# Electronic – Triac Turn-on Stress Calculation

triac

When snubbing a triac, one of the concerns is minimizing turn-on stress as your snubber triac discharges across the newly-shorted terminals of the triac. This paper says that the main concern is keeping \$\frac{dI}{dt}\$ below the datasheet maximums during turn-on.

How exactly do I work out what \$\frac{dI}{dt}\$ will be theoretically given a particular RC snubber?

The datasheet for my triac indicates the "turn-on time" of the triac is \$2 \mbox{ } \mu s\$. Is the switching speed? Is the calculation \$\left(\frac{dI}{dt}\right)_{max}=\frac{80\% \cdot 120V √2}{47\Omega \cdot 2µs} = 1.4 \frac{A}{µs} \$ (for a 47 Ohm snubber resistor) too simplistic somehow? It is based on the assumption that the voltage across the triac terminals drops approximately linearly from 90% to 10% of the peak AC voltage in 2µs. 1.4 A/µs is a good margin under my triac's rated \$\frac{dI}{dt}\$ maximum of 10 A/µs.

I'm doubting my reasoning because the paper I linked to above says that \$47 \mbox{ } \Omega\$ is barely enough to limit \$\frac{dI}{dt}\$ to \$50\frac {A} {\mu s} \$ at turn-on (see figure 6). Am I to understand from figure 6 that the switching time of STM triacs is much less than 2µs (the STM datasheets don't have a "turn-on time" field). If I'm eyeballing figure 6 properly, it looks like the snubber discharge current peaks only ~0.1µs after it begins to rise.