[tisme]

No, but it is inefficient. A larger charger will have a larger 'dead load', the internal consumption of the charger. So if you plug in a very small consumer into a such a charger then percentage wise you're losing a lot of energy to heat.

[kens]

Author here - it's a bit more complicated than that. For a linear power supply (old-fashioned wall wart), all the unused power is converted to heat, so you'd waste a lot of energy. But for switching power supplies (such as USB chargers), theoretically only the power that is needed is used, so the efficiency shouldn't depend much on the load. In practice, larger chargers might have better overall efficiency since they can implement better circuits (for space and price reasons). But larger chargers might not optimize as much for low loads. So it's hard to say offhand whether a large charger or small charger would be more efficient for a smaller load.

I did a quick test to see what really happens, plugging a Samsung phone into a iPhone charger and an iPad charger. In both cases, the charger used 3.0 watts of wall power. (The phone was turned off and charging since if it is turned on, the load fluctuates a whole lot as the phone does random things.) So my conclusion is that the size of the charger doesn't affect efficiency.

[tisme]

For the general case, all other parameters being equal (supply mode, quality and so on): bigger charger -> larger dead load.

An iPad charger is not a 'large' charger, it's a fairly small step up from an iPhone charger, since you are reporting 3.0 watts of 'wall power' but your multiplication of scope measured values does not correct for power factor you are likely off by quite a bit on both measurements.

GP mentioned a HP touchpad charger to charge a phone, I don't have a HP touch pad charger here but the specs are quite terrible [1], you'd have to measure with that specific charger to answer the specific question or you'd have to do a comparison of a large range of chargers with accurate measurement methodology in order to really answer the general question.

As it is your conclusion contradicts practical engineering and I'm afraid it will not hold up in a better test, which would be to try a number of switched mode supplies of various sizes designs with various loads. Plugging in one device and doing a hasty (wrong, ignoring phase shift) measurement does not warrant your conclusion.

To measure efficiency you're going to have to take the power factor[2] into account, this can be quite hard to do, and theoretical efficiency doesn't matter for a practical test (you're measuring, not theorizing).

The wave forms that switched mode chargers [3] output and consequently the kind of load they represent to the grid is so irregular that most non-caloric and power factor corrected measurements will give values that are not accurate. That noise that is present on the output wires will be to some extent visible on the input side.

A normal Watt meter will work best with transformer based supplies or resistive loads, accuracy for small switched mode loads will be anywhere from 'so so' to 'terrible' depending on the make and model power meter. Good brands (for instance Fluke) do most calculations right and will be able to deal with CFLs and other phase shifted loads, bad brands (I won't name them but they're killing it in the domestic watt meter department) will give wildly in-accurate results.

But even a quality meter like a Fluke will still have trouble with this kind of spiky load, especially if it is small.

It would probably be a good idea to (properly) describe your test rig along with the results it says:

"I measured the AC input voltage and current with an oscilloscope. The oscilloscope's math functions multiplied the voltage and current at each instant to compute the instantaneous power, and then computed the average power over time. For safety and to avoid vaporizing the oscilloscope I used an isolation transformer. My measurements are fairly close to Apple's[15], which is reassuring. "

But you can't really do it that way and get accurate results, instantaneous power draw using a switching supply changes several hundred thousand times per second and is likely phase-shifted so a simple multiplication is not going to work.

Accurately measuring (low) power draw from switched mode consumers is a really tricky problem, it's easy enough to read some numbers from a display but I can assure you that this is not a simple problem to work on if you want to get meaningful results.

[1] http://en.wikipedia.org/wiki/HP_TouchPad#Power_adapter

[2] http://en.wikipedia.org/wiki/Power_factor

[3] http://en.wikipedia.org/wiki/Switched-mode_power_supply

[kens]

Thank you for your detailed comment. I went to a great deal of effort in my article to measure the power consumption accurately, accounting for the power factor, but I left out most of the details since most people don't care. I'm not multiplying the average current and average voltage to compute watts, because that obviously would not work due to the power factor. Instead, I'm multiplying the instantaneous voltage and current 50,000 times a second and summing this up, which gives the actual power, corrected for the power factor. (While the internal current changes tens of thousands of times a second, the line current changes slowly due to the input filtering, so this is plenty of resolution. I'm using a Tektronix TDS5104B 1 GHz oscilloscope, so I have a pretty accurate view of the input voltage and current.)

The main sources of error in my measurements are the cheap isolation transformer (which causes a bit of line voltage distortion under load), the current sense resistor, the tolerances of the voltage divider resistors, and noise in the measurements. So I wouldn't claim these measurements to be better than 10%.

You can take a look at one of the oscilloscope power graphs at https://picasaweb.google.com/lh/photo/pbrO8BQz38kDo9xU5ejffd... Yellow is the input voltage, and turquoise is the input current. The non-sinusoidal current shows the non-unity power factor. Note that there's no phase shift, but instead the current flow happens only at the voltage peaks (which is a consequence of the input diode bridge, not of the switching power supply per se.) At the bottom of the image is the instantaneous power, computed from the instantaneous voltage and current.

For the iPad vs iPod measurement above, I didn't have the oscilloscope handy so I used a Kill-A-Watt, which does in fact take the power factor into account.

Going back to your statement that "bigger charger -> larger dead load". By "dead load", do you mean the power consumption under no load, which I call "vampire power" in the article? This varies widely between chargers, having more to do with the design than the size of the charger. But in any case, this wasted power is pretty much irrelevant under load. For instance, 100 mW is a typical vampire power usage. So if a hypothetical larger charger has twice that wasted power, at a 3 watt load, this is only a 3% difference.

[tisme]

The peaks in your scope image clearly show a phase shift, which is kind of logical if you take into account the fact that the main component in a switched mode supply is a coil.

If you look a little more carefully at your scope trace you'll see the coils reactance at work in the lower trace, the peak is where the FET in the supply is closed and is drawing real power, the purple trace past the peak and beyond the 0 crossing is inverted and drops slowly back to 0 before the next peak hits. If you use the controls on your scope to zoom in on the bottom trace by increasing the vertical sensitivity you'll get a much better idea of what I'm getting at here. You'll see '0' voltage and yet current is still flowing.

You can't correct for power factor by simply increasing the resolution and averaging. The base frequency of your oscilloscope does not enter into the discussion here, it could be 500 Hz for all I care and that would be enough.

Furthermore, the power factor of a switched mode supply changes as a function of the load applied and gets (much) worse if that load is also reactive or capacitive. Under some circumstances it is possible to draw negative power from the wall socket if you do a naive measurement, or you'll see wall socket power decrease as output current increases.

All this is possible because voltage and current are more or less out of phase with each other.

The kill-a-watt will work well with some reactive loads (such as CFLs) as long as they're of the ballast type.

A switched mode supply presents challenges that can't be met at the cost constraints of a consumer device like that.

Vampire power is a new term, I'm not familiar with it. Dead load (or simply the losses) is anything that does not end up in your consumer (the live load), I'm not sure if that is an accurate translation of the terms. It normally goes up as a function of the amount of power consumed, the base line (consumption without any load at all) is probably your 'vampire power'.

Total efficiency is 100 * ((output power)/(input power)) and will in practice be anywhere from 60% to 98% depending on how well load and supply are matched, and can vary wildly from one powersupply to another due to component variations.

Finally, classical power factor correction applies to sinusoidal wave forms, as you've already discovered switched mode supplies waveforms on both the input and the output side are anything but sinusoidal further complicating an already hairy problem.