General questions

General topics and discussions about the VESC and its development.
Dor
Posts: 1
Joined: 03 Mar 2018, 08:22
Location: Israel

General questions

Postby Dor » 03 Mar 2018, 09:50

Hi everyone,
Little bit on me:
My name is Dor and I am new here.
Right now I am learning robotics at college, and I decided to bulid an electric skateboard as a hooby.

So I have some basic questions in 3 subjcects.
I hope that the questions are clear and easy to understnad, so let's start:

Subject: VESC
1.what is the latest version of the Vesc?Does it come is several hardware forms?
2.What is the latest firmware version?
3.Except for conncetion a motor, what other inputs does he has?for exaple, can I conncet a small LCD screen?or any sensors?

Subject: Electric skateboard
1. I bought this motor: https://diyelectricskateboard.com/colle ... 6355-190kv
and I also have a 6S lipo battery, does it enoght to bring me to 15 mph?
2.In which speed the motor can provide the highest tourqe?
3.Which factor determines the tourqe?Does it the voltage or does it the current?

Subject:Lipo battery
Let's guess I have a lipo battery with these features: 4200mAh 35C 6S 1P 22.2 v count 35C Brust 70C
1. It capacity is 4200 mAh, right? So as long as I am using it ,the capacity is getting lower and lower, right?
Like a smartphone?
So if the answer is "yes", is any damage will be occur if I finish all her energy?Is it dangerous for the cells?Is it necessary means that at least one cell(or more) are lower thant 3V?

2.There is this formula : CxA=Amax ,right?
So as the mili ampers are getting lower and lower due to using, so also the max current is getting lower and lower?

3.What does it mean "6S2P"?
-On the buttom line, how much cells do I have in serial? how much cells do I have in paralell?

4.what are the differences between connceting cells in serial to parallel?

I am really sorry if I have some grammers errors or something was not wrriten right
I hope I was enught clear

I am realy apreciate your help
Thanks!
Dor :)

arvidb
Posts: 230
Joined: 26 Dec 2015, 14:38
Location: Sweden, Stockholm

Re: General questions

Postby arvidb » 03 Mar 2018, 17:32

Dor wrote:2.In which speed the motor can provide the highest tourqe?
3.Which factor determines the tourqe?Does it the voltage or does it the current?

Torque is directly proportional to current.

In fact, torque is equal to the motor's torque constant Kt [Nm/A] times the motor current I [A]: T = Kt ⋅ I.

The torque constant is the inverse of the motor constant Kv expressed in SI units [radians⋅s⁻¹⋅V⁻¹]. The motor you bought has a motor constant of 190 rpm/V = 190⋅2⋅π/60 ≈ 19.9 radians⋅s⁻¹⋅V⁻¹. Since Kt = 1/Kv, its Kt is 0.05 Nm/A. So e.g. at 10 A it will develop a torque of 0.5 Nm.

You can push the maximum amount of current through a motor at stand-still, so that is also where it can provide the highest torque. The VESC however limits the current to sane levels (so that the motor doesn't burn out), so you will have a constant torque capability between maybe 0 % and 75 % of the motor's top speed.


Dor wrote:Let's guess I have a lipo battery with these features: 4200mAh 35C 6S 1P 22.2 v count 35C Brust 70C
1. It capacity is 4200 mAh, right? So as long as I am using it ,the capacity is getting lower and lower, right?
Like a smartphone?
So if the answer is "yes", is any damage will be occur if I finish all her energy?Is it dangerous for the cells?Is it necessary means that at least one cell(or more) are lower thant 3V?

2.There is this formula : CxA=Amax ,right?
So as the mili ampers are getting lower and lower due to using, so also the max current is getting lower and lower?

The capacity of the battery is the same whether it's fully charged or not. Just like a glass of water: its capacity (the amount of water it can hold) is the same whether it's full or not.

You need to distinguish between amperes and ampere-hours; they are completely different things. Your (milli)amperes aren't getting "lower" while you use your battery; the remaining ampere-hours of charge is.

The battery's maximum current capability is not affected (much) by its charge level. But yes, you need to make sure the cell voltages doesn't dip too low, since that will damage the battery (and possibly cause it to catch fire!).


Dor wrote:3.What does it mean "6S2P"?
-On the buttom line, how much cells do I have in serial? how much cells do I have in paralell?

4.what are the differences between connceting cells in serial to parallel?

