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06-06-2014, 07:30 AM   #16
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QuoteOriginally posted by BillO Quote
With all due respect, this is not true. The actual voltage under load has absolutely nothing to do with the battery's amp-hour rating. Nothing at all.

The voltage under (VL) load of a battery, is determined by 3 parameters

Vo = No load voltage
Ri = Internal resistance of battery
I = load current

Such that VL = Vo - I * R

The voltage drop is then the load current times the internal resistance or I * Ri
Does Internal Resistance Affect Performance? - Battery University

06-07-2014, 05:08 AM   #17
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Right, so how does that negate what I said?
06-07-2014, 05:57 AM   #18
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QuoteOriginally posted by BillO Quote
Right, so how does that negate what I said?
What did you say again? Just kidding...
06-10-2014, 01:24 AM   #19
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QuoteOriginally posted by awaldram Quote
Originally posted by BillO Quote
With all due respect, this is not true. The actual voltage under load has absolutely nothing to do with the battery's amp-hour rating. Nothing at all.
One of the urgent requirements of a battery for digital applications is low internal resistance. Measured in milliohms, the internal resistance is the gatekeeper that, to a large extent, determines the runtime. The lower the resistance, the less restriction the battery encounters in delivering the needed power spikes. A high mW reading can trigger an early 'low battery' indication on a seemingly good battery because the available energy cannot be delivered in the required manner and remains in the battery

W = V*A
therfore your statemnet is 100% wrong and as I said
The mAh rating (if true) will reflect how much the battery voltage sags under load.

Because the maximum current is determined by the internal resistance of the cells.

Read more at: https://www.pentaxforums.com/forums/172-pentax-k-3/253707-k-3-failure-couldnt...#ixzz34Ds55bxQ
Is true

06-11-2014, 07:06 AM   #20
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Okay,

It is painfully obvious that you don't understand the words that you cut and paste into place, nor what I wrote, if indeed you read it and tried to understand it, because those words and I agree. The problem I think is that you just don't seem to be getting it.

That's fine with me my friend. If you wish to go on in your ignorance, that's cool. It's been called bliss before, so be blissful.

But, just on the off chance that you realize that, even though it's the internet, there are people out here that know more than you, here is an article written by some folks that might know something about batteries:

I'm guessing your attention span is a little on the short side, so here is a pertinent excerpt from that article:
The voltage drop of a battery under load is a
function of total effective resistance and current
drain rate.


And below is a picture, taken with my K-3, of a little remembrance they gave me after leaving the University of Toronto in honor of the 4 years I spent in the bowels of the McLennan physics labs earning my double major degree in physics and mathematics. So, yeah, not every one on the internet can be assumed to be full of sh!t.
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06-11-2014, 07:53 AM   #21
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QuoteOriginally posted by BillO Quote
Okay,
[/INDENT]And below is a picture, taken with my K-3, of a little remembrance they gave me after leaving the University of Toronto in honor of the 4 years I spent in the bowels of the McLennan physics labs earning my double major degree in physics and mathematics. So, yeah, not every one on the internet can be assumed to be full of sh!t.
But mines 3 years older than your so I win !!
And oh look mines in Electronics so pehaps explains you complete failure to understand , if only it had been English

From your linked article (did you read it)
The impact of electronic and ionic resistance can be observed using a dual pulse test. This test involves placing a battery on a low background
drain allowing it to first stabilize and then pulsing it with a heavier load for approximately 100 milliseconds.
Using “Ohms Law”, the total effective resistance is subsequently calculated by dividing the change in voltage by the change in current.


Hey guess what 'change in voltage' means .......yep thats right a voltage drop exactly as I said directly realted to the imternal resistance of the battery.



As you article clearly states
Internal resistance can be calculated based on the voltage drop of the battery under a known load.

And finally to answer your highlighted test in your rude and inaccruate reply putting into context your quoted part.

There are two basic components that impact the internal resistance of a battery; they are electronic resistance and ionic resistance. The electronic resistance plus the ionic resistance will be referred to as the total effective resistance

The voltage drop of a battery under load is a function of total effective resistance and current drain rate.

