This is my first post here on electronics stackexchange. I am a hobbyits in electronics, and a professional in programming.

I am working on a inductor circuit to heat a workpiece. I have a working setup @12Vac. In short I have the following elements in the circuit:

  • Microcontroller to generate pulses with a DC of 50% with its own power supply, sharing ground with the transformer powering the solenoid.
  • 2 MOSFETs (100Amps continues drain current, 150Vds) on the low side to switch the direction of the current, through
  • a 3570 nH solenoid of 11 turns, ~5cm Diameter, made of copper pipe with 1 cm diameter. (Planning to apply watercooling through the coil some time later on)
  • a 230Vac to 12Vac transformer that can deliver up to 35 Amps peaks, or 20 Amps for a while.
  • A MOSFET driver (TC4428A) to drive the gates of the MOSFETs
    • a 10K resistor on each MOSFETs Gate to Source.
    • 1000pF ceramic capacitor on each MOSFETs Gate to Source (to reduce some ringing on the gates). Vpkpk is ~17Volts on the gates

Induction heating circuit

Now the circuit shorts when I want to apply 48Vac to the circuit, by using a welding machine, which the MOSFETs would be able to handle (48Vac = ~68Vdc * 2 = ~~136Vpkpk). Nothing explodes, the MOSFETs are in one piece. But the resistance between the pins of the MOSFETS (Gate,Source,Drain <-> Gate,Source,Drain) are all 0 or very low (<20Ohms). So they broke down.

What caused my MOSFETs to break down? Its hard to examine the circuit when components die.

My equipment exists solely of an osscilloscope and a mutlimeter.

Ringing on Gates without C2 and C3, while solenoid was not powered. Sharing common ground

Ringing on Gates without C2 and C3, while solenoid was not powered. Sharing common ground with transformer. The wires from MCU to the TC4428A driver are, say, 5cm. From the driver to the gates, the wires are ~15cm. Does this cause ringing? thich ~2mm wires where used from the TC4428A driver to the gates.

Snubbed ringing on gates with C2 and C3, while solenoid was not powered. Sharing common ground.

Snubbed ringing on gates with C2 and C3, while solenoid was not powered. Sharing common ground. Looks much better than the first picture.

Ringing on Gates while solenoid was powered

Ringing on Gates while solenoid was powered. Why is the ringing increased when the solenoid is powered on, and how to prevent/mimize it while maintaining switching speed?

Measurement on Source to Drain with workpiece in solenoid @ ~150Khz

Measurement on Source to Drain with workpiece in solenoid @ ~150Khz. Shown in the last picture, if the signal was clean, it would yield a Vpkpk of ~41 Volt. But due to the spikes it is around ~63 Volts.

Would the latter of 150% over/undershoort Vpkpk be the problem? Would this result in a (48Vac => 68Vmax => 136Vpkpk * 150% = ) ~203Vpkpk? How would I reduce the noise on the waves measured on the Source -> Drain?

EDIT enter image description here Here I disconnected one MOSFETs gate from the driver. CH1 is the gate, CH2 is the Drain of the MOSFET that was still connected. Now both waves looks fine. No /minimal current was flowing here. When I do connect both MOSFETs to the driver, and measure the resistance between the two Gates, it says 24.2K Ohm. Could it be that if one MOSFET is turned of by the TC4428A driver, that somehow it still picks up a signal from the other MOSFETs Gate when it is turned on by the driver? Is it a meaningful idea to put a diode like so Driver --->|---- Gate to make sure there is no noise? Preferably a diode with low voltage drop of course.

  • 3
    \$\begingroup\$ I'd add a diode in parralel with the coils as your driving from a DC source. The inductive kick is probably larger and sharper than shown by the scope. \$\endgroup\$
    – Spoon
    Jul 8 '13 at 11:53
  • \$\begingroup\$ What are you actually measuring in those graphs, it's not clear what point in the circuit they refer to? If there's ringing on the gate add a small resistor in series with its driver (10-100 ohm) \$\endgroup\$
    – pjc50
    Jul 8 '13 at 12:02
  • \$\begingroup\$ @m.Alin Is that also a viable solution for a parallel (R)LC circuit? I have no experience with snubbers and only finding RLC in series with snubbers examples. \$\endgroup\$ Jul 8 '13 at 12:51
  • 1
    \$\begingroup\$ @Mike Not really; you need to use diodes. \$\endgroup\$
    – m.Alin
    Jul 8 '13 at 13:06
  • 1
    \$\begingroup\$ Please don't add snubber diodes across the coils - this won't work - you can't use simple snubber diodes on a split winding coil. You need a diode and a zener. The zener voltage has to be above the supply voltage or at least no less than it. \$\endgroup\$
    – Andy aka
    Jul 8 '13 at 13:33

From the driver to the gates, the wires are ~15cm. Does this cause rining?

