Question of the Motor and Motor drive selection

[bhl777]

Hi All,

I am new to motor and motor drive and want to ask the questions for the procedure to select the components based on the rough application requirements.

This is my example application:

1. I want to use an electric motor to lift up an object with a 1 kg weight
2. the lift speed can be constant and very slow. Say, 0.1 meter/ second
3. No need to consider the size/noise/weight/cost like the real application

Assuming the mechanical parts (the orange ones in the picture) can be well designed to fit for the application. I assume I can choose a DC motor and an appropriate motor drive to do the job. My questions are:
1. In order to find a potential motor that can be used in this application, what key motor performance spec I should look for?
2. Once the motor is selected, what is the main electrical spec I should look for to find a matched motor drive circuit?
3. If someone can point me to a real product and its drive circuits, it will be perfect.

Thank you!

 

[KlausST]

Hi,

Homework. So it's not useful that we do your job....but for sure we will help you.

What have you done so far?
We don't see your effort.

I at least expect that you
* read about physics, what is mechanical energy, what is power, what is inertia, what is torque.
* read a motor datasheet, what parameter does it show
* go to a motor manufacturer's internet page and use it's motor selection guide. What parameter does it ask for?
* what are the electrical motor parameters?

... now do almost the same as above for the motor driver.
* what function does your driver need to have? What is your electrical energy source? AC mains, battery, ? Direction control, speed control, overload protection ...

Klaus

 

[stenzer]

Hi,

a simple linear actuator may do the job, which can be driven by a DC voltage source e.g. 12 V. By changing the polarity, the actuator's direction can be changed. For such a simple setup used for example for a motorized TV-lift you need an appropriate linear actuator by means of "changable" length and liftable force (weigth). According to the chosen linear actuator you need a suitable DC power supply, by means of the voltage and sourceable current. The polarity may be changed by a three-pole toggle switch.

How does your application look like?

BR

 

[bhl777]

Hi,

a simple linear actuator may do the job, which can be driven by a DC voltage source e.g. 12 V. By changing the polarity, the actuator's direction can be changed. For such a simple setup used for example for a motorized TV-lift you need an appropriate linear actuator by means of "changable" length and liftable force (weigth). According to the chosen linear actuator you need a suitable DC power supply, by means of the voltage and sourceable current. The polarity may be changed by a three-pole toggle switch.

How does your application look like?

BR

Hi stenzer,

Thank you so much for your advice! According to your advice, I have some followed up questions:
1. Do you mean I can use +12V to lift up and a swapped polarity -12V supply to implement "pull down" as the reverse activity of the lift up?
2. If I am using a actuator, does this mean it will function as a "motor drive + motor"? By saying this, I am asking does an actuator need an external drive circuit in addition to the +/-12V power supply?

Thank you!

--- Updated Dec 7, 2020 ---

Hi,

Homework. So it's not useful that we do your job....but for sure we will help you.

What have you done so far?
We don't see your effort.

I at least expect that you
* read about physics, what is mechanical energy, what is power, what is inertia, what is torque.
* read a motor datasheet, what parameter does it show
* go to a motor manufacturer's internet page and use it's motor selection guide. What parameter does it ask for?
* what are the electrical motor parameters?

... now do almost the same as above for the motor driver.
* what function does your driver need to have? What is your electrical energy source? AC mains, battery, ? Direction control, speed control, overload protection ...

Klaus

Hi Klaus,

Thank you for your advice! I did read through some of the tutorials, but still do not have a clue from the application perspective.

For the motor, I only know the torque, the speed, and the power rating as the motor performance specs.
However, if we come to this example, I am trying to lift up a 1kg object, I am not sure if the following procedure is correct?
1. the force to keep a constant movement will be mg=~10N.
2. the speed requirement of the requirement 0.1m/second
3. if the power is the product of the force and the speed, it will be ~10N*0.1m/second=1 Watt mechanical power from the load side
4. Assuming the efficiency from the input electrical power to the load side mechanical power is 90%, the input power of the motor is 1 Watt/0.9=1.11 Watt
5. How can I use this 1.11 Watt information to specify a motor to choose? For example, I saw a datasheet of a servo motor here. But I do not know how can I relate the 1.11 Watt requirement to the torque and the speed spec.
6. Say my electrical source can be either a 3.7V Li-iOn battery or a 12V supply, and I have bucks and boosts that can step up and down the voltages. Regarding the motor drive circuits, my questions are:
6.1 how do I know if I need one or not? If we look at the motor datasheet above, it does not require a drive circuit.
6.2 if I need to use one motor drive circuit, in addition to the obvious voltage and current rating, what other parameters should be considered?

Thank you!

 

[c_mitra]

I can tell you the steps but will not work out the details.

1. I kg being lifted by 0.1m : can be calculate the work done in this step? Say that the work done is W.

2. Work done per unit time is power. So W work done per 1 second is W watts (I hope you have calculated W in joules).

3. The power must be supplied by the motor. The motor must have minimum power of W watts.

4. Allow 100% extra for losses and other inefficiencies. Many motors cannot start at full load and you need to oversize the motor.

5. Select your motor voltage first; next select the motor power.

6. You need a speed control; you select driver accordingly.

7. Select motor speed and next select your gears for connecting the load.

 

[KlausST]

Hi,

Let's just focus on the motor:

The "motor" you refer to is no motor. It is a servo for RC cars. It can just move a "controlled" angle forward and back. It includes electronics. It needs a dedicated pulse signal to control the angle.

Thus I recommended:
* read a motor datasheet, what parameter does it show
* go to a motor manufacturer's internet page and use it's motor selection guide. What parameter does it ask for?
Please do so.

And (if you don't need some soecial features) you should select motor voltage to match your power source.
Example: In a car the nominal voltage is DC 12V. Thus most motors are rated for DC 12V, too. Everything else means additional effort.

Your mechanical calculations are useful.
What I miss is how the motor rotation can lift the object. This determins motor RPM and motor torques (at the gearbox out if you have).
Whether you use a belt, a rope, a big or small wheel, a toothed rack, threaded spindle or anything else ... won't change the mechanical energy and mechanical power....but every solution will have different loss..this means the motor needs to supply higher mechanical power to move the object.

Klaus

 

[stenzer]

Hi,

1. Do you mean I can use +12V to lift up and a swapped polarity -12V supply to implement "pull down" as the reverse activity of the lift up?

somehow. You only have a single +12 V power supply and a BASIC linear actuator has only two connection wires (there may be more connections to control e.g. speed). So there are two possebilities to connect the two wires form the 12 V power supply with the two wires of your (12 V rated) linear actuator. One connection lifts the inner tube and the other one "pulls" it down.

2. If I am using a actuator, does this mean it will function as a "motor drive + motor"? By saying this, I am asking does an actuator need an external drive circuit in addition to the +/-12V power supply?

No you do not need an "external driver" circuitry, a DC power supply with sufficient current driving capability is all you need. The motor and the required gears are already included within the actuator. Please make a quick web search how linear actuators are working. Here is a video you can start with:

BR