If i leave my charger plugged in and the switch is on without any device charging, does it ‘waste’ electricity? Why/why not?
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It used to be that some chargers wasted power that wasn’t going into charging a device, and it used to be with televisions that parts of the circuits that took time to “warm up” (including the cathode-ray tube) were kept on so that TVs would turn “instant on.”
With more modern switch-mode chargers, and LCD/LED TVs, this wasted power has been reduced dramatically. Government regulation has required these more efficient designs to be used, and the market has shifted from CRT-based TVs to flat-panel displays, so the amount of power “wasted” by plugged-in, but turned-off devices is now just a small percentage of the power in regular use.
Yet, we have billions of people on the planet, and multiplying the small power usage of individual devices by the number of people can be an impressively large number. However, compared to the many other kinds of losses in power distribution systems (resistance in power wires, fringing magnetic fields in high voltage lines, transformer conversion losses, etc.), it’s still a small percentage of the “wasted” or “lost” power.
Any time electricity flows through something a bit is dissipated as heat. (Except in something called a superconductor!) The more resistance it experiences along its path, the more energy is lost to heat. The longer, skinnier, or tougher the path the more resistance it experiences.
This is the basis of electric heater function or a coffee maker. They just provide a hard to travel path for the electric.
The wires in your wall are always live, so, when you plug in your charger you are adding more path and a tougher path to travel so more heat is lost.
Inside the charger is a really long spool of wire and some semiconductors and capacitors. They all stay on, traveled by electric, ready for you to plug your phone in.
A simple way to save money if you want to keep your charger plugged in all the time is to use switch before it like a power strip.
The power in your wall is AC. Your device charges with DC. Essentially, there is a rectifier circuit that converts the AC to DC. But now the voltage is much too high. You need 5v and you are at 120v, so they use a device called a buck converter to drop it down. The buck converter uses an inductor to smooth out the choppiness and a high speed switch to pulse the DC through the inductor to end up with a smoothish lower average voltage for the DC.
Yes, the transistors that perform the switching in the buck converter have a quiescent current at all times. Additionally, there is some magnetizing current lost to the inductor and some voltage drop in the rectifier circuit. There is a small amount of power being lost to this at all times it is plugged in.
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