6S2P means that you have six sets of two cells in parallel. These six sets are then connected in series.

When you connect cells in series, the battery pack voltage increase: six 3.7 V, 4.2 Ah cells in series results in a battery voltage of 22.2 V with a capacity of 4.2 Ah. When you connect batteries in parallel, the battery capacity increase instead: six 3.7 V, 4.2 Ah cells in parallel results in a battery voltage of 3.7 V and a capacity of 25.2 Ah.

So twelve cells of 3.7 V and 4.2 Ah connected in 6S2P results in a battery voltage of 22.2 V with a capacity of 8.4 Ah.

devin
Posts: 253
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: General questions

Postby devin » 03 Mar 2018, 20:54

Dor wrote:Subject: Electric skateboard
1. I bought this motor: https://diyelectricskateboard.com/colle ... 6355-190kv
and I also have a 6S lipo battery, does it enoght to bring me to 15 mph?


I will now give you my opinion whether the board you describe can go 15mph on flat ground.

With some basic assumptions (3:1 gear reduction ratio, 90mm tire diameter, 190kv, 0.05ohm winding resistance lead to lead, 24v battery pack, 1.2 drag coefficient standing human, 0.929m^2 standing human estimated frontal area, 1.225kg/m^3 fluid density of air), I conclude your board as described will not reach 15mph on flat ground.

----------------------

The motor is 190 kv. The motor specifications say max torque "2.83" newton meters.

Image
--------------------------------

your battery is 6S 4200 mAh lipo meaning 6 cells in series * 4v nominal cells = 24v nominal 4.2 amp hour capacity

--------------------------

Kt (torque per motor amp) = 60/(2*pi*KV)

Kt (torque per motor amp) = 0.05025945571323010603228 newton meters torque per motor amp = 60/(2*pi*190)

Your motor gets 0.05025... newton meters torque per motor amp

---------------------------------

if the max torque recommended by the manufacturer is 2.83 newton meters, and torque per amp is 0.05025... newton meters, this implies that the max recommended motor amps can be calculated by:

2.83 max recommended newton meters / 0.05025... newton meters per motor amp = 56.31840a maximum recommended motor amps

based on the max recommended torque, the max recommended motor amps is 56.31840a motor amps

----------------------------------

since your battery is 24v nominal, the max duty cycle of a vesc is 95%, and your motor is 190kv or 190 rpm per volt, the maximum rotational speed of the motor (with no load) is:

24v * 0.95max duty * 190kv = 4332rpm

the max rotational speed of your motor with a 24v battery pack and vesc is 4332rpm

---------------------------------
for gearing let's assume you use a 3:1 reduction ratio and your motor pulley is 20 teeth and wheel cog is 60 teeth.

the effect is to increase the torque output of the motor by a factor of 3 while reducing the rotational speed measured at the wheel by a factor 1/3

60 wheel cog teeth / 20 motor cog teeth = 3 reduction ratio

motor torque * 3 reduction ratio = wheel torque

motor rpm / 3 reduction ratio = wheel rpm

4332rpm [max rotational motor rpm w/ 24v pack] / 3 reduction ratio = 1444rpm max wheel rpm at 24v, 190kv, 3:1 reduction, max 95% duty

2.83newton meters [max recommended motor torque] * 3 reduction ratio = 8.49 newton meters max wheel torque at max recommended motor torque, 3:1 reduction

the max rotational speed of the wheel is 1444rpm no load, and the max wheel torque at standstill is 8.49newton meters, assuming 3:1 gear reduction

---------------------------------

let's assume your tires are 90mm in diameter

90mm * pi = 282.7431mm per rotation

a mile is 1609344mm

15 miles per hour is 15mph*1609344mm per mile = 24,140,160mm per hour

24,140,160mm per hour / 60 = 402,336mm per minute

402,336mm per minute / 282.7431mm per rotation = 1422.97372rpm

in order to travel 15 mph, your wheel needs to turn at 1422.97372rpm assuming 90mm tire diameter

--------------------------
at 15mph, the wheel is turning 1422.97372rpm.

assuming 3:1 gear reduction, 90mm tire diameter, at 15mph, the motor is turning:

1422.97372rpm wheel speed at 15mph * 3 gear reduction = 4268.92116rpm motor speed at 15mph assuming 3:1 gear reduction & 90mm tire diameter

at 15mph, your motor will be turning at 4268.92rpm assuming 3:1 gear reduction.

