Electrical – why reduced resistance and increased current result in an increased amount of heat

A circuit diagram of your setup, particularly including your MOSFET driver would be really nice. However, I'll go out on a limb here and suggest that you need to look closely at your drive circuits. It is entirely possible that you are driving your MOSFET gates from 3.3 volt logic. This is perfectly doable - as long as you are using MOSFETs with logic-level gate thresholds. A lot of power MOSFETs need a minimum of 4 volts on the gate to ensure full turn-on. If you're only giving it 3.3 volts, it will only turn on partially, and will dissipate too much power. It's important that you realize that operating a transistor within both voltage and current limits can still kill it if you don't get rid of the heat dissipated, and at high currents it's easy to generate too much heat.

There is a quick check for this (if you want physical proof): Get ready to sacrifice one more MOSFET, but hey, who's counting, right? Drive a heating element full on. Quick like a bunny, measure the voltage across the nichrome and the voltage across the MOSFET. If your MOSFET voltage is not less than about 10% of the nichrome voltage, you're doing something wrong, and less is better in this case. You don't mention your drive voltage, but it has to be less than 25 volts. Let's say it's 20 volts, and let's say the current is 10 amps - this is just to illustrate, OK? Then total power dissipated is 200 watts, and the effective resistance of the total load is 2 ohms. If your MOSFET is fully on, I'd expect an Rds of .1 ohms or less (and this will give a MOSFET voltage about 5% of the nichrome voltage). This would provide a MOSFET dissipation of 10 watts. Without a heat sink, this would kill the MOSFET, so you need a heat sink in any case. And this better not be one of those little U-shaped jobbers, either. You need real heat sink, possibly with a fan. With more airflow you can use a smaller heat sink.

You need to consult the data sheet for your MOSFET to determine both Vgs(th),the threshold gate voltage, and Rds(on), the on-resistance when the gate is properly driven. Then you will need the specs on your heat sink, specifically the thermal resistance to ambient. You will also need to do a little research on how to specify a heat sink.

As for some of your other questions, consider the two voltages you measured. If they both add up to the nominal voltage of your drive converter, you are not drawing too much current, and you don't have a short. You are just fatally abusing your MOSFETs.

1.When you plug an appliance into a 120 V wall outlet, some amount of amperage will run through the wire depending on the amount of resistance in the circuit (I=V/R, where the voltage is a constant 120 V). Resistance in the circuit depends on the material resistivity and the length/shape of both the wire and appliance.

Correct.

2.The difference between a high amperage appliance (like a refrigerator or space heater) and a low amperage appliance (like a light bulb) is the resistance in the circuit. A light bulb draws less amperage because there is more resistance. In other words, resistance is used to control the amperage drawn by an appliance. The appliance is intentionally made to have the right amount of resistance so as to draw the right amperage. •Is this assumption correct?

Yes. A little more complex than that, but essentially correct.

3.More resistance creates more heat.

Incorrect. For a fixed input voltage, LESS resistance creates more heat.

\$Power = Volts^2 / Resistance\$

As the resistance falls you consume more current and hence more power.

This is due to the electrons bumping into the atoms in the material they are moving through. I kind of envision this as being like having more friction, so therefore more heat. This is the reason a frayed wire can heat up and cause an electrical fire.

True, but again, it is dependent on how fast and how many electrons are moving. Higher current = more collisions = more heat.

•Does this mean that since a light bulb has more resistance than a space heater, it is more likely that it can heat up and cause an electrical fire? Are small appliances therefore more dangerous than large appliances due to their higher resistance?

Your invalid assumptions make this a little invalid.

Further, temperature change is also dependent on geometry. The filament in a 100W light bulb gets orders of magnitudes hotter than your 750W refrigerator because the heat is concentrated in a small area. It is important to separate hot from heat here. Your fridge puts out more heat than the bulb, but does not get so hot.

•Does current in itself create heat?

We already covered that.

So when you reduce resistance and therefore increase current, does more heat get produced (although heat due to resistance decreases)? Conversely, does increasing resistance (e.g. fraying a wire) also help it cool down since current is reduced?

Again you got the resistance part backwards.

Fraying a wire is a little more complex. What you end up doing here is increasing the voltage drop across that frayed part in the wire, which will have a higher resistance than the rest of the wire, adding another load¹ in series with the appliance. This new load steals some voltage from the appliance. The total current is reduced a little. That voltage drop times whatever current the combination load continues to take generates heat in the frayed part. Since the frayed part is small, that heat turns into HOT. If it is frayed enough it can actually start a fire.

^{simulate this circuit – Schematic created using CircuitLab}

ADDENDUM

¹ The term "LOAD" can be confusing in EE. A "LOAD" is generally defined as something that consumes power. However, when you add resistive loads in series to a fixed voltage supply, the "load" on the supply goes down not up. Only when you add loads in parallel does the load on the supply go up.