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Current demand versus battery voltage


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So there is fortionately no feedback loop with the battery voltage and battery power.

Just max speed is limited by lower battery voltage (due to high currents

So that being said, if I am going at fixed speed, fixed wind resistance and fixed grade the amount of current through the Motor will be the same no matter if the battery volts is 80 or down to 50.

We can get deep in the brushless motor theory in this thread if we want, but it really isn't necessary to understand how current must vary as a function of input voltage.

Power out = Power In * efficiency.

That's all you really need to know. 

The motors in our wheels are not resistors.  They look inductive, and there are also capacitors in the power train.  So, over very short periods of time, the principle that Power out = Power In * efficiency doesn't really hold.  But, over longer periods of time (consistent with thermal time constants), that relationship must hold.  It's a very basic conservation of energy theory.

It really doesn't matter what system you are considering.  A DC/DC convertor, a motor, a house, that relationship will always hold true.  

So, if a EUC is under a fixed load, like steadily going up a hill at a constant speed, its output power is constant.   It will be the Input Power * efficiency.  How much power it's putting out will be based on what speed is set by the ESC.  The ESC will by turning currents on and off very quickly and the instantaneous currents that result in FETs, wires, and connectors is important.  But, the average currents in these elements are what's important from a thermal analysis.

Assuming the efficiency (Power Out - Power In)  is constant regardless of the input voltage.  The input power must be constant.  Input power is Input Voltage * Input Current.   So, if the input voltage is lower, the average input current must be higher.

The efficiency is NOT totally constant as a function of input voltage.  However, the efficiency change as a function of input voltage is small relative to the battery voltage impact on the input current.

These fundamental principles are used of power and thermal analysis of virtually all power electronic systems.   Let's debate this, before things get confused with the additional details on how the ESC controls motor speed and what that means for associate FET and motor winding currents. 

Probably, the failure to differentiate between current when devices are on, versus their time average current is the source of much of the misunderstanding about what happens when battery voltage gets lower and it might help to dig into those details.  But, unless you want to argue for a perpetual motion machine, Power Out = Power In * Efficiency.

 

 

 

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Also, need to clarify what current we're talking about.  I'm referring to input or battery current. The motor windings that create the magnetic fields that puil the stator around will have pretty much the same current for a given power output, regardless of the battery voltage.  However, the voltage duty cycle of the windings is set to get the speed you want.  A longer duty cycle is required when the battery voltage is lower than when it's higher.  So, a higher average battery current is required.

 

 

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9 hours ago, DaveThomasPilot said:

Also, need to clarify what current we're talking about.  I'm referring to input or battery current. The motor windings that create the magnetic fields that puil the stator around will have pretty much the same current for a given power output, regardless of the battery voltage.  However, the voltage duty cycle of the windings is set to get the speed you want.  A longer duty cycle is required when the battery voltage is lower than when it's higher.  So, a higher average battery current is required.

I just started out to gather my thoughts about  these points - i'm not really finished with it, but since you started here this topic a short summary of what i have till now:

I started out this example with roughly @Marty Backe"settings" once his cables melted: ~100kg moving up with 10 km/h a 27% slope. This should lead to a needed mechanical power of ~730W (https://drive.google.com/open?id=0B6-Hr1aVKjLLOVNCekpwRkFvbTQ).

Now i looked at the following system for the EUC: The battery (20s4p with 3,8V cell voltage) and an internal resistance (37mOhm for an LG MJ1 times 20 / 4) gets "pwmed" by a mosfet with duty_cycle d to the motor. The motor is a series circuit of the coil inductance (ignored for now - should not matter for "static" systems - i.e. once delta I/delta t gets small enough), the coil resistance (Rc = 0,1 Ohm, see http://forum.electricunicycle.org/topic/7530-coil-resistance/) and a voltage source producing U back emf (https://www.precisionmicrodrives.com/application-notes/ab-025-using-spice-to-model-dc-motors - thanks to @esaj for this link!)
The coil resistance has a big "dampening" incluence on the factor between average battery and motor current - so this is one of the main values!

For the motor i assumed a max no load speed of 55 km/h, at which it produces the 20*3,8V=76V as back EMF.

The on resistance of the mosfet and the resistance of all the cables inbetween are not considered. (1)

This leads to the following:

U_back_emf 13,8181818 V    
R_InternalBattery 0,185 Ohm    
         
  Max Average Min  
         
Ubattery 76,00 74,18 68,45 V
Ibattery 40,79 9,84 0,00 A
         
    Average    
Umotor   17,90   V
Im   40,79   A
d 0,24126725

(https://drive.google.com/open?id=0B6-Hr1aVKjLLZmlMb29fUXcyZ2M) (2)
The same with a speed of just 5 km/h lead to a motor current of 57A with a duty cycle of 17%.

