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Okay so my question is multifold 1)Is it possible to run a BLDC motor at lower voltages than its rated voltage. Since I have a constant power source I was wondering if I could run it with lower voltages which will give me higher current to obtain the starting current. 2) can a BLDC motor start if it is not able to obtain it's starting current. For eg if my peak current required is 60 A for 3s and I'm not able to supply that will the motor start however slowly? 3) can somebody help me with the mathematics involved or at least point me in the right direction. Because all the catalogue deals with are peak and full load values. And as far as I know the motor doesn't always operate at full load.

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  • \$\begingroup\$ It would help if you linked to the motor in question. I don't believe that your brushless DC motor needs 60A to start, unless you are designing an electric car or something. "60 A" is probably the maximum rated current. What do you mean by 3S; is this some RC equipment powered by a three cell LiPo battery? I wrote a fairly comprehensive primer on how BLDC motors behave and how they are driven a while ago: electronics.stackexchange.com/questions/262106/… \$\endgroup\$ – jms Oct 13 '16 at 17:23
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    \$\begingroup\$ Keep in mind when asking questions that we have no Idea what you know. We need a lot of information to be able to answer a question. What motor are you using? What kind of load are you trying to drive? What kind of power supply are you using that can't support 60A for 3s? To what catalog are you referring? I'm not necessarily saying I can answer the question if you provide this info, but I need more info to know if I can. \$\endgroup\$ – ambitiose_sed_ineptum Oct 13 '16 at 17:28
  • \$\begingroup\$ Alright. Really sorry on that part. My load is a solar electric car. I'm using a 48V 1500 W BLDC motor. One of the rounds of a competition I'm taking part in requires me to remove the battery and run it directly from solar panels. Due to space restrictions the largest panel I can use is of 400 W and 48v. By 3s I meant 3 seconds. I received a catalogue from the manufacturer stating that the motor requires a continuous current of 32 amps and will give a continuous torque of 4.8Nm and it will require 60A for 3 seconds to supply 14.4 Nm at the start. I'm yet to receive the data sheet. \$\endgroup\$ – user22483 Oct 13 '16 at 18:57
  • \$\begingroup\$ I roughly need 80Nm as a starting torque. Since I have 's reduction ratio of 8 from my transmission I need 8 NM of starting torque. It doesn't matter if my car moves very slowly. Hence the above question ^ \$\endgroup\$ – user22483 Oct 13 '16 at 18:59
  • \$\begingroup\$ You should rewrite the question with this new information \$\endgroup\$ – jms Oct 13 '16 at 20:53
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The torque output of an electric motor is directly proportional to the motor current (not voltage!), and the current (I) is roughly equal to

Where V is the motor supply voltage, R is the winding resistance and ε is the back-electromotive force (back EMF).

The back EMF is the voltage that would be present at the motor terminals as the motor spins without anything being connected to it. This voltage is produced by the motor acting as an alternator, if you will, and it is directly proportional to the rotation speed. The back EMF limits the maximum motor speed at any given battery voltage, because at some rotation speed the back-EMF will "cancel out" the battery voltage. This prevents any more current from flowing to the motor and thus reduces the torque to zero.

I received a catalogue from the manufacturer stating that the motor requires a continuous current of 32 amps and will give a continuous torque of 4.8Nm and it will require 60A for 3 seconds to supply 14.4 Nm at the start.

You are misinterpreting the specifications.

  • The motor doesn't require a continous current of 32 A to spin, It'll draw only a couple of amps without any load. 32 A is the maximum current you can allow the motor to draw indefinitively without risking damage. (32 A * 48 V = 1536 W, which matches with your description of "I'm using a 48V 1500 W BLDC motor").

  • 4.8 Nm is the torque produced at that maximum continous current of 32 A. This implies that the motor will draw one amp for every 0.15 Nm of load. Assuming that the motor is 85% efficient, this also implies a rotation speed of about 2540 rpm at the maximum voltage of 48 V (1500 W * 0.85 / (2 * π * 4.8 Nm) = 42,3 Hz = 2540 rpm). The back EMF will increase by about one volt for every 53 rpm.

  • The motor doesn't require 60 A for three seconds to start spinning. The original specification probably intended that "you can safely overload the motor with 60 A (about twice the rated current) for three seconds when accelearating from rest". I'm not sure how it would develop 14.4 Nm at 60 A while only producing 4.8 Nm at 32 A; the relationship between torque and current is linear so something isn't right with those values.

Is it possible to run a BLDC motor at lower voltages than its rated voltage?

When you first power your motor on, the speed is zero. This means that the back EMF is also zero, so the only things limiting the motor current are the winding resistance and the supply voltage. If the motor controller was to output the full battery voltage (48 V) to the motor at low speeds, the motor and/or controller would just melt down from the excessive current. Consequently every BLDC controller already does what you are asking; the full battery voltage is applied to the motor only at the maximum speed setting.

Can a BLDC motor start if it is not able to obtain it's starting current?

No. The torque output of an electric motor is directly proportional to the motor current, so if your motor needs say 20 A to generate enough torque to overcome the static friction of the vehicle and drivetrain anything less will not do.

That said, your power supply (e.g. solar cells) will not have to provide e.g. 10 A in order for the winding current to be 10 A. This is because when the motor controller outputs a voltage lower than the input voltage it pulse-width modulates the output, effectively acting as a buck converter and trading voltage for current. For example, if the motor draws 60 A at 2.5 V (150 W) your solar panels will only have to provide 3.13 A at 48 V (still 150 W), plus any inefficiencies.

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  • \$\begingroup\$ Alright. So what you're saying is that the motor will provide as much power as I need (considering I never need anything greater than 1500W). My load here is my electric car which weighs around 150kg without batteries( since I have to run it without batteries in one round) and around 220 kh with batteries. So how would I calculate how much power I need from the motor to make it move? As in how do I calculate what my load is? \$\endgroup\$ – user22483 Oct 14 '16 at 8:33
  • \$\begingroup\$ @user22483 Too broad, what does “make it move” really mean? You could probably use a 1W motor to make it move 1mm/hr. Missing many information to design a motor \$\endgroup\$ – PDuarte Jun 13 '18 at 13:40

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