Now to put it all in to simple English for you

the Voltage drop of a battery is a function of the current drain and internal resistance (ionic+electrical) as I states originally and as backed up by you linked article.!!

As you say not everyone on the internet is full of sh!t but you sir protest to loudly !

I would sugest in future you read your posted links before attempting to use them to ridicule others, when you clearly did not understand a word they said nor apreciated they were vindicating my original statement.

If you read the section on flash amps you will understand why I related it to mAh in the context here

Identical batteries rated @2000mAh the other 1860mAh

I said
The mAh rating (if true) will reflect how much the battery voltage sags under load.

Read more at: https://www.pentaxforums.com/forums/172-pentax-k-3/253707-k-3-failure-couldnt...#ixzz34LRyVTxH

note the 'if true' this was to show both mAh rating would be comparable (i.e same load rating)

therfore as v=ir and the rating implies R constant for mAh rating (see flash amps) then V must drop for a lower mAh rating and the difference in mAh rating is directly proportional to the internal resistance of the battery.

Which makes this
QuoteOriginally posted by BillO Quote
With all due respect, this is not true. The actual voltage under load has absolutely nothing to do with the battery's amp-hour rating. Nothing at all.
complete and utter twaddle




.

Last edited by awaldram; 12-23-2015 at 08:32 AM.
06-11-2014, 08:18 AM   #22
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Okay Andrew,

You know my qualifications, so you can be assured I'll understand your explanation. Please explain to be (show me the math) how mAh has an effect on the loaded voltage of a battery. All you really need to do is write down the equation that has mAh somewhere on one side, and the resultant voltage on the other.

Just to remind you, my equation is:

QuoteQuote:
The voltage under (VL) load of a battery, is determined by 3 parameters

Vo = No load voltage
Ri = Internal resistance of battery
I = load current

Such that VL = Vo - I * R
Which seems to be the one everybody else uses too, but I'm all ears (or eyes) an am willing to learn things your way. Please, proceed....

---------- Post added 06-11-14 at 11:44 AM ----------

QuoteOriginally posted by awaldram Quote
I would sugest in future you read your posted links before attempting to use them to ridicule others, when you clearly did not understand a word they said nor apreciated they were vindicating my original statement.
Well, clearly it's your English that needs a little work, especially in the area of spelling.

Listen, now you seem to be quoting exactly what I originally said, that

QuoteOriginally posted by awaldram Quote
the Voltage drop of a battery is a function of the current drain and internal resistance (ionic+electrical) as I states originally and as backed up by you linked article.
Instead of your original statement

QuoteQuote:
The mAh rating (if true) will reflect how much the battery voltage sags under load.

Because the maximum current is determined by the internal resistance of the cells.
therefore
Voltage drop = rating * cell resistance
Now, I began my response to this by using the term "With all due respect" and I had no intention of ridiculing anyone. All I was trying to do was correct an error for the sake of the others on this site. I'll restate that now:

Voltage drop has nothing whatsoever to do with the Ah rating of a battery. Nothing!

It is, as I originally stated and as Andrew himself has just stated, dependent only on the open cell voltage, the current drawn, and the internal resistance (as clearly shown in the equation I have used).

I'm done with this.

Last edited by BillO; 06-11-2014 at 08:53 AM.
06-11-2014, 09:06 AM   #23
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QuoteOriginally posted by BillO Quote
Okay Andrew,

You know my qualifications, so you can be assured I'll understand your explanation. Please explain to be (show me the math) how mAh has an effect on the loaded voltage of a battery. All you really need to do is write down the equation that has mAh somewhere on one side, and the resultant voltage on the other.

Just to remind you, my equation is:

Which seems to be the one everybody else uses too, but I'm all ears (or eyes) an am willing to learn things your way. Please, proceed....

---------- Post added 06-11-14 at 11:44 AM ----------



Well, clearly it's your English that needs a little work, especially in the area of spelling.

Listen, now you seem to be quoting exactly what I originally said, that



Instead of your original statement

Now, I began my response to this by using the term "With all due respect" and I had no intention of ridiculing anyone. All I was trying to do was correct an error for the sake of the others on this site. I'll restate that now:

Voltage drop has nothing whatsoever to do with the Ah rating of a battery. Nothing!

