I am thinking of using a BTA04 type TRIAC in a project to switch a power hungry mains appliance, which could draw up to 12A. I can't find a forward voltage anywhere in the datasheet, or an on-state resistance. I presume, looking at the symbol, and internal construction, that it drops about 1V (silicon diode drop). In which case it would dissipate 12W at this kind of current level and require a good heatsink. Is this correct?
Electronic – TRIAC heatsinking
triac
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Your electric fan speed regulator will have to be controlled by something, what will it be? Because I doubt the heater has something like that inside. Also, a gate resistor will only limit the command current but not the current through the device itself.
The way I see it, if you want to control the heating resistance's current (temperature) with a triac you'd have to resort to a dimmer-like circuit:
Do note, though, that you're dealing with \$ \frac{2kW}{230V_{RMS}} \approx 8.7A_{RMS} \$, which is quite a bit higher than your average light-bulb.
As for the heat-sink, it may well be the case. The dissipated power would be:
\$ P_d = V_{t0} I_{T_{AVG}} + R_d I_{T_{RMS}}^2 \$
With the catalog values and the previously calculated one:
\$ Pd = 0.85V \cdot 8.7A*\frac{2\sqrt{2}}{\pi} + 10m\Omega \cdot 8.7^2 = 7.41W \$
\$ T_j = P_d R_{th_{j-a}}+T_a = 7.41W\cdot 50\frac{^\circ C}{W} + 25^\circ C = 395.5^\circ C > T_{j_{MAX}}=125^\circ C \$
I assumed \$25^\circ C\$ ambient temperature, which means you'll need a heat-sink. After a quick math (involving some graphs), you'll need an \$ R_{th_{h-a}} \approx 13\frac{^\circ C}{W}\$ , which would be an aluminium square of 7.5cm (~3in) with a thickness of 1mm. Of course, conditions vary and this is an informative note, only. What thermal contact you're using, how you're mounting the heat-sink, etc, have an influence on the final result.
Here's an invaluable reference which indicates that your load's on the wrong side of the TRIAC and your gate resistance is way too high.
Best Answer
I'm not familiar with that part so I did a search for "BTA04 triac" and found a datasheet right away.
First, this part is obsolete, so not a good idea for new designs. Second, it's only rated for 4A RMS continuous, so using it with a load that could draw 12A is a bad idea.
The forward drop is in the datasheet, and is listed as 1.65V. That's about as expected since a SCR has two junctions in its conduction path. That's roughly what you'll get from most SCRs and TRIACs.
The method of your power calculation is correct. It will dissipate power equal to the current thru it times the voltage accross it, which would be 12W at 12A and 1V. However, that is considerably out of spec for this part.