I'm not versed in the arcane arts of EE— could anyone give me a basic definition of voltage sag? And would there be any reason to build it as a feature in a charger?
I'm not versed in the arcane arts of EE— could anyone give me a basic definition of voltage sag? And would there be any reason to build it as a feature in a charger?
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.
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.
Afaik USB specs are 5V+-10%, ie 4.5V-5.5V. So the iphone charger is well within the spec and all conforming devices should be able to handle that.
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.
I'm not sure why any end user would care about this, unless they are using a device with no battery.