tl494_inverter_ccfl_lcd_schematic_jpg_nq v1.0.0

TL494 Inverter CCFL LCD Schematic.JPG

TL494 Inverter CCFL LCD Schematic.JPG ->->->-> https://fancli.com/2thVlF

on the secondary side, the transistor q1 acts as a current sink and is also used to regulate the current flowing through the ccfl. the transistor q2 is used to switch off the ccfl when the output voltage from the oscillator/driver is below the required level. the diode d1 is used to prevent the ccfl from lighting up when the inverter is switched on and off.

here is the wiring diagram for the main inverter board. i also soldered the 6 mosfet drivers individually, and added a capacitor on each driver as a bypass circuit. the capacitor values are given in the datasheet of the mosfet. a resistor can be added for an added safety measure.

the second (12v vcc) circuit consists of the egs002 board, the 12v vcc supply and a 5v regulated power supply. the egs002 board is configured to have a 5v supply that may be configured to control another powered device(s) through the 12v input. if you want to use the egs002 with the lm2576 as a non-inverter controller, you can skip this circuit and just connect your 12v line to the +5v pin of the lm2576.

i have a hunch that this is the part that needs a little bit more explanation. check out the inverter circuits site. there is a schematic at the bottom of the page. it's a dual-output inverter with a number of oscillator circuits. the left part of the schematic is what is needed for this application and i'll explain the right side part in more detail.

if you try to connect this output directly to your inverter circuit, you would get oscillations and possibly damage your oscillator. the second transistor, t2, provides a discharge path for the base of the first transistor. the capacitor c2 provides a slow discharge path for the output voltage of the oscillator. this capacitor is not really needed since it is not really used. if you were to connect the output of the oscillator to the base of the second transistor, you would get the same effect as if you were using a capacitor in this circuit. the oscillator output will discharge into t2 through the base of t1. the more base current that t2 receives, the more output voltage will be on the oscillator output. 84d34552a1