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MAX754CSE データシートの表示(PDF) - Maxim Integrated

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MAX754CSE Datasheet PDF : 16 Pages
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CCFL Backlight and
LCD Contrast Controllers
+5V INPUT
C1
0.22µF
2
3
1 15
14
LADJ LON
VDD
BATT
LX
CONTROL ON/OFF
ON-TIME
LOGIC
PRESET
6-BIT COUNTER
CLK
OFF-TIME
LOGIC
PULSE-SKIP
COMPARATOR
BATTERY
INPUT
C2
10µF
L2
33µH
D3
1N5819
LDRV 13
LFB 16
Q3
R3
R4
MAX754
PGND
12
6-BIT DAC
GND
6
VDAC
FULL-SCALE OUTPUT = 1.250V
HALF-SCALE OUTPUT = 0.938V
ZERO-SCALE OUTPUT = 0.635V
POSITIVE
LCD-BIAS
OUTPUT
C6
10µF
35V
Figure 6. MAX754 Positive LCD-Bias Generator
Table 1. CCFL Circuit Component Descriptions
ITEM
C5
R18
R8
D7A, D7B
CCFL
DESCRIPTION
Integrating Capacitor. 1 / (C5 x R18) sets the dominant pole for the feedback loop, which regulates the lamp
current. Set the dominant pole at least two decades below the Royer frequency to eliminate the AC compo-
nent of the voltage on R8. For example, if your Royer is oscillating at 50kHz = 314159rad/s, you should set
1 / (C5 x R18) 3142rad/s.
Integrating Resistor. The output source-current capability of the CC pin (50µA) limits how small R18 can be.
Do not make R18 smaller than 70k, otherwise CC will not be able to servo CFB to the DAC voltage (i.e., the
integrator will not be able to integrate) and the loop will not be able to regulate.
R8 converts the half-wave rectified lamp current into a voltage. The average voltage on R8 is not equal to the
root mean square voltage on R8. The accuracy of R8 is important since it, along with the MAX754 reference,
sets the full-scale lamp current. Use a ±1%-accurate resistor.
D7A and D7B half-wave rectify the CCFL lamp current. Half-wave rectification of the lamp current and then
averaging is a simple way to perform AC-to-DC conversion. D7A and D7B’s forward voltage drop and speed
are unimportant; they do not need to pass currents larger than about 10mA, and their reverse breakdown
voltage can be as low as 10V.
The circuit of Figure 1, with the components shown in the bill of materials (Table 4), will drive a 500VRMS oper-
ating cold-cathode fluorescent lamp at 6W of power with a +12V input voltage. The lower the input voltage,
the less power the circuit can deliver.
8 _______________________________________________________________________________________

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