--------------------------

1/KV = back emf voltage per rpm

1/190kv = 0.00526316v back emf voltage per rpm

4268.92116rpm motor speed [@ 15mph ground speed assuming 3:1 gear reduction, 90mm tire diameter] * 0.00526316v back emf voltage per rpm = 22.4680151v back emf voltage produced by the spinning magnets in the motor at 15mph ground speed

at 15mph, the back emf voltage in the motor is 22.46801v

--------------------------
the maximum duty cycle of the vesc is 95% which in practice means the maximum effective voltage seen by the motor (a result of pulse width modulation) is 95% of the battery voltage

since your battery is 24v, the max voltage to the motor the vesc can apply is:

24v * (95/100) = 22.8v maximum pwm effective voltage with 95% max duty cycle vesc

because the max duty cycle of a vesc is 95% and the battery is 24v, the max effective voltage to the motor is 22.8v.

--------------------------

Detect the winding resistance with the vesc, then double this number because the Vesc detects the resistance from one lead to the center of the motor (virtual ground point), but you want to know the resistance from lead to lead. For example if the vesc detected 0.025ohm, then your lead to lead resistance is 0.05 ohm

let's assume the vesc detects 0.025ohm

0.025ohm vesc detected resistance * 2 = 0.05ohm winding resistance lead to lead

We will assume the winding resistance is 0.05ohm, but the actual resistance of the motor could be different.

--------------------------

since your battery is labeled 35C continuous, and the capacity is 4200mah

4200milli-amp-hours / 1000 = 4.2 amp hours

4.2 amp hours * 35C rating = 147a max amps continuous

147a * 24v nominal pack voltage = 3528w max continuous discharge

The battery is rated to discharge 147a continuously assuming 4200mah and 35C rating.

----------------------------

A=B+(C*D)
(A-B)/C=D

A = pwm effective v = 22.8v maximum pwm effective voltage with 95% max duty cycle vesc & 24v battery [24v*0.95dutymax]
B = bemf v = 22.4680151v back emf voltage produced by the spinning magnets in the motor at 15mph ground speed
C = ohms winding resistance = 0.05ohm winding resistance lead to lead
D = motor current = XX.XXXa

(A-B)/C=D

D motor current = (22.8v-22.4680151v)/0.05ohm = 6.639698a max motor current @ 15mph ground speed assuming 24v pack, 95% max duty, 190kv, 0.05ohm lead to lead, 3:1 gear reduction

A*D = 22.8v maximum effective PWM voltage * 6.639698a max motor current = Electrical Watts = 151.38w = 24v battery voltage * 6.3075a battery amps

The max possible motor current at 15mph is 6.639698a motor amps. At 15mph, 6.639698a motor amps results from drawing 6.3075a battery amps @ 24v battery pack volts.

----------------------------

A=B*C

A = Battery Amps
B = Motor Amps
C = Duty Cycle % (max 95% on a vesc)

6.3075a battery amps = 6.639698a motor amps * (95 max duty %/100)

6.63969a Motor amps at 95% duty cycle equates to 6.3075a battery amps.

----------------------------

A=B*C

A = PWM Effective Volts
B = Battery Pack Voltage
C = Duty Cycle % (max 95% on a vesc)

22.8v pwm effective volts = 24v battery * (95 max duty %/100)

24v battery at 95% max duty cycle equates to 22.8v pwm effective volts

----------------------------
A^2*B = Ohmic Heating of Windings

A = 6.63969a motors amps = Motor amps
B = 0.05ohm = winding resistance

A^2*B = 2.20427w ohmic heating

At 15mph equivalent ground speed, max ohmic heating of the motor should equate to 2.20427w.

----------------------------

6.639698a max motor current * 0.050259nm/a newton meters torque per motor amp = 0.33370458 newton meters max motor torque @ 15mph ground speed assuming 24v pack, 95% max duty, 190kv, 0.05ohm lead to lead, 3:1 gear reduction

The max torque output of the motor at 15mph is 0.33370458 newton meters.

----------------------------

0.33370458 newton meters max motor torque @ 15mph ground speed * 3 gear reduction = 1.00111375 newton meters max wheel torque at 15mph ground speed [1422.97372rpm wheel speed]

The max torque output of the wheel at 15mph is 1.00111375 newton meters assuming 3:1 gear reduction.

