# volts question



## dladd (Jun 1, 2011)

So, I'm still in the very early stages of planning here, I have a basic question on batteries. Say for example I have physical space for 12 batteries. What are the performance differences of voltage choice?

Here are the examples as I see them...

GC2 (6v) 240ah, 66# = 72v battery bank
GC8 (8v) 170ah, 63# = 96v battery bank
group 27/31 (12v) 110ah, 65# = 144v battery bank

So, each of these weighs the same, and takes up the same space. My prior experience with batteries is RV usage, where we basically just compare overall lead weights  So, in my past experience, all of these combo's would be about equal in usage. BUT, in that usage it's all about low amp draw, and making them last a long time. For example, my dual T105's last me about a week when camping. I assume the high current of an EV motor changes the game here.

So, what determines target pack voltage?

fwiw, my targets are fairly modest, I drive ~25 miles/day, maybe 1-2 miles of those are on the freeway (55mph would be fine, there's usually a little traffic anyway). I'm looking at converting a subcompact size car that can seat 4 (I drive my kids to school).


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## taken by aliens (May 26, 2011)

higher the voltage means less current to run... so you can run cheaper wire that doesnt have to be the width of your finger.

Power(watts) = Voltage x Current

1 watt = 0.0013596216 Horsepower
1000 watt = 1.3596216 Horsepower

144V bat pack with 1000A controller = 144000 Watts = 195.7855 Horsepower
96V bat pack with 1000A controller = 96000 Watts = 130.5236 Horsepower
72V bat pack with 1000A controller = 72000 Watts = 97.8927 Horsepower


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## EVfun (Mar 14, 2010)

Group 27 and group 31 batteries are more of a Marine "deep cycle" battery than a traction deep cycle battery. They tend to have a short life as a traction battery.


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## dladd (Jun 1, 2011)

So, simplified down a bit, my question basically is what is the functional difference between these two battery packs:

12x GC2 (6v) 220ah, 66# = 72v battery bank
12x GC8 (8v) 170ah, 63# = 96v battery bank

Both have pretty much the same energy (voltage x Ah) and are the same weight.

I assume the 96v bank would offer a higher top speed at the expense of distance, but what if the higher speed was not required?

If I were to drive both of these cars at a steady 45mph, which would go further? Are they the same?


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## dtbaker (Jan 5, 2008)

dladd said:


> So, simplified down a bit, my question basically is what is the functional difference between these two battery packs:
> 
> 12x GC2 (6v) 220ah, 66# = 72v battery bank
> 12x GC8 (8v) 170ah, 63# = 96v battery bank
> ...



range would be 'about the same', but acceleration and top speed would be better with the 8v pack. With the lower pack voltage, you'll draw more amps to get the same performance, which exacerbates the Peukart effect in floodies, degrading the higher ah rating of the 6v down closer to the 8v.

unless you are building a truck with lots of capacity for carrying the extra space and weight for the 6v, 8v golf cart batteries like the us8vgchc or trojans are not a bad compromise for performance and range.

unless you are only planning a low-speed NEV, I would suggest that a 96v system with 8v batteries is about the minimum you want to operate safely in average traffic even in a small light car for typical suburban driving.

p.s. If you go 96v, I have a couple items in the classifieds, and on ebay right now you might want to snap up. charger, dc-dc, flow-rite watering system, and instrumentation pod.


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## taken by aliens (May 26, 2011)

Amp Hours or ah indicates the capacity of the battery, or the total energy storage it is capable of.

12 bats at 220ah = 2640ah
12 bats at 170ah = 2040ah

the higher the total number of ah your battery pack is the more distance you will be able to drive on a full charge.

for example: if you need 1 Amp per a second to run your motor, and you have 10 batteries they will each supply about 1/10th of an Amp to total up to the 1 Amp needed. If you increase the number of batteries to 100 then each will supply about 1/100th of an Amp to total up to the 1 Amp needed. 1/100th of an Amp discharge is less then 1/10th of an Amp discharge so your batteries wont depleat their stored energy as fast, and therefore will get you farther drive distance.


Now driving 45mph at a Discharge of 1 Coulumb (1 Amp per second)...

3600 seconds = 1 hour

2640/3600 = 0.7333
0.7333 x 45mph = 33 miles

2040/3600 = 0.5666
0.5666 x 45mph = 25.5 miles


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## dtbaker (Jan 5, 2008)

dladd said:


> fwiw, my targets are fairly modest, I drive ~25 miles/day, maybe 1-2 miles of those are on the freeway (55mph would be fine, there's usually a little traffic anyway). I'm looking at converting a subcompact size car that can seat 4 (I drive my kids to school).


your needs sound very similar to mine.... I started w/ a pack of 12 x8v in a Suzuki Swift (Geo Metro). That is physically about all you can fit, and all the suspension will handle. I have upgraded to 120v x 100ah Thunderskies, and have been way happier with performance and range. 

