Do AA batteries equate to a lot more shots, considering the mAh is so much higher?

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So one of the things I like about DSLRs over the mirrorless alternatives is better battery life. I have been trying out my Eneloops in the battery grip though, and wondering. The standard L-ion battery is something like 1800mAh and is rated for close to 600 shots by CIPA on a K3. My Eneloops are 1900mAh each though, for a total of 10,400mAh in the grip alone.

Is it logical to expect the AA mAh to be as effective per mAh as the standard in camera batteries? If 1800mAh will get me near 600 shots, will 10kmAh get me 3000+ shots?
 
You've forgotten the matter of differing voltage. Each Eneloop cell has a nominal voltage of 1.2V whilst for Li-Ion it's about 3.8V. Actually, this depends heavily on load but that's another story.

Watts = Volts * Amps

So, for a given mAh rating, the Li-Ion battery will contain approximately three times the amount of energy (measured in Watt Hours or, more correctly, in Joules).

AAs are still nice though:

http://www.dpreview.com/forums/thread/2744938
 
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So one of the things I like about DSLRs over the mirrorless alternatives is better battery life. I have been trying out my Eneloops in the battery grip though, and wondering. The standard L-ion battery is something like 1800mAh and is rated for close to 600 shots by CIPA on a K3. My Eneloops are 1900mAh each though, for a total of 10,400mAh in the grip alone.
The mAh is per pack, not per cell! (mAh is a measure of how long a cell can deliver a particular current, and since the cells are in series, the mAh figure applies to the pack as a whole.) So,

The 6-cell Eneloop pack is still 1800mAh.
The 2-cell LiIo packs is 1900mAh
So, total battery capacity from the two packs will be 1800 + 1900 = 3700 mAh

However:

The nominal voltage of the Eneloops is 6 x 1.2 = 8.2 V, so energy is 8.2 x 1.8 = 14.8 Wh
The nominal voltage of the LiIon pack is 2 x 3.6 = 7.2 V, so energy is 7.2 x 1.9 = 13.7 Wh

So although the NiMH pack has lower capacity, it can in theory deliver fractionally more energy because of the higher pack voltage. Whether it does so in practice will depend on other factors including the efficiency of the voltage regulator.

--
Mike
http://flickr.com/rc-soar
 
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You've forgotten the matter of differing voltage. Each Eneloop cell has a nominal voltage of 1.2V whilst for Li-Ion it's about 3.8V. Actually, this depends heavily on load but that's another story.

Watts = Volts * Amps

So, for a given mAh rating, the Li-Ion battery will contain approximately three times the amount of energy (measured in Watt Hours or, more correctly, in Joules).

AAs are still nice though:

http://www.dpreview.com/forums/thread/2744938
Actually its worse than that a battery rating is the complete package not the cell voltage

K3 battery = 7.2V * 1800mAh =13W

AA 1.2 * 1900mAH = 2.3W

Even if you count 4 as a pack = 9W

Though in the grip you get 6 and could use 2500mAh batteries giving 15W

Even so you'd only see any advantage using the NimH batteries straight after charging as even a couple days aging or even older batteries will lose any possible advantage.

Li-ion offers greater density giving more power, lower weight and low self discharge as well as better load handling (less lockups.brown outs).
 
So one of the things I like about DSLRs over the mirrorless alternatives is better battery life. I have been trying out my Eneloops in the battery grip though, and wondering. The standard L-ion battery is something like 1800mAh and is rated for close to 600 shots by CIPA on a K3. My Eneloops are 1900mAh each though, for a total of 10,400mAh in the grip alone.

Is it logical to expect the AA mAh to be as effective per mAh as the standard in camera batteries? If 1800mAh will get me near 600 shots, will 10kmAh get me 3000+ shots?
 
