Heating coil over-drawing current

I am currently designing a circuit that includes a nichrome coil used as a heating element. The wall adapter is 25V 5A DC, which leads to different buck converters that takes it down to the appropriate voltages. The Buck converter that leads to the coil is set to have an output of 15A. The Flow to each component including the heating coil is controlled by a number of MOSFETS.

I have noticed that upon testing the heating coil (in testing I have separated the component being tested from the others so in this situation the power goes straight from adapter to buck to coil to MOSFET then back to buck), the MOSFET (which is rated at 30A and 60V) is overheating and melting. I have tried using heat sinks, splitting up the current between multiple MOSFETS, and added PWM to allow the MOSFETS time to cool down, but I can only make the PWM rest periods so long before I can no longer achieve the desired heat. Still with all of this, the MOSFETS heat up and begin smouldering.

Because of the amount of MOSFETS I've ruined and amount of dangerous fumes I've inhaled, I feel it is time to turn to the experts for insight.

My assumption is the circuit is in essence a short circuit, although I would have assumed the coil would have supplied ample impedance.

Does the fact that the buck converter is lowering the voltage mean that excess amperes are being drawn?

On a side but still related note, if the circuit is in essence acting like a short and drawing uncontrolled current, will this risk damaging the adapter?

• What is the actual resistance of the coil? What voltage is your buck converter applying to the coil? How are you driving the gates of the MOSFETs? Show us a circuit diagram. Your question is far too vague to answer as-is. – Dave Tweed Aug 3 '14 at 19:18
• If your heater is creating more heat than desired, reduce the voltage. If it is only doing its job, use a more capable switching circuit. Make sure the MOSFET's are fully "on" (and not already blown), heat sink them, parallel them, add a fan, etc That is a non-trivial amount of power you are working with. – Chris Stratton Aug 3 '14 at 19:19
• The impedance of the inductor is directly proportional to the applied frequency. Since you are driving the coil with DC the reactance of the coil is 0, leaving only the resistance of the windings. impedance = resistance + reactance – sherrellbc Aug 3 '14 at 19:53
• Show your circuit. – Andy aka Aug 3 '14 at 20:34