# Dischargers Diferant Ones



## airconde (Feb 25, 2004)

*Dischargers Differant Ones*

DISCHARGERS differant one's differant ways.Whats your opinion


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## hankster (Jan 1, 1998)

Best discharger made..... your car/truck


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## CClay1282 (Jan 5, 2006)

Honestly, (and im not tryin to be an a$$) i would get a turbo charger. You can get them that do lipo and they will discharge at 35 amps.


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## guver (Jul 31, 2002)

I use the vehicle and think that more batteries have been hurt than helped by "discharging"

Camlight stuff is good. http://www.camlight.com/products/dischargers.html
as is WMR http://www.westmountainradio.com/CBA_ham.htm

Equalizers or discharge trays , well they are just fine if the cells are matched. If the cells are not matched then it is better not to use a tray.


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## airconde (Feb 25, 2004)

Got a Gfx got a Trinity DPD and just brought a Mutch more . Yep car and truck with a 3.5 brushless will suck them down. But i you don't run every weekend some of your batterys go flat.


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## hankster (Jan 1, 1998)

Time to switch to LiPo and avoid that problem


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## airconde (Feb 25, 2004)

This summer offroad lipo and brushless . Saves alot of time .More time for cookin ribs and talkin.


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## hankster (Jan 1, 1998)

Or some of them awesome Bloody Marys!!!!


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## OvalmanPA (Mar 21, 2000)

> Equalizers or discharge trays , well they are just fine if the cells are matched. If the cells are not matched then it is better not to use a tray.


Actually you have that backwards if you ask me. Unmatched cells are even more important to be discharged on an equalization tray so you don't have voltages all over the map when you get ready to charge. This is the main cause of a cell going *BANG*!


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## CClay1282 (Jan 5, 2006)

OvalmanPA said:


> This is the main cause of a cell going *BANG*!


I agree with you there.


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## guver (Jul 31, 2002)

Many people agree with you both. If the cells are eqalized on a tray then they are at the same sod at the discharged state and unequal at the charged state since they are "unmatched"

A better way to equalise them is to do it at the top (when they are full) and when they are at the bottom they will be at thier worst state of imbalance. It is advised to leave them that way so that when they're charged up again they'll be equal.


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## OvalmanPA (Mar 21, 2000)

Huh? Whatchu talkin' 'bout Willis?  Two cells start out at a discharged state. One is 1.12, the other 1.25. You're telling me at full charge they are going to magically both be at the same voltage as a pack? If you're using a lipo and a balancer yes, a NiMH pack.....no. Now discharge that same two cell pack to .9v per cell. Your initial 1.12v cell is going to be going below .9v before the discharger cuts off at 1.80v as a pack. Keep doing this and the pack just continues to get worse. Now take those same 2 cells and equalize them on a tray to .9v. Now both cells rise to about the same "resting" voltage so they start off at the same voltage during charge and in "theory" end up the same voltage at peak charge.


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## guver (Jul 31, 2002)

Only with matched cells. I'm talking about unmatched cells (it is with lipos too)


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## ta_man (Apr 10, 2005)

guver said:


> Many people agree with you both. If the cells are eqalized on a tray then they are at the same sod at the discharged state and unequal at the charged state since they are "unmatched"
> 
> A better way to equalise them is to do it at the top (when they are full) and when they are at the bottom they will be at thier worst state of imbalance. It is advised to leave them that way so that when they're charged up again they'll be equal.


The only real way to equalize them at "the top" is the Spintec ICC (Individual Cell Charger). Even if you try charging each cell one at a time after the pack is almost full, the one you did first will end up with less charge by the time you are done with the fourth or the sixth.

Even "matched" cells aren't matched for the amount of charge they take before they peak. They are matched for discharge. They will be close when new, but after a few charges they diverge and then the cells in a pack don't peak at the same time.

No matter how you equalize them you won't address this problem. And it just keeps getting worse. The cell that takes less charge to peak overheats and deteriorates faster than the other cells in the pack. And then it takes even less charge to peak next time, and so on. I believe this is one reason why the latest NiMH race packs have such a short useful life.

I occasionally rematch old packs and when I do, I group them by charge time, rather then discharge time. That way, once they are equalized, they will all peak at the same time on a charge cycle. Unless they are used for bashing where you draw the pack down all the way, I believe you are better off having all the cells peak at the same time. Of course this technique is rapidly becoming obsolete, but there are lots of threads on that subject.