Almost certainly, and it's a fair bet that this is destroying your MOSFETs, by one or more of these mechanisms:

  1. exceeding \$V_{G(max)}\$ even for the briefest instant
  2. exceeding \$V_{DS(max)}\$
  3. simple overheating due to slow switching and unintended conduction

#3 should be pretty obvious when it occurs, but the other two can be hard to see, since they are transient conditions that may be too brief to be visible on the scope.

C2 and C3 are not decreasing the ringing. You get ringing on the gates because the capacitance of the MOSFET gate (and C2, C3 which add to it) plus the inductance formed by the loop of wire through the driver and the MOSFET gate-source form an LC circuit. The ringing is caused by energy bouncing between this capacitance and inductance.

You should put the driver absolutely as close to the MOSFETS as possible. 1cm is already getting to be too long. Not only does the inductance created by the long trace to the gate cause ringing, but it limits your switching speed, which means more losses in the transistors. This is because the rate of change of current is limited by inductance:

$$ \frac{v}{L} = \frac{di}{dt} $$

Since \$v\$ is the voltage supplied by the gate driver and you can't make that any bigger, the time it takes to increase the current from nothing to something is limited by the inductance \$L\$. You want the current to be as much as possible, as soon as possible, so that you can switch that transistor fast.

In addition to putting the gate driver close to the MOSFETs, you want to minimize the loop area of the path the current through the gate must take:


simulate this circuit – Schematic created using CircuitLab

The inductance is proportional to the area illustrated.

The inductance limits the switching speed, and it also limits how well the gate driver can hold the MOSFET off. As the drain voltage on the MOSFET that just turned off changes (due to the other MOSFET turning on, and the mutual inductance of the coils), the gate driver must source or sink current as the internal capacitances of the MOSFET charge or discharge. Here's an illustration from International Rectifier - Power MOSFET Basics:

schematic of MOSFET capacitances and current

In your case, if the gate traces are long, then \$R_G\$ is also an inductor. Since the inductor limits \$di/dt\$, the gate driver can only respond so quickly to these currents, and then there is significant ringing and overshoot in the resonance between the gate trace inductance and the MOSFET's capacitance. Your C2 and C3 just serve to change the frequency of this resonance.

As the gate voltage is ringing, it sometimes crosses over \$V_{th}\$ of your MOSFETS, and one begins to conduct a little when it should be off. This changes the current and voltage of the connected inductor, which is coupled to the other inductor, which introduces these capacitive currents in the other MOSFET, which can only exacerbate the problem. But, when the coils aren't powered, then the drain voltage is at 0V regardless of the transistor switching, and these capacitive currents (and consequently, the total gate charge that must be moved to switch the transistor) are much less, so you see much less ringing.

This inductance can also be coupled magnetically to other inductances, like your solenoid coils. As the magnetic flux through the loop changes, a voltage is induced (Faraday's law of induction). Minimize the inductance, and you will minimize this voltage.

Get rid of C2 and C3. If you still need to reduce ringing after improving your layout, do that by adding a resistor in series with the gate, between the gate and the gate driver. This will absorb the energy bouncing around which causes the ringing. Of course, it will also limit the gate current, and thus your switching speed, so you don't want this resistance to be any larger than absolutely necessary.

You can also bypass the added resistor with a diode, or with a transistor, to allow for turn-off to be faster than turn-on. So, one of these options (but only if necessary; it's much preferred to simply eliminate the source of the ringing):


simulate this circuit

Especially in the last case with Q3, you have essentially implemented half of a gate driver, so the same concerns of keeping the trace short and the loop area small apply.

  • \$\begingroup\$ I am definitely going to move the MOSFETs Driver (TC4428A) in between the two MOSFETs to be as close as possible. Maybe I thought I could get a way with some longer wires :D \$\endgroup\$ Jul 8 '13 at 12:22
  • 2
    \$\begingroup\$ @MikedeKlerk a diode from source to drain on the MOSFETs adds nothing, because they already effectively have one. I'll edit to add where the diode should go. \$\endgroup\$
    – Phil Frost
    Jul 8 '13 at 12:27
  • 1
    \$\begingroup\$ @MikedeKlerk no, if the diode were going that way, it would just short out the transistor. That is, you could turn the MOSFET off, but the diode would conduct anyway, so it would be pointless. \$\endgroup\$
    – Phil Frost
    Jul 8 '13 at 13:28
  • 2
    \$\begingroup\$ @PhilFrost snubber diodes as shown are not going to work. Think about it - if the two inductors are perfectly coupled when one FET turns on, the voltage on the open circuit fet rises to twice the supply voltage naturally. You'll need diode plus zener where zener returns to positive supply rail and has a Vbr of at least Vsupply. \$\endgroup\$
    – Andy aka
    Jul 8 '13 at 13:37
  • 1
    \$\begingroup\$ @Andyaka good point. I was reading it as two separate solenoids. I don't have time to edit now, though. I'm going to remove that part of my answer, and maybe someone else can address that. \$\endgroup\$
    – Phil Frost
    Jul 8 '13 at 13:40

