| ??? 02/17/03 21:32 Read: times |
#39282 - RE: +5V from battery supply Responding to: ???'s previous message |
"This would be using a 9V battery."
Any particular reason for 9V? Using the term "battery" in its strict technical sense [1], I think you will always get more mAh per unit volume, and a higher % utilisation of the nominal capacity, by usng a single cell instead. Maxim, Micrel, National, Linear et al all do switchers which will boost a single cell - or, if you need to stick with 9V, step-down converters. [1] "battery" Strictly, a "battery" is an arrangement of two or more cells. more mAh per unit volume: With a single cell, you have only 1 casing; With a battery (especially the 9V rectangular ones) you have a casing for each cell, plus an overall casing for the battery; also, the cells tend to be cylindrical, so there is also a lot of wasted space between the cells when they're fitted into the rectangular outline of the battery. higher % utilisation of the nominal capacity: No 2 cells can be made exactly equal, so there will alway be 1 in the battery which discharges first; beyond that point, you are effectively reverse-charging the weakest cells from the remaining cells. This is worse for rechargeables, as reverse-charging is a very bad thing for them. Obviously, this doesn't arise if you have only 1 cell. I'm sure you can find this discussed in Application Notes on the Maxim et al sites |
| Topic | Author | Date |
| +5V from battery supply | 01/01/70 00:00 | |
| RE: +5V from battery supply | 01/01/70 00:00 | |
| RE: +5V from battery supply | 01/01/70 00:00 | |
| RE: +5V from battery supply | 01/01/70 00:00 | |
| RE: +5V from battery supply | 01/01/70 00:00 | |
| RE: +5V from battery supply | 01/01/70 00:00 | |
| DC-DC convertors-Efficiency? | 01/01/70 00:00 | |
| RE: +5V from battery supply | 01/01/70 00:00 | |
| RE: +5V from battery supply | 01/01/70 00:00 | |
RE: +5V from battery supply | 01/01/70 00:00 |



