TRIAC Turns On-Off the Fluorescent Lamp

[(unknown)]

I want to use TRIAC as ordinary ON-OFF switch for a set of fluorescent lamps (the lamps would be sensor-activated). So, there is no dimming of the lamps, only fully ON or fully OFF. Due to large inductive load imposed by ballast, the current and applied mains voltage are not in-phase. To be turned ON, the TRIAC should be triggered immediately after the current crosses the zero, right? That information could be picked off from serial resistor when the current flows, i.e. when the circuit is already ON. But, when the circuit is OFF, no current flows, so how could the current be sensed to enable accurate initial turning on of the TRIAC. Should it be initially be triggered at voltage zero-crossing and than at the current zero-crossing?

Thanks

[Opera House]

Turn off isn't really a concern.  The triac is basically a back to back scr.  When it is triggered on, it stays on till the current stops.  So, you only need to worry about turn on.

[Ungrounded Lightning Rod]

Turn off isn't really a concern.  The triac is basically a back to back scr.  When it is triggered on, it stays on till the current stops.  So, you only need to worry about turn on.

However, note that the current and voltage are out of phase due to the ballast (whether inductive, where the current lags voltage, or electronic, where the current leads it).  The Triac needs to be triggered just after voltage zero-crossing ACROSS THE TRIAC.  Wait too long and you've just tried to dim a fluorescent lamp, a procedure that is not recommended.

Also:  Firing a triac when there's a non-trivial voltage across it isn't an issue if the load is inductive or resistive.  But if it's capacitive (i.e. rectifiers feeding a capacitor in the power supply of an electronic-balasted lamp) you produce a current spike - through both the triac and the rectifiers.  Doing it once when you turn the lamp on may not be an issue but doing it every half-cycle may cause problems.

[(unknown)]

I think that turn-off IS a concern if there is a large inductive load. At the moment when the current aproaches and crosses zero, conduction of triac would stop. However, there is a substantial amount of momentary voltage which results in a large dU/dt across the triac, which could in turn, in conjuction with some other factors, result in the autofiring of the triac. But it could be avoided with a snubber network across the triac.

I agree that the triac should be triggered at the voltage zero-crossing across it, that also means at the current zero-crossing because the triac appears to be purely resistive. But the question is HOW?

When the triac is in a stationary blocked state, there is a full voltage across it, however,when in conduction, a voltage across it is about 2 Volts. Also, there are the transients in between. So, the voltage across the triac could be readily used when it is blocked, i.e. at the initial turn-ON. But than the voltage across it falls to merely a volt or two, which means that the voltage at the sensing point will span over very wide range. However, it could be managed if that is a right way.

Does anybody sees any problem here?

Thanks

[Ungrounded Lightning Rod]

I think that turn-off IS a concern if there is a large inductive load. At the moment when the current aproaches and crosses zero, conduction of triac would stop. However, there is a substantial amount of momentary voltage which results in a large dU/dt across the triac...

Nope.  The substantial voltage across a switch interrupting a conductive load is produced by the current through the inductor.  Because the triac cuts off when the current hits zero there's no current in the inductor (and thus no stored energy) to make a voltage spike.

Think of the inductance as a mass, the current as its motion, and the voltage as the push trying to accellerate or decellerate it.  If you open the circuit you're trying to stop the moving mass RIGHT NOW.  So you have to produce an ENORMOUS force to push it back, and that's the "inductive kick".  You CAN'T stop it in zero time (because the voltage would become infinite, i.e. it WILL rise until the current keeps flowing - at least until the high voltage brings the current to a halt).  But stopping the mass when it's already stopped takes no force at all.

That's why a triac is ideal for switching AC loads to an inductive load.  It automatically turns off when the current it at the right point (zero) to cause no havoc.  And when it turns on the inductance keeps the current from rising abruptly even though the voltage may be non-zero.  It's good for switching pure resistive loads:  Though you get a sudden rise in current (producing radio interference) the resistance limits the current to the nominal value.

The problem comes when you're trying to switch ON a CAPACITIVE load.  If the voltage across the switch is non-zero you get an enormous rush of current when the switch closes (or the triac turns on).  So for capacitive loads (like a compact fluorescent' lamp or other capacitor-charging power supplies) it's necessary to switch your triac at a very low voltage difference or take other measures to limit the inrush current.  Otherwise the enormous current, focussed through the part of the triac that is turning on, can exceed its current-handling capacity and damage or destroy it.

[(unknown)]

Yes, I agree completely with what you said, and the electro-mechanical analogy is OK. But what I ment by "substantial amount of momentary voltage" was not the back-EMF caused by changing the current flow through an inductance; I had in mind the mains voltage which leads the current by almost 90 deg. (with predominantly inductive load) and its momentary value is near the peak at the current zero-crossing. At that very moment the triac opens and the full peak mains voltage is developed across its terminals in a very short time. That high dV/dt can, by my opinion, result in a self triggering of the sensitive triac, which, in turn, could be prevented with a snubber network.

At the turn-on of an inductive load I would prefer the zero-value of the momentary voltage because of unavoidable parasitic capacitances and accompanied inrush current peaks resulting in EMI.