To correctly clamp the voltages on the FET drains to a reasonable value consider this: -

enter image description here

The natural operation of the two coils (if there is any significant magnetic coupling between the two coil halves) is to produce twice the supply voltage on each drain on alternate cycles.

It's like a see-saw with the midpoint (Vs) not moving. Pull one half down and the other rises through transformer action.

This naturally means the FETs have to be rated at least twice the supply voltage or things will fry. Because coupling isn't perfect the zener diodes will catch anything above twice Vsupply.

Recommendations - choose FETs rated at 3 x supply voltage and zener diodes rated at the supply voltage. 5W zener diodes as a minimum too. Get rid of the 330nF capacitor completely - if you think this is going to somehow tune the magnetic filed emitted think again because it just kills the FETs with a current pulse. Maybe 1nF is just about livable-with. Get all connections as short as possible - stray inductance in wires can also be a killer and at the very least give those peculiar gate ringing voltages although, it's likely that these are caused by FET gate drivers with insufficient drive capabilities - in effect the voltage on the drain is coupled back to the gate by internal parasitic capacitance and prevents clean switch-on and switch-off.

  • \$\begingroup\$ Thank you for your input. I have a few questions actually. Why change the value of 330nf to 1nf (C1) or get rid of it at all? This capacitor makes it resonate. And at 150Khz the circuit sources < 1 amp without workpiece. So its quiet efficient. As you can see here calctool.org/CALC/eng/electronics/RLC_circuit if you would change C1 to 1nF its resonance frequency would go up to 2.6Mhz. The MCU isn't able to produce a nice block wave above ~500Khz, and the frequency for inductance heating ussually is < ~250Khz. \$\endgroup\$ Jul 8 '13 at 16:37
  • 1
    \$\begingroup\$ @Mike de Klerk. This type of push-pull bridge drive is absolutely not compatible with a resonant primary - if you drove a parallel resonant circuit with square waves, the harmonics of the square wave get near enough shorted out by the action of the capacitor - you are throwing energy into something that only gets turned to heat. I'd expect this circuit to be drawing in the region of 50mA when unloaded. What are you trying to achieve? \$\endgroup\$
    – Andy aka
    Jul 8 '13 at 16:42
  • \$\begingroup\$ Commercial cooktops use a resonance capacitor like C1. C1 I use is actually taken from one. See openschemes.com/2010/11/11/1800w-induction-cooktop-teardown There is one IGBT inside and the coil is powered from one side, instead of from the middle. By using C1 the circuit is actually more efficient as the energy bounces 'up and down' through the solenoid. If there would be no capacitor. The energy would just leave the coil, instead of being preserved in C1. The energy stored in C1 is reused when the other MOSFET opens. But it has to be at resonance, otherwise the circuit is inefficient. \$\endgroup\$ Jul 8 '13 at 16:56
  • 1
    \$\begingroup\$ I am trying to achieve a high current that switches direction to create a changing magnetic field that switches polarity. This induces heat in the workpiece (core in the solenoid) to (hopefully) reach the curie temperature which is high enough to melt aluminium so I can cast it. \$\endgroup\$ Jul 8 '13 at 17:01
  • \$\begingroup\$ Powering the coil from one side makes all the difference - now you have a primary that can be resonated and effciently too. You just apply a pulse at the right time to push energy into the LC tuned circuit and it keeps it going. Think about the problem with push-pull - one inductor is always tied to Vsupply and grounded - this could never be a sinewave. Single-ended is the same way as high power metal detectors work that I've seen. \$\endgroup\$
    – Andy aka
    Jul 8 '13 at 17:16

I think the problem may be in the gate drive. There is not deadband in the gate drive to allow one fet to turn off and the other to turn on which results in high shoot thru currents. You may have to use two gate drive signals that some deadtime. 1 on, 1 off, wait x time, 2 on, 2 off, wait x time.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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