It is, as I originally stated and as Andrew himself has just stated, dependent only on the open cell voltage, the current drawn, and the internal resistance (as clearly shown in the equation I have used).

I'm done with this.
I don't understand what you dont say or how I can make it simpler
QuoteOriginally posted by BillO Quote
The voltage under (VL) load of a battery, is determined by 3 parameters

Vo = No load voltage
Ri = Internal resistance of battery
I = load current

Such that VL = Vo - I * R
This is true but is true for both internal resistance . mAH rating and Voltage drop

maybe I cut to many corners

Flash amps can also be used to provide an estimate of internal resistance. Flash amps are defined as the maximum current a battery can deliver for a very short period of time. This test is typically performed by electrically shorting a battery with a 0.01 ohm resistor for approximately 0.2 seconds and capturing the closed circuit voltage. The current flow through the resistor can be calculated using Ohms Law and dividing the closed circuit voltage by 0.01ohms

mAh rating is the same thing but at a lot lower resisance to determine run time at it C rating

So given both are exaclt the same ciruit they are cumlative and both determine Voltage at load as per the R in the quoted equation.

so providing Rload is constant and Vbatt is constant then Vload will alter due to Mah rating of the battery becasue the only way that can be different between batteries is because one has lower Rinternal than the other.

06-11-2014, 10:54 AM   #24
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QuoteQuote:
Vload will alter due to Mah rating of the battery becasue the only way that can be different between batteries is because one has lower Rinternal than the other.
Complete rubbish. I asked for the equation that relates the loaded voltage with the mAh rating, and you did not give me one.

mAh is a rating of capacity and has nothing, zero, nada, niets, nichts, rien, nenio, nic, nekas, niente, ingenting to do with the voltage drop due to load, or the internal resistance.

An Energizer E91 alkaline 'AA' cell has a mAh rating of 2700 and an internal resistance of between 150 to 300 milliohms while and an Energizer E94 alkaline 'C' cell has a mAh rating of over 8000, 3 times as much as the E91, yet still has an internal resistance of 150 to 300 milliohms and will experience the same voltage drop at a given current at the E91.

Taking that a step further, the E95 'D' has a mAh rating of 20,000, nearly 7.5 times that of the E91, yet will still experience exactly the same voltage drop as the E91 given that same current draw. Why? because it has the same internal resistance of 150-300 milliohms.

Now please, stop spreading this misinformation. I am sorry you can't get this, but you are wrong and you'll do yourself a favor to face it.

E91, E93, E95

PS, this time I really am done. I obviously can't help you, and I can only hope others here do their own home work on this. It's simple stuff.

Last edited by BillO; 06-11-2014 at 11:05 AM.
06-11-2014, 01:30 PM   #25
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I give up, Your complete failure to understand is staggering

We are talking identical batteries with same chemical family same size same voltage but 1 2000mAh and 1 1860maH
Showing diffent batteries with same IR has nothing at all todo with my statment which sort of shows you either havn't read or are ignoring what I'm saying and are off on some tangent unrelated to the context the statement was made in.

Given this very specific framework the only thing that can alter is IR as an adjuntant to improved load bearing

It's simple stuff
If technology, Voltage and size is identical but 1 batery is rated 10% better than the other then IR must be lower on the higher rated baterey and if Ir is lower is shoudl sag less under the same load

there is no new equation as its just Ohm's law as were not even considering inductive loads here just considering a circuit with voltage, Current and Resistance


Of cause if you want a more complete idea of voltage sad you woudl need to consider randles model and what hold up the capacitance of the battery will deliver
06-11-2014, 03:04 PM - 1 Like   #26
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Enough Already

Since the overlying issue is well addressed in another thread, maybe the mods can just kill this thread and you propeller heads can find an electronics forum somewhere to throw stuff at each other.
06-11-2014, 03:19 PM - 2 Likes   #27
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LOL....

I print-screen the back-and-forth conversations.... it's just too funny when you need to showcase your certificate in an internet argument.

This is just too funny.

I like Pentax Forum!
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