Electronic – Determine efficiency, voltage, current from motor characteristic curves

brushless-dc-motorcurrentdevice-characteristicsefficiencyvoltage

I would to identify efficiency, voltage, current for a particular Brush less DC motor in our lab given characteristic curves.

Electrical specifications are

enter image description here

The operating conditions are

Torque: 0.35 N.m
RPM   : 1200

Based on the operating condition, i drawed one more torque vs speed line (orange color) passing through 0.35 N.m & 1200 rpm and keeping slope constant.

To explain the image 

Horizontal axis (X) : Torque (N.m)
Left Vertical axis  : U(V)     - Applied voltage         - Light Blue color
                      I(A)     - Current                 - Green color
                      P1(w)    - Electrical power input  - Dark Blue color
Right Vertical axis : n(r/min) - Revolution per min      - Orange color
                      P2(w)    - Mechanical power output - Red color
                      Eff(%)   - Efficiency              - Brown color

enter image description here

Unfortunately, i don't have better quality image. Please click on the image or Zoom the browser page to enlarge a bit more.

I have drawn few vertical and horizontal dashed lines(pink in color) passing through the operating point (0.35 N.m and 1200 rpm).

Question:

As the voltage vary the torque vs speed curves, does the efficiency curve vary ? what about the voltage and current ?

What are the approximate values in my case ?

Current = 8 Amps ?
Voltage = 15.00 or 17.50 V ?
Efficiency = 50 % ?

Thanks

Best Answer

As voltage varies torque vs speed, does the efficiency curve vary ?

Yes

  • Motor Specs

  • Max Eff.= 52.2% @ 5.916 A, 2999 RPM , 0.236 N.m

  • Max Power 45.07%@ 9.427 A ,2226 RPM , 0.436 N.m Pe=226.2W, Pm= 101.6W

Your operating specs:

  • Torque: 0.35 N.m
  • RPM : 1200

  • What are V,I,Eff.? :

  • Using Max Eff.,

\$\dfrac {0.236 Nm}{0.436}(=54\%) \cdot 24V = 13V, ~~2999 ~RPM \cdot 54\% =1619~RPM\$

  • \$\dfrac{T_{spec}}{T_{max}}= \dfrac{0.35Nm}{0.436}= 80.3\%\$
    which is much greater than torque ratio 54% at peak eff.
    • 0.35 N.m = 80.3% of max T thus current
      Iss=80.3%*9.427 = 7.57 A and power Pe = 98.4W = 7.57A⋅ 13V
    • yet RPM has dropped to 54% of max power at 80.3% T so mechanical power has dropped to 80% of 54% = 43.2% of 101.6W= Pm=43.9W mechanical power (P2)
  • thus efficiency is 44.7% about the same as the max T eff. (45.07%)

You will need a tach to regulate the RPM if you use this to vary the BLDV voltage or some other servo method.

enter image description here

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