I've been working on variable control for a immersion heater. The problem is that the heater consumes a lot of current, about 10 to 15 Amps. I already have to circuits to control it, the problem is that both circuits overheat to the point that they already burned some components. The heating comes from resistance in the triac circuit while the components support more than 25 Amps (Triac and the Mosfet).

Here are the two circuits that I've tried. I need a way to avoid the overheating of the one or both circuits. How do I avoid overheating in these circuits?

PWM and Dimmer control for immersion heater with triac

PWM and dimmer control for immersion heater with mosfet

  • If a resistor is burning up, get a higher-power resistor. – WhatRoughBeast May 18 '17 at 20:12
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    wondering why you would use such a circuit for that application.... what is wrong with the thermostat? – Trevor_G May 18 '17 at 21:22
  • Its to try a PID control, and it is suppose to be an immersion heater, just that I thought that figure kind of look a like . :) – Laza Alvarez May 19 '17 at 4:35
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    What is the thermal inertia of your heating element? If your time constant is several seconds worth, as seems likely, then you can use a relay. This will be much cheaper than the semiconductor solution for such a high current, and will still allow you to experiment with PI control (Derivative very unlikely to be required here.) The relay lifetime isn't a concern if you are just experimenting. (Putting this as a comment because it doesn't answer your question about using a TRIAC.) – replete May 19 '17 at 4:53
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    @LazaAlvarez. you don't need PID for a slow heater just on off. 15A is not a lot of current ! Ovens & Stovetops do this every day with $0.25 Relays. $1 (1of) – Tony EE rocketscientist May 19 '17 at 6:27

Current draw is determined by the heater power you need. If you can reduce the heater watts, do that. I'll assume that you have already sized the heater.

Both triacs and IGBTs have relatively high losses, and you can expect roughly 1W/A of losses for a triac. The IGBT may have a bit more, and the required bridge will have still more.

Your best bet is to use an adequately rated part (for example a 40A triac for 25A use) and put a large enough heat sink on the part to keep the temperature rise above the highest anticipated ambient temperature reasonable.

For example, if losses are 25W and if you need to allow for 50°C ambient and you think 100°C case temperature is acceptable you will need a heatsink that has a thermal impedance of less than 2°C/W. With natural convection that is a fairly large (and fairly expensive) heatsink. The below photo is one that is 5" x 3" x 1.5".

enter image description here

A smaller heatsink will be possible if you use forced convection (a fan).

Get the maximum power dissipation for a given current from the datasheet as with the typical 40A triac linked:

enter image description here

For this one, as I said above, the dissipation at 25A RMS is maximum 25W.


If your application can deal with a bit of variation, it may be better to use a relay or contactor and switch the heater with a on/off cycle of appropriate times. A mechanical contactor will have little in the way of losses but you have to be concerned about lifetime, as the contact will wear out after perhaps 100,000 operations. It's possible to use a triac to do the switching and short it with a relay (hybrid switching) with appropriate timing.

The BUZ41A is rated at 4.5 A so that circuit is totally inappropriate for your application.

For the Triac based circuit you need a heatsink as the forward voltage drop will be in the 1.3 - 3 V range resulting in a potential power dissipation up to 45 Watts at 15 A.

  • Hello yeah you are totally right , I actuallly change the components to an IRF 640 wich can handle 18 A. I will update the picture, do you think that 'd be? – Laza Alvarez May 19 '17 at 4:33
  • The 1N5408 Diodes must be changed also. The load current flows through the bridge rectifier. – Jack Creasey May 19 '17 at 4:39

In any semiconductor switch the junction temperature rise ΔTj is;

  • a product of two resistances;

    • electrical On resistance sometimes called Rs, Ron, ESR, Rce or RdsOn [Ω]
    • thermal resistance from junction to ambient Rja = Rjc+Rca [°C/W]

Such that the rise above ambient;

> ΔTj = Pd * Rja = I² * ESR * Rja <

Since Triacs are composed of 2 BJT's the drop is 1 P-N jcn + Vce/Ice where the diode drop has an ESR and the Rce is another ESR.

  • All transistors can be modeled as bulk resistors in saturation Vce/Ice=Rce , ~ plus some offset voltage <0.1V)
    • MOSFETS are rated for RdsOn in a similar way.

So you have 3 choices ; lower either R or both.

Always start with a spec:

  • like ΔTj=25°C then choose your design accordingly.
  • let ΔTj=25°C=I² * ESR * Rja
  • so for Imax=25Arms then ESR*Rja=25/25²= 0.04

    • consider cost of Rja=0.1 ESR= 0.4 vs Rja=1 ESR=0.04 then decide.
    • since Rja=0.1°C/W is like a "big $" CPU heatsink, ignore & try ESR=0.01
  • A more complex design of low cost is to use a Triac for transients and a bypass RELAY with no arc current for long life (1e5 cycles) otherwise a good relay can only be expected to operate less life with frequency cycles every few minutes. This can be regulated by hysteresis in set point heat temp. – Tony EE rocketscientist May 18 '17 at 20:41

I have reduced power to a heating element controlled by a PID/SSR by inserting a suitable power rated diode in series with the heater which reduced the average current to the load by half. If your application allows that large a power reduction then it is an easy change.

  • Ill try also your advise, any diode that you recomend me ? my power consumption is 1700 watts and 15 Ams 120V – Laza Alvarez May 19 '17 at 4:46
  • APT30D40BG, APT30D30BG, LQA30T300 are a few part numbers that I found by searching on line for "POWER DIODES", "GENERAL PURPOSE DIODES", and "RECTIFIER DIODES". These 3 examples are in the 30A, 300V reverse peak range. They are all under $5. – Entrepreneur May 19 '17 at 11:10
  • The loss in the triac will still be roughly 1W per A RMS of current to the load, plus additional losses in the diode of maybe 2/3 of that. So the total power dissipated by the diode + triac is not significantly less (might be more), though the maximum heater power is half. If you don't mind half-waving the heater (power will be reduced to 50%, so RMS current will be reduced to 0.707 ) a better scheme might be to put a 1N4006 in series with R2. I think that will work though it might not with the zero crossing circuit. – Spehro Pefhany May 19 '17 at 12:37
  • P.S. Just to be clear- the total power dissipation with diode+load series will be somewhat less- this is the dissipation while the controller is running flat out to reach the final temperature. It will obviously take more time to reach that temperature with about half the heater power- more than double the time (and it may never get there if the power is inadequate). The dissipation in steady state will be somewhat more because a fixed amount of power is required to maintain the temperature at the setpoint, all other things being equal. – Spehro Pefhany May 19 '17 at 12:58

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