I'm familiar with simple dimmer circuit, now I want to build a more powerful triac unit to use with a high power heater.

The resistance of my heater is made of 9\$\Omega\$ 3mm diameter nichrome wire. On 240V(rms) voltage this makes 25A(rms) current. I have an access to a mains line capable of delivering 30A, so in theory I will be able to drive heater at full load. In reality I probably won't, but nevertheless, I want to build a unit that could do that in theory if needed.

My question is - what should I take into account when building a simple optotriac/triac/zerocrossing circuit when moving from 8A triac to 40A triac?

Triac i'm going to use is this: BTB41

My schematics looks like this:

enter image description here


1 Answer 1


Two things I would check:

  • the trigger current, your triac requires 50 mA. If this is more than the triac your circuit was deisgned for you might need some changes

  • cooling! At 25A your triac seems to dissipate in the order of 25 .. 35 W. At a Tc of 75C and an assumed Tamb of 40C this means a heatsink of 1C/W. That's not a small one!

  • \$\begingroup\$ yesss, I already have the biggest heatsink i've seen for this. Good point about triggering -- (360+470)*50mA = 41.5V. This means, I can't fire in the first asin(41.5/340) = 7 degrees? \$\endgroup\$
    – miceuz
    Apr 14, 2013 at 17:10
  • \$\begingroup\$ This is not my specialty, but I think the C2 provides a phase shift of the trigger wrt. the power, to make triggering on the 0-crossing possible. Do you intend to DIM this load? IMH that is not needed for a heater. A simple and slwo on-off should be enough. \$\endgroup\$ Apr 14, 2013 at 17:22
  • 1
    \$\begingroup\$ IV quadrant requires 100mA according to the spec so don't skimp on this part of the circuit \$\endgroup\$
    – Andy aka
    Apr 14, 2013 at 17:22
  • \$\begingroup\$ In fact I want to dim it as I will want to limit the power when the mains line can't supply enough current. Now I've just realised that I don't get R6/C2 at all. I thought it's just a filter to cut off high frequency transients. \$\endgroup\$
    – miceuz
    Apr 14, 2013 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.