USB Charger 2.0

ok, member GoodAtIt wont reply at all, i have double all the parts for his usb charger 2.0 but all i want to know is how to make it into a double thing he did by putting 2 together to make one, someone help out. double the picture i posted thing.

Picture of USB Charger 2.0
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gmoon9 years ago
It is a bit difficult to decypher your post... But you want a dual-output charger (two USB ports), right? All you need is a second USB port, connected in parallel with the existing one. You'll need to change the PCB traces if the orientation of the ports is such that one blocks the other. Also, if the current draw of two USB devices is too much for a 9V / 7805 pair, you could use two 9V batteries together in parallel (two in series would work too, but the 7805 would just dissipate the extra voltage as heat..)
Yerboogieman (author)  gmoon9 years ago
well its like 2 processors if you only have one it works pretty hard, 2 processors that one works less hard, 2 portable chargers integrated into 1 to make the 1 regulator get less hot and not have to work as hard
A couple points:

-- 7805 regulators are inefficient. A 7805 in combination with a 9V battery is at best 60% efficient. So you're loosing 40% of your current to heat, etc.

Using two 7805 regs will double your losses, not reduce them.

-- The estimated max current that a 9V battery can supply through a 7805 is 300mAH (60% of the battery capacity), even though the 7805 can source ~1 amp (1.5 with a heatsink.) So two or three parallel 9V batteries still aren't enough to overtax a 7805.

And at that rate of discharge, a single 9V battery will drain below the 7805 minimum voltage in an hour....

-- Lastly, if you really want to use two separate regulators (meaning you're planning for a larger current source than a 9V battery or two, and you intend to drive both regulators hard), then you need to replicate the entire circuit twice, since the caps are required for reliable operation of the the 7805 (the LED would be optional.)
Yerboogieman (author)  gmoon9 years ago
what would be more efficient?
-- Switching regulators are much more efficient, but expensive.

-- LDO regulators aren't as efficient as switching regs, but you can use a lower input voltage (7.5V or 6V perhaps, 4 or 5 AA cells will last much longer than a 9V batt.) Lower input voltage leads to less power loss (in dropping the input voltage to the output voltage.)

For any approach--if you don't have enough current capacity on the input side, you certainly can't get more on the output (in fact less is guaranteed.) So think about it-- regardless of the the # of regulators, is the battery (or batteries) sufficient to supply what's needed? If not, then two regulators isn't really going to help...
Yerboogieman (author)  gmoon9 years ago
hmm, i saw switching regulators at radioshack so i think i might go with those but would AAA battery's work? i want it to fit inside an altoids can, and sell them at school
Sure, AAA are fine--better than a 9V, anyway.

Check out the chart on [ wiki for batteries], "Capacity (mAh)" specifically. The AAA have twice the capacity (1250 vs 625 mAh) of a 9V.

You'll need decide how many batteries are needed. Use too low an input voltage and the batteries drain and quickly drop below the Vin minimum of the regulator. Use too high an input voltage, and you're wasting current through heat, which is just as bad (I'd suggest 5 batteries for 7.5V, especially if they are rechargeables.)

Do they make a battery holder for 5 AAA? :P -- I dunno. But they make 'em for 1, 2, 3 and 4 batteries, so some combination...

If you could cram in AA batts, you'd more than double the capacity again.
Yerboogieman (author)  gmoon9 years ago
yea they make 3 AAA battery holders so i can do 6 batterys
Yerboogieman (author)  gmoon9 years ago
thank you
tech-king gmoon9 years ago
now i think i get it. the laptop charges off the usb device right?
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