Unmatched cells are a crapshoot. All you can do is equalize them before charge (NiMH, not LiPo) and hope for the best. That's why those stick packs of the latest NiMH cells are such a joke. The cells self-discharge at different rates and keep getting further apart in capacity over time.


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## guver (Jul 31, 2002)

Charging cells individually will also work, but is time consuming. It is customary to charge at a c/10 rate and rely on a "forced overcharge" to equalise the cells at the top. This is a rate that is good for ensuring all cells are "full" yet not damage the ones that get full first.

It is possible we have a different definition of "matched" since the only "match" I am concerned about is a capacity match. The IR,voltage,ect is irrelevent when talking about cells being full and empty at the same time (exactly the same with lipos) If a cell had severe voltage depresion and averaged only 1 volt were paired up with cells that delivered 1.2 volts and were "matched" by my definition they would all be empty and full at same time since they are in a series.


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## ScottH (Nov 24, 2005)

guver said:


> Only with matched cells. I'm talking about unmatched cells (it is with lipos too)


Re-read Ovalman's post. I think you missed his point.

If you simply charge and discharge unmatched cells, they will be out from one another, hence the term UNMATCHED. In theory, the MATCHED cells charge and discharge more like each other making a better pack.

to keep UNMATCHED cells from being so far out of one another as to cause deep discharging or over charging it is even more important to equalize those cells.

If you discharge an UNMATCHED pack say at 35amps, there are going to be those cells that discharge further than others raising your odds for damaging the cells. On the charge side, some of those cells may peak way before some others thus raising the risk of damage/explosion of those cells that over charge.


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## guver (Jul 31, 2002)

It doesn't work that way, other than varying rates of self-discharge (which doesn't apply here since there's not a long time spent between testing)

As I posted earlier, regardless of type, voltage, resistance, or current thru the pack any cells in aseries will discharge/charge at exactly the same "rate" and thier capacity will be changed at exactly the same amount. This is why a grossly unbalanced/unmatched pack (that has all cells full when charged) will recharge right back to it's balanced state without balancing or equalising. If the pack were to be equalised on a tray at the discharged state and recharged it would be grossly imabalanced when it is charged again.

It is really the same principal as here http://www.hobbytalk.com/bbs1/showthread.php?p=2457087#post2457087


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## OvalmanPA (Mar 21, 2000)

Once again, I'll quote myself this time....



> Two cells start out at a discharged state. One is 1.12, the other 1.25. You're telling me at full charge they are going to magically both be at the same voltage as a pack?


Measure temps at the end of the charge on this hypothetical 2 cell pack and let me know which cell is the hottest. Do this enough with today's 42/4600s and watch it go POP.



> It is customary to charge at a c/10 rate and rely on a "forced overcharge" to equalise the cells at the top. This is a rate that is good for ensuring all cells are "full" yet not damage the ones that get full first.


Do this and you are almost CERTAIN to have the cell that started out at 1.25 in my quote above go BANG.

Ok we're talking NiMH packs here but I'll give you an example I ran into with my Reedy 3200 lipo pack since you say a pack will charge/disharge at exactly the same rate. I discharged the pack shortly after I got it and watched the individual cells. Cell one reached 3v and below before the entire pack reached 6v. So I decided to charge this same pack before I got a balancer. At the end of the charge, cell 2 was at 4.25 and cell one was at 4.10. Now with your theory even without the balancer the packs cells should have been equalized at the end of the charge.


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## King Dork (Nov 23, 2008)

I'm totally with Ovalman on this one. Unmatched cells will not charge equally no matter what. Any unmatched pack I've charged (whether stick or side by side, discharged either as a pack or on a tray), the individual cell temps usually vary from 30 to 40 degrees, coolest to warmest, when peaked. Cells with higher internal resistance do not charge or discharge the same as cells with lower internal resistance, and these cells also heat up faster. No cells magically peak with the same voltage or capacity, even high end matched packs (they just stay within closer numbers). It is what it is.


