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Can some one help me on this??

We are using two different make PMDC motors with different gear ratios(motor to gear box to lead screw to load) in one of our application.Below are the details. Application:Solar Tracking systems. Load:Around 1 ton solar structure.

***At No load condition:***
Motor 1:                               Motor 2:       
current: 0.2A                          current: 0.6A
speed:2100 rpm                         speed:3515 rpm
input voltage: 24V                     input voltage: 24V
Gear Ratio:67                          Gear Ratio:57.
Gear working:smooth                    Gear Working:smooth to hard.
Max current rating: 6A                 Max current rating: 12A
Load current: 1.2 to 2.3A(max)         Load current: 2.5 to 7A(max). 

We are using the both motors for same application.

  1. We have used motor 1 and it's working fine.But we have chosen motor 2 by thinking it has high no load and full current rating so that it can produce high torque.is it correct??

  2. For lifting the same load motor 2 taking high current. Why? and How to reduce load current??

  3. We need to reduce the load current around 3A when we are using motor2.How to achieve this for same load?

  4. is there any possibility to increase the load torque and reduce the load current when we are using motor 2??

  5. And finally where we have done wrong?and what are the changes need to do to get same performance as motor 1 connected to load???

Thanks

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All the current ratings, no load, rated, max, of motor 2 are about 2x that of motor 1. The fact that it's taking twice the current of motor 1 when driving a load sounds consistent with all it's trying to tell you on the data sheet.

Note that motor 1 has about half the speed at the same voltage, which means its torque constant is about 2x that of motor 2. So you would expect half the current in motor 1 for the same torque.

Sounds like motor 1 is the one for the job.

So now the most difficult part is political. Having chosen motor 2 for the wrong reasons, does anybody have to lose face to choose motor 1 instead? I do not lie, after 4 decades of lab politics, this is often the most difficult part of any project, when you're halfway through and it suddenly dawns on the team that someone's made a wrong choice. Can you swallow pride and fix it, or do you do a Trump (other politicians are available), say nothing's wrong, and soldier on? Perhaps sneak into the workshop in the middle of the night and remove one of the brushes of motor 2, 'oh look, it's stopped working, now we'll have to use motor 1!'

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  • \$\begingroup\$ Presently we have ordered around 4000 No's of motors(2).we don't have any option right now,we need to use those motor only.Can you please suggest what changes need to adopt in our system to use motor 2 for same application?? \$\endgroup\$ – Siva Pratap Nov 4 '17 at 6:41
  • \$\begingroup\$ Is half the present motor speed permitted? If you could change the order so they wind the motors with twice as many turns of wire of half the area, then that might work. At the same voltage, this will halve the speed and halve the current for the same output torque. Another way to get much the same effect is to supply the motors through a DC to DC converter to halve the supply voltage to the motor, which halves the supply current to the converter compared to the motor. Again the motor works at half speed. Otherwise, you're screwed. \$\endgroup\$ – Neil_UK Nov 4 '17 at 6:53
  • \$\begingroup\$ An order for 4000 is quite a big political problem. What are the penalties for cancelling? Sit down and do the sums. It may be cheaper to cancel, and do the project right, than to soldier on with a lead turkey. Remember, even if there's no time or will do to it right early on, failure will persuade you to do it right eventually. The earlier you start to do it right, the cheaper it will be, even if that seems to be very expensive right now. \$\endgroup\$ – Neil_UK Nov 4 '17 at 6:55
  • \$\begingroup\$ Due to our site requirements ,our ultimate goal is to reduce the load current.Previously we have tried PWM technique to reduce the speed of the motor.At that time speed reduced but load current is same. if we reduce PWM to 50% What are the parameters are going to be effected?.If we do like that is it possible to reduce the load current?And can you please help me to reduce load current ?? \$\endgroup\$ – Siva Pratap Nov 4 '17 at 7:28
  • \$\begingroup\$ PWM is not quite the same as a DC-DC converter. The latter will reduce the current draw. \$\endgroup\$ – Neil_UK Nov 4 '17 at 7:39

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