# [EVDL] battery sag question



## EVDL List (Jul 27, 2007)

In a message dated 11/19/2007 6:35:54 AM US Mountain Standard Time, 
[email protected] writes: 
> EVDL] battery sag question 
> Date:11/19/2007 6:35:54 AM US Mountain Standard Time
> From:[email protected]
> Reply-to:[email protected]
> To:[email protected]
> Received from Internet: 
> 
> 
> 
> I've been reading about the huge battery voltage sags seen by the racers 
> (ex. 50% by Killacycle). Does this ever lead to cell-reversal issues or is that 
> non an issue with lithium? How about the racers using AGM's? With massive 
> sags are they risking cell reversal?
> 
> TIA
> 
Bracket racing the truck (AGM pack with 125 timeslips)I have had no reversals 
and the mph keeps going up each week(by about .35mph) indicating more hp.I 
cycle the hawkers out of the dragster with 500+ cycles.I do however use BMS. 
Dennis Berube 
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## EVDL List (Jul 27, 2007)

For "Maximum Power Transfer" (from the battery pack to
the controller) you need to match your source
impedance to your load impedance, in which case, you
end up with half of your open circuit voltage coming
from your source. In other words, if your battery
pack isn't sagging to 1/2 of the open circuit voltage,
you are not getting the max power (H.P.) out of it, at
that instant. This might mean that your pack is
larger (heavier) than what is optimal. 

There might be good reasons why your pack doesn't sag
to 1/2 voltage (for racing applications). For
example, you might have to parallel more cells to get
enough amp-hours to get down the track. Boulder Cells
come to mind. If I remember correctly, they are about
the stiffest cell out there (or were, since it's not
possible to get the same boulder cells I'm referring
to) but did not have that many amp-hours. Their watts
per kg was great, but there watt-hours per kg was
pretty poor. Tons of current (or power), but not for
very long. So to get enough amp-hours to make it down
the track meant that you had to parallel a lot of
cells, which also gave you more current capacity than
you could use, so they didn't sag as much. Until the
pack was almost empty, like at the end of the 1/4
mile, which is where you need the HP the most.

Yes, you are very correct in that by sagging this
much, you are greatly increasing the risk of reversing
a cell. This is why a BMS is _so_ important for
racing! If your pack is out of balance, then your
weakest cell has the greatest risk of reversal. If it
_doesn't_ reverse, then that means all your other
cells are stronger than they have to be, and you might
be carrying more battery weight down the track than
you have to! On the old Boulder cells, racers were
always reversing cells and rebuilding the pack. With
the new A123 cells, not only does the Killacycle have
a lighter pack then the lead acid Boulder cell pack
(allowing the bike to go faster), but it has several
times more amp-hours. The increased amp-hours allows
for a burnout, keeps the pack from sagging as much
(especially at the end of the strip, which is where
you need the power the most), and greatly reduces the
chances of reversing a cell.

There are always exceptions to pretty much everything,
but in general the above is true (to the first order
of approximation). Something similar can be said for
battery interconnects in racing. Sure, smaller
interconnects drop more voltage, loose more power. 
You would think that bigger is better, right? Would a
4" diameter solid copper buss bar be better? Nope, it
would weigh too much. So a balancing act must be made
between weight and power loss. Bigger interconnects
will loose less power, which will make you go faster. 
But they will weigh more, which will make you go
slower. Smaller interconnects will weigh less, which
will make you go faster. But they will drop more
voltage, which will make you go slower. You need to
find that balancing point to be "optimal".

A quick example on why you need your pack to sag to
1/2 voltage for optimal power transfer. Let's say you
have an exactly 1V battery, with exactly 1 ohm of
internal impedance. With no load, you get 1V out, but
no current, so you get no power out of the battery. A
direct super-conducting short circuit will get the
most current out of this battery, which will get you
one amp (1V divided by one ohm). But here again, the
battery isn't delivering any power, because the
voltage across your zero ohm load is zero. But if you
match your load to the source impedance of your
battery (one ohm in this case) by placing a one ohm
load across this battery, you get (one volt divided by
two ohms equals) one half amp out of your battery,
into your load. And one half amp times one ohm is one
half volt across your load. Notice how this is 1/2
your open circuit battery voltage? The power
delivered by this battery is volts times amps, or one
half times one half which is a quarter watt.