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## ScottH (Nov 24, 2005)

guver said:


> It doesn't work that way, other than varying rates of self-discharge (which doesn't apply here since there's not a long time spent between testing)
> 
> As I posted earlier, regardless of type, voltage, resistance, or current thru the pack any cells in aseries will discharge/charge at exactly the same "rate" and thier capacity will be changed at exactly the same amount. This is why a grossly unbalanced/unmatched pack (that has all cells full when charged) will recharge right back to it's balanced state without balancing or equalising. If the pack were to be equalised on a tray at the discharged state and recharged it would be grossly imabalanced when it is charged again.
> 
> It is really the same principal as here http://www.hobbytalk.com/bbs1/showthread.php?p=2457087#post2457087


We are talking NiMh.

So you are telling me that if in a pack of two cells. One starts out at 1.1v and the other at .9v, when they peak they will be the same?

Ummm NO, they may well indeed end up the same but if you watch that cell that started at 1.1 it will have peaked much sooner than its .9v counterpart and will likely let you know so by going BOOM.

But hey, do it your way.


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## King Dork (Nov 23, 2008)

Might not want to monitor that battery TOO close. Wouldn't want your face next to it when it goes BOOM! Despite what's been said about the potential dangers with lithiums, those NiMH packs can be quite violent when they decide to die (think Kamikaze pilots).


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## guver (Jul 31, 2002)

OvalmanPA said:


> Once again, I'll quote myself this time....
> 
> 
> 
> ...


Lets start out with the two different cells at a charged state since that is where I am advocating them to be equalised (not at the discharged state) Lets also exgerate this a lot so it shows up very clearly.

Cell A is an old 1500 nicad , cell B is a newer IB4200 and both are full and the voltage is irrelevent , but may be around 1.4 - 1.5

Discharge them any amount less than 1500 mah and the cells will be grossly unbalanced (the more we pull out of them the worse theunbalance gets) , however they both have been drained exactly the same CAPACITY and when charged will both be full at the same time regardless of discharge/charge rate (within reason)

Let's do it your way and tray them both to same level (choose voltage or sod) they are both discharged/dead/empty, ect. and are "equalised or balanced" When they are charged they will again get exactly the same amount of capacity delivered to them , but the 1500 is the cell that will be oveheated OR the 4200 will never get full (it will be less than half full)

The type, quantity,IR,voltage is completely irrelevant. In fact we can make a pack from any combo of these and it'll alway repeat these findings if the cells are in series.


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## ScottH (Nov 24, 2005)

You just made the argument for running matched cells.


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## guver (Jul 31, 2002)

As far as your lipo pack without a balancer goes the principal is exactly the same , had you started out with a balanced pack (equal voltage at the top) then it is correct that when recharged they would come right back to perfect balance whether or not they are in balance at the discharged state since they are in series and are pulled down together and are charged up together.

It is possible we are in agreement, yet I start out equal at the top yet you start out equal at the bottom. My way will tend to be equal at the top and not the bottom. Your way will tend to be equal at the bottom yet not the top. I think we also agree that "matched by capacity" packs will be equal at both ends.


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## guver (Jul 31, 2002)

King Dork said:


> I'm totally with Ovalman on this one. Unmatched cells will not charge equally no matter what. Any unmatched pack I've charged (whether stick or side by side, discharged either as a pack or on a tray), the individual cell temps usually vary from 30 to 40 degrees, coolest to warmest, when peaked. Cells with higher internal resistance do not charge or discharge the same as cells with lower internal resistance, and these cells also heat up faster. No cells magically peak with the same voltage or capacity, even high end matched packs (they just stay within closer numbers). It is what it is.


Unmatched IR will vary in temp and voltge during the charge , they do not peak with the same voltage, but they do peak at the same time (if they all started out full before use) and they also peak with the same capacity.


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## guver (Jul 31, 2002)

ScottH said:


> We are talking NiMh.
> 
> So you are telling me that if in a pack of two cells. One starts out at 1.1v and the other at .9v, when they peak they will be the same?
> 
> ...


I'm saying that in a pack of two (different) cells that the capacity discharged is what's important and the voltage is completely irrelevent as long as it remains above 0)


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## guver (Jul 31, 2002)

ScottH said:


> You just made the argument for running matched cells.


 
Yes, I would run matched (capacity) cells. That probably is a good argument for them, and I was not making a case for or against matched cells since all of us are in agreement that cells run together are better off being matched.

The case is cell balance of unmatched cells 

My position is to balance them at the charged state and they will tend to stay that way if a tray is not used.

The other position seems to be to tray them on a tray at the discharged state.

They will also tend to stay that way too , but will be further towards unbalance when full charged.

Read post 1 , then read my answer in post4 .


