9.0 TUTORIAL SELECTION OF MICROMOTOR APPLICATION PROBLEMS

9.1 Two Problems In Angular Motion

(A)  It is required that a system reach a rotational speed of 2000 RPM in 20 MS. Initial velocity is zero. What is the angular acceleration required?
   
(B) It is required that a system have an angular displacement of 3 radians in 20 MS. Initial velocity is zero. What is the angular acceleration? 

9.2 Drive Motor Application Problem

Customer has a "sniffer" drive problem. It is to be battery operated and the carbon brush motor he has selected consumes excessive current. This motor has a 12 VDC rating, 14,000 RPM no load speed, a 5 ohm rotor, and 150 ma no load current. The operating conditions under load are a terminal voltage of 6 volts and 300 ma.

    a) What is the load torque?
          
M_L = K(l - l_NL) = 76.7 x 10^-4(0.3 - 0.15) = 11.5 x 10^-4Nm = 0.16 oz-in

  b)  What is the motor speed?
      
   C)  What is the efficiency?
        P_out = M_Lω = 11.5 x 10^-4 x 586.7 = 0.675 watt
        P_in = Vl - 6 x 0.3 = 1.8 watt
        η = P_out/P_in = 37.5%

9.3 Drive Motor Selection Problem

This customer powers the above sniffer by Nickel-Cadmium rechargeable batteries of 1.25 - 1.30 volts per cell output. He would like to use the least number of cells consistent with optimum efficiency and reasonable motor life. Small size is desirable.

a)  What is your motor recommendation?
      
Try 22C11-216, R_a = 6.0 ohms @ 90ºF
       
       l = l_NL + lL = 0.119 + 0.007 = 0.126A
    
      V = 6.36 Volts (5 cells = 6.25 - 6.50 Volts)

b) What is the efficiency?

      P_out = M_Lω = 11.5 x 10^-4 x 586.7 = 0.675 watt
      P_in = Vl = 6.36 + 0.126 = 0.801 watt
     
c) How does this efficiency compare to the motor's peak efficiency?
     
       l_NL = 0.007A
     
       η_max = 84.4% (0.998 effective)

d) What is temperature of rotor above case?
      P_a = l²R_a = (0.126)² x 6 = 0.095 watt
      ΔT = RP = 9 x 0.095 = 0.86C negligible

9.4 Gearbox Application Problem

A customer has a machine tool application where he needs exactly one revolution of a shaft in 1.0 - 1.5 seconds. He has a 24 VDC supply available and load friction is 5 oz-in, inertia is 0.05 oz-in-s². Position of the output shaft will be sensed externally and on-off control applied to the motor.

	a)	What would your first try be using a B24? initially assume start-up and shut down times negligible relative to one second. Speed therefore is one revolution in approximately one second or 60 RPM. Gear ratio is needed of approximately 100. Selecting the smallest 24 VDC Motor, try the 26P11-210-100 with B24 ratio 128.
	b)	What would the motor and gearbox output speeds be under load? Convert friction load to metric: MLf = 5 oz-in x 70.62 x 10-4Nm/oz-in = 353 x 10-4 Nm Motor speed loaded:        output speed:
	c)	What time to speed (95%)? Convert Inertia to metric: JL = 0.05 oz-in-s² x 7.062 x 10-3 Kgm²/oz-in-s² = 3.53 x 10-4 Kgm² JM = 5.3 x 10-7 Kgm² 3τ = 0.057 Sec
	d)	Assuming equal times to start and stop, how long does it take from start to rest with a total angle of 1.0 revolution? Ref: Figure 15 For 3τ = 2.05ωτ = 2.05 x 5.6 x 0.019 = 0.22 Rad

	Start Slow Stop	0.22 Rad 5.84 Rad 0.22 Rad	0.057 Sec 1.043 Sec 0.057 Sec
	Total	6.28 Rad	1.16 Sec

9.5 Case Study: The "DC 300" Cartridge Drive

9.5.1 "DC 300" cartridge specifications (CARTRIDGE FOR ILLUSTRATION PURPOSES ONLY - NOT AVAILABLE FOR SALE )

Excerpts from the "DC 300" cartridge specifications per American National Standard's ANSI X3.55-1977 and ANSI X3.56-1977 necessary for our purposes are the following:

-	Friction force = Ff The worse case tangential force required at the outer driving radius (ra = .445 in) of the belt capstan to maintain a constant operating speed. Ff = 4.5 oz   (1.25N)
-	Total inertia = mra2 The total equivalent inertial mass of all cartridge elements must not exceed m, given in linear units referred to the external radius (ra) of the belt capstan. m = .002 oz s2/in   (.022 kg)
-	Inter block gap (lBG) The minimum length of the inter block gap should be: lBG = 1.2 in   (30.5 x 10-3)
-	Drive ratio: The ratio of the tape velocity, vt, to the tangential velocity of the external driving radius of the belt capstan, v, (which is also equal but opposite to the tangential velocity of the drive roller) shall be: vt/v = .76

9.5.2 System requirements
  - Density = 6400 bits/in (DC300XL)
  - Tape read-write speed: v_to
     The common speed for most applications is:
            v _to = 30 in/s
     therefore v_o = 30/.76 = 39.5 in/s       (lm/s)
  - Tape rewind speed =v_trw
     The common speed for most applications is:
           v_trw = 90 in/s
     therefore: v_rw = 90/.76 = 118 in/s (3m/s)
  - Acceleration or deceleration time on read-write mode (t_a; t_d):
            (lBG - .15 - .3)/v_to

   Where;
   .15 in., block integrity.
   Block integrity is a safety factor where a portion of the tape is DC erased before information block.

   .3 in., distance between read head and write head

  - Acceleration or deceleration on rewind mode can vary from one system to another,
     however, it is greater than 75 ms.
  - External drive roller radius:
           r_c = .25 in      (6.35. 10^-3 m)

9.5.3 Model of Drive

with vo=39.5 in/s M = .002 oz s2/in r = rc = .25 in. Ff = 4.5 oz.

9.5.4 Solution

Because the greatest average power dissipation in the motor armature is when reading or writing on start-stop mode with the smallest increment (or block) of information (128 bits = .02 in.), this will be used for an example of the worst case. The following results were obtained with a 23D21-216E Escap motor.

-- Torque Tm1 = (Jm + mr2)vo/rta + Ffr Tm2 = Ffr Tm3 = (Jm + mr2) vo/rtd + Ffr
-- Current l1 = Tm1/k l2 = Tm2/k l3 = lm3/k
-- Voltage U1 = Rl1  U2 = Rl1 + kvo/r U3 = Rl2 + kvo/r U4 = Rl3 + kvo/r U5 = Rl3
-- Dissipated power Pd1 = Rl12 Pd2 = Rl22 Pd3 = Rl32

Average dissipated power =
                                        
                               when    
where:
l₁ = A + B            
l₂ = B 
l₃ = -A + B            
Therefore,

in our case this will be:
                                                              2.32 W 
-- Optimal drive roller radius r_o, which minimizes the average dissipate power, can be obtained by:
      r optimum which minimizes
     This results in the following optimal radius:
    
if F_f = 0, no friction force, "matching inertia"
r_o = ( (m²/J_m²) + (t_a² + .5t_at_o)F_f²/J_m²v_o²)^-1/4  
r_o= .15 in

The above results can be tabulated in the following curve:



Comparision of Power Dissipaion Versus Drive Roller Diameter
Curves Using Various Escap D.C. Micromotors





page 6 - Dynamic Braking of a DC Micromotor