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I have a bluetooth headset that charges at direct current 5 Volt and 450 Milliampere. I charge it with its original AC-DC-converter and I want to be able to charge it over USB (e.g. from a portable pre-charged battery/charger or a solar USB charger).
I could easily cut the cable and solder the one end to a charging USB cable, but with this I would almost always give 5 Volt and 1 Ampere since that is the standard, right?
- Are there modified USB "cables" with extra electrical components on the market that use 500mA from 1A?
- Is there a DIY way to take out those 650mA or 500mA?
scjorge
Posted 2016-10-18T02:09:29.853
Reputation: 244
6The device will only draw 500ma – Ramhound – 2016-10-18T02:14:30.257
really?! by closing the circuit I would only get as much as the battery of the headset demands? this means the 450mA on the AC-DC-adapter are printed to advice users that at more mA it might burn the internal circuit of the charger?? – scjorge – 2016-10-18T02:21:49.943
“this means the 450mA on the AC-DC-adapter are printed to advice users that at more mA it might burn the internal circuit of the charger??” Utterly no idea what this means. But past that, ever think about how lightbulbs work? I mean a 40 Watt bulb and a 60 Watt light bulb and a 100 Watt lightbulb all work in the same socket, right? How can that be based on your logic. – JakeGould – 2016-10-18T02:56:44.757
@scjorge Yes; Really! – Ramhound – 2016-10-18T03:06:55.467
1This logic must be based on Grassman Algebra or something... – Ale..chenski – 2016-10-18T03:07:23.783
I wonder why someone downvote my question... – scjorge – 2016-10-18T05:58:23.400
https://en.wikipedia.org/wiki/Ohm%27s_law – Daniel B – 2016-10-18T06:29:15.687
@DanielB Kirchhoff's Rules are more accurate – scjorge – 2016-10-18T06:37:03.483
1@DanielB, The simple Ohm or Kirchhoff formulas are applicable to circuit components with constant parameters. The AD-DC adapter is not a linear voltage source, and it will either abruptly reduce the voltage output when connected load exceeds its rated DC current capability (either along some artificially defined cut-off function, or just turn off). The device is also a non-linear load, starting with constant current, which will be reduced when the internal battery enters a constant voltage stage of charging. So these simple formulas do not apply. – Ale..chenski – 2016-10-18T06:55:58.753
@AliChen Ohm’s law provides a suitable abstraction for most people. Of course you can always go into further detail, but why bother? – Daniel B – 2016-10-18T07:06:22.403
1@DanielB, I believe that the details are necessary because otherwise some curious people get confused with the discrepancy between their expectations and measured results. – Ale..chenski – 2016-10-18T07:20:23.140
@AliChen, the adapter may be non-linear, but the load is defined by Ohm's law. I would add a comma to your first comment: Kirchoff might seem relevant after some grass, man. – fixer1234 – 2016-10-18T20:45:20.450
1@fixer1234, certainly the load is defined by Ohm's Law, if you mean that to consume 450mA from 5V it must have about 11 Ohms of effective resistance. However, there is a bunch of active elements that make this "impression", and this "resistor" will change in time as charging mode changes. So, given an unknown state of battery in the device when you plug it in first, you never know what its resistance is. How does the Ohm Law help here? So I am unsure what the objection is. – Ale..chenski – 2016-10-18T22:07:36.467
1@AliChen, you're correct. Where I was going is that we aren't designing the system. The question was whether a 1 A charger would damage the headset. While the effective resistance isn't fixed, there always is one, and it will be at least the value that defines the maximum current the headset draws. So Ohm's law is enough to explain that the charger won't damage the headset. But actually, my previous Ohm's law comment was just a setup for a lame joke. :-) – fixer1234 – 2016-10-19T01:07:09.203