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## OvalmanPA (Mar 21, 2000)

> As far as your lipo pack without a balancer goes the principal is exactly the same , had you started out with a balanced pack (equal voltage at the top) then it is correct that when recharged they would come right back to perfect balance whether or not they are in balance at the discharged state since they are in series and are pulled down together and are charged up together.


The problem is your wrong. I later charged the pack with a balancer to full capacity (same voltages on cells), discharged to 6v, recharged without the balancer, and cell one was less voltage than cell two at full capacity. Now according to your assumption, if the pack starts off balanced, is discharged and recharged it should be in perfect balance again, correct?

In your analogy above with cells of two different MAH of course your theory is going to prove true since you are dealing with capacity. Chargers typically don't look at the capacity of a cell during charge but at the overall voltage. Hence my theory of if you start off with all cells at the same voltage you will have all cells closer to the same voltage at peak charge. Say our 2 cell pack (from my original example) is going to peak at 3.5v. When the charger looks at the voltage, since the original 1.12v cell is lower than the 1.25v cell, the 1.25 cell is going to get _overcharged_, possibly going bang before the pack would actually reach peak charge. Now start off those same cells at about 1.10 after traying and they should peak out to about the same voltages preventing overcharge of one cell.


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## guver (Jul 31, 2002)

Your balancer probably has some margin of error and is not as close as some balanced cells. I balance within 1 mv and can do a D<C and the balance is still within that same tolerance. There are some abnormal circumstances that can cause an imbalance , but are not normal and would be considered abuse or misuse.

Yes, my analogy works since it deals with capacity ONLY. I have said all along that the voltage is irrelevent. The voltage during charge of nimh batts is also irrelevent as it applies to charging and peak voltage. Only the drop is the only thing most chargers look for. The voltage could be 4,5,6,7,8 volts and the charger wouldn't care as far as find the full mark. Voltage is also a poor indication of sod , it is better than nothing suppose.

Let me make sure we understand. Are you saying that a unmatched pack should be trayed to equal voltage and then when charged the cells will be full at same time? I say no. There's nothing wrong with traying them to equal voltage, but if they are not equal cells then there's no way they will b full at same time. If they were mached cells (which we aren't debating) then they would be full at same time.

Shall we continue or drop it?


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## OvalmanPA (Mar 21, 2000)

> Your balancer probably has some margin of error and is not as close as some balanced cells. I balance within 1 mv and can do a D<C and the balance is still within that same tolerance. There are some abnormal circumstances that can cause an imbalance , but are not normal and would be considered abuse or misuse.


I didn't just depend on the balancer. I checked the voltages through the balance connector when it was done balancing, at discharge, and then again at charge end. So my pack has been "abused"?

How can the voltage be irrelevant when that is what the charger is looking at during the charge cycle (NiMH)?? Even lipo chargers look at the voltage of the pack before trying to "top off" the battery to a certain portion of it's overall capacity (at least with the GFX).



> Let me make sure we understand. Are you saying that a unmatched pack should be trayed to equal voltage and then when charged the cells will be full at same time?


Full _capacity_ of each cell? Possibly not but closer voltages at peak which "should" avoid overcharging of certain cells in the pack which could lead to failure. (BOOM)

I dunno, shall we continue or drop it? I'm enjoying the spirited debate.


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## ScottH (Nov 24, 2005)

Just thinking about this.

Using the example of one 1500mah cell and one 4200 mah cell in the two cell pack.

A load of 20amps is placed across the pack for say 300 seconds. Now we all know that the 4200 cell can withstand this load without a problem, but the 1500mah cell cannot. Once the 1500mah cell is fully discharged, the load will be totally up to the 4200mah cell. But the voltage output will be way down. Unless you are on a piece of test equipment there is no way to predict this, only when the car/truck/plane slows down will you be able to tell this. 

Now at this point the cells are WAY OUT OF BALANCE, at the bottom so to speak, yes?

With the 1500 being so far discharged, probably to zero and the 4200 cell no where near zero. What will happen when you charge them? Once allowing the pack to cool and placing it on a charger how are these two cells going to act in the same pack?

Now in a vaccum we will have theoretically pulled 1500mah from both cells and in theory once each cell took 1500mah worth of charge they should both peak. I think that the 1500 is going to peak way before the 4200 and start going the other way sooner than the 4200 causing the pack to false peak or possibly the 1500 to vent.

