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

Electrical specifications are

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
```

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

Yes

Motor Specs

Max Eff.= 52.2% @ 5.916 A,

2999 RPM,0.236 N.m9.427 A,2226 RPM ,0.436 N.mPe=226.2W, Pm= 101.6WYour operating specs:

What are V,I,Eff.? :

Using Max Eff.,

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

which is much greater than torque ratio 54% at peak eff.

Iss=80.3%*9.427 =

7.57 Aand powerPe = 98.4W= 7.57A⋅ 13VPm=43.9Wmechanical power (P2)You will need a tach to regulate the RPM if you use this to vary the BLDV voltage or some other servo method.