The "formula set" i used in the excel sheet is by now not "finished/?determined?" - it still relies on iteration to find the solution...;(

Ubattery max is for the open circuit (mosfet disconnects) for (1-d) and Ubattery min for the closed circuit (mosfet conducts) while d (24,1% in this example) and by this U battery average = U battery max * (1-d) + U battery min * d.

Umotor = Im*Rcoil + Uback_emf which has to equal Ubattery average min * d

For Imotor exists only the average value, since i assumed that the LR-circuit averages the current so it does not change anymore for this calculation (L can be ignored). (In regard to this @esaj provided a nice link: http://i.imgur.com/DGAEEJ5.png - with that one can maybe calculate/estimate the motor current ripple and by this the torque ripple producing the whining sound?!)

further Questions/Conclusion in my mind from this results are by now:

- How do LiIon cells cope with pulsed load in general and additionaly with loads above the max continous current - the average current is well within the specification (10A continous per cell), but the max load exceeds this. For packs with only 2 cells in parallel this should be quite an horrible burden... EUCs with 1p should be out of question?! The cells in such an EUC maybe have a chance since these are normally low power EUCs with much higher coil resistance, which dampens the current multiplier?

- measurements till now showed, that KS/GW wheels report in the app more current than measured at the battery side - could be that they report the motor current as calculated here?!

- Driving up a steep incline slowly normaly is not a real "steady" constand driving - speed changes and accelerations (additional torque spikes) happen - so the worst case could be these above numbers for even lower speeds + the current needed for additional acceleration! So the average current could be even higher!

Edit: (1) also "magnetic losses" are not considered as internal friction of the motor. But this could be considered as part of the rolling resistance...

Maybe @lizardmech, @electric_vehicle_lover, @zlymex or others could chime in with their expierience and knowledge - they could maybe also make an empirical proof with their equipment? As with the numbers for an idling "low" power EUC (16s2p, 10W, v max no load 30 km/h, v 5 km/h, Rcoil=0,3 Ohm) still lead (theoretical) to an I battery average of 0,16A vs. I motor average of 0,96A (https://drive.google.com/open?id=0B6-Hr1aVKjLLZjdVQkVOWmhBLUE)

Edit2: cleaned up the excel sheet for the needed power... And recognized by this, that i just took the "potential power" instead of the total for the example... but since it is just an example, it doesn't matter too much :ph34r:

Edit3: (2)

Just to avoid misunderstandings: UMax, Imax and Umin, IMin are each in the same column one under another but Imax happens at Umin and Imin at Umax... ;)

As the sheet shows is I battery max equal to I motor average. As the "power provided" by the battery is U battery min * I battery max * d (time the mosfet conducts) and the motor is all the time supplied by Umotor average and has I motor average flowing this U motor * I motor = U battery min * I Battery max * d. Since U motor = U battery min *d we get I motor = I battery max.

Seems like i found the last relation to get rid of the iteration in the excel sheet. So they way could be paved to produce some I_motor over speed graphs for fixed power, max motor power, etc... and by this the dissipated power for the mosfets / ~30cm cable AWG14 vs AWG 16.

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41 minutes ago, Chriull said:

- How do LiIon cells cope with pulsed load in general and additionaly with loads above the max continous current - the average current is well within the specification (10A continous per cell), but the max load exceeds this. For packs with only 2 cells in parallel this should be quite an horrible burden... EUCs with 1p should be out of question?! The cells in such an EUC maybe have a chance since these are normally low power EUCs with much higher coil resistance, which dampens the curren

They normally react with a massive voltdrop to loads above the max continuous current and if the "pulse" is to long or to often they generate a lot of heat, which then can lead to chemical cell destroyment,....because of this (serious) cell producer only give a max continuous current....but no "pulse" max current.....as pulse in generell is not defined in time....

Btw....as you speak from Martys example...his Acm is 20s6p....

Eucs with 1p (GOOD ones like luffy or Ips I5) normally using cells with at least 20Amp (lg h2) or 30amp (sony vtc) or the new 20700 in the I5

 

And: Fantastic report....its hard for me to follow!

