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?

  • \$\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
  • 5
    \$\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

4 Answers 4


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%).


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:


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)


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.


Since noone seemed to directly answer your actual underlying question yet: The actual reason they are efficient is because the transform the voltage instead of just resisting the voltage or current.

for example 5V at 1A is equal to 1V at 5A in power use both 5W. but if we would have a device which needs 1V and 1A of power, for example a electromagnetic coil with 1 ohm to move a needle to show it is powered on or of. then in the first case we would need to waste those 4 volts so would add a 4ohm resitor, still using 5W but wasting 4W. in the second case we would have 1V as output voltage and it can handle 5A but the device would only draw 1A without the need for extra limiting, this means 4A of energy isn't used so we save 80%, 4W of the 5W, making it 5 times as efficient.

that is basically it, transforming a voltage can theoretically be 100% efficient, a lower voltage will automatically limit the current due to resistance, even "magical" components such as leds do have a internal resistance based on voltage and so will limit their power usage if the voltage is right. on the other hand limiting Current while on a high voltage is a waste since that requires increasing resitance

2 formullas and 2 extra wissdom

  1. Power use in Watts=Volt*Amps
  2. Amps=Volt/Resistance
  3. Resistance causes loss of energy through for example heat or light
  4. if you have fixed maximum wattage by for example having a fixed load which requires a certain voltage, and so won't get the amps going to infinity. then trying to limit the amps will in general when using a resistor to limit it cause it to still use the exact same amount of power, if you lower the voltage however you can theoretically get down to what the exact power usage of the component you are powering is.

as we can see, if the voltage won't change then to change the amperage we need to reduce the current by adding resistance, this will waste the energy with a ratio similar to that of how far we need to reduce the voltage and then multiply that with the current we want to end up with.

so if the input is 100V and we need 1V 1A on the output(again a 1ohm 1W component), we would need to add a 99ohm resistor, the component would have 1V 1A over it which is 1W, the other resistor would have 99V 1A over it which is 99W. this totals the power usage at 100W

but if we have the same setup having 100V input and needing 1V 1A as a output again then if we transform the voltage down to 1V then on the output if it would draw 1W, 1V at 1A, that would make it only draw 0.01A at the input thus only using 1W of power in total if the conversion rate is 100%

actually to make it more easy to remember, the way we use electronics and physics on earth in general makes it so that:

  1. Voltage is a absolute value, meaning it will act the way it is.
  2. Current is a relative value, meaning it will act depending on the other parts of the cirquit and the voltage.

this is also where the voltage current resistance triangle comes from. since the relative value shares a link to voltage and the other relative value which is in this case resistance which means energy loss, this means that to prevent wasting energy you should only change the absolute value which is voltage(please not not official terms, just for explaining).

now converters/transformers do have a energy loss in the real world, this is caused by things like resistance, eddy currents/induction, having to much or to little conduction in the wrong places, or energy loss through a IC, transistor, fet, diode, etc.

the main 2 types of AC to DC adapters are

  1. the linear power suply, this is the most common one, which used to be used a lot in the past, they are also typically less efficient sometimes even getting as low as 40% to 60% in some cases, they are just 2 normal transformer coils, they are very safe in general since the electronics are electrically isolated, while transformer coils can be made very efficiently for AC to AC or for high voltages or frequencies, they have problems on low voltage AC-DC conversions, this is why these power adapters used to get hot even when not seriously loaded. the main issue is that they need a rectifier on the low voltage DC side, low voltage means high current, diodes become less efficient at high currents, and diodes have a voltage drop, while not digital this voltage drop wastes energy, so lets say we have a 4V 2A adapter, then the rectifier is made of diodes which have a voltage drop of 0.7V, since it needs to go through 2 at least,this gives a voltage drop of 1.4V meaning that to get 4V on the output the transformer needs to generate at 5.4V and so for that 8W output 2.8W of extra energy would be wasted. Next to that we also have energy loss due to coil resistance on both ends, and currents caused in the iron induction core. optimally a transformer coil should have 0ohm resistance but many turns, so be made of a superconducter, the core should take over the magnetic field or empower it but shoul not create currents inside of itself unless it is also a super conductor or does not exist. this also gives them a problem in efficiency which is why it also doesn't work that well on somewhat higher voltages often, in general they work better if the voltage is much higher than the current and if they are tuned to perfect resonance where the resistance comes from the impedance. at the end there is often some capacitors and a voltage regulator to make the output constant
  2. Switching mode power suply, these are way more common these days, mostly due to their much higher efficiency they actually can get to almost 100% in some cases and can also handle both very low voltages, and very high currents. the have a voltage rectifier in front of the voltage conversion part, so essentially they are a DC-DC converter, they also have a transformer, but this one is driven by a IC and at a very high frequency, a higher frequency means less of a magnetic core is needed and also less currents are caused in the core, at the end they often some filters, so for example capacitors and coils around a core to make sure the typically 10khz to 200khz ADC turns into DC, after that there is also a part where it measures the voltage as part of a feedback loop to the frequency generator back on the AC side directly after the rectifier. to put it simple, a rectifier turns AC into DC, this is at high voltage low current still, if the voltage on the output is to low the frequency generator on the input will start generating a frequency or a increased PWM duty cycle, in the middle is still a transformer, but since it is driven at a much higher frequency it can be really small and insanely efficient, since the microtransformer is driven by a ADC(pulsating DC), the output will also be a ADC, some filters are added to turn the ADC into a DC.

EDIT---- found this, here: https://www.circuitbread.com/ee-faq/what-are-the-efficiency-levels-of-different-power-supplies you can see some basics about the mayor transformer types in a easy way and with some graphical schematics, for the physics based stuff from the actual answer well you have to visualize it yourself based on what I said.


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