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" Since the designers of the Apple charger went to a great deal of effort to build a high quality charger, I conclude they must not consider voltage sag worth w

by GuiA 14y ago

" Since the designers of the Apple charger went to a great deal of effort to build a high quality charger, I conclude they must not consider voltage sag worth worrying about. Or, more interestingly, maybe they built this sag as a feature for some reason. In any case, the chargers lose points on this." 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?

nostromo 14y ago
It's exactly what it sounds like: a short-term dip in voltage. I'm not sure why any end user would care about this, unless they are using a device with no battery.
- jopt 14y ago
  Right on. Perhaps the company we deride for not letting us remove the battery could be excused for not optimising for devices with no battery.
kens 14y ago
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.
- sliverstorm 14y ago
  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.
  - zokier 14y ago
    > USB specifies 5V, not the iPhone 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.
- tisme 14y ago
  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.
- rzimmerman 14y ago
  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.

by GuiA 14y ago

nostromo 14y ago

It's exactly what it sounds like: a short-term dip in voltage. I'm not sure why any end user would care about this, unless they are using a device with no battery.

jopt 14y ago
Right on. Perhaps the company we deride for not letting us remove the battery could be excused for not optimising for devices with no battery.

kens 14y ago

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.

sliverstorm 14y ago
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.
- zokier 14y ago
  > USB specifies 5V, not the iPhone 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.
tisme 14y ago
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.
rzimmerman 14y ago
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.