So one of the things I like about DSLRs over the mirrorless alternatives is better battery life. I have been trying out my Eneloops in the battery grip though, and wondering. The standard L-ion battery is something like 1800mAh and is rated for close to 600 shots by CIPA on a K3. My Eneloops are 1900mAh each though, for a total of 10,400mAh in the grip alone.
The mAh is per pack, not per cell! (mAh is a measure of how long a cell can deliver a particular current, and since the cells are in series, the mAh figure applies to the pack as a whole.) So,

The 6-cell Eneloop pack is still 1800mAh.
The 2-cell LiIo packs is 1900mAh
So, total battery capacity from the two packs will be 1800 + 1900 = 3700 mAh

However:

The nominal voltage of the Eneloops is 6 x 1.2 = 8.2 V, so energy is 8.2 x 1.8 = 14.8 Wh
The nominal voltage of the LiIon pack is 2 x 3.6 = 7.2 V, so energy is 7.2 x 1.9 = 13.7 Wh

So although the NiMH pack has lower capacity, it can in theory deliver fractionally more energy because of the higher pack voltage. Whether it does so in practice will depend on other factors including health of the battery, and how efficiently the voltage regular handles the higher voltage.
 
You beat me to it.

But I make 6 *1.2 =7.2v !!!!
Doh! well spotted:) And I got the capacities the wrong way round.

So...

The nominal voltage of the Eneloops is 6 x 1.2 = 7.2 V, so energy is 7.2 x 1.9 = 13.7 Wh
The nominal voltage of the LiIon pack is 2 x 3.6 = 7.2 V, so energy is 7.2 x 1.8 = 13.0 Wh

Nothing much in it, though of course you can get higher capacity Eneloops these days which would tip things further in their favour.

--
Mike
http://flickr.com/rc-soar
 
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You beat me to it.

But I make 6 *1.2 =7.2v !!!!
Doh! well spotted:) And I got the capacities the wrong way round.

So...

The nominal voltage of the Eneloops is 6 x 1.2 = 7.2 V, so energy is 7.2 x 1.9 = 13.7 Wh
The nominal voltage of the LiIon pack is 2 x 3.6 = 7.2 V, so energy is 7.2 x 1.8 = 13.0 Wh

Nothing much in it, though of course you can get higher capacity Eneloops these days which would tip things further in their favour.
Thanks all so far for the comments, this electricity stuff is hard to conceptualize. So let me get this straight, voltage is the resistance, so that's why they make us stack 6xAA so the voltage is 7.2, otherwise it would be too much juice at a time?

And both the voltage and mAh stack with all 6 AAs, offering a decent W and 7.2v? One thing I am noticing is the AAs seem to be lasting longer, but I haven't charged my L-ion pack but twice I think and the AAs have been cycled a handful of times. It's not a scientific test but just keeping an eye on how much I have shot with each the AAs seem to be doing better, is that because I haven't cycled my battery pack enough to "break it in"?
 
And, if unused, the Eneloops will still have 70% of their charge after 5 years:


Unlike the Li-Ion battery which will likely be flat because of the drain from its internal protection circuit - there to stop it catching fire.

And the Li-Ion will permanently lose a significant proportion of its capacity while it is sat there fully charged. Top Tip: never charge a Li-Ion more than 80% (some say 40%) if it's just going to sit there for some time.

And, if want ultimate performance, I can pop out the Eneloops and replace them with Energizer L91s. Compared with the D-LI90P I get over 10% more energy capacity in 70% of the volume and 67% of the weight and they'll keep working down to -40 degrees and they'll happily sit there for years.

 
So one of the things I like about DSLRs over the mirrorless alternatives is better battery life. I have been trying out my Eneloops in the battery grip though, and wondering. The standard L-ion battery is something like 1800mAh and is rated for close to 600 shots by CIPA on a K3. My Eneloops are 1900mAh each though, for a total of 10,400mAh in the grip alone.

Is it logical to expect the AA mAh to be as effective per mAh as the standard in camera batteries? If 1800mAh will get me near 600 shots, will 10kmAh get me 3000+ shots?
 
And, if unused, the Eneloops will still have 70% of their charge after 5 years:

https://en.wikipedia.org/wiki/Eneloop

Unlike the Li-Ion battery which will likely be flat because of the drain from its internal protection circuit - there to stop it catching fire.

And the Li-Ion will permanently lose a significant proportion of its capacity while it is sat there fully charged. Top Tip: never charge a Li-Ion more than 80% (some say 40%) if it's just going to sit there for some time.

