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As I've repeatedly said, the ONLY difference is the DC level the signal is sat on - most electronic uses of AC wouldn't be AC by that theory - amplifiers, radios, almost everything.
The EMF will still be there but the current will flow in only one direction since impendance of a positive polarised dieode is much much smaller than imp. of the negative polarised diode. that's just it, but the emf will be there.
Regardless of the quality of diode used (meaning low low leakage current) you'll not get any DC buildup because winding is almost a short circuit for any DC component.
If it appears then You've got a problem elsewhere.
But then again if the pulses are short and far away allowing the atmosphere in the lamp to cool down before next cycle then you'll get cold lamp on each positive half-cycle, which means it will have a higher current rush later in the cycle. That will certainly damage one of the electrodes.
Another thing. and i'm on the edge of shooting my foot now:
-Since you do intend to get the best of the transformer, you'll use the frequency as high as convenient, too low the freq. and you'll waste money, too high and you'll heat up the core thus wasting energy.
-If you follow this, you'll see that usually voltage rise-up and fall-down times of the oscillator are much smaller than current rise up and fall off times. (Which is of course governed by the Z of the tansformer.) And current rise/fall times of the transformer are equal, that's it's characteristic. (The strange thing is that I'm lying right now, there is a slight difference due to hysteresis of the core, since we're driving dhe flux in only one direction, but believe me in these cases you'll not see it.)
So the point is JUST TUNE THE DAMN THING WELL AND IT WILL WORK.
P.S.
Thanks guys, just for laughs I've checked my cellar and found there something beautiful: The MAGNETIC AMPLIFIER, I myself would never have had heard of it if not for some old centrifuges, but that's another story now.
How can this simple circuit power the tube without blackening one of the ends?
How can it deliver equal voltage positive and negitive cycles?
I don't see how you can make the current in the primary grow at the same rate as it will decay when the current is interrupted.
I don't think you understood the point I made earlier on about the fly back effect. This circuit works like an ignition coil driver, it's the current being interrupted that generates the high voltage which causes the tube to glow - the highest voltage occurs when the transistor switches off not when it's being turned on!
Simulate it on TLSpice and you'll see what I mean!
Hi all
I've used circuit#2 (with 7.2V constant supply for charging) for emerg.Light. I used a new battery 6v 4.5Ah . The problem is when the battery voltage drop to 5.8 volt circuit stop working the time of glowing light is 30 minutes. I want to increase the lighting Time what should I do ?
Obviously, if I said circuit one is the best then implication was for you to use it.
However, I don't think this it the problem here, although circuit 1 is likely to power the tube at lower voltages as both of the cathodes will be warmer.
Perhaps you need to decrease the number of turns on the primary, you could also hook up the base of the transistor to a scope to see if it's oscillating at the lower voltage. Can you power a neon lamp via a resistor when the input is only 5.8V?
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