DC Motor Stall Current

[NaikD]

I have following specification for a DC Motor:

Operating Voltage: 28 VDC ± 0.56 VDC power bus
Number of Phases: 3 (connected in Delta)
Number of Poles: 24
Resistance: 15 ohms
Torque Sensitivity 275 ounce-inch/amp
Power: 3.4 Watts
RPM: 31.6

Maximum current then would be Operating Voltage/Resistance = 28/15 = 1.86 amps
Maximum/Stall Torque = Torque Sensitivity * Maximum current = 275*1.86 = 511.5 Oz-In.

Shouldn’t I get the same result by using Power = I * V to compute the Max amps?
i.e. Stall current = Power/V = (3.4/28) = 0.12 amps?

Thanks!

 

[Skogsgurra]

NaikD,

Are you sure that you have three phases? A DC motor only has a positive (+) and a negative (-) armature connection and (if separately excited) a field winding with a + and a - connection. Sometimes the field and the armature windings are connected at the minus end, which will get you three connections, but definitely not a three-phase motor. Most of your data point to a DC motor.

It could also be a BLDC motor since you are talking about 24 poles. Anyhow, the discussion below is valid in either case - even if the concept of armature resistance is hard to visualize in the BLDC (and three phase) case.

There is a big difference between operation at rated speed/rated torque and operation with zero speed and full torque. At rated speed and rated torque, you get output power in W by multiplying speed in rad/s with torque in Nm.

When the motor stalls, it develops a torque that is proportional to the current that you get when dividing supply voltage and armature resistance. This torque is normally many times higher than the rated torque and it cannot be sustained for more than seconds or minutes before the armature gets hot.

In the stalled condition, the motor produces no output power. It is like you lifting a heavy wheight. Once you got it up, you do not do any work. You can probably lift 100 pounds without difficulty. It is only when you start climbing stairs with this wheight that you do any real work and you will notice that you do since you will start sweating hevily after only four or five floors with hundred pounds of extra wheight.

So, the short answer to your questions is: No. You shall not get the same current in these two situations.

 

[NaikD]

Skogsgurra,

Thanks for clarifying. Yes as you guessed, it is a BLDC motor.

Regards,

 

[eemotor]

If your DC motor is stalled, then the stall current will be approximately V/R since it's just an RL circuit with no change in voltage making your CEMF very low. So you'd be looking at something around 1.87 A.

 

[aolalde]

NaikD:

The nameplate Power of an electric motor refers to the mechanical output delivered in the shaft at full load.
Then the 3.4 watts refer to mechanical power output, this results in full load Torque =145.4 Oz-in at 31.6 rpm. Considering the torque sensitivity, the current at full load, Ifl =145.4/275 =0.529 Amps.
The full load current will be related to the electric power input to the motor and is proportional to the power output divided by the motor efficiency. It should be reported on the nameplate data.

The inrush current as view from the 28 VDC power bus (input to the BLDC motor control) is a resistance resulting from the combination of the motor and the control reactance’s.

Apparently it will be the result of V/R (1.86 Amps) as you calculated.

 

[NaikD]

aolalde, Skogsgurra, eemotor,

Thank you very much for contributing. After Slogsgurra clarified the details, I calculated the amperes and the torque at 31.6 rpm and numbers match with those that you reported.

Thanks again,
Regards,