_More to ponder_


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## guver (Jul 31, 2002)

ScottH , I think you've just about got it. But anyways to wade thru all the side roads here makes the issue more complex to talk about since some terms are not quite clear and may mean different things to some of us.

Lets try to simplify this or seperate the issues to understand. See if you'all can agree with this statement and others and then we'll see exactly where we part company.

First statement is this:

A parts of a series circuit (regardless of voltage and/or resistance) all gets the exact same current and capacity thru all it's parts. It is a complete circuit.

All parts of a series circuit all gets the exact same resistance regardless of where that resistance is in the circuit. The resistance gets added up and totaled.


If you can agree with these two statements then it is impossible for any one cell to get a different current or a different total charge than the rest of the cells in a series. 

If you can't agree , then an argument for all (equal/balanced/matched) cells in a series being full at the same time falls apart .


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## guver (Jul 31, 2002)

Let me address the charger and capacity too: 1+ nimh cells

Even though the charger doesn't care about capacity, voltage,current, ect. It does look at and for voltage , but it cares not what that voltage is and merely looks for the "peak" or "drop"

There are some parameters on some chargers that may check cell count vs. voltage , but it matters not to what we are discussing. In otherwords if we have a typical 6 cell pack and it shows 7,8,9,10 volts it won't matter. All that matters is the peak or drop as it relates to finding out whether the battery is full or not. The voltage can be ran way up or way down , yet the charger will still look for it's programmed mv drop to terminate charge.

There are a couple of other instances where voltage "matters" , but doesn't relate at all to our debate. Some chargers compare the cell count to the voltage and must be in a range perhaps of 1-1.9 volts/cell and if voltage is under 1 volt/cell it may give low voltage error and if it's over 1.9 it may give a high voltage error. Another instance would be an automatic charger where the mv programmed is multiplied by the estimated number of cells and if the cell count happens to be off by 1 or 2 it still makes no difference and sure has nothing to do with our debate.

A couple of other things can also hide or affect the effective current/capacity taken by the cells. Self-discharge or varying degrees of self discharge , but TIME is involved and is difficult to predict unless it is known. I have left this variable out since that is a different discussion. The heat and overcharge characteristics of NI cells can also affect the capacity of the cells in a pack, but again this is something different and results normally AFTER the cells are already full.


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## guver (Jul 31, 2002)

Let me quickly address varying IR, and voltage of individual cells in a pack . To repeat what I stated those values are completey irrelevant when they are in a series circuit and will affect only the voltage/current/capacity or the pack as a whole.

To illustrate allow me to construct another frankenstein pack 6 cells.
1 3000 nimh cell (1.2 volts)
2 1500 nicad (1.4 volts)
3 2200 lipo cell (3.7 volts)
4 AA nimh 2700
5 piece of coat hanger
6 D sized Alkaline cell

Every cell in this 6 cell pack (as long as it is between dead and full) during charge/discharge will get exactly the same capacity and current. The resistance of the pack will be the total of all cells added up even though each cell may vary tremendously.


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## ScottH (Nov 24, 2005)

guver said:


> First statement is this:
> 
> A parts of a series circuit (regardless of voltage and/or resistance) all gets the exact same current and capacity thru all it's parts. It is a complete circuit.
> 
> All parts of a series circuit all gets the exact same resistance regardless of where that resistance is in the circuit. The resistance gets added up and totaled.


The first statement is partly true. In a series circuit the current is the same, it is the voltage that is divided. I am not so sure about the capacity part.

The second statement, I do not agree with. That is how the voltage is divided, by the resistance being different.


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## ta_man (Apr 10, 2005)

guver said:


> Lets try to simplify this or seperate the issues to understand. See if you'all can agree with this statement and others and then we'll see exactly where we part company.
> 
> First statement is this:
> 
> ...





ScottH said:


> The first statement is partly true. In a series circuit the current is the same, it is the voltage that is divided. I am not so sure about the capacity part.
> 
> The second statement, I do not agree with. That is how the voltage is divided, by the resistance being different.


First statement isn't true either.

The same current may pass through all the cells in series, but if the cells have different internal resistance, different amounts of energy will be converted into heat in the different cells, leaving them with a different state of charge.

That is one reason why a LiPo of a specific capacity has more runtime than a NiMH of the same capacity. Less energy is converted into heat in the LiPo so more is available to run the car.


