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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?

Discussions

0
whitres
whitres

3 months ago

These circuits usually use simple converters operating at high frequency and using high voltage miniature transformer. Output current is limited capacitively using a high voltage (few kV) capacitor with a small capacitance (tens of pF). Cold cathode fluorescent lamp (CCFL) has only two outlets, because it has no preheating. Capacitive current limiting is possible because the output voltage is sinusoidal. The inverter operates in resonant mode. It is an current fed inverter. This is ensured by L1 choke. Current of the fluorescent tubes (and thus power) can be adjusted by changing C1 and C2. Operating frequency is about 50kHz. It can be adjusted by C3. By changing the frequency the power is also changed. The brightness of fluorescent lamps can be controlled by changing the supply voltage in range from 5V to 12V. Control is possible in the range from about 5 or 10% to 100% brightness. With less voltage than 5V, it may happen that only a part of the lamps will be lit. One end of the lamp starts to glow at cca 1.5 V and the entire CCFL lamp is lit from 5V. When connecting CCFLs to Tr1 transformer observe the cold and live end (the cold end of the transformers and fluorescent tubes have thinner insulation than those live). The circuit I tested successfully with transistors 2SC5707, 2SC1384 and BD139. Tr1 can be obtained from the LCD monitor, or other device with CCFL. Winding your own would be difficult due the large number of turns of a thin wire at secondary.

0
whitres
whitres

3 months ago


this is almost your drawing

0
RobertK3
RobertK3

5 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.

0
orksecurity
orksecurity

8 years ago

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.

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steveastrouk
steveastrouk

8 years ago

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

Steve

0
steveastrouk
steveastrouk

8 years ago

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

0
lemonie
lemonie

8 years ago


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

L


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orksecurity
orksecurity

8 years ago

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

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steveastrouk
steveastrouk

8 years ago

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

0
steveastrouk
steveastrouk

8 years ago

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

Steve

0
orksecurity
orksecurity

Answer 8 years ago

+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.

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orksecurity
orksecurity

8 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.