And, if want ultimate performance, I can pop out the Eneloops and replace them with Energizer L91s. Compared with the D-LI90P I get over 10% more energy capacity in 70% of the volume and 67% of the weight and they'll keep working down to -40 degrees and they'll happily sit there for years.

http://data.energizer.com/PDFs/l91.pdf
This will give you another viewpoint


Note the maintenance requirements of Nimh and hybrids.

How many AA users actually have a charger than can run a maintenance cycle (mine can).

Without correct maintenance a Nimh/hybrid set of 6 batteries will have at least one mis-behaving with 12 months .

My battery technologies in order of preference

1 Li-ion (propriety)

2 Hybrid (AA)

3 Nimh (AA)

4 NiZn (AA)

5 Lithium (AA)

6 Alkaline (AA)

The propriety wins as it just works charge , store and use

I still use 3rd party DLi90 that were brought in 2010 ... Non of the Nimh are still working from that period even though I look after them properly.
 
You beat me to it.

But I make 6 *1.2 =7.2v !!!!
Doh! well spotted:) And I got the capacities the wrong way round.

So...

The nominal voltage of the Eneloops is 6 x 1.2 = 7.2 V, so energy is 7.2 x 1.9 = 13.7 Wh
The nominal voltage of the LiIon pack is 2 x 3.6 = 7.2 V, so energy is 7.2 x 1.8 = 13.0 Wh

Nothing much in it, though of course you can get higher capacity Eneloops these days which would tip things further in their favour.
Thanks all so far for the comments, this electricity stuff is hard to conceptualize. So let me get this straight, voltage is the resistance, so that's why they make us stack 6xAA so the voltage is 7.2, otherwise it would be too much juice at a time?

And both the voltage and mAh stack with all 6 AAs, offering a decent W and 7.2v? One thing I am noticing is the AAs seem to be lasting longer, but I haven't charged my L-ion pack but twice I think and the AAs have been cycled a handful of times. It's not a scientific test but just keeping an eye on how much I have shot with each the AAs seem to be doing better, is that because I haven't cycled my battery pack enough to "break it in"
Voltage is very roughly speaking the push/force applied to the electrons. Wiring the batteries in series increases the voltage (push). It might help to think of a battery as something like a little pump, and voltage is the pressure it produces. Putting two pumps in series allows you to get twice the pressure.

Current (amps) is a measure of how much electricity is flowing (or how fast it goes). It can be defined as a rate of electrons passing a given point.

Amp-hours, and milliamp-hours, are measures of charge—a quantity of electrons. If we were talking about water rather than electrons, they'd be something like gallons.

Watts are a measure of power, and watt-hours a measure of energy. Wattage is found by multiplying voltage times current, so one watt is the power of one amp at a voltage of one volt. Watt-hours, the measure of energy, can be found by multiplying amp-hours by voltage.

Resistance is a separate thing altogether; it's a measure of the opposition to an electric current. The resistance in a theoretical circuit represents the energy being consumed by something, whether usefully or otherwise. By definition, a current of one amp will flow when there's a voltage of one volt across a resistance of one ohm—that's the essence of Ohm's law. A camera does not represent a fixed resistance; its sometimes uses a lot of power (high current), and sometimes very little power (low current). The camera does need a more or less constant voltage to operate, hence the use of six AAs or two Li-ion cells to produce a nominal 7.2 V supply. (Internally, this is regulated and manipulated to produce a bunch of different voltages that the different camera parts need—possibly 3.3V or something similar for much of the digital electronics, possibly 5V for the LCD backlighting, several hundred volts for the flash strobe, and various others.)
 
I had power problems with my first Pentax, an IstDL. I tried NiCads, NiMHD, and disposable batteries. The disposable Lithium worked great, but cost too much. The NiCads and the NiMHD didn't work very well. They would discharge sitting in the camera and quit just as I needed a picture. I was directed to try Eneloops by some members of the Pentax forums. The power problems were much improved when I switched to Eneloops. I bought the starter pack of Eneloops that contained AA and AAA batteries and a charger. Still though my battery life was not as long as people were reporting on Pentax forums. I posted about this and was told about chargers. I bought a charger that charged each cell individually instead of cells in series and started to get the full benefit of the Eneloops. I used those cells though a IstDL then a K100Ds then a K200D and then a K-R and finally now with a K-5 and a 360 flash. None of those Eneloops or Duraloops have failed me, they still work good. I really can't remember when I got them, but it has been years ago.
 
