I have this old power adapter with a rating of 6V 2.4 Amp. I used a voltage divider circuit made of two 1000 ohm and one 10000 ohm resistors. when i measured the voltage across the 10k resistance, I got an exact drop of 1 volt, i.e., it was reading 5 V. i connected the ends to the micro USB and connected to my windows phone. The phone was not charging. When i read the voltage with charger, the voltage was only 2.5 V instead of 5 V. What just happened here? the adapter was a 6 V adapter that was used for a router, why did stepping down voltage result in drop in voltage on connecting load. thanks for helping me out. How would i get a 5V to charge my cell phone from this adapter?
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2\$\begingroup\$ 1: hackingmajenkoblog.wordpress.com/2016/08/30/… 2: Get a 5V UBEC. Alternatively do it right and use the proper charger. (although I'd class a windows phone as a disposable brick) \$\endgroup\$– MajenkoCommented Jun 22, 2017 at 14:13
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\$\begingroup\$ Your resistors are far too high in value to deliver the current you require. \$\endgroup\$– Leon HellerCommented Jun 22, 2017 at 14:21
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\$\begingroup\$ Sorry, I wanted to add.. since my 6V is regulated, do i simply use the L7805 with the 6V adapter and not use any capacitors? will the 2.4 AMP maintain in the output? \$\endgroup\$– marc KatzCommented Jun 22, 2017 at 14:29
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3\$\begingroup\$ No, a 7805 needs more than 1 V of overhead, it won't work with 6 V input. \$\endgroup\$– ColinCommented Jun 22, 2017 at 14:46
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The basic assumption of a resistor divider is that the current flowing through both resistors is the same. When the current is not the same, the resistor divider equation does not work.
6V is pretty close to 5V. You almost don't need a regulator. Instead of a normal resistive divider, make a divider with a diode "on top" and a 1k resistor "on the bottom." An ordinary silicon diode. A nice big one that can dissipate lots of heat without a separate heatsink. Look for something rated at 10A or something.
simulate this circuit – Schematic created using CircuitLab
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\$\begingroup\$ thank you. will the current output be the same in this case? i am passing 2 amp, will i get 2 amp from this circuit? \$\endgroup\$ Commented Jun 27, 2017 at 16:52
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\$\begingroup\$ You should be able to get 2A if you follow my advice on choosing a diode. The only problem is that the voltage might be a bit too high under no-load conditions. I guess you can try it, but measure with a volt meter before connecting anything expensive. \$\endgroup\$ Commented Jun 28, 2017 at 5:54