 

from a feeling i would say that KS and GW work with total different drive firmwares and ways to generate speed/torque. On some measurements on a treadmill at electro-sport it has shown that KS wheels on low voltage are loosing a lot of their max speed, while GW is able to hold this high speed down to a much lower voltage level...

but also it has shown that the max lift cut off speed on KS is nearly the same as with a load on the treadmill, while the GW wheels have a high max lift cut-off, but are not able to do this speed under a load on the treadmill.....(speak: torque seams to be higher on the KS's)

so i guess they are working a complete different way, but thats just guessing and feelings and has not the substance your mathematics have....

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6 minutes ago, KingSong69 said:

They normally react with a massive voltdrop to loads above the max continuous current and if the "pulse" is to long or to often they generate a lot of heat, which then can lead to chemical cell destroyment,....because of this (serious) cell producer only give a max continuous current....but no "pulse" max current.....as pulse in generell is not defined in time....

but pulses can be defined by repetition and duration - that's quite common for other electrical components. Since i did not see till now such specifications for li ion cells, i got the feeling that "they just don't like this at all"...

6 minutes ago, KingSong69 said:

Btw....as you speak from Martys example...his Acm is 20s6p....

Eucs with 1p (GOOD ones like luffy or Ips I5) normally using cells with at least 20Amp (lg h2) or 30amp (sony vtc) or the new 20700 in the I5

also this high current batteries show a high voltage sag if under such burdens (from what i remember from the dampfakku measurements) - but at least they "survive" it!

6 minutes ago, KingSong69 said:

And: Fantastic report....its hard for me to follow!

Thanks - i am in the process of writing together the formulas, so there is the possibility to follow the numbers - also it's maybe not really making it simpler ;)

6 minutes ago, KingSong69 said:

from a feeling i would say that KS and GW work with total different drive firmwares and ways to generate speed/torque. On some measurements on a treadmill at electro-sport it has shown that KS wheels on low voltage are loosing a lot of their max speed, while GW is able to hold this high speed down to a much lower voltage level...

Should be the speed restriction of KS on low battery implemented in firmware?

 

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9 hours ago, DaveThomasPilot said:

The motor windings that create the magnetic fields that puil the stator around will have pretty much the same current for a given power output, regardless of the battery voltage.  However, the voltage duty cycle of the windings is set to get the speed you want.  A longer duty cycle is required when the battery voltage is lower than when it's higher.  So, a higher average battery current is required.

2 hours ago, Chriull said:

Ubattery max is for the open circuit (mosfet disconnects) for (1-d) and Ubattery min for the closed circuit (mosfet conducts) while d (24,1% in this example) and by this U battery average = U battery max * (1-d) + U battery min * d.

Umotor = Im*Rcoil + Uback_emf which has to equal Ubattery average min * d

I assume thats both the same content? The motor defines by its Back EMF (which is a motor constant times speed) and the Coil resistance the Voltage wich has to be applied (Umotor). By the duty cycle d this is "down scaled" from the Battery Voltage.

The motor current is defined by Imotor = (Umotor - Uback emf) / Rcoil).

So with every Battery Voltage (1) above Umotor this state of the motor (accomplishing a certain output power while rotating with a certain speed with Umotor and Imotor) can be accomplished by different duty_cycles.

 

Edit: (1) Minimal Battery Voltage under maximum load while the "circuit is closed - mosfet conducts"

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1 minute ago, Chriull said:

Should be the speed restriction of KS on low battery implemented in firmware?

yes, but i guess they do these speed restrictions because of their different algorythm.....and not that the speed restrictions are the reason for the missing speed....there algorithms seam to not allow high speeds on low batterie voltage....while torque is always there...

 

Perhaps thats also the reason why they didn't jump the 84volt train, as it doesn't match their way of doing firmware/motor driver?

I have read somewhere they (Boss KS) said they would never do a 84volt wheel.......

I guess i will have to ask Chris from 1rad when i catch him in a good mood :-)

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22 minutes ago, Chriull said:

but pulses can be defined by repetition and duration - that's quite common for other electrical components. Since i did not see till now such specifications for li ion cells, i got the feeling that "they just don't like this at all"...

Thats a long discussion in the vaping world :-) yes, those pulses are not defined by any batterie "regulators"...

as the 18650's are used in high power vaping gear, there are always "rebranders" using for example a Lg h2 (20 Amp max cont), put them in a nice yellow new wrap, new nice name and sell them as 20 max, 40amp pulse.....

Then the "batterie experts" and tester always have to explain that this is dangerous game...as in vaping a "draw" can go from 2-10 seconds and if it works out really bad, that can go up to an blow out of the 18650....several videos of exploding e-cigarette can be found :-)

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56 minutes ago, KingSong69 said:

Thats a long discussion in the vaping world :-) yes, those pulses are not defined by any batterie "regulators"...

as the 18650's are used in high power vaping gear, there are always "rebranders" using for example a Lg h2 (20 Amp max cont), put them in a nice yellow new wrap, new nice name and sell them as 20 max, 40amp pulse.....

