How many (V) per (1000w) Output to you drop?

DogDipstick

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How many volts should you drop with the average ebike amperage draw?

I drop about 1.25v for a 1000w draw on my ebike. I have no idea, but is this good? What does an average 16Ah 18650 pack draw? It will drop ~2v for 2000w, 3v for 3000, and so on...

Mine is 20s, 82.5v, Chevy Volt cells.

Is this good? How many volts do you ( on average) drop for every 1000w output? Curious here. Never had an 18650 pack.
 

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I bet those are very healthy numbers there!

But to compare apples, best to use a **percentage** of the resting voltage, rather than an absolute voltage delta.

And the load as well, use C-rate, since a very high capacity Ah pack will suffer a radically lower V sag.

Where you measure may bring wire gauge and connection quality issues in as well, best right at the pack.

 
a few threads with related info
https://endless-sphere.com/forums/search.php?keywords=voltage+drop&terms=all&author=&sc=1&sf=titleonly&sr=topics&sk=t&sd=d&st=0&ch=300&t=0&submit=Search
 
Wow that is some signifigant data. Seems like they drop alot.


Someone told me my "1000w/V was weak.

I guess average IR read on the CA3 is 50-100mOh? Ranging around there? Where is the formula the CA3 uses to calculate that exactly? Anybody know? How the CA3 calculates that? Its important to me.
 
john61ct said:
I bet those are very healthy numbers there!

But to compare apples, best to use a **percentage** of the resting voltage, rather than an absolute voltage delta.

And the load as well, use C-rate, since a very high capacity Ah pack will suffer a radically lower V sag.

Where you measure may bring wire gauge and connection quality issues in as well, best right at the pack.


I kinda figured voltage was directly proportional to power.

Power is important to me. Im not quite sure what a "percentage " of resting voltage is, do you mean SOC? I am not trying to figure SOC here, I am figuring how many volts the average ebike pack ( 36-84v) drops, droops, sags, when a certain ( directly proportional to the v.drop ) power is output.... Based on an average ebike ( 30-60A)...

This way you can figure the average, given ebiker "A" runs at 48v and 10C discharge, or "B" whom is at 80v and 6.5 C.. whatever, regardless of Crate, or of certain installation size in Ah.. these two things are great variables in the ebiking world, given the cells that are out there..

Power is a standard I understand. All of us are subject to similar loads, on ebikes, round these voltages, and speeds. This requires a congruent power to move them. This is the similitude between any two ebikers, and the drop they are having will show the average power loss in average cells.
 
V sag under load per 1000w is meaningless information unless you know the size in Ah or wh of the battery. I'm not familiar with what size those chevy volt cells are. is your pack 16 ah?

1v sag for 1000w load would be very good for a 500wh pack, like some of the typical shark packs of 18650's. The cheapo packs can sag 6v or more, in that size, with that load. The cheap one that burned my house was 1000wh, and sagged 6-8v under a 1000w load. this is new charged, half discharged, it would sag even more.

But if the pack was 2000wh, some pretty shitty cells would be able to sag only 1v per 1000w. lower the c rate enough, and sag pretty much vanishes under a moderate 1000w,
 
dogman dan said:
V sag under load per 1000w is meaningless information unless you know the size in Ah or wh of the battery. I'm not familiar with what size those chevy volt cells are. is your pack 16 ah?

1v sag for 1000w load would be very good for a 500wh pack, like some of the typical shark packs of 18650's. The cheapo packs can sag 6v or more, in that size, with that load. The cheap one that burned my house was 1000wh, and sagged 6-8v under a 1000w load. this is new charged, half discharged, it would sag even more.

But if the pack was 2000wh, some pretty shitty cells would be able to sag only 1v per 1000w. lower the c rate enough, and sag pretty much vanishes under a moderate 1000w,

I disagree completely. It is very proportional when voltage is directly related to power? We could do it with weight. That is W/Kg, a common known number to some... in batteries.


Yeah well, I asked someone the same question, and they said the have a 64Ah pack that sags 5-7v on a 60A draw. That is 1.75 volts dropped(per 1000w pulled). And they think its good? My homemade battery ( of howevermany ah) tromps their homebuild.

You said a Cheapo 1000wh pack sags 6-8v/1000W? Thats 8X worse than mine! Good comparison... same size, same Ah approx, ect.

Mine only sags 1v, for aprosximatly the same energy density that you had probably ( or close to it, within 5-6 lbs).... 8v/1000w I horrid! I couldn't imagine pulling 7000w.....

.....that would sag 56v by the time you hit it with the same power as I! Lol No wonder it was the center of a bad thing happening.... Those cells probally heated up on any good amp draw!

