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If I wanted to drive, say, a 1V 0.1A device from mains, I could step down the voltage with a transformer, but if the device was basically just a resistor (overall), then the resistor would have to be fairly large to draw only 0.1A and it would heat up. This heat would indicate that the device is not efficient.

However, my USB wall charger does not heat up, even when left plugged in all day.

I assume something clever is being done with semiconductors, but what?

How can a low-voltage, low-current device be efficiently supplied by mains?

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  • \$\begingroup\$ Just about every electronic device has a switchedmode psu these days. righto.com/2012/10/a-dozen-usb-chargers-in-lab-apple-is.html \$\endgroup\$
    – Kartman
    Jul 21, 2021 at 10:30
  • \$\begingroup\$ It's a good question. I wondered, myself. You can see here where I discussed my own questions like this. In any case, good question and best wishes! \$\endgroup\$
    – jonk
    Jul 21, 2021 at 10:31
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    \$\begingroup\$ You seem to refer to your load as a device then the "voltage dropper" as the device. Please clear this up. \$\endgroup\$
    – Andy aka
    Jul 21, 2021 at 13:13

3 Answers 3

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Transformers can have very high efficiency- higher than switch mode supplies. Transformers are quite different from resistors; transformers can convert one voltage to another at high efficiency (high 90s percent) while resistors efficiently convert electrical energy into heat (close to 100%).

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Don't forget about the Capacitor-Dropper supply circuit.

It's very popular in ultra-cheap, low power supplies, where life-and-limb may not be first in the design-concerns list:

https://www.eetimes.com/cap-drop-supply-odd-interesting-useful-and-somewhat-dangerous/

enter image description here

In this capacitive transformerless power supply, the voltage at the load will remain constant as long as current out (IOUT) is less than or equal to current in (IIN). IIN is limited by R1 and the reactance of C1. R1 limits inrush current; its value is chosen so that it does not dissipate too much power yet is large enough to limit inrush current. (Source: Microchip Technology)

and to continue:

But behind the cap-drop there are lurking challenges: the dropping capacitor is subject to full AC-line stress and spikes, and so can fail if a low-grade unit is used. Most vendors strongly suggest you use a capacitor which is “X-rated” meaning that if it fails due to voltage spikes or overloads, it will still maintain galvanic isolation rather than “fail-to-short-circuit” mode which would put users in danger. Further, since the design is not isolated by a transformer, there’s a potential hazard to users (we mean “potential” in both senses of the word!) Since the cap-drop circuit has a so-called ground wire and is not floating, there can be serious consequences if the line-AC plug or socket are miswired and the hot, neutral, and ground wires get re-arranged; if it’s a two-wire AC-line connector without a formal ground — or the ground is not connected — then large risks are also present.

(The same EE Times article)

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Your USB wall charger is using a switched-mode power supply (SMPS):

an SMPS transfers power from a DC or AC source [...] to DC loads, such as a personal computer, while converting voltage and current characteristics. Unlike a linear power supply, the pass transistor of a switching-mode supply continually switches between low-dissipation, full-on and full-off states, and spends very little time in the high dissipation transitions, which minimizes wasted energy. A hypothetical ideal switched-mode power supply dissipates no power. Voltage regulation is achieved by varying the ratio of on-to-off time (also known as duty cycles).

A capacitor on the output provides smoothing of the rectified signal to produce a dc waveform.

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