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I am doing my best to make a portable charger. Here is the link with the schematic of the circuit I am working on:Voltage regulator question .I was told that the 9 V battery I was using didn't have enough capacity to charge any device,so I decided to connect 5 1,2 rechargeable batteries(NiMh and one NiCd) is series in its place,the lowest capacity one having 800 mAh.Here are the questions:
1)Will these batteries perform better?If not,what kind of battery would I need(mAh and voltage)?
2)I suspect that the device,when connected to my circuit may be losing energy at the same time while it charges.Would I need a diode somewhere to prevent this?
3)I would appreciate a few examples of how to make this charger better(without changing the battery's parameters).

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  • \$\begingroup\$ Please let me know if you need more information or if I should clarify something. \$\endgroup\$ Aug 29 '15 at 10:56
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Frankly, if you want to use rechargeable batteries, I'd suggest to go for Li-ion batteries. A single-cell (3.7V nominal) or a double-cell (7.4V nominal) sound like good choices. For the single-cell version, you'd need a DC-DC boost step-up converter to get 5V, for the double-cell you can either use a "normal" dissipative power regulator (something similar to the 7805, but with a lower dropout voltage, since the minimum output voltage of a double-cell Li-ion battery is about 6V, and that's not enough for the 7805), or, if power loss is an important factor (I'd guess it is), you should use a DC-DC buck step-down converter. Also, if you don't want to prematurely kill your Li-ion batteries, you should use a power converter which shuts down when the minimum input voltage is reached (3V for a single-cell, 6V for a double-cell).

So I'd look for a battery with the proper capacity (you should do a rough calculation for the expected minimum milliamp-hours: Pout / efficiency / battery voltage * run time), and a circuit designed for converting from that battery into 5V. Ebay is for example full of these components (both Li-ion batteries and circuits designed for this purpose).

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  • \$\begingroup\$ It seems I found a better solution.After googling a bit I found out that AA batteries (non-rechargeable) can have quite a big capacity.google.ro/… enough of these in parallel I could make the 7805 work.Am I right?Also,I'm a bit confused by the voltage regulator's voltage drop.I heard about 2 volts and 1,5 volts.Which is correct? \$\endgroup\$ Aug 31 '15 at 12:14
  • \$\begingroup\$ @DanielTork: You download the data sheet of your 7805 from the manufacturer's site, and look for "Dropout voltage". There, you'll see what the manufacturer guarantees (STM, for example says that the worst-case dropout voltage is 2.5V). AA batteries are nice, but you'll need at least 5 or rather 6 of them in series to have high enough voltage. So it'll be large and heavy. And because of the linear regulator, you'll lose about 1/3 total capacity to heating your 7805. \$\endgroup\$ Aug 31 '15 at 12:29
  • \$\begingroup\$ That's true,the datasheet always has the exact answer.The minimum I saw(in fact it's the typical) was 2 volts. \$\endgroup\$ Aug 31 '15 at 12:48
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mixing battery chemistry is bad. mixing new and used is bad. mixing capacities and brand also bad...

6V into a 7805, also bad. Minimum drop out is 2V over Vout, so you need atleast 7V. And thats before you take battery drain or Voltage droop from a high current use into account.

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