Hi, the author here. I'm using voltage sag to describe how the voltage drops as the load increases. (This has nothing to do with the presence of a battery or not.) Chargers normally are designed to output a constant voltage until the charger hits maximum power. But with the iPhone charger, the voltage is 5.2 volts if there's no load, and steadily drops to 4.6 volts as the load increases. If a device is expecting 5 volts, this could cause problems. Also, charging times could increase since the voltage is lower.
Generally, if the voltage drops under load like this, it's a sign of a poorly designed charger that doesn't regulate well. But since Apple's charger appears otherwise well-designed, it's a bit of a puzzle. One possibility is that maybe the exact voltage input to the iPhone doesn't matter, and the designers knowing this didn't care about the voltage sag. Another possibility is for some unknown reason they deliberately designed the charger this way.
It's almost always a balance between cost of manufacture and performance. I would bet money it was not designed that way as a feature.
The interesting points to remember here are:
- USB specifies 5V, not the iPhone
- Voltage drop across the iPhone's internal regulators and charging circuits are probably less than 0.5V
- Most cell phone LiIon batteries are 3.7V
So, presumably they made a charger that allows droop to 4.6V because this is a voltage still capable of charging the battery, yet also allows a less expensive charger design.
That is pretty interesting. I guess that's still within spec for USB. The iPhone probably doesn't care as long as it's a few hundred mV above the battery voltage (which would max out at about 4.2V for a lithium ion cell). Also, it looks like the big drop occurs when you go over the 1A rating of the charger.
While it's an interesting measurement, I'm not sure how relevant the current sag is to overall quality. Devices really shouldn't be using the charger in the constant current range. I guess it does reflect the overall workmanship and thought put into the design.
Also, nice job doing those ripple measurements - that was interesting stuff! It's not super surprising that the cheap knockoffs fail miserably there. Probably cheap capacitors that are way too small.
The correct term is internal resistance, and you can calculate it easily using ohms laws. The internal resistance is easiest to think of as a series resistor in line with one of the output terminals.
Generally, if the voltage drops under load like this, it's a sign of a poorly designed charger that doesn't regulate well. But since Apple's charger appears otherwise well-designed, it's a bit of a puzzle. One possibility is that maybe the exact voltage input to the iPhone doesn't matter, and the designers knowing this didn't care about the voltage sag. Another possibility is for some unknown reason they deliberately designed the charger this way.