Battery Question

[giordano]

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A 9600mAh battery is capable of dumping a hell of a lot of current in a fault condition so could cause carnage in your camera should a fault or short develop on the main board in the camera.

 

That's what fuses are for.

 

Also it would take many more hours to charge the battery....

 

Just use a charger appropriate for the size of the battery.

[giordano]

A fully-charged Li-ion cell will read about 4.2V no-load, but the voltage falls under load (but not by much) and as the battery discharges. The 3.7v is a nominal voltage. Probably you'll find that the two-cell Canon battery reads about 8.4V fully charged with no load.

 

I tried with a 12V adjustable battery eliminator and it worked fine, so i'm also making a plug for this one for both charging and the possibility of connecting to a wall socket

 

Be very careful with the charging arrangements. Li-ion cells have specific requirements - especially if you want fast charging, lots of charge/discharge cycles, and no explosions:eek:. It's not safe to use an ordinary battery eliminator to charge them.

 

If the main reason for wanting extra battery power is cold weather, why not build your new housing (a) with insulation and ( to hold two batteries. One battery to drive the camera, the other to discharge through a resistor inside the housing to keep both batteries warm enough to deliver near full-capacity?

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[atufte]

A fully-charged Li-ion cell will read about 4.2V no-load, but the voltage falls under load (but not by much) and as the battery discharges. The 3.7v is a nominal voltage. Probably you'll find that the two-cell Canon battery reads about 8.4V fully charged with no load.

 

 

 

Be very careful with the charging arrangements. Li-ion cells have specific requirements - especially if you want fast charging, lots of charge/discharge cycles, and no explosions:eek:. It's not safe to use an ordinary battery eliminator to charge them.

 

If the main reason for wanting extra battery power is cold weather, why not build your new housing (a) with insulation and ( to hold two batteries. One battery to drive the camera, the other to discharge through a resistor inside the housing to keep both batteries warm enough to deliver near full-capacity?

 

 

I will not use a eliminator as a charger, only for test purposes...and it works ;-)

 

I'm using a eneloop hand warmer, (electric) which is custom fitted under the baseplate, and although it works, the M8/9 batteries are still a joke compared to my 5D MKII, and i do agree that for most work they are "ok", but for what i do, it's not, i can't change batteries every 30 minutes, and will of course only use this "battery grip" for these occasions, for everything else i will only use regular batteries like the rest the you...i like to have the option, like we have with C&N and like the Leicavit and motordrive, so i can't see why not a battery grip for a camera with bad battery capacity would be such a bad idea...i think not having one is a bad idea, i feel there is much more need for this for the M8/9 than for my 5D MKII which i can use for days without charging without even using a battery grip...

[atufte]

That's what fuses are for.

Just use a charger appropriate for the size of the battery.

 

Can you guide me to the right fuse for this battery:

 

Your Online Source For 3.7V 6600 mAh Li-Ion Battery Pack - With Protection IC

 

It also says "Protection IC" and "cut off voltage 4.35V... is this a built

in fuse...?

 

Thanks

[farnz]

Can you guide me to the right fuse for this battery:

...

Alexander,

 

A fuse won't help you here because it will activate too slowly; any semi-conductor device (diode, transistor etc) will be dead before a fuse will operate. To explain, a fuse is no more than a thermal link which will continue to pass current until its thermal rating value, determined by i2t, is exceeded. i2t (read: "i squared t") is the value of current in amps, squared, and multiplied by time in seconds.

 

So, the thermal rating for a small current with a long time period will be the same as a large current with a small period. This means that if there is a sudden, large in-rush of current then the fuse may operate quickly (but never quicker than a semi-conductor device because the fuse has to heat up to its thermal rating, which takes time) but it also means that a slightly smaller in-rush current will continue to flow for some time. This current is likely to also kill your M8 although it *might* prevent the Li-Ion battery from exploding.

 

The recommended method for quickly stopping the flow of a large in-rush current is to use PTC (positive temperature coefficient) thermistor, whose resistance will increase with heat caused by the in-rush current. Since this will be the first semi-conductor device between the battery and the camera, it will blow first and protect all the components in your M8.

 

Sorry about the lengthy post and sounding like a lecture but I think it's important that you understand this.

 

Pete.

[atufte]

Thanks again Pete

 

But what about the build in IC chip which according to manufacturer "Cut's off voltage before 4.35 V", will not this do the exact same thing...?

