Now the real question still is, how is current affected by the “effective voltage” when the BEMF/RPM changes, is it still linear?
The power has to be higher at higher speeds. The current being the same is due to having to overcome all the resistances, losses…
And your right, Power = Current x Effective Voltage is wrong. It needs to be Power = (Current x Effective Voltage) + initial power. Or Power needed to be added to reach desired RPM = Current x Effective Voltage
The “effective voltage” can be handled as the speed difference that the motor wants to achieve and it therefore pulls current to try to achieve that speed, If there is no “effective voltage”, then the motor is already at the speed it wants to spin and it won’t pull any current and therefore power = 0.
That is true if you want to neglect ALL losses. But then you can’t compare it to the videos of a skateboard…
Yes the statement is correct, but factoring the real life losses, it will never spin at the speed it wants to spin (as you said), so there is always a little bit of “effective voltage” giving current to combat the losses and it then settles somewhere below the “desired speed” where the used power counteracts the losses.
I’m glad we agree. You might want to think about what that means for your implementation of current/torque control. That is the whole point of the thread, isn’t it?
I still need to know how the current relates to the “effective voltage” and BEMF. Is it linear to “effective voltage” no matter what the BEMF/RPM is or does it change with it. AKA 2nd question.
`The back emf limiting the torque though…you only need a very small voltage across the motor, the difference between the battery voltage and back emf, to have potential for a huge amp flow. lets take this motor as an example because it’s similar in size to what we’ve been working in equations just with a 5x higher kv
http://www.rapidtables.com/calc/electric/watt-volt-amp-calculator.htm With only two volt potential and using my motors first in the equation with its resistance , I can draw 17 amps per motor. Simos
And with this 5x higher kv motor run it through ohm’s law and with the 2 volt potential
I can get 80 amps!
And here’s my question: the wattage number above is dismal at 166 watts even though 80amps. Of course u can point to the low voltage…but what good is voltage anyway!? I still can’t figure out the basic of how voltage is power if a motor only runs on amps
On another note about speed and power, this is how I see speed in the power equation: say I’m coasting at 20mph down a hill without a motor it would be fair to say I have power. I have power to get from a to b as I finally slow but it’s not power in the typical sense I think about it. Torque is power to me. Without torque there will be no speed. Speed is just torque’s dregs Simosmcmuffin I’m probably far from ur project talking about this but it’s a fundamental that I still can’t get my mind around after years
The effective voltage values I have been using are not based on any real values, as I have not measured it ever, yet. So they are just there to say: “So we have some voltage that is doing work, and let’s say X amount of voltage causes Y amount of current.” point being, that they might not be close to actual values we get from motors we use.
The calculator you use is just a static power calculator, but as it doesn’t have the RPM/BEMF element to it, is this what the motor would see at 0 RPM? What about if the motor is spinning, as an example: the BEMF is 10 Volts and then we supply 12 Volts to the motor, giving the effective voltage of 2 V. Will the motor still pull 83.3 Amps? And then our power would be 12 V * 83 A = ~1000 Watts. This makes sense with the Power = Torque x RPM.
Also, if your assumption that current increases linearly with voltage across the motor winding resistance. Than that would answer the 1st question:
I would like add a point a couple of things about about the performance graph.
The RPM, as depicted in the graph in percentage, is a relative measurement.
Not necessarily at full throttle, but at 85 % BEMF voltage of the constant voltage.
as an example: Motor has 100 kV, we have a lab power supply and a controller which runs constantly at 100% duty. If our power supply voltage is 10 V, the peak efficiency is at 850 RPM supply voltage 15 V, peak efficiency at 1275 RPM supply voltage 20 V, peak efficiency at 1700 RPM
This supply voltage can be also directly controlled with a battery voltage and the duty cycle:
Motor 100 kV, 20 V battery voltage Duty cycle 50 % → voltage 10 V, peak efficiency at 850 RPM Duty cycle 75 % → voltage 15 V, peak efficiency at 1275 RPM Duty cycle 100 % → voltage 20 V, peak efficiency at 1700 RPM
Point being that the RPM axis is always 100% given the voltage applied to the motors kV value. And you’re running at peak efficiency when the BEMF is 85 % of the supplied voltage.
“The DC Motor block represents the electrical and torque characteristics of a DC motor using the following equivalent circuit model:”
http://www.mathworks.com/help/physmod/elec/ref/dc_motor_diag.png Source: http://www.mathworks.com/help/physmod/elec/ref/dcmotor.html
As seen in the model, BEMF (Vb) is just a counter voltage against the applied voltage. And the BEMF is a function of the motor’s kV and RPM.
Not really contributing here…but this thread is total mind F@#&
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No time to read all this because it reminds me of physics class too much, but where are you going to school? my eboard would have been a cool senior design project…at least till the advisors suck all the fun out of it…
I’m Finnish and I go to the Tampere’s Polytechnic University. 24 Years old, studying in the embedded systems department (microcontrollers). I plan on keeping the project under my own wing as much as I can. I’ll of course let other people give their opinions on things, but how far they go depend on me. AKA we will do as we see fit.
I’m interviewing the last 2 guys on Monday and then assembling the team of 3 out of the 5 interested volunteers.
You don’t have to have full throttle (100% duty) to get peak performance.
Yes, but then we can use the duty % to change the no load 100% speed. On my controller this is possible, for example I can use a 50 V battery, but limit my duty cycle at full throttle to 50 %, then my “constant voltage” would be just 25 V and my no load is 2500 RPM, then the peak efficiency is at ~2100 RPM.
Then when you factor in the dynamics of accelerating from standstill. It causes the graph to sort of stretch on X axis (power and efficiency) as the duty cycle increases, while the torque & amps curve doesn’t get stretched, it gets offset horizontally.
that sounds about right, good luck on this project, hopefully you can get all this theory into something we can see in the real world, like more torque and/or efficiency.