Micromotors are smaller fractional-hp designs frequently found in medical equipment. Selecting one for a particular application can be a rather involved process and should be done in consultation with application engineers from the motor manufacturer. Still, it is useful to “ballpark” a motor selection on your own. To that end, keep in mind these few rules and equations relating to the physics and the practical application of such motors.

A few equations

For example, the major constraint on motor operation is thermal. The heat a motor must dissipate can be calculated from:

P_dis = I²R

where P_dis = heat dissipated, W; I = current through the motor, A; and R = resistance, ohm.

The current through a motor is solely determined by the torque it produces and is found with:

I = M/k_m

where M = torque; and k_m = torque constant.

Also consider:

• The constant current operation of a dc motor. It produces a constant output torque regardless of speed.

• Given a constant load (i.e. torque) a motor's speed is solely dependent on the applied voltage.

• Power is the product of speed and torque. A dc motor produces maximum power at an operating point defined by operation at half the no-load speed and half the stall torque. A motor seldom operates at maximum output due to thermal considerations.

• One rule of thumb says run a dc micromotor at about 90% of its no-load speed and from 10% to 30% of its stall torque. This is where the motor is most efficient.

• When used with gearing, select a motor for a practical minimum speed by choosing one with higher voltage ratings than the available voltage supply. This results in lower noise and better life characteristics.

• For a dc motor that will work at a constant voltage, its speed and torque are inversely related. The higher the torque produced, the lower the speed of the motor.

• Other factors may include size, environmental conditions, weight, and required life.

An example shows how to use the selection sequence. Suppose a medical device needs a motor as small as possible to run at these conditions:

Available voltage = 20 Vdc

Output torque required = 0.425 oz-in

Output speed required = 5,000 rpm

Ambient temperature= 22°C

It is unlikely a standard catalog motor will fulfill all parameters because they are not independent. The selection process finds a best fit.

For example, first find the required output power from:

P_o = nM/1,350

where n = speed, rpm; and 1,350 is a unit-conversion factor.

P_o = (5,000 T 0.425)/1,350

=1.57 W

The motor should be rated at least 1.5 to 2 times the required output power in relation to its maximum output power (at nominal voltage). Therefore, a motor with about 2.4 to 3.2 W maximum output should suffice. Referring to the MicroMo catalog, the smallest motor with this power rating is the 1331 series (13-mm dia., 31-mm long). The 24 V version is closest to the required operating voltage of 20 V.

To find the no-load speed, a good first approximation is simply to ratio the voltages and speeds with:

n₂₀ = (20n₀)/24

= (20 × 10,400) / 24

= 8,666 rpm

The required speed of 5,000 rpm represents only 58% of the no-load speed (at 20 V). Unless size is of paramount importance, this motor is not a great choice even though it can provide the power required.

The next selection from the catalog which meets the power requirements is the series 2230 motor (22-mm dia., 30-mm long). The selection would again be the 24V version. Pertinent data for this motor includes:

n₀ = 9,000 rpm (at 24 V),

I₀ = 0.005 A

R_term = 50 Ohms

k_m = 3.59 oz-in/A

P₀ = 2.88 W

T_amb = 22°C

where R_term = resistance at the terminal, ohms.

Hence, the approximate no-load speed at 20V would be:

n₂₀ = (20 × 9,000)/24

= 7,500 rpm

The required speed is 67% of the no-load speed, and even though it's not quite 70%, it is worth continuing the selection process. The motor current will be the sum of the load current and the no-load current.

I_m = I + I₀

Where I_m = current through the motor, A; I = current due to load, A; and I₀ = no-load current, A.

Because I = M/k_m

where M = required torque

Therefore, I_m = (M/k_m) + I₀

Then: I_m = 0.425/3.59 + 0.005

= 0.123 A

To calculate the speed at the required load torque, use:

n= 7,500-1,350×0.123∞50/3.59

= 5,187rpm

This speed is close to the required value so the selection of this motor appears reasonable.

Check the motor's temperature rise to confirm the selection. The power dissipated by the motor is found with:

P_dis = I²R_term

Hence:

P_dis = (0.123)² × 50

= 0.76 W

Find the heat rise under steady-state conditions with:

³T = P_dis (R_th1 + R_th2)

where:

³T = motor temperature rise, °C

R_th1 = thermal resistance from rotor to case, °C/W, and

R_th2 = thermal resistance from case to ambient, °C/W.

(These latter two values come from the catalog)

³T = 0.76(4 + 28)

= 24.3°C

The motor temperature under steady state is found with:

T_m = T_amb + ³T

where:T_m= operating temperature, °C; T_amb= ambient temperature, °C.

Then:

T_m= 22 + 24.3

= 46.3°C

This value is satisfactory because it is well below the maximum operating temperature for this type of motor, which is 125°C

Make contact
MicroMo, (727) 572-0131, micromo.com

Comparing the required to selected values

Parameter	Required Value	Selected Value
Voltage	20 Vdc	20 Vdc
Torque	0.425 oz-in.	0.425 oz-in.
Speed	5,000 rpm	5,187 rpm
The required values versus the selected values for the 2230T024S motor show good agreement.