Now what if you want this battery to sag to less than
half voltage? You can do this by putting more than a
one ohm load. For example, a 1.1 ohm load. There,
this battery will put out (one volt divided by 2.1
ohms equals) 0.476 amps. The voltage across the 1.1
ohm load is 1.1 ohms times 0.476 amps equals 0.5238
volts (less than 1/2 voltage of sag). But the power
is 0.2493 watts, less than the 0.25 watts above.

A similar thing happens if you sag this battery too
much, by putting a 0.9 ohm load across it. You get
0.5263 amps out of the battery, but only 0.474 volts
across this resistor. This gives you 0.249 watts
again.

You may notice that with a one ohm load and a one ohm
source resistance, the power being delivered to your
load is the same as the power that's being dissipated
in your battery. When the Killacyle is putting out
hundreds of HP, if the pack is being sagged to half
voltage, it's dissipating hundreds of HP internally as
well! 

I'm sure others can explain it better. I hope that
helped more than it confused...

- Steven Ciciora



> --- Frank John <[email protected]> wrote:
> 
> > I've been reading about the huge battery voltage
> > sags seen by the racers (ex. 50% by Killacycle).
> ...


----------



## EVDL List (Jul 27, 2007)

Nice work Steve.

This one belongs in the FAQ's!!!

