Electronic – Why is voltage gradually increasing from the source of a power MOSFET

Lose the diode. I can't even guess what you think it does, but it will prevent the FET from being turned off quickly.

The FET driver will drive the gate high quickly, which turns on the motor. However, when it tries to drive the gate low, the diode prevents it from doing so. That means the gate just floats. It probably slowly drifts low, running the FET in the intermediate region where it dissipates significant power.

You need to switch the FET quickly to prevent significant dissipation. The power the FET dissipates is the voltage across it times the current thru it. When it is off, there is no current, so the power is 0. When it is on, the voltage drop is very small, so the power is also small. In the middle of its operation, it would have 18 V across it and 5.7 A thru it, for a dissipation of over 100 W. Poof!

The job of the FET driver is to slew the gate voltage quickly to have the FET only spend a few ns at a time in the high dissipation region. The diode is preventing that from happening.

Added:

You now mention that the reverse diode across the motor and the FET are bolted to the same aluminum heat sink. That could be a problem, depending on what pin of each part is connected to the mounting tab. This is, of course, all clearly spelled out in the datasheets. If it's not the FET drain and the diode anode, then that is very bad. At least one of the two then needs to be insulated. Or, use two separate heat sinks.

Another problem you may have is that the FET or diode aren't connected properly. Again, you have to actually read the datasheets, then double check that you have things wired right. This could explain why the FET driver blew out.

Also, do what Tom Carpenter suggested, which is to replace the motor and diode with a resistor for debugging. However, I'd use different values. Use a 1 kΩ resistor between the drain and the 12 V supply. Until the drain is switching crisply and opposite of the gate drive signal there is no point going further. With 12 V and 1 kΩ nothing else can get hurt, including the FET driver even if you flipped some pins on the FET.

Don't forget the bypass cap across the FET driver power and ground pins, and a 10 Ω or so resistor between the FET driver output and the FET gate.

If the MOSFET doesn't open at logic level of your MCU you either need:

• a MOSFET that opens at lower VG (and gives you enough drain amps at that VG), or
• you need to use a MOSFET gate driver that does the logic-level translation.

What you have there with that NPN is a workable albeit basic MOSFET driver.

Which of these two approaches is better depends on various factors... with the former, you need new MOSFET[s], with the latter you can use your existing ones, but the circuit becomes more complicated and driver isn't free either.

There is also catch with lower Vgs MOSFETs: those that are both low Vgs and high current tend to come in packages that are nasty to solder by hand, e.g. NVTFS4823N is similar enough to BUZ11 in Rds and max Id terms, but will give you more Id current at lower Vgs. The catch? Difficult to solder by hand. They aren't all that bad though; BUK9219-55A comes in DPAK and does 50A with 3V gate drive; also has good enough Rds on (compared to BUZ11):

Mosfet drivers can be realized in many, many ways. A bipolar transistor is used often enough because it is cheap. Sometimes a pair of push-pull transistors are used; these allow both transitions (on and off) to happen fast. You can find such transistor pairs as a single package e.g. MCH6541.

Additionally, if you need level shifting (and here you do) add another transistor in front) as in this answer:

^{simulate this circuit – Schematic created using CircuitLab}

(The gate resistor depends whether you have any ringing problems with a given MOSFET.) However, for slow switching requirements, you'd be hard pressed to tell the difference between the 1-transistor (just level shiftier) and faster driver (i.e. with extra push-pull transistors):

Actually, if you also do level shifting, you can optimize away one of the transistors by using a twin-NPN totem pole instead of a push pull (but you need to add a diode):

^{simulate this circuit}

There are also more sophisticated IC drivers that use custom-designed gate driving curves. These will usually reduce switching losses and prolong the life of the MOSFET. Some don't look much more than a transistor in size and connectivity requirements (4-pin devices) but basically have something like the above inside:

The catch? They cost more.