Then the "batterie experts" and tester always have to explain that this is dangerous game...as in vaping a "draw" can go from 2-10 seconds and if it works out really bad, that can go up to an blow out of the 18650....several videos of exploding e-cigarette can be found :-)

"Exploding batteries" in vaping gear are usually the sign of someone shorting a battery. Putting it naked in the pocket with keys and change, or in a bag. Or screwing on a tank on your hybrid mech, without checking that the positive pole actually protrudes enough to avoid having the negative pole touching the battery with the positive... boom!

Try to find a video of an exploding vape battery, where the explosion were not triggered by sheer and obvious user stupidity. I've tried, and found exactly zero such videos or reports. I did that when another vaper challenged me to do so.

Safe chemistry batteries usually vent, and then that's the end of the very uninteresting story. As long as the gasses and temps has somewhere to go, that is. But with all the stupid rebrands and clones that are floating around, the likelihood of having the wrap saying 40A, while the cell has an actual CDR of 15A or less is far too high. And the likelihood of getting an ICR-battery in place of the INR you thought you had is also high. The end result could well be a Christmas Sparkler on steroids, as the cell goes thermal and the lithium ignites.

I won't use rebrands, and I make sure the batteries I buy are genuine Sony's, Samsung's or LG's.

I routinely ask my VTC5As for 40A+ pulses, which means a voltage sag, and maybe 38A effective at an effective Voltage of 3.8 instead of 4.2. The risk for an explosion is infinitesimal even at those powers, as long as you keep your pulses under 5 seconds and let the mod cool down for 20-30 seconds between pulses. I usually take two drags of a couple of seconds, then put the mod down for a while. The batteries I use right now has about 250 cycles on them, and will be good for maybe 200 more. They will never do the 1K cycles they could have, that is the price I pay for overtaxing them.

For EUCs the parameters in play are slightly different. One thing are the sheer number of cells involved, and the limitations on venting and temperature dissipation that introduces. While I can push my single 18650 more than just a little without fear, I really, really wouldn't want to do that in a EUC power-pack. A temp runaway there, could well end in utter disaster, as the runaway cell heats up its neighbours and takes them with it into oblivion... The thought of 16 cells in a pack playing domino venting and then NOT cooling down, is quite scary.

Metal fire is brutal, and I'm quite happy to have two powder-based extinguishers at hand in my home. I bought those a few years ago because of all the battery fed electronics we have here. Two iPads, five phones, seven laptops, four of them being used daily, and plenty of vape gear.

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5 hours ago, Chriull said:

 

I just started out to gather my thoughts about  these points - i'm not really finished with it, but since you started here this topic a short summary of what i have till now:

I started out this example with roughly @Marty Backe"settings" once his cables melted: ~100kg moving up with 10 km/h a 27% slope. This should lead to a needed mechanical power of ~730W (https://drive.google.com/open?id=0B6-Hr1aVKjLLOVNCekpwRkFvbTQ).

Now i looked at the following system for the EUC: The battery (20s4p with 3,8V cell voltage) and an internal resistance (37mOhm for an LG MJ1 times 20 / 4) gets "pwmed" by a mosfet with duty_cycle d to the motor. The motor is a series circuit of the coil inductance (ignored for now - should not matter for "static" systems - i.e. once delta I/delta t gets small enough), the coil resistance (Rc = 0,1 Ohm, see http://forum.electricunicycle.org/topic/7530-coil-resistance/) and a voltage source producing U back emf (https://www.precisionmicrodrives.com/application-notes/ab-025-using-spice-to-model-dc-motors - thanks to @esaj for this link!)
The coil resistance has a big "dampening" incluence on the factor between average battery and motor current - so this is one of the main values!

For the motor i assumed a max no load speed of 55 km/h, at which it produces the 20*3,8V=76V as back EMF.

The on resistance of the mosfet and the resistance of all the cables inbetween are not considered. (1)

This leads to the following:

U_back_emf 13,8181818 V    
R_InternalBattery 0,185 Ohm    
         
  Max Average Min  
         
Ubattery 76,00 74,18 68,45 V
Ibattery 40,79 9,84 0,00 A
         
    Average    
Umotor   17,90   V
Im   40,79   A
d 0,24126725

(https://drive.google.com/open?id=0B6-Hr1aVKjLLZmlMb29fUXcyZ2M) (2)
The same with a speed of just 5 km/h lead to a motor current of 57A with a duty cycle of 17%.

