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Arouse1973
Adam
What about a diode or two?
Cheers
Adam
Cheers
Adam
Thought about that - the only question is if i need a special kind of diode for 0.5A ~10W or can i use these small ones that i have here, without frying them.
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They are 1N4007 and if i'm reading correctly the datasheet, they can handle a device of ~20V 0.5A. The problem is that the battery wire is so thick, while the diode wire is so thin. It makes me think if i am missing some kind of essential data here..i guess it's a kind of a stupid question, but the battery was pricey and it took it 3 weeks to arrive, so i'd better ask even stupid questions..
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OK, your main problem is that the battery voltage will vary with charge level.
It will vary from some multiple of (maybe) 3V (discharged) to some multiple of 4.2V (charged).
When you say 25.2V fully charged, this probably means it has 6 cells in series. This will give you 25.2V when fully charged, dropping to 18V when approaching discharge.
You have three problems:
It will vary from some multiple of (maybe) 3V (discharged) to some multiple of 4.2V (charged).
When you say 25.2V fully charged, this probably means it has 6 cells in series. This will give you 25.2V when fully charged, dropping to 18V when approaching discharge.
You have three problems:
- The higher voltage when fully charged. 25.2 vs 24V is pretty trivial, and whilst it's not guaranteed, most devices rated to operate at 24VDC would hardly even notice the small increase in voltage.
- The lower voltage (18V) when the battery approaches discharge. If the device is normally battery powered, it may operate on the reduced voltage. However it is posible it won't. Your need to boost voltage may be an issue.
- Limiting discharge. Unless the battery pack does this itself you will need to turn off the load when the battery approaches discharge or you will very quickly destroy the battery. A similar problem exists when charging -- overcharging is also dangerous.
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