vinod chandran
- Jun 21, 2011
- 192
- Joined
- Jun 21, 2011
- Messages
- 192
I can only see one real issue, i would have opted for a higher miliamp hour battery, say four x AA battery's or bigger, four x D, or C, size in a battery pack. other than that, fine.
Dave.
It will work down to approximately 7V battery voltage (depending on current draw), then the 5V will start to drop. Charging will slowly diminish.
There is only very little current available from non-alkaline batteries like in your picture, better to use alkaline, and preferably 5-6 AA cells.
Hi BobK,What are you planning on charging with this? If it is a cell phone, you will exhaust the 9V without fully charging the phone. Perhaps it is OK for an MP3 player.
Bob
The 5v will just simply start to drop, and the charging current will start to drop also, simple as that, nothing more.
This does not mean the current will start flowing in the opposite direction. You do not need a diode.
There is a reason those 9V batteries are cheap. A 9V non-alkaline battery have about 100mAh, while an alkaline have about 500mAh.
Alkaline AA batteries have about 2000mAh. So you figure out what will actually be the cheapest (Rs/mAh) batteries. It's a costly recharge method anyway.
With 4.5V input there's obviously no (or rather a negative) headroom for the 7805 to work. It needs 2-3V of headroom, hence 7.5V is the minimum input requirement for its use.Hi Resqueline,
Thanks for the reply. So what if i use 3 AA size batteries ?. In that case do i need the 7805 regulator?.
-vinod
With 4.5V input there's obviously no (or rather a negative) headroom for the 7805 to work. It needs 2-3V of headroom, hence 7.5V is the minimum input requirement for its use.
It might work for a while (3 fresh AA batteries, no 7805), depending on the phone charging circuit, but it'll certainly cease to function long before the AA batteries are depleted.