# simple triac current limiter

I have a 20 liter electric storage water heater which has a 230V AC, 2KW electric heating element. The heating element is failing frequently primarily due to bad quality of water inducing faster corrosion of heating element. I decided to experiment with the current to increase the life of heating element. I want to limit the heating to about 80% of normal so that heating element runs at slightly cooler temperatures.

I have BTA41-600B 41Amp Triac available to me. This triac has sufficiently high current rating for a 2KW coil. Two options come to my mind.

1. Buy a electric fan speed regulator. These are available cheap and in plenty. Open the enclosure, identify the triac in the circuit and just replace the triac with BTA41-600B. The question is will this work? I am an electrical Engineer and I do build small electronic circuits on my own, but I have never built a circuit involving Triac.

2. The other option is to build a simple circuit with a preset resistance connected to the gate. But what should be the rating of this preset resistance?

There is one more question. Should I need to use a heat sink for this application?

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.

• Very good explanation! You must be a great teacher! Are You? Best regards! Ing Eftim Stoyanov – user78274 Jun 8 '15 at 16:25