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Old May 30, 2003, 7:43 AM   #11
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Folks, I know the 7.2V battery works...however talking to the Minolta Techies 7.2V is the maximum it can handle. Talking to some battery guys the initial charge on the batteries might be a little more when it is freshly charged...somewhere around 8.5V. I was using this battery and I think it may have been the cause of my camera dying on me, $370 later it is workign again. So to be on the safe side I took the 7.2V 3600mAH battery and took out one of the cells in it, reducing it to a 6.0V...which is what is recommended for it. Yes it will not last as long as 7.2V battery but it beats coughing up another $370...just my two cents. I haven't tested it yet but I think it should still give something like 400 pictures.
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Old May 30, 2003, 11:52 AM   #12
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Rather than pulling one of the cells, wouldn't it be better to place a voltage regulator in between the batteries and the camera? I'm not an electrical engineer, but I would think that a switching regulator would provide safer and steadier voltage and current to the camera, and maybe even give you close to the full number of pictures the pack would allow, at least from things I've read over the years.
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Old May 30, 2003, 2:12 PM   #13
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There's already a regulator in the camera... I guess it wouln't hurt, but a better idea would be a simple limiting resistor, a diode drop, or just a long cable!

I don't know why Minolta says this, but if you look at Minolta own external (and expensive) battery pack (single or dual) they are also rated at 7.2V (probably higher as well when initially charged!) :? :? :?



http://www.bhphotovideo.com/bh4.sph/...ID=F5B3688EE90
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Old May 30, 2003, 9:02 PM   #14
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KCan wrote:
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You still have a theoric 3600mAh rating , because each of the cell in series have a 3600 mAh rating itself. The mAh rating has nothing to do with the voltage.

But the battery pack with a higher voltage (within the camera's limitation) will last longer just because when the voltage is higher, the camera will draw less curent.
I beg to differ on that point. If the D7i uses a linear power supply, the camera will draw more current and dissipate the extra voltage/current off as heat to get the voltage to the required level.

I = V/R (Current = Voltage over Resistance) The resistance being the camera, if you increase the voltage you increase the current through it. P = IV (Power = Current * Voltage) , a higher voltage with higher current through it leads to higher power thus shorter operating time.

Now there is also a "minimum" voltage that the camera will take and still operate. If you start with a higher voltage, even though you may dissipate more power, it will be longer before the cells drop to below this minimum operating voltage.

Some cells like alkaline drop off rapidly, others like Li and NiMH hold nominal voltage over a long part of the discharge curve.
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Old May 31, 2003, 12:40 AM   #15
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Quote:
Originally Posted by Binji7
I beg to differ on that point. If the D7i uses a linear power supply, the camera will draw more current and dissipate the extra voltage/current off as heat to get the voltage to the required level.

I = V/R (Current = Voltage over Resistance) The resistance being the camera, if you increase the voltage you increase the current through it. P = IV (Power = Current * Voltage) , a higher voltage with higher current through it leads to higher power thus shorter operating time.
Binji7,
If -- if the D7i uses a linear regulator ( like a 7805, 7912 etc… which is very unlikely the case) , it does not mean the current will increase with the supplied voltage level.

Lets take an example :
let say the load ( the D7) requires 2 Amps at 6 Volts, and the external battery is a 10 Volts pack. The regulator , which is an equivalent of a “7806” regulator acts as a ballast resistor to dissipate the 4 Volts difference at the imposed current , 2 Amps, thus throwing away 4x2=8 watts of heat.
The current (2Amps) can not be higher because the function and purpose of the linear regulator is precisely to maintain a fixed voltage at the load . Since the load “sees” the same voltage always (6 volts) , it will behave just the same way, ie demanding for a 2 Amps current, not higher ; the load have no idea what is connected at the other side of the regulator (ie the external battery side).
Of course, in this case , the more you add cells in series, the more the regulator work hard to dissipate the ballast heat, but the camera operation time will not be shorter in anyway. You got the same mAH rating of 1 cell , whenever you use 4 cells or 100 cells in series.

(In fact , … it will be shorter because the regulator, which have a max heat dissipation rating, will fry as a BBQ chicken because of heat :shock: )
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Old May 31, 2003, 6:21 AM   #16
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This is an interesting discussion, but IMO if I was a designer of this camera: It'll be suicidal for me to put in a linear voltage regulator with batteries as power source and expect the camera to perform (the hungry fast processor needs the power and not the "ballast"). The competition will eat it alive!
http://www.stevesforums.com/phpBB2/v...137&highlight=

Beside a linear regulator usually requires large voltage differential to work properly (ie guaranteeing the heat effect). A low-drop out type is all the more expensive offsetting any cost of a single chip switcher (did someone posted early on of hissing sound coming from his D7's as well)... Have you guys try to remove the 4 screws on the left side of the camera?
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Old Jun 12, 2003, 8:02 PM   #17
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Well....I opted to play it safe warranty wise and bought the Minolta external AC supply. Was very handy doing lots of critical shoots of our company's products.

BTW...Going on a long trip with the new MaHa MH-DPB140LI at my side. I'll let you know how it performs.
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