Can anyone explain to me exactly what the ballast capacitors (cap in series with a lamp) function is in this circuit? I have done some searching and found: "The ballast capacitor controls current amplitude through the lamp's negative impedance by dropping an approximately equal voltage across its positive impedance." What does this mean? What are the implications of not using any ballast cap at all? TIA.
If you applied enough voltage to strike the lamp the current would increase until the lamp destroyed itself. Like any impedance, the capacitor causes a decrease in lamp voltage as current increases and stabilizes the circuit. An inductor or resistor could be used but presumably a capacitor is more cost effective and wastes less power.
I found the Fairchild app note, AN4136, useful for a description of what happens when a resonant Royer circuit is used to drive a CCFL. They give a slightly different slant on the purpose and value of the ballast capacitor.
They state that the cap is there in order to try and isolate the non-linear CCFL load from the parallel resonant tuned circuit. The intention is to reduce the harmonics in the tank and to obtain a better sine current. Ideally the voltage across the ballast cap should be many times that across the CCFL, but practical considerations limited them to 2x.
-- Tony Williams.
Oh dammit... I'd forget my head if it wasn't screwed on.
Note that the voltages across the cap and CCFL are 90 degrees out of phase. So the voltage from the inverter is the square-root of the sum of the squares. ie, If V(cap) = 2xV(CCFL) then V(Inverter) = 2.2xV(CCFL).
However, someone seeing (say) a 500V V(CCFL) and an 1100 volt V(Inverter) might unthinkingly assume "roughly equal voltage drops across the cap and CCFL".
-- Tony Williams.
You seem to be talking about using the Baxandall class-D oscillator circuit to drive a cold-cathode back-lights. This was popularised by Jim Williams of Linear Technology.
Reposting from an earlier thread.
As has already been mentioned by others, the capacitor provides a positive (if non-dissipative) impedance in series with the lamp to limit the current despite the negative impedance characteristic of these lamps. The implication of not using any cap at all would be that you might blow up the lamp if the driver circuit didn't blow up first. If you were really lucky, the oscillating inverter might stop oscillating before it blew itself up, or blew up the lamp, but Sod's Law would suggest that you would lose at least one component during the experiment.
-- Bill Sloman, Nijmegen
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