The lead 96v pack was fine for almost 18 months, but then the range starts fading pretty fast. If you look at replacing a lead pack every 18 months to retain a safe 20-25 miles, saving up for Lithium the first time looks better. 

have a read thru my site (link in sig....)


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## dtbaker (Jan 5, 2008)

taken by aliens said:


> Amp Hours or ah indicates the capacity of the battery, or the total energy storage it is capable of.
> 
> 12 bats at 220ah = 2640ah
> 12 bats at 170ah = 2040ah


these ratings only apply at a specific discharge rate (way below normal EV needs). You will find that when discharging at a 1hr rate rather than the published 20hr rate that peukarts effect with floodies is SIGNIFICANT.

for example. us8vgchc 8v floodies have a 20hr rating of 185ah. HOWEVER, at 100 amp discharge (aprox pull required to hold 45 mph with a 96v system) the effective rating is closer to maybe 100ah.

hence... at a lower system voltage, more ah in a 6v system may not get you any further. you cannot use the 20hr ah ratings to estimate range in an EV.


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## taken by aliens (May 26, 2011)

opps... yea please take his consideration into account, my numbers are high estimations.


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## tomofreno (Mar 3, 2009)

Just want to clarify a couple things:



> Amp Hours or ah indicates the capacity of the battery, or the total energy storage it is capable of.


 Ah is a unit of charge, Wh, product of Ah and V, is a unit of energy. The Ah rating of a battery is typically it's "20 hr rate", meaning you will get that much Ah of charge out of it if you discharge it at a current level at which it will take 20 hours to fully discharge the battery. If you discharge it at a higher current you will get less Ah out of it. That's the Peukert effect dt mentioned.



> if you need 1 Amp per a second to run your motor...


 Amp per second? You just need charge per time, or current, through the motor to run it, which requires voltage applied to the input. 

The torque the motor produces is roughly proportional to the current. The power output at the shaft depends on current and motor rpm. The current you can "push" through the motor at higher rpm, depends on voltage, V. Higher voltage gives higher peak power using a given controller, so higher max vehicle speed as was already said.

Since the originator of the thread is used to using batteries at low discharge currents, it would be good to review the Peukert effect mentioned to understand the impact it has on "effective" battery capacity as dt mentioned, since it is reduces capacity considerably at high discharge currents. This effect is extremely small for LiFePO4 cells, but they are more expensive than lead acid batts - much longer cycle life though.


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## major (Apr 4, 2008)

Hello Mr. taken by aliens,

Your comments here are a bit off the wall. I'll try to explain:



taken by aliens said:


> Amp Hours or ah indicates the capacity of the battery, or the total energy storage it is capable of.


Amp Hours (Ah) does indicate the capacity of the battery. It is the charge. It can also be expressed as Coulombs. 1 Coulomb = 1 Ampere times 1 second. 1C = 1A * 1s. 

But, contrary to your statement, it is not the "total energy". Energy must include the potential, or voltage. It is generally expressed in terms of watt hours or kilowatt hours, kWh. The basic unit is the Joule. 1J = 1V * 1A * 1s. And 1J = 1Ws.



> 12 bats at 220ah = 2640ah
> 12 bats at 170ah = 2040ah


This is incorrect because the dladd is obviously connecting the batteries in series. 



dladd said:


> 12x GC2 (6v) 220ah, 66# = 72v battery bank,
> 12x GC8 (8v) 170ah, 63# = 96v battery bank


When batteries are connected in series, the capacity (Ah) does not change. The twelve 220Ah batteries in series have a total of 220Ah.



> for example: if you need 1 Amp per a second to run your motor,


What physical quantity is represented by a "Amp per second"?

Your example is way off base and makes no sense.

Regards,

major


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## major (Apr 4, 2008)

taken by aliens said:


> for example: if you need 1 Amp per a second to run your motor, and you have 10 batteries they will each supply about 1/10th of an Amp to total up to the 1 Amp needed. If you increase the number of batteries to 100 then each will supply about 1/100th of an Amp to total up to the 1 Amp needed. 1/100th of an Amp discharge is less then 1/10th of an Amp discharge so your batteries wont depleat their stored energy as fast, and therefore will get you farther drive distance.
> 
> 
> Now driving 45mph at a Discharge of *1 Coulumb (1 Amp per second)...*
> ...


Your edit does not help much  One Ampere is one Coulomb per second, not the other way around.


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## dtbaker (Jan 5, 2008)

no need to get lost in the numbers.... 

in a given amount of physical space and weight of lead between a 6v and 8v pack, the higher voltage will give better performance and very close to the same range.

if you have a truck and can have MORE space and weight for the same VOLTAGE with 6v, you'll have better range than same voltage 8v pack.

the same volume of prismatic Li cells will be 60% lighter, 30% higher voltage, yield 20% better range, cost 3x and last probably 5x as long as the lead.

ballpark....