----------------------------

1422.97372rpm wheel speed at 15mph

(1422.97372rpm*2*pi)/60=149.01333 radians per second

1.00111375 newton meters max wheel torque at 15mph ground speed * 149.01333 radians per second = 149.179w max mechanical watts at the wheel at 15mph ground speed

The max mechanical watts at the wheel at 15mph is 149.179w mechanical watts.

---------------------------

A+B=C=D*E=F*G

A = Mechanical Watts = 149.179w mechanical watts
B = Ohmic Heating Watts = 2.20427w ohmic heating
C = Electrical Watts = 151.38327w electrical watts
D = PWM Effective Volts = 22.8v pwm effective volts
E = Motor Amps = 6.63969a motor amps
F = Battery Voltage = 24v battery pack voltage
G = Battery Amps = 6.3075a battery amps

A+B=C=D*E=F*G

149.179w mechanical watts + 2.20427w ohmic heating = 151.38327 electrical watts = 22.8v max pwm effective volts * 6.63969a Motor amps = 24v battery pack voltage * 6.3075a

6.63969a motor amps at 22.8v pwm effective voltage equates to 6.3075a battery amps from a 24v battery pack, 151.38w electrical watts drawn from the battery at 15mph ground speed, 149.179w mechanical watts and 2.20427w ohmic heating of the windings.

---------------------------

A = 1.225kg/m^3 = Fluid Density of Air

B = 1.2 = Standing Human Estimated Drag Coefficient | source: http://www.taylors.edu.my/EURECA/2014/downloads/02.pdf

C = 0.929m^2 = Standing Human Estimated Frontal Area

D = 15mph [= 0.149129086 seconds per meter = 6.7056 meters per second]

E = 0.149129086 seconds per meter [ = 15mph]

F = 6.7056 meters per second [ = 15mph]

G = XX.XXX = Air Drag in Newtons

H = XX.XXX = Wind Resistance Mechanical Watts

(1/2)*A*C*(F*F)*B = G = Air Drag in Newtons

(1/2)*1.225*0.929*(6.7056*6.7056)*1.2 = 30.7028252006784 newtons = G = air drag force @ 15mph standing human = Air Drag in Newtons

G = 30.7028252006784 newtons = Air Drag in Newtons of Standing Human at 15mph

G / E = G * F = H = Wind Resistance Mechanical Watts

30.7028252006784 newtons [air drag force @ 15mph standing human] / 0.149129086 seconds per meter = 205.8808648547500653226w Mechanical Watts Wind Resistance of standing human at 15mph

H = 205.8808648547500653226w = Mechanical Watts Wind Resistance Applied to Standing Human @ 15mph

The power applied to a standing human by wind resistance at 15mph is 205.88086w mechanical watts, which equates to a force of 30.7028252006784 newtons.

----------------------------

149.179w max mechanical watts at the wheel at 15mph ground speed <[is less than] 205.8808648547500653226w mechanical watts wind resistance standing human 15mph

The max mechanical power applied by the skateboard at 15mph (149.179w mechanical watts) is less than than the opposing power applied to the rider by wind at 15mph (205.88086w mechanical watts).

-------------------------

Dor wrote:Subject: Electric skateboard
1. I bought this motor: https://diyelectricskateboard.com/colle ... 6355-190kv
and I also have a 6S lipo battery, does it enoght to bring me to 15 mph?
2.In which speed the motor can provide the highest tourqe?
3.Which factor determines the tourqe?Does it the voltage or does it the current?


conclusion

since the max mechanical watts at the wheel at 15mph ground speed (149.179w) is less than the mechanical watts applied by the wind at 15mph to a standing human (205.88w), I believe your setup as described will not be able to reach 15mph under its own power on flat ground assuming 3:1 gear reduction, 90mm tire, 190kv motor, 0.05ohm lead to lead winding resistance, 24v battery, 57a motor amp limit, 147a battery amp limit.

------------------------

for illustration, notice in the second chart that the mechanical watts are less than the wind watts at 15mph.

Image

rew
Posts: 937
Joined: 25 Mar 2016, 12:29
Location: Delft, Netherlands.

Re: General questions

Postby rew » 04 Mar 2018, 10:12

Dor, please ignore Devin. He's got some non-scientific ways to get some results that are not scientifically valid.

He starts out with assuming a gear ratio. That's the last thing you need.

