Please help me understand Samsung mob phone battery chager

[peterlonz]

I found an unused older style Samsung mobile phone battery charger, the type that is connected to vehicle 12 VDC cig lighter.
I was interested to see if I could use it in a project so I checked & found:

1) Output voltage (no load) is 6.55 VDC.
2) The unit carries a 2 amp, vehicle style plug in two prong fuse, can't tell if its a fast or slow blow.
3) The unit looks well made with an IC present on the small CB.

I am pretty sure the phone used a Li-iron cell(s). But I now wonder how a 6.55V charger could ever have been adequate?
Also at nominal 13 watts it seems substantially over powered for the task.
Why on earth is a relatively high 2 amp fuse employed, surely 1A or maybe 1.5A would have been sufficient & provided real protection.

Great to hear any comment that would shed some light on the possible reasons for this design & enlighten me in the process.
Thanks

 

[poorchava]

A photo of the intestines of the charger would help.

Are there some magnetics inside (I am almost sure there is some smps transformer or at least a relatively large choke). I guess that 6.55V voltage is beacuse most regulators (both linear and switching) need some load to work properly. Theoretically it could be a linear regulator, but droping from 12V to 5V @ 2A would dissipate 15W of heat, so the device would get seriously HOT, that's why I would bet on smps.

Current rating on the fuse may be adequate because you should always allow some gap between circuit nominal power draw and fuse rating (to avoid blowing at startup for example)

 

[peterlonz]

I thought the means of accomplishing the voltage drop would be well known since almost all mobile phones use car cig lighters for "in vehicle charging".
6.55V is the measured no load output voltage.
So what phone battery needs 6.55V to charge @ 2 amps?
Surely Li-iron would require 3.6 or 7.2V depending on whether one or two cells were used?