--
Stay Charged!
Hump


On Mon, 19 Nov 2007 07:00:46 -0800 (PST), Steven Ciciora <[email protected]> wrote:
> For "Maximum Power Transfer" (from the battery pack to
> the controller) you need to match your source
> impedance to your load impedance, in which case, you
> end up with half of your open circuit voltage coming
> from your source. In other words, if your battery
> pack isn't sagging to 1/2 of the open circuit voltage,
> you are not getting the max power (H.P.) out of it, at
> that instant. This might mean that your pack is
> larger (heavier) than what is optimal.
> 
> There might be good reasons why your pack doesn't sag
> to 1/2 voltage (for racing applications). For
> example, you might have to parallel more cells to get
> enough amp-hours to get down the track. Boulder Cells
> come to mind. If I remember correctly, they are about
> the stiffest cell out there (or were, since it's not
> possible to get the same boulder cells I'm referring
> to) but did not have that many amp-hours. Their watts
> per kg was great, but there watt-hours per kg was
> pretty poor. Tons of current (or power), but not for
> very long. So to get enough amp-hours to make it down
> the track meant that you had to parallel a lot of
> cells, which also gave you more current capacity than
> you could use, so they didn't sag as much. Until the
> pack was almost empty, like at the end of the 1/4
> mile, which is where you need the HP the most.
> 
> Yes, you are very correct in that by sagging this
> much, you are greatly increasing the risk of reversing
> a cell. This is why a BMS is _so_ important for
> racing! If your pack is out of balance, then your
> weakest cell has the greatest risk of reversal. If it
> _doesn't_ reverse, then that means all your other
> cells are stronger than they have to be, and you might
> be carrying more battery weight down the track than
> you have to! On the old Boulder cells, racers were
> always reversing cells and rebuilding the pack. With
> the new A123 cells, not only does the Killacycle have
> a lighter pack then the lead acid Boulder cell pack
> (allowing the bike to go faster), but it has several
> times more amp-hours. The increased amp-hours allows
> for a burnout, keeps the pack from sagging as much
> (especially at the end of the strip, which is where
> you need the power the most), and greatly reduces the
> chances of reversing a cell.
> 
> There are always exceptions to pretty much everything,
> but in general the above is true (to the first order
> of approximation). Something similar can be said for
> battery interconnects in racing. Sure, smaller
> interconnects drop more voltage, loose more power.
> You would think that bigger is better, right? Would a
> 4" diameter solid copper buss bar be better? Nope, it
> would weigh too much. So a balancing act must be made
> between weight and power loss. Bigger interconnects
> will loose less power, which will make you go faster.
> But they will weigh more, which will make you go
> slower. Smaller interconnects will weigh less, which
> will make you go faster. But they will drop more
> voltage, which will make you go slower. You need to
> find that balancing point to be "optimal".
> 
> A quick example on why you need your pack to sag to
> 1/2 voltage for optimal power transfer. Let's say you
> have an exactly 1V battery, with exactly 1 ohm of
> internal impedance. With no load, you get 1V out, but
> no current, so you get no power out of the battery. A
> direct super-conducting short circuit will get the
> most current out of this battery, which will get you
> one amp (1V divided by one ohm). But here again, the
> battery isn't delivering any power, because the
> voltage across your zero ohm load is zero. But if you
> match your load to the source impedance of your
> battery (one ohm in this case) by placing a one ohm
> load across this battery, you get (one volt divided by
> two ohms equals) one half amp out of your battery,
> into your load. And one half amp times one ohm is one
> half volt across your load. Notice how this is 1/2
> your open circuit battery voltage? The power
> delivered by this battery is volts times amps, or one
> half times one half which is a quarter watt.
> 
> Now what if you want this battery to sag to less than
> half voltage? You can do this by putting more than a
> one ohm load. For example, a 1.1 ohm load. There,
> this battery will put out (one volt divided by 2.1
> ohms equals) 0.476 amps. The voltage across the 1.1
> ohm load is 1.1 ohms times 0.476 amps equals 0.5238
> volts (less than 1/2 voltage of sag). But the power
> is 0.2493 watts, less than the 0.25 watts above.
> 
> A similar thing happens if you sag this battery too
> much, by putting a 0.9 ohm load across it. You get
> 0.5263 amps out of the battery, but only 0.474 volts
> across this resistor. This gives you 0.249 watts
> again.
> 
> You may notice that with a one ohm load and a one ohm
> source resistance, the power being delivered to your
> load is the same as the power that's being dissipated
> in your battery. When the Killacyle is putting out
> hundreds of HP, if the pack is being sagged to half
> voltage, it's dissipating hundreds of HP internally as
> well!
> 
> I'm sure others can explain it better. I hope that
> helped more than it confused...
> 
> - Steven Ciciora
> 
> --- Frank John <[email protected]> wrote:
> 
>> I've been reading about the huge battery voltage
>> sags seen by the racers (ex. 50% by Killacycle).
>> Does this ever lead to cell-reversal issues or is
>> that non an issue with lithium? How about the
>> racers using AGM's? With massive sags are they
>> risking cell reversal?
>>
>> TIA
>>
>>
>>
>>
>>
>>
> ____________________________________________________________________________________
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>>
> 
> 
> 
> 
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## EVDL List (Jul 27, 2007)

not really 1/1 =1 1amp x 1ohm = 1volt amp x res = voltage still 
----- Original Message ----- 
From: Tim Humphrey<mailto:[email protected]> 
To: Electric Vehicle Discussion List<mailto:[email protected]> 
Sent: Monday, November 19, 2007 9:15 AM
Subject: Re: [EVDL] battery sag question



Nice work Steve.

This one belongs in the FAQ's!!!