[farnz]

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Thanks again Pete

 

But what about the build in IC chip which according to manufacturer "Cut's off voltage before 4.35 V", will not this do the exact same thing...?

Alexander,

 

Not necessarily because it's only looking for a trigger voltage of 4.35 Volts and isn't monitoring the current. It would be quite feasible with such a 'large' capacity battery for a short-circuit to cause an in-rush current in the order of Amps for a short time and for the voltage not to rise beyond 4.35 Volts. If the IC also monitors for current and clamps it at a maximum pre-set level than you might be okay.

 

It's unlikely that this situation would occur but in light of the M8's susceptibility to moisture you'd be better to be on the safe side.

 

The precise parts you'd need is beyond my field of expertise so I'm going to stop offering advice now.

 

Pete.

[atufte]

Thanks Pete...

 

I asked to vendor if i needed a "security fuse/circuit/thermistor" but he told me that the build in IC did just that and actually is a Thermistor, only with more advanced control system, and is even safer than a Thermistor alone, so i feel this is pretty safe, but the only way to find out is to try for myself...and i will

 

Thanks again for all the help...

[SJP]

Maybe repeating the obvious, but anyway, it is worth distinguishing current, voltage and energy content.

 

The voltage is your primary concern, it should be 3.8 V (like for the M8) but anything in that ballpark is OK regardless of current or power (so anything from 3.4 - 4.2V or so is fine, it may throw the M8 charge meter into a state of confusion).

 

Current, if at all listed, is only a peak curent value and will not be a factor as the M8 is relatively lean on current usage.

 

Energy content (= mAh) is the measure for the total amount of energy you can collect in terms of current & hours of use. Note that in fact the energy should be V.I.t, V.I is the power = energy/time, so the mAh meaure is somewhat ambiguous without mentioning the voltage.

[atufte]

Maybe repeating the obvious, but anyway, it is worth distinguishing current, voltage and energy content.

 

The voltage is your primary concern, it should be 3.8 V (like for the M8) but anything in that ballpark is OK regardless of current or power (so anything from 3.4 - 4.2V or so is fine, it may throw the M8 charge meter into a state of confusion).

 

Current, if at all listed, is only a peak curent value and will not be a factor as the M8 is relatively lean on current usage.

 

Energy content (= mAh) is the measure for the total amount of energy you can collect in terms of current & hours of use. Note that in fact the energy should be V.I.t, V.I is the power = energy/time, so the mAh meaure is somewhat ambiguous without mentioning the voltage.

 

Thanks...but this was a bit greek to me, is there a easier way to say this?, it get's a bit hard with all these tech stuff for my english skillz..

[SJP]

Sorry, I hope this helps, some examples:

 

3.7V, 1800 mAh versus 3.7V, 3600 mAh - same voltage but twice as much energy i.e. 2x more shooting time

 

3.7V, 1800 mAh versus 7.4V, 1800 mAh - twice as much voltage so twice as much energy content.

 

7.4V, 1800 mAh versus 3.8V 3600 mAh - voltage 2x lower, current 2x higher, same energy content

 

The overall energy content is V x mAh however the M8 can deal with more energy content (more mAh) but not with more voltage. So you need to buy the same voltage (3.7 V or so)and more mAh for your power pack.

[don_panko]

Thanks for the feedback guys, but i eventually think of converting a 5D MKII battery to M8, but do you know if it's the battery that's bad or the power control in the M8?, since the M8/9 battery is 3.5V 1900 mAh and the 5D MKII battery is 7.2V 1800mAh, i get if i'm lucky 200 shots with my M8 and the other day i got 1787 shots from a single charge on my 5D MKII, no that's more like it...

 

Rereading this suggests that you may be chasing the wrong problem. If a 5D battery has twice the capacity of an M8 battery yet it takes 8 times as many shots, then either the M8 battery is defective or your M8 is draining the battery faster. Doubling the capacity with a newly designed M8 battery external to the camera will only give you 400 exposures.

 

I assume that mechanical actions like cocking the shutter or lifting a mirror take lots of battery power but perhaps the internal computer processing of each shot and writing it to the card can also be different enough to cause the symptom that you have. The M8 is made of discrete cards that are cabled together like early computers while a totally integrated design like you might see in a modern cell phone are much more efficient.

 

Before you spend a lot more time building a battery pack maybe getting tech insight from others on this site who have opened up an M8 would help.

 

My thoughts for your consideration.