The motor can put out 2500 Watts. If we take that to be correct, the main question is: Can you do 15mph on 2500 W?

On my bike I measured the amount of energy I put out to bike 5m/s or 18km/h to be about 70W. If we estimate that this becomes 2.5 times more at 24km/h (15mph, 6.6m/s) we get about 175W to go 15mph.

Now a bike is more efficient (but the air resistance becomes dominant somewhere in the region 20-25km/h) so you might assume a skateboard to be twice as inefficient as the bike. So 350W.

So... Can your 2500W motor deliver 350W? Yes it can.

Next... If you gear it wrong, it won't be able to achieve the theoretical performance. So you need to gear it in such a way that it achieves max RPM at the desired speed. With the current margin (350vs 2500), I would want it to run 15mph at the lowest possible battery voltage. So 6*3V = 18V. Times 190 Kv means it will be running at 3400 RPM. NOW the assumptions made by devin can come in to play. (3:1 gear reduction ratio, 90mm tire diameter) means that your tire will be doing about 1100RPM, with a 282mm tire circumference, that means you'll be doing 5.1 m/s, about 19km/h or below. the target speed. If you want to go 1.5 times faster, you need a 1.5 times lower gearing ratio (or bigger wheels). So... with a 2:1 gearing ratio you'll easily do 15mph, close to 20mph on fully charged battery.

If I cherry-pick a bit more from Devin, 2.8 Nm of torque means that you get about 5.5Nm at the wheel. This gives a force of about 120N of thrust. Should be enough. If you weigh 80kg, you'll reach 5m/s in about 3 seconds.

devin
Posts: 253
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: General questions

Postby devin » 04 Mar 2018, 14:23

i agree with rew if you lower the gear reduction to 2:1 or increase the battery voltage, then 15mph should be possible, but using 3:1 reduction, 90mm tire, 24v battery, then I believe (as rew seems to confirm) 15mph would not be possible.

I do have a question for rew...

what happens if he uses 3:1 gear reduction so that the board won’t do 15mph on its own on flat ground, but then on a downhill he coasts it up to 20mph, then while at 20mph, he uses the throttle. my understanding is the back emf voltage at this speed will exceed the battery voltage. in this case, will using the throttle act as a brake? if the answer is yes there is a real danger to riders of the board who exceed the no load speed of the motor while coasting downhill.
Last edited by devin on 04 Mar 2018, 17:04, edited 1 time in total.

devin
Posts: 253
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: General questions

Postby devin » 04 Mar 2018, 14:52

& as rew says, if you change the assumptions I chose from 3:1 to 2:1 gear reduction, then by my calculations and factoring the wind you can reach somewhere between 20-25 mph (32.1-40.2kph) before the wind drag force exceeds the thrust force from looking at the middle, top row chart.

Image

further reducing the gear reduction to 1.5:1, it would appear you can reach about 25mph before the wind drag force exceeds the thrust force. on a test bench with no wind drag load the wheel would turn slightly above 30mph equivalent ground speed (middle, top row chart), assuming 1.5:1 gear reduction-- for example with 20 motor pulley teeth and 30 wheel cog teeth...

Image

devin
Posts: 253
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: General questions

Postby devin » 05 Mar 2018, 17:11

i made a comparison chart of 1 motor set to 57a max motors amps as previously described [24v battery, 0.05ohm winding, 1.5:1 gear reduction, 90mm tire, 190kv], and 4 identical motors set to 14.25a max motor amps. in theory the riding performance should be almost indistinguishable, but with 4 motors at 14.25a max motor amps, the overall electrical to mechanical conversion efficiency is increased and total battery amps drawn at all rpms is decreased (in theory, improving range).

Image

^notice in the left chart, black line (battery amps-4 motors), battery amps are decreased across the chart, while in the right chart, total mechanical watts at most rpms is identical.

a huge benefit to this increase in efficiency would be a reduction in ohmic heating of the windings per motor. with 1 motor at 57a motor amps we get 57a*57a*0.05ohm=162.45w average ohmic heating per motor. by contrast with 4 motors at 14.25a we get identical riding performance but only 14.25a*14.25a*0.05ohm=10.15w average ohmic heating per motor. in other words achieving identical overall performance with 4 motors instead of 1 motor reduces average ohmic heating of the windings by 152.3w per motor to a very manageable 10.15w heating per motor.

devin
Posts: 253
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: General questions

Postby devin » 11 Mar 2018, 16:24

rew wrote:Dor, please ignore Devin. He's got some non-scientific ways to get some results that are not scientifically valid.