The "formula set" i used in the excel sheet is by now not "finished/?determined?" - it still relies on iteration to find the solution...;(

Ubattery max is for the open circuit (mosfet disconnects) for (1-d) and Ubattery min for the closed circuit (mosfet conducts) while d (24,1% in this example) and by this U battery average = U battery max * (1-d) + U battery min * d.

Umotor = Im*Rcoil + Uback_emf which has to equal Ubattery average min * d

For Imotor exists only the average value, since i assumed that the LR-circuit averages the current so it does not change anymore for this calculation (L can be ignored). (In regard to this @esaj provided a nice link: http://i.imgur.com/DGAEEJ5.png - with that one can maybe calculate/estimate the motor current ripple and by this the torque ripple producing the whining sound?!)

further Questions/Conclusion in my mind from this results are by now:

- How do LiIon cells cope with pulsed load in general and additionaly with loads above the max continous current - the average current is well within the specification (10A continous per cell), but the max load exceeds this. For packs with only 2 cells in parallel this should be quite an horrible burden... EUCs with 1p should be out of question?! The cells in such an EUC maybe have a chance since these are normally low power EUCs with much higher coil resistance, which dampens the current multiplier?

- measurements till now showed, that KS/GW wheels report in the app more current than measured at the battery side - could be that they report the motor current as calculated here?!

- Driving up a steep incline slowly normaly is not a real "steady" constand driving - speed changes and accelerations (additional torque spikes) happen - so the worst case could be these above numbers for even lower speeds + the current needed for additional acceleration! So the average current could be even higher!

Edit: (1) also "magnetic losses" are not considered as internal friction of the motor. But this could be considered as part of the rolling resistance...

Maybe @lizardmech, @electric_vehicle_lover, @zlymex or others could chime in with their expierience and knowledge - they could maybe also make an empirical proof with their equipment? As with the numbers for an idling "low" power EUC (16s2p, 10W, v max no load 30 km/h, v 5 km/h, Rcoil=0,3 Ohm) still lead (theoretical) to an I battery average of 0,16A vs. I motor average of 0,96A (https://drive.google.com/open?id=0B6-Hr1aVKjLLZjdVQkVOWmhBLUE)

Edit2: cleaned up the excel sheet for the needed power... And recognized by this, that i just took the "potential power" instead of the total for the example... but since it is just an example, it doesn't matter too much :ph34r:

Edit3: (2)

Just to avoid misunderstandings: UMax, Imax and Umin, IMin are each in the same column one under another but Imax happens at Umin and Imin at Umax... ;)

As the sheet shows is I battery max equal to I motor average. As the "power provided" by the battery is U battery min * I battery max * d (time the mosfet conducts) and the motor is all the time supplied by Umotor average and has I motor average flowing this U motor * I motor = U battery min * I Battery max * d. Since U motor = U battery min *d we get I motor = I battery max.

Seems like i found the last relation to get rid of the iteration in the excel sheet. So they way could be paved to produce some I_motor over speed graphs for fixed power, max motor power, etc... and by this the dissipated power for the mosfets / ~30cm cable AWG14 vs AWG 16.

 

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So, how can use the spreadsheet do compare two different battery voltages with the same motor load?

I assume I change the UBattery average cell, until cell C20 is closed to zero (like the text says).  When I do that, PMotor is higher.  So, something else needs to be tweaked to get that.

Or did you do that already and I missed an associated conclusion about whether battery current is (roughly) inversely proportional to battery voltage if motor load is held constant?

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34 minutes ago, DaveThomasPilot said:

So, how can use the spreadsheet do compare two different battery voltages with the same motor load?

I assume I change the UBattery average cell, until cell C20 is closed to zero (like the text says).  When I do that, PMotor is higher.  So, something else needs to be tweaked to get that.

Or did you do that already and I missed an associated conclusion about whether battery current is (roughly) inversely proportional to battery voltage if motor load is held constant?

The sheet uses the kv ( motor constant for U back_emf = kv * rotations) by calculating U back_emf = Ubattery max / v max no load speed * v. And by this changing U battery max in the sheet changes the whole motor characteristic (kv) ;(

I changed here the sheet by inserting a seperate U battery max (cell B 10), so that U back emf (kv) gets calculated from this, and the U battery max in call b 13 can be changed as one whishes: https://drive.google.com/open?id=0B6-Hr1aVKjLLU0ZpbVV4TGhsRmM

ps.: don't forget to use excels "iteration" on cell c 21 (should get 0) by changing cell b22 after each change...

pps.: I'm not really sure if you intented your topic to go in this direction - if you feel that i captured this thread to somewhere else, i can split it to somewhere new...

