I have the schematic diagram in front of me.
Let me know, but I believe I’m not wrong if I say that the power section is made up of 3 elements when the circuit is closed with mains :
- The Triac,
- The 1.5 mH inductance,
- The external load (e.g. incandescent bulb(s), or other resistive load).
In normal operation, the Triac controls the output voltage (and current too, according to Ohm’s law).
If there is no load connected to the output, any attempt to drive the Triac from 0% to 100% will not ruin anything.
But, if the Triac is delivering, let’s say 75%, to an existing load, and if you open the circuit (as a blowing bulb does), what happens ?
What appears between the terminals of the inductance ? (Not really a resistive load).
Don’t you believe that a VDR, able to short-circuit 6 500 A during 25 µs (0.70 EUR VAT included), won’t protect the Triac from blowing, if it’s mounted between A1/A2 ? (According to Ohm’s law : I = U/R between A1/A2).
Anyway, as it’s usual to protect the switching circuit of a relay in a DC domain by a diode, it’s a recommended practise, in the AC domain, to protect a triac with an appropriate VDR, either if it just switches a load with zero-crossing detector, or if it acts as light dimmer.
Since 1976, when I started to engineer light dimmers for amateur theatre lightning, driving up to 3 500 W each, I never lost a Triac when a spotlight bulb broke down on duty.
Trying this solution by soldering a VDR on the PCB kit will not affect, in any way, the behaviour of the original circuit.
Philippe J. BOUCHON
Audio Engineering Society, Inc.
French section Elected Officer