40W Magnetic Ballast With Electronic Ignition

By Robert Dvoracek

The ballast itself is unlabeled, but the chassis sticker calls for 25, 32, or 40 Watt 48" type T12 bulbs.

WARNING: This circuit is AC line powered and contains voltages and currents which can KILL if you are not careful. Charged capacitors will SURPRISE YOU! In the event of a bleeder resistor failure, they can hold a lethal charge for hours! If you don't know much about working with line voltages or if you aren't crazy (like me) then DO NOT attempt to construct this circuit. I CANNOT BE RESPONSIBLE if you electrocute yourself to death! That said, let's have some fun!

40W Magnetic Ballast With Electronic Ignition circuit

Description:

The circuit is essentially a line frequency operated magnetic ballast with an electronic igniter circuit in place of the standard glow switch.  Unfortunately, I could not measure the inductance of L1 as I cannot afford an LCR meter.  Does anyone want to donate one to my cause?  I can tell you the size though which really means nothing if you were to try and build this thing, because your core may have a different permeability and Q factor than the one used here, and getting it wrong will damage your bulbs.  At any rate, L1 is a laminated iron core inductor with x number of turns to give a reactance limiting the current flowing through the fluorescent bulb FL1.  This is necessary because when gas (in this case mercury vapor) becomes ionized, its resistance drops to near zero.  Higher temperatures reduce the resistance even further.  Lower resistances draw more current and more current makes more heat.  Without some kind of external current limitation, this would end in an explosion, and we all know how dangerous mercury is.  Current can also be limited with a resistor, but inductors are preferred because a resistor converts the energy into heat, whereas an inductor turns it mostly into magnetism.  The 4u7 capacitor is matched with L1 and helps balance out its reactance to make the load appear more resistive.  Across it is a 330K resistor to drain any residual voltage when the circuit is switched off for safety reasons.

In the starter circuit, current is flowing through the filaments and through the 1M5 resistor and charging a 22N capacitor.  When it reaches around 32V, the DIAC, a 32V DB3, breaks over and dumps it into the TRIAC, momentarily causing a greater current flow.  This happens some 10s of times a second, warming the filaments.  At the same time, when the TRIAC switches off, as in a standard glow switch circuit, the decrease in current causes the excess flux in L1 to collapse, creating a high voltage transient, or "flyback" pulse.  This high voltage is enough to ionize the mercury and get the current flowing end-to-end.  Once that happens, the starter circuit is basically shunted and no current flows through the filaments.

Notes:

No filament current isn't necessarily always a bad thing.  It helps keep overall power consumption down and in some cases prolongs filament life.  Not all tubes may like this though, but this was the standard for many years.

I'm not sure about the function of the diode and the 27K resistor.  It may be there to shunt away some current from the 22N capacitor and slow down the pulsing on the filaments, or it may be there to protect the TRIAC from some of the transients from L1.  It may even be possible to omit it.  Who knows, give it a try and see what happens.

The starter circuit can be used in place of a glow switch in any fluorescent light that uses one.  It's a pretty neat concept since those glow switches are always in need of replacement.  The TRIAC circuit will last forever and gives a more rapid start.  All you would need to do is take an old glow switch and bend the tabs on the bottom to separate the can from the little plastic disc with the terminals on it.  Then just cut off the glow switch and compensation capacitor and solder in the TRIAC circuit.  Then it's just a simple matter of putting the can back on and putting it back in the light.

*UPDATE 3/23/2006*

I've done some thinking about the way this circuit works and listened some more to the sound it makes when it starts up.  It sounds as though it only fires once every few cycles, which would be due to the diode.  When the current is flowing in the direction of forward biasing the diode, it acts like a shunt since the 27kR has much lower resistance than the 1M5.  The 27k resistor is there to keep the diode from damping the HV pulses from the inductor too much.  The TRIAC cannot fire and so very little current flows through the filaments.

When it's flowing in the other direction, the 22n cap resumes charging.  After enough cycles go by, the cap reaches the breakover voltage of the DIAC and the TRIAC turns on and warms the filaments for the rest of that 1/2 cycle.

Eventually the bulb will light and the remaining charge in the 22n will bleed off through the 1M5 and what is now a very low resistance across the bulb.

It may be advantageous to add another resistor between the gate of the TRIAC and the DIAC to keep the TRIAC on a little longer.  I noticed that this circuit has a bit of trouble starting older bulbs out in the garage when they're really cold.

Parts Info:

MAC97A6 400V TRIAC in TO-92 case
DB3 32V DIAC
 

http://kickme.to/lightningstalker 3677 08 December 2007

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