User manual

1.24V x R18
V
TURN-OFF
- 1.24V
= 15.8 k:
R11 = =
1.24V x 750 k:
60V - 1.24V
V
HYSO
= R18 x 20 PA = 750 k: x 20 PA = 15V
V
HYSO
20 PA
= 750 k:
R18 =
15V
=
20 PA
R4 = 16.9 k:
R13 = 10 k:
R5 = 1.43 k:
V
HYS
=
R5
20 PA x 16.9 k: x (1.43 k: + 10 k:)
1.43 k:
V
HYS
=
+ 20 PA x R
UV2
20 PA x R4 x (R5 + R13)
+ 20 PA x 10 k: = 2.9V
R4 =
R5 x (V
HYS
- 20 PA x R13)
20 PA x (R5 + R13)
= 16.9 k:
R4 =
1.43 k: x (2.9V - 20 PA x 10 k:)
20 PA x (1.43 k: + 10 k:)
V
TURN-ON
=
R5
1.24V x (1.43 k: + 10 k:)
1.43 k:
1.24V x (R5 + R13)
= 9.91V
V
TURN-ON
=
1.24V x R13
V
TURN-ON
- 1.24V
= 1.42 k:
R5 = =
1.24V x 10 k:
10V - 1.24V
D1 o 12A, 100V, DPAK
P
D
= I
D
x V
FD
= 1A x 600 mV = 600 mW
Design Procedure www.ti.com
(43)
The chosen component from step 10 is:
(44)
6.12 Input UVLO
Since PWM dimming will be evaluated, a three resistor network will be used. Assume R13 = 10 kΩ and
solve for R5:
(45)
The closest standard resistor is 1.43 kΩ therefore V
TURN-ON
is:
(46)
Solve for R4:
(47)
The closest standard resistor is 16.9 kΩ making V
HYS
:
(48)
The chosen components from step 11 are:
(49)
6.13 Output OVLO
Solve for R18:
(50)
The closest standard resistor is 750 kΩ therefore V
HYSO
is:
(51)
Solve for R11:
(52)
The closest standard resistor is 15.8 kΩ making V
TURN-OFF
:
10
AN-1985 LM3429 Buck-Boost Evaluation Board SNVA403C–July 2009–Revised May 2013
Submit Documentation Feedback
Copyright © 2009–2013, Texas Instruments Incorporated