Vtyjkyuk

I've been wondering for a while now, why do some electrical appliances overheat and burn out when you lower their incoming voltage? Shouldn't the less voltage mean they receive less current, and less current should mean less heat?

For example, I have a LED lamp, which when run through a dimmer burns out its LEDs.

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.

Non-linear loads (like switchmode power supplies - a phone charger is an example of one) behave differently than might be expected.

Assuming 85% efficiency through all conditions, 5V output and I'll assume 1 amp phone charging current. Input power to the converter is 5V * 1A / 0.85 = 5.88 Watts

Ignoring the fact that dimmer does interesting things to reduce the voltage, I'll apply a voltage of 85Vac, 120Vac, and 240Vac to this imaginary converter.

5.88 watts / 85V = 69.2mA

5.88 watts / 120V = 49mA

5.88 watts / 240V = 24.5mA

For this circuit, as the input voltage is reduced, the input current rises. This is why it is called a non-linear load.

The most common sort of appliance where this can occur is those using conventional refrigeration cycles with a compressor. If the voltage goes low the compressor can stall which causes overheating (the back-EMF from the motor drops and thus the current increases).

Compressors are used in refrigerators, freezers, dehumidifiers, air conditioners, and heat pumps.

There are 3rd party devices sold to protect compressors. Here is a brochure from one such supplier.

For example, I have a LED lamp, which when run through a dimmer burns out its LEDs.

LED lamps usually use a capacitive dropper - which does not work properly with the high frequencies introduced by triac based dimmers.

The voltage behind the dropper tends to be higher instead of lower - killing the LED or supporting electronics.

A dimmer doesn't lower the voltage that's going to the attached lamp - instead, it turns the lamp on and off about a hundred times each second.

If you just turned a LED lamp on and off really quickly manually, using the switch, it would probably die, too. They're just not designed to be switched that often.

I have a LED lamp, which when run through a dimmer burns out its LEDs.

I have a AC 220V to DC 12V converter, AFTER it a dimmer and after it
some LEDs

actually LED strips inside with no other components

It's advertised as a dc 12v dimmer.

There is a piece missing to this puzzle.

When you say the LED strip has just LEDs and "no other components", this would be a problematic design. If powered with 12V (no dimmer) and no current limiting resistors this in itself would be enough to burn up the LEDs.
There must be some component(s) between the wires coming out of the power supply/dimmer and before the LEDs.

The term "burn out" is a bit ambiguous. Does this mean stopped illuminating or burned up in smoke leaving a charred LED?

This is what I consider burnt LEDs:

which when run through a dimmer

This means to me the 12V dimmer is optional and not an integral part of the lamp design.

The max current of the 12V supply is an issue as well. Strip LEDs (with no thermal management) cannot take but about 100 mA max or they get too hot.

You say the lamp worked fine without being dimmed. I find this very difficult to believe. Unless the "dimmer" can boost the output voltage to greater than its 12V input.
You need to measure the voltage of the dimmer's output.