Hell, If I do a current draw on datalog, with a single 4.2v cell.... It will still drop 1v/1000W output? It is a 1000w capable cell. I coudl datalog that. You could do the same test to an 18650 ...

Dont tell me power output and voltage are not what a cell is designed around. This is an important number. I see how you guys want me to liken it to Ah capacity, but I am just figurine the average that affects e bikers, given we all have many diffent designs and use many diffent cells. .

Yeah My cells are 15.5Ah on the dot... average ebike size, be it 48, 52, or 72v....
 
You voltage drop is proportional to your power out.

On any battery, any size, anywhere, anytime. Always, and Forever, it has been this way, since Ohm wrote the law. Period.

Interesting read: https://en.wikipedia.org/wiki/Maximum_power_transfer_theorem
 
DogDipstick said:
john61ct said:
But to compare apples, best to use a **percentage** of the resting voltage, rather than an absolute voltage delta.
Im not quite sure what a "percentage " of resting voltage is, do you mean SOC? I am not trying to figure SOC here, I am figuring how many volts the average ebike pack ( 36-84v) drops, droops, sags, when a certain ( directly proportional to the v.drop ) power is output.... Based on an average ebike ( 30-60A)...
Averages aren't helpful in this case, precision is.

Say a 2V drop, if a 36V battery system that is a 6% decline, but for an 84V one it's only 2%.

And power is not helpful as a causal factor.

Amps is better, but to get proportional need to use C-rate.

So a 1000W draw for the 36V battery is 28A, or 1.75C for a 16Ah pack.

But a 1000W draw for the 84V battery is 11.9A, or 0.74C for a 16Ah pack.

V sag is a function of resistance, the lower the resistance at a given current rate, SoC and temperature, the lower the percentage V drop.

This is how members with good testing gear arrive at the "real life" C Rating for different model batteries, look at V sag and rise in temperature

a vendor might claim 30C but in reality the battery is only good for say 10C keeping voltage reasonable and not overheating.

 
Say a 2V drop, if a 36V battery system that is a 6% decline, but for an 84V one it's only 2%.

And power is not helpful as a causal factor

EXACTLY!

Same amp draw, ( average for ebikers,)( say, 10A or 100A) and the POWER comes into play...

10A, with a 2v sag on a 84v pack is losing so much...2%? Right? So it is 840w- 20w sagging to 82v... outputs 820w... It is sagging 20w...

10A, with a 2v sag, on a 36v pack, is 6%, or 20w also.

Both packs sag . However, the 36v pack sags ( 2v) 20w for every 360 watts output ( about 6v sag for every 1000w out) , where the 84v sags 2v for 840w power... It is a better battery. ( Only 2.2 v/1000w)..... The power is in the voltage... Yes the amp draw is relative to the size of the pack, and the resulting droop, but this is a good way to level the playing feild... we all have 10-20Ah batteries in the 36-84v range, I think it is a good average number to look at. When judging power density of a cell.
 
You voltage drop is proportional to your power out.

With no regard to the actual number of volts, or regard to energy transferred, in Wh... or regard to "crates".... or capacity....

Your voltage drop will ALWAYS be proportional to your power out. No matter what, period.

On any battery, any size, anywhere, anytime. Always, and Forever, it has been this way, since Ohm wrote the law. Period.

IDk just gauging the power of an average ebike pack.
 
It must be horrible if your pack drops 8 volts per 1000w ( as stated above for a 18650 pack).... I guess you hit LVC everytime when you try a "high power" pull. Lol "Shark"...

Like I said, I had a nanotec/multistar lip0 guy tell me his 64Ah 72v pack sags 7v with a 60A hit.. or 1.75v/1000w... Very less power dense than my pack if you look at it.

I also have heard the "wolf pack" sags 4-6v on a 2000w pull ... AKA loses 3v per every 1000w output... pull 4000W or more and you hit LVC...

Anyone have actual data on a 'wolfpack " or 14s4p of a good ( lol15moh) cell (4p) pack?

I bet you couldnt even do a 100A hit. You would have to limit it to 20-40A max lol. Or have 500$/Kwh cells and a huge... er... unit .. to compensate for your ..... er, uh, shortfalls.
 
You know, I know it is not a constant ratio... But I think it is a good comparison. Without knowing what the impedance, size, or voltage of the pack. I d know if it is anything rational... I bet out packs all kinda weigh the same too.... 10-15 lbs.?

I know it is not linear, looking at it. I think that's what you guys are trying to point out. Thanks for the input for sure.
 
