¨§1+
©
1
12
x
¨¨©§414.052mAA¸¸¹·2¸¸¹·
IT-RMS = 830 mA
PT
=
IT-
2
RMS
x
RDSON
=
830
mA2
x190
m:
=
129
mW
The chosen component from step 6 is:
Q1o 5.7A,70V, DPAK
7. DIODE
Determine minimum D1 voltage rating and current rating:
VD-MAX = VIN-MAX = 42V
ID
=
(1-
D)
x
ILED
=
¨¨©§1-
VO
VIN x
K
¸¸¹·
x
ILED
ID =
¨§1-
©
14V ¸·
24V x 0.90¹
x
1.02A
=
358
mA
A 60V, 5A diode is chosen with VD = 750mV. Determine PD:
PD = ID x VD = 358 mA x 750 mV = 268 mW
The chosen component from step 7 is:
D1 o 5A,60V, SMC
8. INPUT UVLO
Solve for RUV2:
RUV2
=
VHYS
22 PA
=
1.1V =
22 PA
50 k:
The closest 1% tolerance resistor is 49.9 k⦠therefore VHYS is:
VHYS = RUV2 x 22 PA = 49.9 k: x 22 PA = 1.1V
Solve for RUV1:
RUV1 =
1.24V x RUV2
VTURN- ON -1.24V
=
1.24V x 49.9 k:
10V -1.24V
= 7.06 k:
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