On Sun, 30 Dec 2007 15:59:04 -0800 keith_w <[EMAIL PROTECTED]> wrote:
> Derby Chang wrote: > > Derby Chang wrote: > >> Roman Melihhov wrote: > >>> I'm using 2 CR-V3 Li rechargeables in my K100D Super and Albinar > >>> cherger for 1 battery, so I charge them in sequence and they stay > >>> charged when unused. That's one good thing about them. Although they > >>> last longer too I'd noticed Li batteries tend to die fast when camera > >>> indicate half the charge or less. Just my 2ยข... > > >> Roman, > >> > >> I am pretty sure I remember Pentax recommending lithium rechargeables > >> *not* be used in the *ist DS, and presumably that applies for all of > >> the other AA-type cameras as well. They work, but the higher current > >> output isn't healthy for the AF motor IIRC. > >> > >> Another vote for me for the Eneloops in my *ist DS, since I don't use > >> it much. It is nice to know the charge will be there when I want to > >> use it occasionally. > >> > >> D > > > Checking my facts again, the Pentax manuals state which types of > > batteries can be used (of which lithium rechargeables are not included). > > Here is a pretty detailed explanation why, not from Pentax, but a very > > clear nonetheless. > > > > "Now consider a 4.8-6V circuit divided into 1.2-1.5V, typical 4x AA, or > > even 2x CR-V3 (Li-Mn). You'll probably be fine with slight variances in > > your AA like 1.6V that even push it up to 6.4V total. Now consider using > > 2x RCR-V3 (Li-Ion), which is chemically impossible of delivering > > anything less than 6V, and has a nominal operation of 7.2V! That's 1.8V. > > And if you take it off the charger, and it's not clamped, it could be as > > over 8V high as 8.4V total, a whopping 2.1V divided!" > > > > http://thebs413.blogspot.com/2007/07/li-ion-is-not-lithium-eg-li-mn-li-fe.html > > > > D > > All of which goes to say it's the higher *voltage* that's the possible > harm-causing culprit, not the current. > Put a higher voltage battery in, and the current becomes higher. Not really true. Infact it will be the oposite. Higher voltage means less current. > Current isn't "delivered" per se, voltage is. Current 'happens', as a > function of the resistance the load presents, up to the limit the > supply is capable of delivering. > > Remember I = E/R ? > > If the load resistance (R) stays the same, and the voltage (E) rises, > the current (I) also rises. > > In simple DC circuits, control current by adjusting and controlling the > voltage impressed on the load. Your forgetting the fact that no modern circut is without its own dc-dc converter of some sort. Some with more input voltage range than others. Without it, you simple cannot pull all the power from the battery. This is why damage can occur. Running a dc-dc converter at the edges of its range constantly damages them eventually, unless said dc-dc converter is designed for this. > IMMHO, keith whaley > > > -- > PDML Pentax-Discuss Mail List > [email protected] > http://pdml.net/mailman/listinfo/pdml_pdml.net > to UNSUBSCRIBE from the PDML, please visit the link directly above and follow > the directions. -- Ben 'Polyhead' Smith KE7GAL -- PDML Pentax-Discuss Mail List [email protected] http://pdml.net/mailman/listinfo/pdml_pdml.net to UNSUBSCRIBE from the PDML, please visit the link directly above and follow the directions.
