# Triac Turn-on Stress Calculation

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.

## 1 Answer

There sure is something wrong with your reasoning. With 120Vac, the peak voltage across the snubber network will be 120V x 1.414: 169.68V. this gives you a peak current of 3.6A. (The app note you linked to starts from 230Vac. Peak voltage: 325V.) Now you have to calculate the rise time but I think it's better to measure it after you build a proto-type. But you can make a guess on where to start based on the calculations you did.

It's difficult to see the current rise time in the app. note you linked to. It looks like 25A/µS to me. You should also take into consideration that the peak voltage appearing across the triac might become higher if you're switching an inductive load like a motor for example.

• Good catch: I now know that "x VAC" means "AC with $V_{rms} = x$". However, this still leaves my $\frac{dI}{dt}$ unexpectedly well below the max $\frac{dV}{dt}$ of my triac. I'm still worried about the discrepancy between this result and the app note's assertion that $47 \Omega$ is barely enough to limit $\frac{dI}{dt}$ to $50 \frac{A}{\mu s}$. – Isaac Sutherland Jun 3 '11 at 0:39
• Furthermore, I'm trying to design this circuit to be robust to different loads (varying inductance and resistance) so I still need to understand the theoretical calculation for peak $\frac{dI}{dt}$. – Isaac Sutherland Jun 3 '11 at 0:43
• I think it's almost impossible to calculate the slope of the current peak right after turn on. It's the sum of the load current and the current trough the snubber. Fig 6 in the app note shows an unloaded triac while the di/dt is made up of the load current and the current trough the discharging snubber capacitor. The snubber alone, as shown in fig.6 of the appnote, will produce a di/dt of 21A/µS. The slope will be very close to 50A/µS when the triac has to switch 4.5A. – Hendrik Jun 3 '11 at 10:53
• Also, switching speed and and rise time are two different things. Switching speed means the time it takes to switch between 10% and 90%. What happens in between these two values is what it's all about. In fig6 of the appnote, you will see that the rise time of the current trough the triac is 1.5µS. But the double arrow indicates a portion of the rise time that's very close to 250nS. (bit difficult to see in the app note.) – Hendrik Jun 3 '11 at 11:16
• The $\frac{dI}{dt}$ contribution of my inductive load is trivial (on the order of 0.0012 A/µS). So, basically it boils down to the question: Can I assume a linear drop from 90% to 10% voltage (between the triac terminals) in 2µS? i.e. Is "switching speed" equivalent to the "turn-on time" in the datasheet? – Isaac Sutherland Jun 3 '11 at 17:23