Electrical apliances overheating when receiving LESS voltage
It depends a bit on the type of appliance, as to why they burn out (or not).

There is the kind of appliance that - to put it simple - will try really hard to keep working. Imagine a buck switching voltage regulator for instance. (Like the charger of your phone is one big buck converter on the inside.)



The supply voltage is always higher then the voltage at the load. When the switch closes, power flows through the coil to the load. The coil prevents the current from becoming too large, as it acts as a resistor for high frequencies. But instead of turning the excess energy into heat, as a normal resistor would, it stores the energy in it's magnetic field (ready to be re-used when the switch opens). When the voltage at the load becomes too high, the switch turns off, and the coil powers the load (through the diode) with the energy it stored when the switch was still on. That way the voltage at the load doesn't just drop immediately to 0 when the switch opens. Now, when the voltage at the load becomes too low, the switch closes again, and the whole process repeats. (See picture below.)

Now the higher the input voltage, the shorter the switch needs to be on, to achieve the desired voltage at the load. The lower the input voltage, the longer the switch needs to be on.


At nominal input voltage, the switch might be on for, say 30% of the time.
Now as the input voltage drops, the switch needs to be on for a longer time, to be able to keep the load voltage at the same level. So at minimum input voltage, the switch might be on for 90% of the time.
Now if the input voltage drops even more - and there is no undervoltage protection (usually there is) - the switch will be on for 100% of the time. The coil can only store a limited amount of energy, after which it will start to act as a normal resistor (they call that saturation). The saturated coil will become really hot, because that's what resistors do when current flows through them. It wasn't made to become this hot. Also, because the coil isn't acting like a coil anymore, the current through the switch will become much higher. All of this will cause big losses in the circuit and in the end either the switch itself, or the coil (or some other parts) will burn. To put it simply, in order for the circuit to work properly, the switch needs to keep switching. When it stops for too long, things will start to go very wrong.




Blue/green = switch on time   |   Black = voltage at the load



Now in your example, with the LED and the dimmer, something else is happening. There are dimmable LED's, that don't have this problem, but taking an LED that is not dimmable can indeed lead to a burned LED.

This is how (simply put) a dimmer works. Normal AC voltage is shown in the top waveform - it's a sine wave. This is also the input voltage of the dimmer. Now when you start dimming the lamp, like in the second waveform, the dimmer turns the lamp off every time the input voltage reaches 0. Then it waits a set time (the time depends on where the knob of the dimmer is), and then suddenly turns the lamp on. Basically it 'cuts off' the first part of the sine wave.

Now as you dim the lamp even further (turn the knob), the lamp voltage will look like the third or the fourth waveform. You can see that the time the dimmer waits before it turns on the lamp is longer and longer, leaving less of the sine wave for the lamp to 'use'. With less of the sine wave to 'use', less power is available to the lamp, and thus it will burn less bright.



Now you connect an LED lamp. Inside, it looks a bit like this (simplified, DON'T build this schematic!)



Imagine you dim the LED at 50%. The voltage on the LED will look like the third waveform. That means the dimmer will turn on the power to the LED, when the voltage is at it's HIGHEST peak! A normal lamp doesn't care about that, but the electronics inside your LED do! All of a sudden, the big capacitors in the LED need to charge, and they don't get the ususal time and relatively slowly rising voltage to do it. Instead they get the full voltage, and need to charged themselves in a very short time. This causes the current in the LED to be really high spikes, right at the moment when the dimmer switches your LED on. The components and the PCB itself were not designed to handle those high spikes, and your LED will 'burn'.

Now I simplified things a bit here, but I hope you get the idea.

As you can see, different appliances will burn for different reasons...

Hope this helps.