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## ScottH (Nov 24, 2005)

guver said:


> Let me quickly address varying IR, and voltage of individual cells in a pack . To repeat what I stated those values are completey irrelevant when they are in a series circuit and will affect only the voltage/current/capacity or the pack as a whole.
> 
> To illustrate allow me to construct another frankenstein pack 6 cells.
> 1 3000 nimh cell (1.2 volts)
> ...


Sorry but I really think you are off base here. If you put a 35amp load on that pack there are two things that are going to happen. The AA is going to breakdown and pop, the D Alkaline is going to do the same thing.

I really do not think that you get it. The AA and the D cell CAN NOT PROVIDE 35amps of current so they WILL NOT provide the same current as those cells that can provide that amount of current.

Now charge that pack at say 6amps, you are still going to get BOOM.

The only thing that I agree with is that the total IR will be the sum of al the IR's.


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## OvalmanPA (Mar 21, 2000)

I just have one thing to say. I ran my 4600 packs last Saturday. Monday I discharged them to 6.6v (6 cell pack) on my GFX. I put each of them on a Novak tray to .9v per cell. Resting voltage after an hour of sitting was 1.16v on each cell. Recharged at 6 amps and periodically checked each cells voltage. All during charge the individual cell voltages remained the same until a peak charge of 9.16v at which time all cells were at the same resting voltage. So....voltages at the equalized bottom portion of the charge ended up equalized voltages at the top of the charge.


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## guver (Jul 31, 2002)

It looks like there is some agreement with the fact of the same current passing thru the series wired cells. And Scott agrees that the IR is the total of all parts added together.

As far as the charger and pack are concerned that's a good start.

I can go along with differing cells,IR, "converting different amounts into heat" (ta man)
I also agree with "the voltage being divided" (depending on what Scott means)

To answer both lets remember that the imaginary pack I construct is grossly exaggerated to illustrate the point. Lets go back to a real (unmatched pack) with varying capacity,IR,voltage and see that the very real condition described by Scott makes an imeasurable amount of difference in retained capacity as long as the currents are within some reasonable tolerances. To get specific let's use a 6 cell 3000-5000 mah pack and a tolerance of 1-8 amps charge and 1-50 amps discharge.

This is also an answer to the fact that a AA cell and/or a D alkaline may indeed get hurt by a 35 amp discharge or with 6 amps charge. Again please keep in mind that the frankenstien pack is illustrative only and if we'd rather remove the 2 cells or replace them with a C nimh and D nimh that's fine. The tolerances for such a pack need to be within the tightest recommended tolerance for the "worst" cell. An example may be a 1 amp charge and a 3 amp discharge. 

It's too bad I can't think of a nice easy test to do to illustrate that "capacity added is equal to every cell in a pack" We all see and witness it at nearly every discharge and at nearly every re-charge of nearly every pack.


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## guver (Jul 31, 2002)

Perhaps I should describe how to do this in a couple of different ways , but I would like suggestions as to how to graphically illustrate it. In otherwords what kind of demo would it take to convince you that it is true?

A pack of perfectly matched cells starts out charged full (every cell) We pull out 2000 mah and then replace 2000 mah (every cell is exactly full again)

A pack of perfectly matched cells (with a 4500 actual capacity) starts out empty and we add 4500 mah to the pack and every cell is exactly full) 

A pack of un-matched cells starts out charged full (every cell) We pull out 2000 mah and then replace 2000 mah (every cell is exactly full again)

A pack of un-matched cells (with a 4500 capacity) starts out empty and we add 4500 mah to the pack and every cell is not full. Some are over , some are under.

A pack of perfectly matched cells starts out full and we pull 1000 mah from only one cell. We pull out 2000 mah from the pack. Upon recharge 5 cells will be full when about 2000 mah is added , but the 6th cell is still down by 1000 mah. We simply add 1000 mah to the 6th cell and all cells are exactly full again.

A pack of un-matched cells starts out full and we pull 1000 mah from only one cell. We pull out 2000 mah from the pack. Upon recharge , 5 cells will be full when about 2000 mah is added , but the 6th cell is still down by 1000 mah. We simply add 1000 mah to the 6th cell and all cells are exactly full again.

I can easily do these tests and prove to anyone watching what's going on. It's actually quite simple, easy to understand. Perhaps 6 glasses glued to a 2by4 would illustrate?