Edit: i just got now that you changed U battery average - that's a calculated value and not to be changed. Ubattery max in cell b13 is the one to adopt. I should start to user different fill colors or so to mark input cells... Done

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Good thread, and good points made.

I'll put it in a different way, ignoring the details(such as PWM, 3 phases) and make a block diagram below.
EUC-block.gif.71dbe057412b7ce585a6b96d2de1468e.gif

The battery output voltage is Ub, current Ib and power Pb, these are also the input of the board.
The board output voltage is Ud, current Id and power Pd, these are also the input of the Motor.
And the output mechanical power of the motor is Pm.
That equation(of Power out = Power In * efficiency) is universal, so we have
Pd = Pb * Eb, where Eb is the efficiency of the board,
Pm = Pd * Ed, where Ed is the efficiency of the motor.

I agree very much that for certain driving conditions, Pm, Ed and Eb are more or less constant, so, if Ub drops, Ib must be increase, and  is roughly inversely proportional to Ub.

In real life, what are those parameters especially Eb and Ed?
Well, I can monitor Ub and Ib very precisely and I did measured them in many riding conditions. An app(such as Wheellog) can give Id and log it.
Generally speaking, the larger the Id, the less the Eb. If the current is doubled, the power will be quadrupled for certain resistor. This is true both for motor wires and on-resistance of the MOSFET.
In normal riding conditions(such as no wind flat road, speed of 25kmph), Id is around 10A, I can give the best estimate of Eb to be around 97%.
However, if Id is large as happened in steep slope climbing, Eb is dropping dramatically. Not only due to the on-resistance of the MOSFET, the transit state also consume a lot power, and doubling the number of MOSFET(from 6 to 12) seems not very useful(transit time may be double too because the input capacitance is also doubled). 

Yesterday, I rode up a very steep hill, I monitored Id by Wheellog that even reached to 90A, and the temperature has gone up to 70 degree C.

SkiHill.thumb.gif.5e44fecaccd9fbc20c07ad7d24dd68e2.gif


In most cases, Id is larger than Ib, sometimes twice or even larger. However, the battery is not bare this PWM spikes because there is a capacitor(two for GW) paralleled with the battery input.

As for Em, I estimate to be around 65% in normal riding conditions(for my Gotway Msuper3s+) but dropping fast at low speed.
 

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16 hours ago, zlymex said:

Yesterday, I rode up a very steep hill, I monitored Id by Wheellog that even reached to 90A, and the temperature has gone up to 70 degree C.

...

As for Em, I estimate to be around 65% in normal riding conditions(for my Gotway Msuper3s+) but dropping fast at low speed.
 

With 90A and a coil resistance of ~0,1 Ohm that would make a motor loss of 810W only from this part - not taking into account friction/"magnetic"/etc losses...

Edit: and a bit more is burned within the battery packs?! Msuper3s+ also has a 20s6p configuration? Would make ~0,12 Ohm? (1)

Even with the average ~50A you reached going uphill that's around 250W and 300W (1) burned in the motor and in the batteries - still a small but nice cooking plate... ;)

---

I finshed my formula set (so everything can be calculated in excel without iteration): https://drive.google.com/open?id=0B6-Hr1aVKjLLaE4tbDhubGVpdzQ (1)

I renamed the one U battery max to U back emf max to avoid misunderstandings - this value is used to calculate the actual U back emf

U battery max ist by default set to number of cells in series * Voltage per cell, but can be changed as wanted. (Pay attention - it can be set too low, so the duty cycle becomes >1 :wacko:) - i did not thouroghly check it, but it seems quite ok...

The used formulas:

BCD92X0.png

and a first graph: (again) (1)

Thats for a fixed Power of 800W, resulting in a Average Battery voltage of 74,68V and average Battery current of 10,71A. The max battery voltage equals to I motor shown in the graph. The graph is made over Speed (0-55 km/h). Max U Back Emf was set to 70V (at 55 km/h)

gK4aTRn.png

Edit (1): ups - 6 cells in parallel don't have 1/6 of the resistance of one cell... :wacko: ... i'll change that...

Edit2: or it is - pfff - i have to make a break.... ;)

Edit3: Ok - it is 1/6 - don't know what happened yesterday... :wacko:

 

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11 minutes ago, Chriull said:

With 90A and a coil resistance of ~0,1 Ohm that would make a motor loss of 810W only from this part - not taking into account friction/"magnetic"/etc losses.