He starts out with assuming a gear ratio. That's the last thing you need.

The motor can put out 2500 Watts. If we take that to be correct, the main question is: Can you do 15mph on 2500 W?

On my bike I measured the amount of energy I put out to bike 5m/s or 18km/h to be about 70W. If we estimate that this becomes 2.5 times more at 24km/h (15mph, 6.6m/s) we get about 175W to go 15mph.

Now a bike is more efficient (but the air resistance becomes dominant somewhere in the region 20-25km/h) so you might assume a skateboard to be twice as inefficient as the bike. So 350W.

So... Can your 2500W motor deliver 350W? Yes it can.

Next... If you gear it wrong, it won't be able to achieve the theoretical performance. So you need to gear it in such a way that it achieves max RPM at the desired speed. With the current margin (350vs 2500), I would want it to run 15mph at the lowest possible battery voltage. So 6*3V = 18V. Times 190 Kv means it will be running at 3400 RPM. NOW the assumptions made by devin can come in to play. (3:1 gear reduction ratio, 90mm tire diameter) means that your tire will be doing about 1100RPM, with a 282mm tire circumference, that means you'll be doing 5.1 m/s, about 19km/h or below. the target speed. If you want to go 1.5 times faster, you need a 1.5 times lower gearing ratio (or bigger wheels). So... with a 2:1 gearing ratio you'll easily do 15mph, close to 20mph on fully charged battery.

If I cherry-pick a bit more from Devin, 2.8 Nm of torque means that you get about 5.5Nm at the wheel. This gives a force of about 120N of thrust. Should be enough. If you weigh 80kg, you'll reach 5m/s in about 3 seconds.


@rew -- out of curiosity, I wondered how fast he could go with 1 motor if he switched from a 6S (~24v) to 12S (~48v) battery pack...

if we revert back to 3:1 gear reduction ratio (20 teeth motor pulley & 60 teeth wheel cog), change to 97mm tire diameter (same as 97mm abec 11's), 190kv, 0.05ohm measured winding resistance lead to lead, 48v nominal battery pack (12S - 12 series lipo), 57a motor amp limit, 147a battery amp limit, 95% duty cycle limit, 1.2 drag coefficient standing human, 0.929m^2 standing human estimated frontal area, 1.225kg/m^3 fluid density of air)

using these assumptions with a single motor I find that the rider can in theory travel faster than 30mph on flat ground:

Image

rew
Posts: 937
Joined: 25 Mar 2016, 12:29
Location: Delft, Netherlands.

Re: General questions

Postby rew » 14 Mar 2018, 07:22

devin wrote:my understanding is the back emf voltage at this speed will exceed the battery voltage.\
When the voltage from the motor exceeds the battery voltage, the parasitic diodes in the mosfets will function as a three-phase-rectifier. Thus even without using the throttle, the motor will start working as a generator.

The problem is that the FETs are cooled so that they can handle 50A and dissipate P = I^2 * R = 50A * 50A * 1.5mOhm = 3.7W. But now the diode forward drop is going to be 0.6V, So now you're going to hit the 3.7W at about 6A of motor/battery current! So, once your motor starts delivering about 10A, the fets will start overheating....

devin
Posts: 253
Joined: 08 May 2017, 01:55
Location: San Francisco, California, US

Re: General questions

Postby devin » 22 Mar 2018, 16:46

rew wrote:
devin wrote:my understanding is the back emf voltage at this speed will exceed the battery voltage.\
When the voltage from the motor exceeds the battery voltage, the parasitic diodes in the mosfets will function as a three-phase-rectifier. Thus even without using the throttle, the motor will start working as a generator.

The problem is that the FETs are cooled so that they can handle 50A and dissipate P = I^2 * R = 50A * 50A * 1.5mOhm = 3.7W. But now the diode forward drop is going to be 0.6V, So now you're going to hit the 3.7W at about 6A of motor/battery current! So, once your motor starts delivering about 10A, the fets will start overheating....


@rew - just curious-- by your calculations, why does 6A of "reverse current" (when motor bemf voltage exceeds pack voltage) deliver similar heating to the MOSFETs as 50A of standard "forward current?"


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