--
Stay Charged!
Hump


On Mon, 19 Nov 2007 07:00:46 -0800 (PST), Steven Ciciora <[email protected]<mailto:[email protected]>> wrote:
> For "Maximum Power Transfer" (from the battery pack to
> the controller) you need to match your source
> impedance to your load impedance, in which case, you
> end up with half of your open circuit voltage coming
> from your source. In other words, if your battery
> pack isn't sagging to 1/2 of the open circuit voltage,
> you are not getting the max power (H.P.) out of it, at
> that instant. This might mean that your pack is
> larger (heavier) than what is optimal.
> 
> There might be good reasons why your pack doesn't sag
> to 1/2 voltage (for racing applications). For
> example, you might have to parallel more cells to get
> enough amp-hours to get down the track. Boulder Cells
> come to mind. If I remember correctly, they are about
> the stiffest cell out there (or were, since it's not
> possible to get the same boulder cells I'm referring
> to) but did not have that many amp-hours. Their watts
> per kg was great, but there watt-hours per kg was
> pretty poor. Tons of current (or power), but not for
> very long. So to get enough amp-hours to make it down
> the track meant that you had to parallel a lot of
> cells, which also gave you more current capacity than
> you could use, so they didn't sag as much. Until the
> pack was almost empty, like at the end of the 1/4
> mile, which is where you need the HP the most.
> 
> Yes, you are very correct in that by sagging this
> much, you are greatly increasing the risk of reversing
> a cell. This is why a BMS is _so_ important for
> racing! If your pack is out of balance, then your
> weakest cell has the greatest risk of reversal. If it
> _doesn't_ reverse, then that means all your other
> cells are stronger than they have to be, and you might
> be carrying more battery weight down the track than
> you have to! On the old Boulder cells, racers were
> always reversing cells and rebuilding the pack. With
> the new A123 cells, not only does the Killacycle have
> a lighter pack then the lead acid Boulder cell pack
> (allowing the bike to go faster), but it has several
> times more amp-hours. The increased amp-hours allows
> for a burnout, keeps the pack from sagging as much
> (especially at the end of the strip, which is where
> you need the power the most), and greatly reduces the
> chances of reversing a cell.
> 
> There are always exceptions to pretty much everything,
> but in general the above is true (to the first order
> of approximation). Something similar can be said for
> battery interconnects in racing. Sure, smaller
> interconnects drop more voltage, loose more power.
> You would think that bigger is better, right? Would a
> 4" diameter solid copper buss bar be better? Nope, it
> would weigh too much. So a balancing act must be made
> between weight and power loss. Bigger interconnects
> will loose less power, which will make you go faster.
> But they will weigh more, which will make you go
> slower. Smaller interconnects will weigh less, which
> will make you go faster. But they will drop more
> voltage, which will make you go slower. You need to
> find that balancing point to be "optimal".
> 
> A quick example on why you need your pack to sag to
> 1/2 voltage for optimal power transfer. Let's say you
> have an exactly 1V battery, with exactly 1 ohm of
> internal impedance. With no load, you get 1V out, but
> no current, so you get no power out of the battery. A
> direct super-conducting short circuit will get the
> most current out of this battery, which will get you
> one amp (1V divided by one ohm). But here again, the
> battery isn't delivering any power, because the
> voltage across your zero ohm load is zero. But if you
> match your load to the source impedance of your
> battery (one ohm in this case) by placing a one ohm
> load across this battery, you get (one volt divided by
> two ohms equals) one half amp out of your battery,
> into your load. And one half amp times one ohm is one
> half volt across your load. Notice how this is 1/2
> your open circuit battery voltage? The power
> delivered by this battery is volts times amps, or one
> half times one half which is a quarter watt.
> 
> Now what if you want this battery to sag to less than
> half voltage? You can do this by putting more than a
> one ohm load. For example, a 1.1 ohm load. There,
> this battery will put out (one volt divided by 2.1
> ohms equals) 0.476 amps. The voltage across the 1.1
> ohm load is 1.1 ohms times 0.476 amps equals 0.5238
> volts (less than 1/2 voltage of sag). But the power
> is 0.2493 watts, less than the 0.25 watts above.
> 
> A similar thing happens if you sag this battery too
> much, by putting a 0.9 ohm load across it. You get
> 0.5263 amps out of the battery, but only 0.474 volts
> across this resistor. This gives you 0.249 watts
> again.
> 
> You may notice that with a one ohm load and a one ohm
> source resistance, the power being delivered to your
> load is the same as the power that's being dissipated
> in your battery. When the Killacyle is putting out
> hundreds of HP, if the pack is being sagged to half
> voltage, it's dissipating hundreds of HP internally as
> well!
> 
> I'm sure others can explain it better. I hope that
> helped more than it confused...
> 
> - Steven Ciciora
> 
> --- Frank John <[email protected]<mailto:[email protected]>> wrote:
> 
>> I've been reading about the huge battery voltage
>> sags seen by the racers (ex. 50% by Killacycle).
>> Does this ever lead to cell-reversal issues or is
>> that non an issue with lithium? How about the
>> racers using AGM's? With massive sags are they
>> risking cell reversal?
>>
>> TIA
>>
>>
>>
>>
>>
>>
> ____________________________________________________________________________________
>> Be a better pen pal.
>> Text or chat with friends inside Yahoo! Mail. See
>> how. http://overview.mail.yahoo.com/<http://overview.mail.yahoo.com/>
>>
>> _______________________________________________
>> For subscription options, see
>> http://lists.sjsu.edu/mailman/listinfo/ev<http://lists.sjsu.edu/mailman/listinfo/ev>
>>
> 
> 
> 
> 
> ____________________________________________________________________________________
> Be a better sports nut! Let your teams follow you
> with Yahoo Mobile. Try it now. 
> http://mobile.yahoo.com/sports;_ylt=At9_qDKvtAbMuh1G1SQtBI7ntAcJ<http://mobile.yahoo.com/sports;_ylt=At9_qDKvtAbMuh1G1SQtBI7ntAcJ>
> 
> _______________________________________________
> For subscription options, see
> http://lists.sjsu.edu/mailman/listinfo/ev<http://lists.sjsu.edu/mailman/listinfo/ev>


