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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.

Motor 1:

No load:
current: 0.2A
speed:2100 rpm
input voltage: 24V
Gear Ratio:67

Loaded:
Gear working:smooth .
Max current rating of motor: 6A
Load current: 1.2 to 2.3A(max)

Motor 2:

No load:
current: 0.6A
speed:3515 rpm
input voltage: 24V
Gear Ratio:57.

Loaded:
Gear Working:smooth to hard.
Max current rating of motor: 12A
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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    \$\begingroup\$ Possible duplicate of Regarding PMDC motors attached to two different gear boxes \$\endgroup\$ – John D Nov 4 '17 at 9:14
  • \$\begingroup\$ Please use the buttons on the editor toolbar to format your question properly. Use the bullet point for the specifications and the numbered list for the questions. You may also need to edit the post for content as it is already attracting "close" votes. \$\endgroup\$ – Transistor Nov 4 '17 at 9:46
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have chosen motor 2 by thinking it has high no load and full current rating so that it can produce high torque

High no-load current compared to full-load current indicates the motor is not very efficient. Efficiency an be estimated as (full-load current minus no-load current) / full-load current.

Motor output power is full load input current X voltage x efficiency. Torque (Newton-meters) = Output power (Watts) X 9549 / RPM

In determining the power to drive a mechanical load, you must not forget to consider the efficiency of the gear. A small, inexpensive speed reduction gear can be quite inefficient particularly if it is a worm gear. Half the motor torque could be lost in the gear.

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

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

For the same motor, torque is proportional to current. So it makes no sense to reduce the current if you need a certain torque. If you cannot deliver more current, you have to use a higher reduction gear, thus increasing the output torque at the same current, and live with the slower movement of your tracking platform.

For your application, I think 1:10000 is still okay.

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