NO!

Do an internet search of performance curves for Alkaline, NiCad, NiMH and Lithium cells. You will find that Alakaline have a much lower peak current output than all the others. They do have a high total over time, but much lower peak current. Sort of like cold cranking power vs total mAh. The very high peak demands of a DSLR and other high current electronics will deplete an alkaline very fast as they cannot recover from the really high short peak demands.

An Alakaline at 3000 mAh may have a higher milliAmpHour rating, but it cannot handle a high current load. For a low current load it could deliver a total of 3000 MilliAmps over an hours time.
 
Voltage is very roughly speaking the push/force applied to the electrons. Wiring the batteries in series increases the voltage (push). It might help to think of a battery as something like a little pump, and voltage is the pressure it produces. Putting two pumps in series allows you to get twice the pressure.

Current (amps) is a measure of how much electricity is flowing (or how fast it goes). It can be defined as a rate of electrons passing a given point.

Amp-hours, and milliamp-hours, are measures of charge—a quantity of electrons. If we were talking about water rather than electrons, they'd be something like gallons.

Watts are a measure of power, and watt-hours a measure of energy. Wattage is found by multiplying voltage times current, so one watt is the power of one amp at a voltage of one volt. Watt-hours, the measure of energy, can be found by multiplying amp-hours by voltage.

Resistance is a separate thing altogether; it's a measure of the opposition to an electric current. The resistance in a theoretical circuit represents the energy being consumed by something, whether usefully or otherwise. By definition, a current of one amp will flow when there's a voltage of one volt across a resistance of one ohm—that's the essence of Ohm's law. A camera does not represent a fixed resistance; its sometimes uses a lot of power (high current), and sometimes very little power (low current). The camera does need a more or less constant voltage to operate, hence the use of six AAs or two Li-ion cells to produce a nominal 7.2 V supply. (Internally, this is regulated and manipulated to produce a bunch of different voltages that the different camera parts need—possibly 3.3V or something similar for much of the digital electronics, possibly 5V for the LCD backlighting, several hundred volts for the flash strobe, and various others.)
 
NO!

Do an internet search of performance curves for Alkaline, NiCad, NiMH and Lithium cells. You will find that Alakaline have a much lower peak current output than all the others. They do have a high total over time, but much lower peak current. Sort of like cold cranking power vs total mAh. The very high peak demands of a DSLR and other high current electronics will deplete an alkaline very fast as they cannot recover from the really high short peak demands.

An Alakaline at 3000 mAh may have a higher milliAmpHour rating, but it cannot handle a high current load. For a low current load it could deliver a total of 3000 MilliAmps over an hours time.
Maybe a simpler way to put is, the cutoff voltage for a typcal application using AA cells with an alakline cell arrives a lot sooner under high loads than it does with a NiCad, NiMH and Lithium cell. The latter will hold at 1.2 volts long after the AA has dropped below 1.2 volts.
 
So one of the things I like about DSLRs over the mirrorless alternatives is better battery life. I have been trying out my Eneloops in the battery grip though, and wondering. The standard L-ion battery is something like 1800mAh and is rated for close to 600 shots by CIPA on a K3. My Eneloops are 1900mAh each though, for a total of 10,400mAh in the grip alone.

Is it logical to expect the AA mAh to be as effective per mAh as the standard in camera batteries? If 1800mAh will get me near 600 shots, will 10kmAh get me 3000+ shots?
 
I still use 3rd party DLi90 that were brought in 2010 ... Non of the Nimh are still working from that period even though I look after them properly.

--
My experience has been different. I have 6+ year old Eneloops that behave as they did new and I have rarely done anything but a charge cycle with any of them. Granted, before I had a Maha PowerEx my Nimhs died in about a year, but those were Energizers and the 15 minute charger was cooking them to death.

Now my luck with immedions and other non-eneloop low discharge batteries has been dismal. Some are saying the new Panasonic branded eneloops aren't as good and I hope that isn't the case. I'm rather sick of everything that simply works right being replaced by utter junk.
 

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