Technical

The first step in selecting a motor/generator for your application is to come up with any two of the following three parameters: Available Power, Speed, and Torque. Once you know any two of these, you can calculate the third by using the following equation: P (watts)=(Torque (in-oz) x Speed (rpm))/1352. Then comes the amount of voltage required. By knowing the required voltage and the available speed, you can determine the Kv value of the motor/generator that will fulfill your application's needs.

***NOTE: by selecting the Kv of the motor, the sizing process has just begun. The most important factor in selecting which motor/generator will work in your application is to make certain that the amount of heat which is generated in the motor can be safely dissipated.

Required

   

Regulator Output Power

 

This application requires that the selected motor have

Kv =7000 rpm/12V
=583.3 rpm/V

With this value in mind, you can search our standard product list and select a motor which has a Kv close to the one you have calculated. If one can not be found we can build a motor with a custom number of turns to specifically meet your needs.

P=

150 watts

Output Voltage

 

V=

12 v

Output Current

 

I=

12.5 A

Input Spee

 

S=

7000 rpm

Total Heat Dissipation Capability =

28.24 watts

 
     

Motor Specifications

   

Motor Model

36/38/4

As we have calculated, this motor can safely dissipate 28.24 watts of heat. You now have to calculate the losses in the motor and see if they are less than or equal to the amount the motor can safely dissipate.

In very simplified terms, a DC brushless motor's losses can be broken down into two major categories:

1) losses due to hysterisis and drag torque
(V x Io)
2) losses due to the resistance in the motor's windings (I2 x R)

Voltage Constant (Kv) =

600 rpm/V

Torque Constant (Kt) =

2.24 in-oz/A

No Load Current (Io) =

1.2 A

Resistance (R) =

0.085 ohm

Length (L) =

3.58 in

Diameter (D) =

1.4 in

Total Surface Area =

18.82 inˆ2

Total Heat Dissipation Capability =

28.24 watts

 
     

Losses

   

Power Loss at Speed (No Load)

36/38/4

The losses in the motor are 32.58 watts which are slightly more than 28.24 watts the motor can safely
dissipate if it were running coninuously with no cooling. At this point the generator's duty cycle, mounting location, and cooling become important factors. You may also select a different motor and work through the calculations once again.

P1 =

Io x Vg

=

14.00 watts

Torque Loss at Speed (No Load)

36/38/4

T1 =

P1 x 1352/S

=

2.70 in-oz

Power Loss due to Current

36/38/4

P2 =

Ig2 x R

=

18.58 watts

Torque Loss due to Current

36/38/4

T2 =

P2 x 1352/S

=

3.59 in-oz

Power Loss in Regulator

36/38/4

Regulator Efficiency (RgE) =

85.00%

P3 =

22.50 watts

Heat Lost in Generator

36/38/4

 

P4 =

P1 + P2

=

32.58 watts

Total Torque Losses

36/38/4

=

T1 + T2

=

6.29 in-oz

     

Requriemtents After Losses

   

Total Power Required into Regulator

 

If you do select to use this motor as a generator, a 39.61 in-oz input of torque power will be required to spin the motor's shaft and generate the desired torque

Ptot =

P + P3

=

172.50 watts

No Motor Loss Torque Required

 

T =

Ptot * 1352/S

=

33.32 in-oz

Current Required

 

Ig =

Ptot/Vg

=

14.79 A

Total Torque Required

 

 

Ttot =

T + T1 + T2

=

39.61 in-oz

 

*** NOTE: The above calculations are merely for demonstration purposes..