Triac problems in turning CFL on/off

[Flavini]

I have made a circuit to turn on and off 2 CFLs remotely. For the remote control part, I used RC5 protocol. For turning the CFLs on and off, I used triac. I have a few questions here.

Firstly what I did was:
To turn the CFL on, the microcontroller kept the triac turned on as signal was applied to gate continuously until the command to turn the CFL off was sent. After a few days, the triac was burnt. Was it due to the fact that I kept the triac on?

Then, I changed the driving to:
The microcontroller turns the triac on at a zero-crossing (for the first time), then keeps on sending that signal to the gate for 2ms, turns the gate off (but the triac is still on) for 6ms, then turns triac gate on for 4ms. This worked for a few weeks, but then, the triac got burnt again.

So I put a snubber across the triac, but this caused the CFL to turn on and off every few seconds. I think this is because the leakage through the snubber charges the input capacitor in the CFL circuit.

So what could be the solution to the problem? Is it really a problem if I keep the gate of the triac turned on? How can I prevent the triac from burning?

 

[betwixt]

More information needed:

1. what is the triac type and rating?
2. what are the CFL ratings (voltage, power) ?
3. what power source are you using (voltage, AC/DC, generator)?

Under normal circumstances it is quite safe to leave the gate voltage on all the time, either you are using mismatched CFLs and triac or you are driving the gate with far too much current.

Brian.

 

[KerimF]

Please note that some CFLs take their current at the mains voltage peak (not having PFC) and their starting current could be relatively high.
Brian already put the important questions that help finding the cause.

Kerim

 

[Flavini]

More information needed:

1. what is the triac type and rating?
2. what are the CFL ratings (voltage, power) ?
3. what power source are you using (voltage, AC/DC, generator)?

Under normal circumstances it is quite safe to leave the gate voltage on all the time, either you are using mismatched CFLs and triac or you are driving the gate with far too much current.

Brian.

The triac is BT134 rated for 4A RMS on state current and 25A non-repetitive peak state current.

The CFL is rated at 30W, 220V AC, 50Hz.

It is running from 220V mains.

I'm driving the triacs with 5v with a 1k-ohm gate resistor.

I hope this can help identify the problem.

---------- Post added at 08:17 ---------- Previous post was at 08:16 ----------

Please note that some CFLs take their current at the mains voltage peak (not having PFC) and their starting current could be relatively high.
Brian already put the important questions that help finding the cause.

Kerim

How could the triac be protected from that high starting current? Would placing an NTC help?

 

[KerimF]

So your gate current is:

Ig = (Vcc-Vg) / Rg
Where
Vcc = 5V
Vg = 0.7V (like the forward voltage of a diode)
Rg = 1K

Ig = (5-0.7) / 1 = 4.3 mA

I guess your Ig is rather small. You can check this by using as a load an incandescent bulb (about 40W to 100W) and see how it lights.

The gate sensitivity (Ig to turn the triac on for the full cycle) differs with the gate current polarity.
The triggering gate current is lower if it is negative (its direction out of the gate, from T1 to G). This could be done by connecting T1 to Vcc and the gate resistor between gate and ground (usually a small PNP transistor as BC547 is used as a driver; emitter to ground and collector to Rg).

I think for a negative gate current, 50mA will surely turn on the triac in full. Use an incandescent bulb for test. But you can try lower values till you find out at which value the bulb starts flickering or being off.

If you need to supply a positive gate current, you can do the same but perhaps even 50mA might not be enough.

If you are using a transistor to switch the gate current on and off, you can use the following formula to calculate Rg (you may need taking the nearest standard value, preferably lower):

Rg = (Vcc – Vg – Vc ) / Ig
Where
Vcc = 5V
Vg = 0.7V (like the forward voltage of a diode)
Vc = 0.3V (estimated for Vce saturated when Ic/Ib < 30)
Ig = your value (in mA)
Rg in Kohm

Let us see if your problem will be solved after driving the triac properly.

Good Luck (try not to adjust anything on your test circuit before disconnecting it from the two mains lines).

Kerim