Today I did a voltage sag test on 2 different Chevy Volt module packs...a 21s 2015 from a low mileage car that I've had in service for about a year with no notable signs of diminished capacity. At 210A the voltage sagged from 80.2V to 75.2V. The other pack is a 16s module from a 2017 Volt version2 pack that had less than 10 miles on it and I've used it very little. At 115A its voltage sagged from 64.2V to 62.0V . Note that these packs are made from complete modules, 3 cells in parallel and 2 larger cells in parallel respectively, though I'm still unclear as to what the exact nominal capacity is for the v1 and v2 Volt modules since I never run the low enough to find out and usually charge only to 4.05V/cell. I think 44ah and 50ah is about right.
 
A better measurement might be V drop per C (amps) of pack capacity. This would compensate for pack size and voltage but give you a good measurement of internal resistance.

The measurement you get will depend on all sorts of other things like temperature and state of charge.
 
Anybody know how the CA3 calculates IR from its inputs? I mean exactly? I got more to input based on your guys thoughts.. but I'll get back to that. Ive really been wondering how the CA3 does its extrapolation.
 
John in CR said:
Today I did a voltage sag test on 2 different Chevy Volt module packs...a 21s 2015 from a low mileage car that I've had in service for about a year with no notable signs of diminished capacity. At 210A the voltage sagged from 80.2V to 75.2V. The other pack is a 16s module from a 2017 Volt version2 pack that had less than 10 miles on it and I've used it very little. At 115A its voltage sagged from 64.2V to 62.0V . Note that these packs are made from complete modules, 3 cells in parallel and 2 larger cells in parallel respectively, though I'm still unclear as to what the exact nominal capacity is for the v1 and v2 Volt modules since I never run the low enough to find out and usually charge only to 4.05V/cell. I think 44ah and 50ah is about right.


See? Thats awesome. Now I know your power is kinda out there..( lets say 88v, or a blunt 21s or so) on that 21s3p Volt module.. but anyway, looking at it like this:

210A x 88v? =18.48Kw

Sag? = 5v.

V/1000w dropped? Approx 0.27V /1000w is dropped. What does this tell me? This tell me ...

.........THIS IS A STRONG frocking BATT ERY.............that straps on the bike and adds weight...

Now lets take a typical "Shark" 18650 pack of 14s4p of some cells in the battery that straps on the bike and adds (good ol negligible) weight.. Specs above... "6v or more" dropped/1000w.. This tell me...

.................THIS IS A WEAK EFFEN BATTERY.....................

Or the guy with the Lipo bricks of Nanoted, and Multi, for 64Ah of 72v... 60A is what he designed it for.... Sags 5-7v, for 4Kw, and this is ( in my wackjob formula here that is not mathematically relevant, right, or provable) 1.75V Dropped / 1000w OUTPUT, and this is..

...............A WEAKER THAN THE FIRST BUT STRONGER THAN THE ONE LISTED ABOVE...EFFIN BATTERY.......

... and finally, take my ebike, with the (X) cells in teh (Y) capacity and the (Z) weight.... cells... and its 1v/1000w sag.... 100A design system....

.................MINE IS BETTER THAN THE LAST TWO BUT NOT AS GOOD AS THE FIRST EITHER ( cause 3p of Chevy Volt cells > 1p, proven ) ...........FARGIN ( It is)....


There. My system works. We took 4 batteries, all that can be used on some sort "Ebike"... and for some (weight) rider to go some ( speed).... Some where in the world, at some temperatures ( preferably the range that supports life on earth too) .... And figured out, through my "How many V/1000W" baseless, not mathamatically correct, formula, which batteries were strongest and which were weakest, based only on how many volts are dropped per a certain power output.

haha Tolt you it works.

Anyway, Thankyou for that Volt data, John in Ct. That stuff is priceless to me, even if it doesn't come toan fruition in my strange way of loooking at this, I know your data is empirical and accurate. I appreciate taking the time to look at this for me.

So.. Another question... I said above ( and repeated, like I knew for sure) that " Voltage is proportional to power"...

Is that true? I mean, its not linear.... I think am wrong, in thinking that....
 
Power is Watts

Watts are Amps (flow rate) x Volts (pressure)

So yes **if** the Amps rate is held constant

aka CC control

then increasing Volts will proportionally increase the resulting power.
 
Wow, really set you off dog. I never said its not proportional. I said if you simply state I drop 1v per 1000w, that's not all the data you need for comparison. One volt under load is great for a 500wh pack, but not exceptional for a 5000wh pack.

Comparisons based on incomplete data is junk science. So I was asking for your data, what size battery is your pack?

Yeah,, 8v sag for 1000wh size pack did suck.. Sucked even more when that pack burned my house down. And yeah, I totally agree it could not have pulled any more than that 1000w. But,,, Its typical of 18650 packs made from the cheapest, shit cells.

The 18650 pack I have now, from luna, does waaaaaaaay better.
 
