1
$\begingroup$

I have a project where I need a motor that can turn some number of rotations which will spool up a cable attached to a spring-closed device to open it up. When power is disconnected, the spring closure will cause the spool to unwind and the device to close.

In the closed position, no power is available. (i.e. The closure mechanism needs to be 100% passive.)

In order to keep this open for some time, I will need a motor that is capable of being stalled for long periods without having the windings burn up. I know some motors can do this, such as the motors they use on spring closed HVAC dampers, but I don't know how to find them or if there's a particular name I should be using to find them. I know I could probably do this with a stepper motor, but that seems overkill for the application.

The only requirements are higher torque to open this mechanism, no gearing that prevents the motor from spinning when power is disconnected, and the ability to be stalled.

$\endgroup$
  • $\begingroup$ Why not use something like a gear stopper while "stalled"? $\endgroup$ – fibonatic Mar 10 '15 at 3:19
1
$\begingroup$

The current required to open the device is greater than the current required to keep the device from closing.

A motor is a piece of wire conducting commutated DC. While it is stalled, it is just DC. A reduction gear can increase the torque applied (further reducing the force and current required to open and hold open) without interfering with the automatic close function, because a standard DC motor should provide little, or no, "Detent Torque" like a stepper.

Try this:
-Calculate Iholding using Ohm's law to limit the power dissipated by the stalled motor.
-I("RMS")lifting will be less since the DC is being commutated.
-The power limit is arbitrary; the wires will un-solder at >300F.
-The power limit should allow the motor enough current to open the device.
-If it doesn't, add a reduction gear.

An example using a motor's datasheet:
1.0V < Vin < 3.0V (This motor can supposedly tolerate 12V, no problem)
"locked-rotor current" = stall current

Working backwards with V = 3.0V and I = 0.8A, using Ohm's law:

E = I * R  
R = E / I  
R = 3.0V / 0.8A = approximately 3.75ohms

With Vin = 3.0V, using Ohm's law:

E = I * R  
I = E / R  
I = 3.0V / 3.75ohm = 0.8A
P = I * E  
P = 0.8A * 3.0V = 2.4W

Same motor with Vin = 1.0V, using Ohm's law: (it will turn slower)

E = I * R  
I = E / R  
I = 1.0V / 3.75ohm = 0.26A
P = I * E  
P = 0.26A * 1.0V = 0.26W

Same motor with Vin = 3.0V and 5ohm series resistor, using Ohm's law:

E = I * R  
I = E / R  
I = 3.0V / (3.75ohm + 5ohm) = 0.34A
EdropAcrossR = 0.34A * 5ohm = 1.7V
EdropAcrossMotor = 0.34A * 3.75ohm = 1.28V  <= above no-load start voltage
P = I * E  
Pr = 0.34A * 1.7V = 0.58W  
Pmotor = 0.34A * 1.28V = 0.44W  

Double-check:
Ptotal = 0.58W + 0.44W = 1.02W  
Ptotal = Itotal * Etotal  
Etotal = Ptotal / Itotal  
Etotal = 1.02W / 0.34A = 3.0V
$\endgroup$
  • $\begingroup$ This doesn't answer the question. Even with less load, the motor still needs to be designed to stall or the windings will burn up. $\endgroup$ – David Pfeffer Feb 7 '15 at 16:01
  • $\begingroup$ moved comments to answer, added detail $\endgroup$ – Jon Feb 7 '15 at 20:46
  • $\begingroup$ Most motors don't have a stated stall torque or current. Also, the formulas in this answer assume that the motor is an Ohmic load. Motors are inductive loads (except when stalled), not resistive. $\endgroup$ – David Pfeffer Feb 8 '15 at 1:57
  • $\begingroup$ All of the calculations above are with the motor stalled $\endgroup$ – Jon Feb 8 '15 at 5:37
2
$\begingroup$

One simple alternative is to attach the spring in such a way that it doesn't pull the device closed when it is all the way open (see diagram below). In this fashion there is no torque acting on the device, so the motor will not need to be running at this point.

Note that this will require a small torque back in the original direction in order to close.

enter image description here

$\endgroup$
1
$\begingroup$

One possible option would be a linear solenoid. I would get a push type (Push solenoid's will push a rod out when energized and the builtin spring will auto retract it when the power is removed.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.