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## EVDL List (Jul 27, 2007)

> "FRED JEANETTE MERTENS" <[email protected]> wrote:
> 
> > not really 1/1 =1 1amp x 1ohm = 1volt amp x res = voltage still
> > ----- Original Message -----
> ...


----------



## EVDL List (Jul 27, 2007)

Hi Steven,

Pretty good explanation. The other approach is:

Battery terminal voltage = Vb = Voc - I * Rint.
Where:
Voc = open circuit battery voltage
I = current
Rint = internal resistance of the battery.

To get power, use P = Vb * I,
Then substitute previous equation into this,

P = (Voc - I * Rint) * I
or

P = Voc * I - I^2 * Rint

Differentiate this equation with respect to I and set
equal to zero to find the current at which maximum
power will occur. Result is

0 = Voc - 2 * I * Rint
or

I = Voc / (2 * Rint) 

Which is the current at which maximum power is
delivered. Since I = Voc / (Rint + Rload), max power
occurs when the load resistance equals the internal
resistance, Rload = Rint.

To find the maximum power, just multiply Vb at max
power times the current at max power. Takes a few
substitutions, but you end up with

Pmax = (Voc^2) / (4 * Rint).

If you look at the power as a function of current,
you'll see it to be a parabola. As you increase
current nearing maximum power, the increase in power
per increase in current diminishes. So, in my
opinion, it is best to stay 15 percent or so below the
current at which max power occurs. Avoids drawing a
hundred amps more for just a few extra watts.

Jeff M



> --- Steven Ciciora <[email protected]> wrote:
> 
> > For "Maximum Power Transfer" (from the battery pack
> > to
> ...


----------



## EVDL List (Jul 27, 2007)

his amperage out of the battery with 1 ohm resistance and 1 volt ==1 amp
----- Original Message ----- 
From: Tim Humphrey<mailto:[email protected]> 
To: Electric Vehicle Discussion List<mailto:[email protected]> 
Sent: Monday, November 19, 2007 12:00 PM
Subject: Re: [EVDL] battery sag question







> "FRED JEANETTE MERTENS" <[email protected]<mailto:[email protected]>> wrote:
> 
> > not really 1/1 =1 1amp x 1ohm = 1volt amp x res = voltage still
> > ----- Original Message -----
> ...


----------



## EVDL List (Jul 27, 2007)

OK, I believe your issue is in the following cut....


"Let's say you have an exactly 1V battery, with exactly 1 ohm of internal impedance. With no load, you get 1V out, but no current, so you get no power out of the battery. A direct super-conducting short circuit will get the most current out of this battery, which will get you one amp (1V divided by one ohm). But here again, the battery isn't delivering any power, because the voltage across your zero ohm load is zero."

"Let's say you have an exactly 1V battery, with exactly 1 ohm of internal impedance. With no load, you get 1V out, but no current, so you get no power out of the battery."
This part is just giving us the open circuit voltage of the battery. No connections, therefore no current flow and no power out.


