Electronic – Why is the N-channel MOSFET getting very hot and the power it provides to device increases with temperature

If you provide the wrong circuit we can oblige with the wrong answer :-)

If the supply voltage being used is the same or lower than before then mu answer dos not explain what is happening.

If the supply voltage is greater than before then the zener may be not providing the isolation intended.

What is the old supply voltage ?
What is the new MEASURED in circuit running supply voltage?
What is the zener voltage?

If Vzener is < V_supply_new_actual then what I describe below will be happening to some extent.

The problem is that you are shorting the windings with the internal diodes in the ULN2003.

As you can see from your drawing (even though it tends not to be intuitive at first glance) - each centre tapped winding is like two magnetically coupled inductors or two halves of a transformer winding. When you connect the centre tap to V+ and ground one end the other end rises to 2 x V+ - or tries to. BUT each driven output is connected via a diode to com (anode to driver, cathode to com). When you ground one end of the winding and the other end is connected to V+ via a diode you are trying to drive the supply with 2 x supply (less a diode drop). Something has to give. As you have discovered.

The internal "catch diodes" are intended to return energy in eg inductive spikes from isolated coils but are not suited to this role.

With a stepper you may not get substantial inductive kicks so the com diodes may not be needed. YMMV.

Fix:

Remove the battery connection to "com" and one of:

• In the unlikely event that you have a 2 x V+ rail, connect com to that. That would be a near perfect solution. If you connect com to a capacitor you will get a 2 x V+ supply :-).

• Leave it floating (check with oscilloscope or magic smoke)

• or connect com via a resistor to supply

• Connect a zener from com to ground (Vzener > 2 x Vsupply) or com to V+ (Vzener > V+). Zener cathode to com in each case so com can rise to 2 x V+ without zener conducting.

• or connect com via a resistor to a capacitor with other terminal grounded, with a second resistor from capacitor to ground.

Just leaving COM open circuit MAY be OK.

The above schemes with capacitor and resistor provide a load for inductive spikes. They also load the transformer formed by the two halves so the resistor to the capacitor is to reduce the unwanted loading. The resistor to ground drains the cap. Dimension as required.

Doing it right:

MOST circuits on the internet which show a ULN200x driving a stepper motor with centre tapped windings show com (incorrectly) connected to V+.

The easy practical test of my assertion is to either disconnect com (slight risk of ULN2003 dying) or connect to V+ with a zener as above, the monitor com with an oscilloscope. Or connect a capacitor with voltage rating > 2 x V+ from com to ground, operate stepper and measure capacitor voltage. Voltages of ~=2 x V+ should appear.

__

Here is one circuit which almost gets it right - except he has the zener diode polarity reversed. As shown the zener acts like a low grade diode with the same polarity as the ULN200x internal diodes. Reverse it and it lets com rise to V+ + Vzener.

[The above diagram is from here]( http://ssecganesh.blogspot.com/2008/05/driving-stepper-motor-using-uln2003.html)

Hooray hooray ! - here is somebody who has got it right ! :-)

The above circuit is from here - he doesn't explain the use of the zener - see my comments above.

While I was writing this answer, @Connor posted an answer covering most of it. In any case...

There are a few things that need to be addressed in the presented circuit.

1. Eliminate C2 entirely: The MOSFET is being used in a switching topology, not for linear amplification, C2 completely subverts the sharp switching desired for minimal power loss. The IRFZ44N needs to be switched as rapidly as possible between fully conducting and fully blocking states, for least power wastage i.e. heat.
2. The maximum current available to charge the gate at the high-going edge of the gate input (from PWM signal) is limited by R3 = 4.7k ==> Ig < 5.1 mA. This current charges up the substantial gate capacitance at each rising edge for V_gs to rise, and is way too low. This will cause R_ds to rise very slowly, and while in this rising part of the graph, the MOSFET will waste a lot of power as heat.
   • Reduce R3 as far as the optocoupler's collector current rating will allow, or better yet:
   • Use the optocoupler to drive a BJT or smaller FET with very low gate capacitance as a switch to allow far higher gate charging current to the power MOSFET.
3. Apply the same rationale to the discharging of the gate capacitance on each falling edge of gate input. For the very high PWM frequencies indicated, a push-pull gate driver, either an integrated device or made of discrete components, is typically used instead of a passive gate drive such as shown.
4. If the rather high PWM frequency mentioned is not really needed, consider moving to a far lower PWM frequency: 500+ Hertz is often good enough, but 20-30 KHz is typical, so as to be beyond human hearing and therefore PWM noise from the motor. The higher the frequency, the greater the percentage of time the power MOSFET will be in its intermediate transition stage, rather than on or off. Therefore, more heat.
   Edit: 244 Hz as updated by OP is much more realistic.
5. The higher temperature at low duty cycles is again due to capacitor C2: It is being unable to charge up to the gate's switching voltage during the too-brief high pulses of the PWM signal. The V_gs to aim for is not the V_gs(th) of 2 to 4 Volts, but 6+ Volts, where the curve begins to flatten out in Figure 3 of the datasheet. With higher duty cycles, the capacitor does manage to breach the desired V_gs most of the time.