That 90A is only brief in time. With motor power peaked at >3000W, a loss of 810W is not surprisingly high, given the efficiency of about 65%.. In one extreme test, I even 'achieve' the limit(120A).

However, I cannot access the Google page, nor the 2 photos attached.

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32 minutes ago, Chriull said:

With 90A and a coil resistance of ~0,1 Ohm that would make a motor loss of 810W only from this part - not taking into account friction/"magnetic"/etc losses...

Edit: and a bit more is burned within the battery packs?! Msuper3s+ also has a 20s6p configuration? Would make ~0,12 Ohm?

Even with the average ~50A you reached going uphill that's around 250W and 300W burned in the motor and in the batteries - still a small but nice cooking plate... ;)

---

I finshed my formula set (so everything can be calculated in excel without iteration): https://drive.google.com/open?id=0B6-Hr1aVKjLLaE4tbDhubGVpdzQ

I renamed the one U battery max to U back emf max to avoid misunderstandings - this value is used to calculate the actual U back emf

U battery max ist by default set to number of cells in series * Voltage per cell, but can be changed as wanted. (Pay attention - it can be set too low, so the duty cycle becomes >1 :wacko:) - i did not thouroghly check it, but it seems quite ok...

The used formulas:

BCD92X0.png

and a first graph:

Thats for a fixed Power of 800W, resulting in a Average Battery voltage of 74,68V and average Battery current of 10,71A. The max battery voltage equals to I motor shown in the graph. The graph is made over Speed (0-55 km/h). Max U Back Emf was set to 70V (at 55 km/h)

gK4aTRn.png

Could you explain this plot?  What are the axes?  Seems motor and current are on the same axis?  so, its volts or amps depending on the curve?  What is the horizontal axis?

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16 minutes ago, zlymex said:

That 90A is only brief in time. With motor power peaked at >3000W, a loss of 810W is not surprisingly high, given the efficiency of about 65%.. In one extreme test, I even 'achieve' the limit(120A).

However, I cannot access the Google page, nor the 2 photos attached.

Strange - @DaveThomasPilotseems to had no prob? the pictures where at imgur.com and till now seemed to work wo prob?

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Ah, I see now from your text in the post, thehorizontal axis is speed.  Sorry

Just now, Chriull said:

Strange - @DaveThomasPilotseems to had no prob? the pictures where at imgur.com and till now seemed to work wo prob?

I was able to download a spreadsheet with no problems.  That's all I tried to do.  I could check something else, if you want.

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1 minute ago, DaveThomasPilot said:

Could you explain this plot?  What are the axes?  Seems motor and current are on the same axis?  so, its volts or amps depending on the curve?  What is the horizontal axis?

yes - everything is on the same axis, so volt,ampere and km/h depending on the curve. Horizontal is the speed in km/h

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1 minute ago, DaveThomasPilot said:

Can this spreadsheet be used to get battery current versus battery voltage at given motor output power?

Since this are all "direct" formulas now without iteration needed, one should be able to plot everything.

At a given (constant) motor output power average battery current and average battery voltage should be constant - Just the ohmic resistance from the motor should be a loss and not motor output power - so some small changes would be needed in the sheet. Also the internal resistance of the battteries should be considered as loss and so only the "real" battery output power equaled to the motor power without ohmic coil losses - seems like a next iteration for the formulas has to come...

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2 hours ago, zlymex said:

Good thread, and good points made.

I'll put it in a different way, ignoring the details(such as PWM, 3 phases) and make a block diagram below.
EUC-block.gif.71dbe057412b7ce585a6b96d2de1468e.gif

The battery output voltage is Ub, current Ib and power Pb, these are also the input of the board.
The board output voltage is Ud, current Id and power Pd, these are also the input of the Motor.
And the output mechanical power of the motor is Pm.
That equation(of Power out = Power In * efficiency) is universal, so we have
Pd = Pb * Eb, where Eb is the efficiency of the board,
Pm = Pd * Ed, where Ed is the efficiency of the motor.

I agree very much that for certain driving conditions, Pm, Ed and Eb are more or less constant, so, if Ub drops, Ib must be increase, and  is roughly inversely proportional to Ub.