"A direct super-conducting short circuit will get the most current out of this battery, which will get you one amp (1V divided by one ohm)."
This gives us a direct short circuit across the battery, so 1 amp flows. Since there is no resistance in the load, all of the power, 1 watt, is dissipated inside the battery by it's internal resistance.

"But here again, the battery isn't delivering any power, because the voltage across your zero ohm load is zero."
I suppose you could say that the battery IS delivering 1 watt of power. BUT the load isn't consuming it, the battery is. So what good does it do for us. 

HTH



--
Stay Charged!
Hump
I-5, Blossvale NY





> "FRED JEANETTE MERTENS" <[email protected]> wrote:
> > his amperage out of the battery with 1 ohm resistance and 1 volt ==1 amp
> > ----- Original Message -----
> > From: Tim Humphrey<mailto:[email protected]>
> ...


----------



## EVDL List (Jul 27, 2007)

[No message]


----------



## EVDL List (Jul 27, 2007)

Hi Andre',

I think the point is that you do have control of the
source impedance when choosing your energy storage
system. Yes, zero source impedance would be nice, but
who can afford one of those? So it becomes a
compromise of cost, mass and performance. Being able
to calculate the maximum power transfer is important
so you do not end up with a energy storage system (or
power source) incapable of delivering your required
power.

My thoughts on the subject.

Jeff M



> --- Andre' Blanchard <[email protected]> wrote:
> 
> > "match your source impedance to your load impedance"
> >
> ...


----------



## EVDL List (Jul 27, 2007)

>You may notice that with a one ohm load and a one ohm
source resistance, the power being delivered to your
load is the same as the power that's being dissipated
in your battery. When the Killacyle is putting out
hundreds of HP, if the pack is being sagged to half
voltage, it's dissipating hundreds of HP internally as
well!

So that's why my batteries get hot!

storm

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## EVDL List (Jul 27, 2007)

For fixed internal impedances, the maximum power delivery always occurs
when the load resistance is equal to the source impedance and efficiency
is 50%.

For a 1V batt with 1 ohm impedance, it delivers max power to a 1 ohm
load, 0.5A @ 0.5V = 0.25W. 0.9 ohm or 1.1 ohm load gets less power.

Danny

----- Original Message -----
From: Tim Humphrey <[email protected]>
Date: Tuesday, November 20, 2007 8:46 am
Subject: Re: [EVDL] battery sag question
To: Electric Vehicle Discussion List <[email protected]>

> 
> OK, I believe your issue is in the following cut....
> 
> 
> "Let's say you have an exactly 1V battery, with exactly 1 ohm of 
> internal impedance. With no load, you get 1V out, but no current, 
> so you get no power out of the battery. A direct super-conducting 
> short circuit will get the most current out of this battery, which 
> will get you one amp (1V divided by one ohm). But here again, the 
> battery isn't delivering any power, because the voltage across your 
> zero ohm load is zero."
> 
> "Let's say you have an exactly 1V battery, with exactly 1 ohm of 
> internal impedance. With no load, you get 1V out, but no current, 
> so you get no power out of the battery."
> This part is just giving us the open circuit voltage of the 
> battery. No connections, therefore no current flow and no power out.
> 
> 
> "A direct super-conducting short circuit will get the most current 
> out of this battery, which will get you one amp (1V divided by one 
> ohm)."This gives us a direct short circuit across the battery, so 1 
> amp flows. Since there is no resistance in the load, all of the 
> power, 1 watt, is dissipated inside the battery by it's internal 
> resistance.
> "But here again, the battery isn't delivering any power, because 
> the voltage across your zero ohm load is zero."
> I suppose you could say that the battery IS delivering 1 watt of 
> power. BUT the load isn't consuming it, the battery is. So what 
> good does it do for us. 
> 
> HTH
> 
> 
> 
> --
> Stay Charged!
> Hump
> I-5, Blossvale NY
> 
> 
> 
> On Mon, 19 Nov 2007 15:13:08 -0600, "FRED JEANETTE MERTENS" 


> > <[email protected]> wrote:
> > > his amperage out of the battery with 1 ohm resistance and 1 volt
> > ==1 amp
> > > ----- Original Message -----
> ...


----------