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## ta_man (Apr 10, 2005)

guver said:


> Perhaps I should describe how to do this in a couple of different ways , but I would like suggestions as to how to graphically illustrate it. In otherwords what kind of demo would it take to convince you that it is true?
> 
> A pack of perfectly matched cells starts out charged full (every cell) We pull out 2000 mah and then replace 2000 mah (every cell is exactly full again)
> 
> ...


You couldn't really do the tests you describe for the simple reason that it always takes more to recharge a NiMH cell than you took out of it. Minute variances in internal properties, (beyond the detection level of the most accurate matcher) will cause each cell to require a slightly different amount to recharge. That includes variance in self-discharge rate which is not a parameter checked/measuerd by any matcher.

People have years of experience that NiMH cells in a single pack differ in amounts of charge over many cycles and need to be equalized. Your denial (or whatever it is) of this, or proposal of hypothetical test cases, doesn't countermand those years of experience.


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## guver (Jul 31, 2002)

minute variances including self-discharge are very real. They will not affect any test of "capacity added or removed" Varying amounts of self-discharge will not have time enough to come into play. If we did tests over days or weeks self-discharge will affect the results to the degree that they vary from cell to cell in the same pack.

Any other differences internal or otherwise are the differences I am saying that do not matter in a series wired pack. I'm afraid I am repeating myself and perhaps "a man convinced against his will is of the same opinion still"

Perhaps we can go about another way , Anyone is welcome to convince me that anyone of my above "tests" is untrue. Give me 1% tolerance on capacity. 

I might agree with this statement "NiMH cells in a single pack differ in amounts of charge over many cycles and need to be equalized" What do you mean?

If you are saying that a pack of 4200 actual capacity cells over the lifetime do not all stay at 4200 or that they do not lose capacity at the same rate , then I would agree. The "matched" pack becomes unmatched.


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## guver (Jul 31, 2002)

How about someone put 2 (somewhat different capacity) of the same cells together , tray them to equalise and show how they will get full at the same time. I say that they cannot/will not. I can do the same thing and start out with them full , discharge in series and upon re-charge in series will show by temp and/or voltage that they'll be full at the same time.

The only problem is I can't figure out how to "capture" this event graphically or visually so that one would be convinced.


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## ta_man (Apr 10, 2005)

guver said:


> Any other differences internal or otherwise are the differences I am saying that do not matter in a series wired pack. I'm afraid I am repeating myself and perhaps "a man convinced against his will is of the same opinion still"


I know that is what you are saying. But the differences do matter. When one cell in the series pack has more resistance then another, that cell heats up more during charge and discharge. That heat comes from mAHr that was converted to heat instead of chemical energy to charge the cell. During discharge, that heat comes from chemical energy that ends up as heat instead of mAHr coming out of the cell. So even though the cells are in series, they differ in the amount of mAHr that gets converted to chemical energy, and thus the level of charge.

Your theory seems to imply the heat comes from the air or something and none of the mAhr put into the cell is what is heating it up. If so, the cells would heat up spontaneously on there own (I could use them to heat my house!). They don't. Excluding the cell going into runaway because of internal damage, they only heat up when being charged and discharged.



guver said:


> Perhaps we can go about another way , Anyone is welcome to convince me that anyone of my above "tests" is untrue. Give me 1% tolerance on capacity.


I doubt anyone will convince you.



guver said:


> I might agree with this statement "NiMH cells in a single pack differ in amounts of charge over many cycles and need to be equalized" What do you mean?


I mean that, since the cells differ in internal properties, different amounts of mAHr get converted into heat. So taking charge out and adding it back to the pack as a whole without equalizing the cells means there ends up being different amounts of charge in different cells in the pack. Thus they become unbalanced. (That's kind of the definition of "unbalanced", at least as regard to battery packs.)


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## guver (Jul 31, 2002)

It seems that your focus is on the varying IR or other properties of cells that may make them spec out differently than just capacity and I am focused on capacity only. If this is true then we can agree on these?

A pack of perfectly matched cells starts out charged full (every cell) We pull out 2000 mah and then replace (I relise this is going to be higher, but all cells are getting same amount for this sake) 2000 mah (every cell is exactly full again)

A pack of perfectly matched cells (with a 4500 actual capacity) starts out empty and we add (I relise this is going to be higher, but all cells are getting same amount for this sake) 4500 mah to the pack and every cell is exactly full) 

and possibly even this?