In real life, what are those parameters especially Eb and Ed?
Well, I can monitor Ub and Ib very precisely and I did measured them in many riding conditions. An app(such as Wheellog) can give Id and log it.
Generally speaking, the larger the Id, the less the Eb. If the current is doubled, the power will be quadrupled for certain resistor. This is true both for motor wires and on-resistance of the MOSFET.
In normal riding conditions(such as no wind flat road, speed of 25kmph), Id is around 10A, I can give the best estimate of Eb to be around 97%.
However, if Id is large as happened in steep slope climbing, Eb is dropping dramatically. Not only due to the on-resistance of the MOSFET, the transit state also consume a lot power, and doubling the number of MOSFET(from 6 to 12) seems not very useful(transit time may be double too because the input capacitance is also doubled). 

Yesterday, I rode up a very steep hill, I monitored Id by Wheellog that even reached to 90A, and the temperature has gone up to 70 degree C.

SkiHill.thumb.gif.5e44fecaccd9fbc20c07ad7d24dd68e2.gif


In most cases, Id is larger than Ib, sometimes twice or even larger. However, the battery is not bare this PWM spikes because there is a capacitor(two for GW) paralleled with the battery input.

As for Em, I estimate to be around 65% in normal riding conditions(for my Gotway Msuper3s+) but dropping fast at low speed.
 

Great way to clarify things!

I've been referring to battery voltage (Ub) and battery current (Ib).  It's important to differentiate from Ud and Id.

It's harder to talk about Id, since there are multiple conductors at different phases.  So, there's not just one current to consider, but three (or more)?  That's why I was trying to stick with just input current (or Ib) and battery voltage (Ub).

I think that the average current in each winding will be (first order approximation) proportional to the torque, or at fixed speed, proportional to the mechanical output power.  I think this is what many are thinking about when they assert the input voltage doesn't matter for motor current.

But, as I've tried to point out, battery current (Ib) MUST be inversely proportional to battery voltage (Ub) at a fixed motor speed and load.

That principal is good to use for sanity checking a detailed spreadsheet model.  If it doesn't hold up, something is probably wrong somewhere.

 

 

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9 minutes ago, DaveThomasPilot said:

It's harder to talk about Id, since there are multiple conductors at different phases.  So, there's not just one current to consider, but three (or more)?  That's why I was trying to stick with just input current (or Ib) and battery voltage (Ub).

Quite true. If we read the current from an app(such as Wheellog), it is only one value. I suppose they just converted to single phase motor. There are two current sensor ICs on the board in series connection with 2 of the 3 motor wires, the 3rd current may be calculated from the two.

And yes, the torque is proportional to the motor current, and the back EMF is proportional to the speed.

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6 hours ago, DaveThomasPilot said:

Great way to clarify things!

I've been referring to battery voltage (Ub) and battery current (Ib).  It's important to differentiate from Ud and Id.

It's harder to talk about Id, since there are multiple conductors at different phases.  So, there's not just one current to consider, but three (or more)?  That's why I was trying to stick with just input current (or Ib) and battery voltage (Ub).

I think that the average current in each winding will be (first order approximation) proportional to the torque, or at fixed speed, proportional to the mechanical output power.  I think this is what many are thinking about when they assert the input voltage doesn't matter for motor current.

But, as I've tried to point out, battery current (Ib) MUST be inversely proportional to battery voltage (Ub) at a fixed motor speed and load.

That principal is good to use for sanity checking a detailed spreadsheet model.  If it doesn't hold up, something is probably wrong somewhere.

 

 

I will try to simplify.

  A motor is running at 10mph, constant grade, no change in rolling resistance. The battery is 80 and the current shows that to maintain this speed is 10AMPS.

  Now lets say same conditions but the battery is at 40 volts.  The current will still be 10AMPS to make the wheel go 10mph.

The numbers are academic and not real but only to demonstrate the relationship.  

So why is this so!  Well at the time when the battery is 80volts, the computer was putting let say 10% duty cycle to generate 10amps to produce the mechanical torque required to maintain 10mph.  When the battery is 40volts the computer then increases the duty cycle to generate 10amps to produce the same mechanical torque required to keep the wheel spinning at 10mph. To produce a mechanical torque on a motor only current is the parameter that determines how much torque is produced. Inductance, Resistance, Capacitance, Voltage are also important but as secondary items on the complete design.

Battery volts is only relevant to the point that the voltage is not capable of producing the minimum AMPS to produce a mechanical torque required. And yes it is also important because of the secondary issue of BackEMF. At some point the 80 and 40 volts example dont work because if you hit the region when RPM is high enough to produce a backemp high enough to prevent enough delta volts to maintain the minimum current needed.

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From energy point of view, if the load is fixed say at 800W, the battery must provide at least the same power at 80V * 10A. If the voltage has dropped to 40V, the current must be increased to 20A in order to maintain that 800W of power.

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