watt hours in series vs parallel - discuss

I have seen some discussion on this in old threads, but nothing definitive....

excuse the unrealistic numbers but metric seems to work well here.......

If I have 10 x 10v 10ah cells, I can wire them as 100v 10ah.....or as 50v 20ah...either way I have 1000 watt hours available.

for either configuration, the same work will be done in theory say 10amps @ 100v will give the same output as 20 amps @ 50v - but in reality, what happens?

Obviously the speed & winding needs to be considered.....the motors most efficient "sweet spot", presumably right before it reaches saturation.

If the motor's most efficient speed matches the voltage correctly then I would assume that would be best, but if you double the voltage then you have to work with 50% duty cycle and it's consequent losses......I don't see any motors rated at 100v for example, so this has to be considered.....

The controller also has a sweet spot - 100% duty cycle I presume. So for most scenarios this would seem to favor the lower voltage. But as I understand it higher volt / lower amp situations have less heating losses....so what wins out?

What Math ???

Range won't vary much at all if you run at the same speed with both. With 100V, you have the option of shortening your range for increased speed and power. That's the option I chose, and I'm perfectly happy with it.

You hit the sweet spot. It's all going to be the same wh capacity. The question is, for that particular duty, using a particular motor, controller, and voltage, are you in the sweet spot.

Wh/mi on the CA tells you if you are, or a thermometer in the motor will let you know. Make less heat, you are in the sweet spot. I have two bikes set up for riding in dirt. Both mongoose blackcombs. One is a 48v 20 amp controller running a 2810 9c. It gets around 48 wh/mi. My other bike running 72v 20 amp controller and 2812 motor tends to get about 43 wh/ mi. So for sure, the 72v motor and battery are in the sweet spot a bit better for those particular dirt trails. Change the ride to a different type of trail, or maybe a road instead of a trail, and you have every possiblity the 48v bike gets better wh/mi.

All other things being equal, I prefer running at a higher voltage. Doing so I can use smaller/thinner wires. Using your numbers: 1,000W. At 50V the current is 20A, at 100V it's only 10A, thus I can use wires half the size.

Perfect logic if your speed is ideal at 100v.

I forget about stuff like that. I pick a speed and type of duty. Such as 5 mph up steep dirt, vs 25 mph on flat pavement. What ever the need is for that particular bike. Then select a motor and controller to do the job. Then select a battery c rate, Ah size, and voltage that meets the needs that have been detremined by the motor and controller you selected. At that point you can look at wires and see if they need to be fatter than what came with the motor or not. Selecting a voltage just to use a particular wire makes no sense at all to me. You want to select a voltage that gets what you want done. Then wire with whatever the set up needs.

dogman said:

Selecting a voltage just to use a particular wire makes no sense at all to me.

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That's why I began with "All other things being equal,"

For a given amount of power you need from the pack, the cells feel exactly the same load if the pack is the same amount of energy.

So if you have a 40ah 25v pack or a 10ah 100v pack (both identical energy packs), if you need to dear 1000w, its 1C loading on the cells either way. So each cell has no idea if its in the 100v pack or the 25v pack.


Interestingly, the phase current only depends on the rate the motor is spinning and the power entering it. So, your 100v pack and your 25v pack both dumping 500w into the motor at the same speed both are placing exactly the same phase current on the motor and need the same size wires. The higher voltage pack can use smaller battery cables, but places much more load on the caps in the controller who have to Buck down the voltage to send the motor the current it needs.


Best efficiency for the whole system comes from picking the Max speed you wish to travel, multiply it by roughly 1.2x, and divide it by the KV of the motor to end up with ideal pack voltage. Higher pack voltage would be worse, lower pack voltage would be worse.

Anyone know the kv of the 9c 9x7?
I prefer to series the battery for higher voltage and dial down the amps on the CA for commuting and long rides.
1.) I can go faster or slower depending on the weather as sometimes a strong headwind can really eat up your wh's
2.) Less wires to hook/unhook.
I seem to get good wh/miles at 68v 20ah dialed down to 16-18 amps or 1000-1200 watts.
8) 8) 8) 8) 8)

999zip999 said:

What Math ???

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The math that I thought might be forthcoming in a discussion about switching losses, efficiency, heating etc.