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## dladd (Jun 1, 2011)

Thanks for the responses, lets see if I'm getting it... 

from what I gather, in a hypothetical lead acid system where one pack is 72v and one is 96v, but both have identical Ah x V numbers, the 96v pack would go a little further, since the higher voltage means a lower current, so there is less of a Peukert effect. Is that right?


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## major (Apr 4, 2008)

dladd said:


> Thanks for the responses, lets see if I'm getting it...
> 
> from what I gather, in a hypothetical lead acid system where one pack is 72v and one is 96v, but both have identical Ah x V numbers, the 96v pack would go a little further, since the higher voltage means a lower current, so there is less of a Peukert effect. Is that right?


Hi dladd,

I've seen this stated a number of times, but do not agree. I think the Peukert effect would be very close. It is true for a battery of a given Ah the due to Peukert effect, you will get less Ah at the higher current. But because your lower voltage battery will have a higher Ah rating, it will have a different value of C used in the Peukert calculation and therefore should deliver the same energy at the same power as the higher V lower Ah battery. In other words, both batteries would be working at the same C rates for equal power, and that means Peukert effect would be the same for each.

Regards,

major


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## dtbaker (Jan 5, 2008)

dladd said:


> Thanks for the responses, lets see if I'm getting it...
> 
> from what I gather, in a hypothetical lead acid system where one pack is 72v and one is 96v, but both have identical Ah x V numbers, the 96v pack would go a little further, since the higher voltage means a lower current, so there is less of a Peukert effect. Is that right?


I think you've got it.
If the (20-hr rating)Ah x V were identical, I would say that the EFFECTIVE 1-hr ah rating for the 6v system would suffer slightly more from peukarts (internal resistance) because it would be operating at a higher amp draw.

what you have to keep in mind is that the range of a lead system starts degrading pretty quickly after about 500 cycles, and that the last 30% of the lead pack will have significant voltage sag, which is very noticeable in performance and exacerbates the peukarts as the voltage drops the amps must go up to maintain same output. The last quarter of the charge can sag so bad that the amps needed to maintain moderate performance exceed your controller rating and the car just feels really limp.

save up your pennies and go with Lithium! CALB is probably the best value at this point, especially since the supply of Thundersky/Winston seems limited with less than reliable delivery from several wholesalers.


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## GerhardRP (Nov 17, 2009)

major said:


> Hi dladd,
> 
> I've seen this stated a number of times, but do not agree. I think the Peukert effect would be very close. It is true for a battery of a given Ah the due to Peukert effect, you will get less Ah at the higher current. But because your lower voltage battery will have a higher Ah rating, it will have a different value of C used in the Peukert calculation and therefore should deliver the same energy at the same power as the higher V lower Ah battery. In other words, both batteries would be working at the same C rates for equal power, and that means Peukert effect would be the same for each.
> 
> ...


A year or so ago the question of range came up in this thread: http://www.diyelectriccar.com/forums/showthread.php/trying-diagnose-my-poor-range-49667p2.html
I did some modelling of instantaneous Peukert effect on the data dump provided to me. I then did "what-ifs" using different batteries.
Range when going to 70% DOD.
8 Volt energizers ... Peukert factor 1.28 ... range 20.2
8 volt US battery 8VGCHC ... Peukert 1.21 ... range 28.2 [best 8V]
6 volt Us battery 145-xc ... Peukert 1.15 ... range 35.4
It actually seems that the improvement in Peukert factor within a fixed volume of the battery favors 6V over 8V.
Overall, look for the best Peukert factor for the battery.
Gerhard


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## tomofreno (Mar 3, 2009)

I made up this graph back in 2008 when trying to see if I could achieve the range I desired with lead in the conversion I was planning. It is based on the equations given at http://www.smartgauge.co.uk/peukert.html
I bought LiFePO4 cells.
View attachment Graph, batt capacity vs discharge current.pdf


Edit: Sorry, I noticed I cut off some of the legend. After T145 is T125 (the line just below the T145 line), T890 (the line just below the T125 line), and Trojan31-Gel (the line coincident with with the Odessey 2150 line at bottom).

Look for low Peukert value with high capacity - the 6V batts.


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## Gas Free (May 31, 2011)

dtbaker said:


> range would be 'about the same', but acceleration and top speed would be better with the 8v pack. With the lower pack voltage, you'll draw more amps to get the same performance, which exacerbates the Peukart effect in floodies, degrading the higher ah rating of the 6v down closer to the 8v.
> 
> unless you are building a truck with lots of capacity for carrying the extra space and weight for the 6v, 8v golf cart batteries like the us8vgchc or trojans are not a bad compromise for performance and range.
> 
> ...


I am looking for a reasonable 96v charger. Did you say you have one for sale?
Tom
[email protected]


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