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How does this CCFL inverter work?


Hello,


I have this CCFL (Cold Cathode Fluorescent Light) inverter from a scanner. it powers the scanner light.

That's the schematic, but I can't see all the components very good, I'm sure about everything that's noted in the schematic, but I don't know the value of that capacitor between both collectors. It's that brown one from the picture.

So the question is:
How does this circuit work?

I'm still trying to figure out how the transistors switch, it's really weird, because when I try recreating this on my breadbord it doesn't work.

And why is the first coil shorted out?

What I do know about this is that it has an output voltage (very low current, serveral micro amps) of 2 to 3kV. It also works at a very high frequency, about 30kHz I think, and it has a ferrite core transformer. The primary windings are very thick, and there are about 10 turns. the secondary has Many windings, a few thousand.

The primary current is limited by the frequency (Xl = 2*Pi*f*L) so high frequency means high resistance of the coil, and that means a low current.


now  how does the switching process work? and what's the use of the brown capacitor?

Picture of How does this CCFL inverter work?
lemonie6 years ago

This page shows a build and explains how the inverter works (in detail). The design isn't exactly the same, but similar enough.
http://www.talkingelectronics.com/projects/FluorescentInverter/FluorescentInverter.html

L


RobertK32 years ago

The only connections that are right in the diagram are the emitters, and the transformer high voltage side.

The thing that you show as a shorted winding is the center tap of the input, the other end of the input windings hooks to the collectors of the transistors.

The drive winding hooks to the bases of the transistors and is biased by the resistor.

The brown capacitor is the tank circuit capacitor,usually about .022uf at 250v.

orksecurity6 years ago
I'm not convinced your schematic is correct, since connecting both ends of a coil to a single wire (at the top of the transformer on your page) does absolutely nothing.

Analog isn't my forte' these days, but conceptually this is straightforward. DC is used to run an oscillator, which is fed into a transformer to step up the voltage. Of course the current available at the output of the transformer goes down correspondingly, since even if this was perfectly efficient the amount of power can't be increased and watts (power) is volts times amps -- the transformer just trades off one against the other.