[atufte]

Sorry, I hope this helps, some examples:

 

3.7V, 1800 mAh versus 3.7V, 3600 mAh - same voltage but twice as much energy i.e. 2x more shooting time

 

3.7V, 1800 mAh versus 7.4V, 1800 mAh - twice as much voltage so twice as much energy content.

 

7.4V, 1800 mAh versus 3.8V 3600 mAh - voltage 2x lower, current 2x higher, same energy content

 

The overall energy content is V x mAh however the M8 can deal with more energy content (more mAh) but not with more voltage. So you need to buy the same voltage (3.7 V or so)and more mAh for your power pack.

 

Thanks, that's was what i thought you meant

[atufte]

Rereading this suggests that you may be chasing the wrong problem. If a 5D battery has twice the capacity of an M8 battery yet it takes 8 times as many shots, then either the M8 battery is defective or your M8 is draining the battery faster. Doubling the capacity with a newly designed M8 battery external to the camera will only give you 400 exposures.

 

I assume that mechanical actions like cocking the shutter or lifting a mirror take lots of battery power but perhaps the internal computer processing of each shot and writing it to the card can also be different enough to cause the symptom that you have. The M8 is made of discrete cards that are cabled together like early computers while a totally integrated design like you might see in a modern cell phone are much more efficient.

 

Before you spend a lot more time building a battery pack maybe getting tech insight from others on this site who have opened up an M8 would help.

 

My thoughts for your consideration.

 

Thanks, but after using my 3 M8's since december 2006 i know that it is the batteries/M8 that's the problem, i also tried at least 10 different batteries, and the last two years only used original ones, after

seeing even worse results with cheap fleabay batteries... so i will try this, it will only cost me 70$ (with a dedicated charger) so no big loss compared with a original M8 battery (or a Canon battery grip + batteries for that matter, which is really what this will be, a battery grip for the M8/9)...

 

But thanks anyway...

[atufte]

Fired up a M8 today with 3 AA 1.5V batteries (2400mAh X3), worked prefect outside for a whole day in -7C and still show half a charge 700 + exposures later, this really show how bad the original M8/9 batteries are and i think this solution will work great...

[jaapv]

If you are in this to actually market some kind of battery pack for the M8/M9 I'm interested.

[atufte]

If you are in this to actually market some kind of battery pack for the M8/M9 I'm interested.

 

I will let you know, but this works great, and i'm working on a solution for easy swopping of the 3 AA 1.2V batteries, so you can buy batteries anywhere if you run empty, although i'm on 1400 + shots now on one single charge, and it's still more juice in them, so it will probably not be a problem, but the option to use batteries from the local supermarket is very appealing...(when you forget to charge your batteries, etc)

 

I have now come to rest with 3 AA 1.2V 2700mAh, which seem to work wonders with the M8...

[giordano]

The recommended method for quickly stopping the flow of a large in-rush current is to use PTC (positive temperature coefficient) thermistor, whose resistance will increase with heat caused by the in-rush current. Since this will be the first semi-conductor device between the battery and the camera, it will blow first and protect all the components in your M8.

 

Good point - but what about the cold resistance of the thermistor. I looked up Mark Norton's http://www.l-camera-forum.com/leica-forum/leica-m8-forum/80260-anatomy-leica-m8s-power-consumption.html thread and he found the M8 draws 1.8A as the motor starts and 1.35A while cocking the shutter, so with only 3.7V to play with even deciohms count.

 

I feel it's a safe assumption that if the M8 is designed to have currents like that flowing in the electromechanical parts, the electronics will be fairly well protected against anything that a 3.7V cell could put out. And the battery Alexander cited Your Online Source For 3.7V 6600 mAh Li-Ion Battery Pack - With Protection IC claims to limit its output current to 6A.

 

Alexander, you asked about the cut-off voltage. That's a different part of the circuitry inside the battery package. It's meant to cut off the charging current when the voltage across the cells reaches 4.3 or whatever, i.e. the voltage of a fully-charged cell. That's a safety measure to reduce the chance of overcharging and explosion. There's also a device to limit the charging current to 3A, again to protect the battery and its surroundings.

[atufte]

If someone's interested i got 1678 exposures on one single charge with my new setup...this is great

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Hallo,

Look here: Leica M

[farnz]

Congratulations, Alexander. I feel sure that there are plenty besides me who'd also like to see some pictures of your M8 with its AA battery pack.

 

Pete