A pack of perfectly matched cells starts out full and we pull 1000 mah from only one cell. We pull out 2000 mah from the pack. Upon recharge 5 cells will be full when about 2000 mah is added , but the 6th cell is still down by 1000 mah. We simply add 1000 mah to the 6th cell and all cells are exactly full again.


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## OvalmanPA (Mar 21, 2000)

Anybody else figure they are doing this







? 

I







, your right, the rest of us are all wrong.


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## ta_man (Apr 10, 2005)

guver said:


> It seems that your focus is on the varying IR or other properties of cells that may make them spec out differently than just capacity and I am focused on capacity only. If this is true then we can agree on these?
> 
> A pack of perfectly matched cells starts out charged full (every cell) We pull out 2000 mah and then replace (I relise this is going to be higher, but all cells are getting same amount for this sake) 2000 mah (*every cell is exactly full again*)
> 
> ...


The major point you don't get is if you have a 4500 mAHr capacity cell that is full, you take out 2000 mAHr and then put back 2000 mAHR, it is *not* full again. There are losses in this process both in the charging and discharging.


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## Minreg (Jan 1, 1970)

airconde said:


> DISCHARGERS differant one's differant ways.Whats your opinion



airconde > You getting all this right? :freak:

PS. Good discussion guys.


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## guver (Jul 31, 2002)

Let's drop it, I hope the OP isn't offfended.


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## hankster (Jan 1, 1998)

guver is not completely off the mark but there are other factors at work. Even if a cell has the same mAh capacity as shown on a matcher, the time that it takes each cell to fully charge will not be the same. Just to throw some numbers out, it may take one cell 1800 sec. to accept 2000mAh to fully change and it may take the next cell 1900 sec. to accept the same 2000mAh to fully charge. Look at the labels on your matched NiMh pack and see the charge time on each cell.

So even you have a pack that has all cells with the same capacity (tested via matching) and the pack is fully equalized on a discharge tray (or via a part discharge), some cells will reach peak before others. This is unavoidable.

For years I have said that to get the most performance from a pack that each cell needs to be charged individually at the same time. There was only one high priced charger that would allow this and everyone also said I was nuts.

Guess what? Now that we have 2 cell LiPo packs and chargers are available to everyone that will "equalize" cells, everyone now says this is the best way to charge cells. What changed? Did the laws of physics change? Did how batteries react to charging change?

Nope. All that changed is that is now practical to do it *and* racers would never admit that the way they are doing something is not the optimal way to do it.


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## ScottH (Nov 24, 2005)

hankster said:


> *...Even if a cell has the same mAh capacity as shown on a matcher, the time that it takes each cell to fully charge will not be the same.* Just to throw some numbers out, it may take one cell 1800 sec. to accept 2000mAh to fully change and it may take the next cell 1900 sec. to accept the same 2000mAh to fully charge. Look at the labels on your matched NiMh pack and see the charge time on each cell.
> 
> So even you have a pack that has all cells with the same capacity (tested via matching) and the pack is fully equalized on a discharge tray (or via a part discharge), some cells will reach peak before others. This is unavoidable...


This is what I was trying to get across but did not put it well.

A 2000Mah battery will take on 1800mah worth of charge at a different rate than a 4000Mah battery.

My head hurts.  I love the smell of Nitro at the track!


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## hankster (Jan 1, 1998)

Two 2000mAh cells that match out at the same capacity and voltage will charge at different rates... they don't even need to be different size cells.


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## ta_man (Apr 10, 2005)

That's why, when I rematch old packs (or at least I used to rematch old packs), I matched them by charge time from an equalized state. Then the cells reached peak closer together. I did not consider ultimate capacity the most important parameter because I never ran the pack down to empty in a race anyway.


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## guver (Jul 31, 2002)

The difference between non equalising chargers for NIMH and equalising or "balancing" chargers is the type and specifically how the cells deal with overcharge. The NI cells can dissipate overcharge as heat long before damage occurs. The lipo doesn't , it will be damaged rather than dealing with any overcharge. This is why it is important that they both be equal and under the max voltage.

I suppose I liked that 6 cell monitoring NI charger (can't remember what it was from England) but maybe it just didn't sell well. I apparently am in the minority again on the best way to charge lipos , as everyone else seems to want balancers and even balancing chargers.

The exact same principal (that I said "lets drop") is exactly the same for lipos , in fact it is even more predictable and repeatable it isn't funny.


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