Luke, where does the 1.2 come into it, is this taking into account losses throughout the whole system?

I have seen somewhere that losses through switching are something like 15-20%, but I don't know if that is an average across the full range of duty cycle or what.

So bearing in mind that the wattage hitting the motor is the same whether 50v full throttle or 100v 50% throttle, then any losses on the motor side of the controller will be the same? and the only difference will be the switching losses in the controller?

liveforphysics said:

Interestingly, the phase current only depends on the rate the motor is spinning and the power entering it. So, your 100v pack and your 25v pack both dumping 500w into the motor at the same speed both are placing exactly the same phase current on the motor and need the same size wires. The higher voltage pack can use smaller battery cables, but places much more load on the caps in the controller who have to Buck down the voltage to send the motor the current it needs.

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How do you figure this one ? For the same speed, if the back EMF of one motor is 25V and the back EMF of the other one is 100V then the current through the first motor will be 20 A and only 5A for second one....? Difference is a factor 4 in amount of windings...

jonathanm said:

Luke, where does the 1.2 come into it, is this taking into account losses throughout the whole system?

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Because at the RPM of KV*V, the motor is capable of zero torque, yet to cruise at your intended top speed requires torque. The 20% shifts you back into a range which you're capable of outputting torque.

Lebowski said:

liveforphysics said:

Interestingly, the phase current only depends on the rate the motor is spinning and the power entering it. So, your 100v pack and your 25v pack both dumping 500w into the motor at the same speed both are placing exactly the same phase current on the motor and need the same size wires. The higher voltage pack can use smaller battery cables, but places much more load on the caps in the controller who have to Buck down the voltage to send the motor the current it needs.

Click to expand...

How do you figure this one ? For the same speed, if the back EMF of one motor is 25V and the back EMF of the other one is 100V then the current through the first motor will be 20 A and only 5A for second one....? Difference is a factor 4 in amount of windings...

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The BEMF is generated by the motor for a given speed is in no way effected by the battery voltage.

Re-read what I said. The same motor at the same speed makes the same BEMF, and if you're feeding the same power into it, it makes no difference if you started from a 10,000v pack or a 50v pack, if the same motor is at the same speed and being fed the same power, it's phase current will be identical, and it makes no difference what the battery current differences happened to be.

Ok, I see your point, my reply was based on two different motors (the 100V one having 4 times more windings as the 25V one to
achieve same rpm, based on the simple PWM controllers everyone seems to use)

I never understand half of this stuff. But I am just barely smart enough to get this. Say you have two combinations. Maybe different motors, but similar wattages delivered by batteries at different voltages.

If one runs hotter than the other on a particular kind of riding, you want to run the cooler one. If you are making heat, you are not making motion. The more volts is better idea may work at first, but eventually you get to where you are just making motors hotter.

Excellent thread

I find I'll be in a position soon to debate going 20s2p with 5Ah lipo packs or 10s4p. At the moment 10s2p has worked remarkably well with a BMC V3, Lyen 12FET.

I find at the moment, my voltage and capacity drain very closely together... in other words, when my capacity is at 35%, my low voltage limit is almost reached, perfect... time to recharge. But, if I double my Ah, will I reach low voltage long before I get close to 35% capacity, thereby negating the reason for going parallel. Or will the battery hold its voltage due to the larger capacity?

Battery will hit the same voltage at 35%. But that's 35% of twice as much battery. So for a given distance, say 10 miles for example, you'd see a higher voltage at 10 miles traveled than you used to.

Put more simply, your range will double at least.

Perhaps more than double because the c rate gets lower. So at a lower c rate your battery capacity should increase a bit. And the portion of the ride that is at the highest voltage will last longer too.

liveforphysics said:

Best efficiency for the whole system comes from picking the Max speed you wish to travel, multiply it by roughly 1.2x, and divide it by the KV of the motor to end up with ideal pack voltage. Higher pack voltage would be worse, lower pack voltage would be worse.

Click to expand...

What's the KV of a 9C 6x10?

Thank you,

Snath

That calculation doesn't work..... Say I want to go 30mph..... 30*1.2=36 36/170kv= .2117 ??????? What?? Why would I want to use .2117 volts to run at 30mph. I would need 10,000 amps!!!! To make the 2117 watts of power...

What am I missing in the calculation?