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TRIAC Heat Generation & Dissipation

Discussion in 'Power Electronics' started by adamq, Oct 4, 2018.

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  1. adamq


    Dec 17, 2013
    Hello all,

    I am putting together a TRIAC power controller that will control a 4500-watt electric heater: [email protected]

    The TRIACs I am considering for this circuit are 40-amp TRIACS:
    NTE56028 in TO-220:
    NTE5679 in TO-3:

    My concern is heat, and how to manage it effectively.

    I am not familiar with how to calculate the heat generated by the TRIAC at a specific load.

    I'm not even sure the load itself impacts the heat generated, to begin with, or if heat is generated strictly by the gate switching.

    Maximum and average gate power of these TRIACS are:
    NTE 56028: 20W Max / 0.5W Avg.
    NTE 5679: 40W Max / 0.8W Avg.

    I've tried searching online for explanations on the matter, but something is being lost in translation. I'm still not sure how to calculate how many watts of heat will be generated at the specified load.

    Consider the NTE 56028. If the maximum gate power is 20W, and the max current rating is 40A, then I should think a ~19A load should produce ~10W heat. Please correct me if I am wrong!

    Active cooling via fans is not ideal. For a variety of reasons, I'd like to use the TO-220 package, if managing heat inside an enclosure is an option. I'd like the project to be relatively dust and splash resistant as it'll be in my shop.

    If heat build-up is going to be an issue, then a TO-3 TRIAC seems easier to mount on an external heat sink if need be.

    My unguided (misguided?) direction at this point is this:
    • Use an aluminum project box. Ground the body of the project box.
    • Mount the TO-220 NTE56028 against the wall of the project box (and electrically isolate it).
    • On the other side of the aluminum wall, install a heat sink using a pad or thermal compound.
    • Alternatively to using a heatsink, use a project box with built-in aluminum fins.
    Hopefully someone here has some insight into this type of project, and can offer corrections or advice where needed.

    Thank you for any input!

  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    Those are max. ratings that must not be exceeded otherwise the triac may be damaged. These values are not what you should operate the device with.

    Gate power and load current are not linearly dependent. In fact, you could drive the triac with a high gate power at low load current.

    The datasheet for the NTE 56028 states a max. gate trigger current IGT = 75 mA and a max. gate trigger voltage VGT = 2.5 V. Your control circuit needs to deliver these values. The resulting gate power dissipation is 2.5 V * 75 mA = 0.19 W only. Note that these values are for continuous gate control with DC. If you restrict the gate control to short trigger pulses only, gate power dissipation will be way less.

    The main power dissipation comes from the load current. The peak on state voltae of the NTE 56028 is VTM = 1.85 V. A load current of 19 A will lead to a power dissipation of PTM = 19 A * 1.85 V = 35 W.

    The thermal resistance junction-ambient of the NTE 56028 is 60 °C/W which amounts to a temperature drop of Delta_T = 35 W * 60 °C/W = 2100 ° (above ambient) which will clearly explode the triac in a matter of milliseconds.

    The thermal resistance junction-case of the NTE 56028 is 1 °C/W.
    The maximum junction temperature of the NTE 56028 is 125 °C.
    Assuming a max. ambient temperature of e.g. 40 °C, the max. allowable temperature difference junction-ambient is 125 °C - 40 °C = 85 °C.
    The max. allowable thermal resistance junction-ambient is therefore Rth_max = 85 °C/35 W = 2.4 K/W.
    As the intrinsic thermal resistance junction-case is 1 °C/W, a heatsink with 1.4 °C/W or less thermal resistance is required.
    This assumes the heatsink is directly mounted to the triac and can be effectively cooled to max. 40 °C. Mounting the gtriac to a project box and the a heatsink to the other side of the box is not a good idea as it will add thermal resistance from triac to box and from box to heatsink. Ideally the heatsink is part of the box, as you can see in many amplifier designs (example). Electrical isolation of the triac from the heatsink and grounding of the heatsink for protection against electric shock is mandatory.
  3. adamq


    Dec 17, 2013

    Thank you for the thorough reply, and for clarifying that the switching is in fact not the primary source of the generated heat.

    A 150mm * 60mm * 60mm, 24-fin aluminium heat sink has been ordered. Per a calculator found online, this heat sink should provide a thermal resistance of under 0.5°C/W (it was calculated at 0.41). Even with a margin of error, it is well within the 1.4°C/W range you calculated.

    An opening will be machined in the wall of the metal project box to allow direct mounting of the TRIAC to the heat sink, which will be mounted to the side of the enclosure. The enclosure will be grounded, and the heatsink will be bonded. The TRIAC will be isolated.

    Regarding the calculations--I'm still not 100% sure I understand, but working backwards, this is what I am understanding:

    40°C ambient temperature; 125°C Maximum Junction (TRIAC) temperature. The difference between those is 85°C. At 35W generated heat, this allows a combined maximum thermal resistance from the TRIAC to the heat sink of of 2.4°C/W. 1°C/W of that is already lost between the TRIAC case to the TRIAC junction, leaving 1.4°C/W to the heatsink.

    If the heat sink has a thermal resistance of 0.5°C/W, then the total should be 1.5°C/W.

    125°C * ([1.5°C/W] / [2.4°C/W]) = ~78°C

    So, at peak heat generation, using this heat sink, at an ambient temperature not exceeding 40°C, the TRIAC junction should operate at a temperature <= 78°C...well within operating range.

    Of course, these are all "ideal conditions" without dust built up, good heat transfer between the TRIAC and the heatsink, etc.

    Am I understanding that correctly?

    Last edited: Oct 4, 2018
  4. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    ??? I don't know how you arrive at this equation.

    Delta_T = Rth * P = 1.5 °C/W * 35 W = 52.5 °C
    Tamb = 40 °C -> Tjunction = Tamb + Delta_T = 40 °C + 52.45 °C = 92,5 °C
    Which is nevertheless well within its limits.

    On the one hand you are right, the conditions you mentin are ideal.
    On the other hand the calculation of power dissipation assumes worst case values.
    Both effects will compensate in reality (at least to some degree). Keep the heatsink clean and well aerated and everything should run well for a long time.
  5. cat o strafic

    cat o strafic

    Dec 3, 2018
    i think you can measure the voltage drop across the triac under load .figure what the equil ohm resistence would be for an equil voltage drop at the same load currant you should get your wattage than add your gate currant voltage for total watts of heat generated by the device at 100 percent on time
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