20368Views11Replies

Author Options:

How does this CCFL inverter work? Answered


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?

11 Replies

user
lemonieBest Answer (author)2011-03-27


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


Select as Best AnswerUndo Best Answer

user
RobertK3 (author)2014-08-17

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.

Select as Best AnswerUndo Best Answer

user
orksecurity (author)2011-03-28

Remember that transformers operate on CHANGING amounts of current. If you feed DC to a transformer, the output will give you a pulse when you turn the input DC on and a pulse the other way when you turn it off, but that's it.

Which is why you need an oscillator before the transformer -- essentially, a DC-to-DC step-up converter always has to be a DC-to-AC-to-AC-to-DC converter.

Select as Best AnswerUndo Best Answer

user
steveastrouk (author)2011-03-27

Well you need a proper drive, because the core will saturate very easily.

Steve

Select as Best AnswerUndo Best Answer

user
steveastrouk (author)2011-03-27

Metal doesn't work well at these frequencies, which is why ferrite it used. There is no intrinsic reason why you need to use HF to power CFL backlight AFAIK, so you can use E I stampings, but your transformer will be very big.

Steve

Select as Best AnswerUndo Best Answer

user
lemonie (author)2011-03-27


It does like you show it. Magnetic fields go through air too.

L


Select as Best AnswerUndo Best Answer

user
orksecurity (author)2011-03-26

In that case, they're definitely doing something I don't understand, so I'll drop out.

Select as Best AnswerUndo Best Answer

user
steveastrouk (author)2011-03-26

Also, your analysis is wrong, a transformer is NOT an inductor, it possesses something called leakage inductance, which is an artefact caused by not all the flux created by the primary links the secondary winding, but any even slightly decent transfomer has a very low leakage inductance.

Steve

Select as Best AnswerUndo Best Answer

user
steveastrouk (author)2011-03-26

Shorting a coil is a very bad thing. It makes the transformer not work.

Steve

Select as Best AnswerUndo Best Answer

user
orksecurity (author)steveastrouk2011-03-26

+1. The shorted coil will draw power into that short circuit. Definitely a bad thing.

Check the circuit traces, and the transformer's actual pinout, again.

Select as Best AnswerUndo Best Answer

user
orksecurity (author)2011-03-26

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.

Select as Best AnswerUndo Best Answer