DogDipkid?
what Volt cells do you use in your pack?
there are first and second GEN cells used in Volt.
I ride on 11S2P secondGen cells.
 
miro13car said:
DogDipkid?
what Volt cells do you use in your pack?
there are first and second GEN cells used in Volt.
I ride on 11S2P secondGen cells.

I have seen ( dissembled myself) Gen 1 cells. I have found, LGXP1. P1.5, and P1.7 cells in these. The P1 datalogged at 13Ah repeatable ( old? Idk.) 2012-13 car.

The P1.5 ('14-'15) and the P1.7 ('15? ) datalogged 15.4Ah-15.6 Ah, every one.

I have 20 ( soon to be 21) LGX P1.5B cells in my bike right now, from '14-15. Ir 1.2-1.4 mOh and great output. 15 mile range at 100Wh/mile, 40 mile range at 30Wh/mile. Ish.

I have never had, tested, or datalogged a LGX P2.5B, (? correct cell name?) from the Gen 2. These are ( your Gen2) are 24.5Ah from datasheet.

I got em from Art at greentec.
 
Good sag then, for 15 ah or so. I figured they were a quality cell with low IR, but I had no idea if they were 10 ah, or 50ah.

Everybody else reading knew, but I didn't .
 
dogman dan said:
Good sag then, for 15 ah or so. I figured they were a quality cell with low IR, but I had no idea if they were 10 ah, or 50ah.

Everybody else reading knew, but I didn't .

No prob I know Im a hack and get rammy, sorry I dint mention it. I am about to go 21s and see if there is much differences.

So... Exactly how many V per 1000w Output does a Luna pack do of good cells, in your average?
 
No pun intended, but this thread appears to be a dog's breakfast. The concept is extremely simple, but this thread has confused me through REALLY weird expressions, like "voltage drop per 1000w".

When it boils down to it, the "voltage drop per 1000w", is nothing more than internal resistance. You can do all the maths you want behind it to convert the units, but the universally accepted one is:

DCIR=(V1-V2)/(I2-I1)

That means, as long as you know any four of the five values in that formula, you can calculate the fifth. And yes, since you have both voltage and current, if you want to convert half the equation to watts instead, you can calculate the "voltage drop per 1000w" :shock:

Since the Cycle Analyst has access to the four values on the right of the equation, it can calculate the DCIR.

As for comparing apples with apples, batteries follow Ohm's law in terms of series and parallel resistors as well, so:

1. As you put cells in series, the internal resistance of the battery will increase. This means that higher voltage batteries will drop more voltage on the same load - But only as a real number - As a percentage, it will drop the same (In a perfect resistor type situation, not real life)

2. As you put cells in parallel, the internal resistance of the battery will decrease. As an extension to point 2 - The larger capacity the cell, all else being equal, the lower the internal resistance (After all, a bigger cell is really just more smaller cells in parallel...)

Given the two above, a 100v, 1Ah battery should sag twice the number of volts, than a 50v, 2Ah battery built with the same cells. But as a percentage, they should be identical (But again, it won't be, because there are connectors, changes in temp, etc).

Anyway, in Li-Ion/LiPo:
3mOhms in a 10Ah cell would be decent, but nothing to write home about.
3mOhms in a 1Ah cell would be pretty impressive.
3mOhms in a 100Ah cell would be somewhat disappointing.

The Multistars were about that (3mOhms on a 10Ah). The Nanotechs from memory were closer to high 1s per cell. You can probably calculate your own DCIR to see how you compare.

Hope that makes any confusion clearer. Because I'm still confused about why someone would want DCIR expressed as volts per 1000W... Especially since you'd have to convert back to compare battery packs of different voltages... But anyway... Maybe I'm the crazy one and this one makes sense to everyone else.
 
Right. Now that I am no longer at work, let me give you a much simpler and easier explanation:

Imagine you have some 5v imaginary cells of 10mOhm internal resistance. Imaginary because round numbers are easier to work with. If you had 20 of these cells, your options are:

A. 10S2P for a 50v battery; or
B. 20S1P for a 100v battery.

The internal resistance of the 50v battery is:
10S1P = 10mOhm x 10 = 100mOhm or 0.1Ohm
2P = 1/0.1 + 1/0.1 = 0.05Ohm, or 50mOhm

The internal resistance of the 100v battery is:
20S1p = 10mOhm x 20 = 200mOhm

Using the formula for DCIR:

On the 50v battery
0.05 = (50v - ?)/(20)
1 = (50v - ?)
? = 49v, or a 1 volt sag

On the 100v battery
0.2 = (100v -?)/10
2 = (100v - ?)
? = 98, or a 2 volt sag.

So expressed in the voltage drop per 1000w, the 50v battery is apparently superior. However, they both sag... 2%.

Can you see why expressing voltage sag as volts per 1000w is misleading?
 
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