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question about question about motors, speed controllers and power supply

George Higgs

Sep 18, 2012
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Sep 18, 2012
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Hello,
I'm looking to buy a motor for a machine I'm building.
I'm looking at this motor: http://radionics.rs-online.com/web/...54D424552267573743D67706D392677633D424F544826
It's rated at 41 watts, 14.5 volts and 6.9 amps. It goes from 0 – 6000 rpm. I can't figure out why that is, if power = volts x current, I figure the amps should be much lower, but obviously I missing something in my thinking.
In any case, the thing is, and the reason the amps are pertinent, is that the only speed regulator I can find has a continuous limit of 3 amps (http://radionics.rs-online.com/web/...46F722C362D3135562673633D592677633D4E4F4E4526), and a peak of 5. The guys at radionics (the online shop) said this wouldn't do, but I'm thinking that because my shaft speed is going to be set at a constant 1920 rpm(I'll never go above this), I am using under a third of the 6.9 amps current the motor could potentially demand, therefore keeping well under the 3 amp limit of the speed controller.
The other issue is a 14.5 volt power supply (the motor is rated at 14.5 volts) which I'm having trouble finding. The larger question that I really had is about how the motor/speed regulator/power supply relationship basically works. Is the motor connected to the speed controller, which is connected to the power supply, so essentially the motor only gets what the speed controller gives it? In that case, does the power supply have to be 14.5 volts, or can't it be below that.
Thanks very much.
-George Higgs
 

duke37

Jan 9, 2011
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The data I saw said 3000 rpm not 6000.
The voltage on a DC motor is approximately proportional to speed so the voltage at 1920 rpm will be 14.5 * 1920 / 3000 plus a bit for drop across the brushes.
To adjust the speed of the motor, you need to adjust the input voltage.

The current taken will depend on the torque which you are extracting, the data gives the current at full torque.

There will be some drop in speed as the torque is increased and extra power lost in the motor which will show up as heat.
 

George Higgs

Sep 18, 2012
3
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Sep 18, 2012
Messages
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George Higgs

Hello duke37,
Thanks for your reply.
The webpage does give 3000 rpm, but the spec sheet I downloaded says it's 0-6000 rpm. I don't understand this, and perhaps I've misinterpreted it.
Also on that sheet there's a graph which shows that the torque at 1920 rpm is 16 ncm.
I'm trying to calculate current from that torque but I'm having a bit of trouble understanding how that's done.
As well as that, I'm confused as to how the load on the shaft will affect the torque. Does the load on the shaft that the motor is driving increase the torque in some manner?
Please forgive my elementary questions, but I am struggling a bit with this.
Thanks,
-George
 

duke37

Jan 9, 2011
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The torque will be from zero to 16Ncm (not 16ncm). It is better to use the SI fundamental units i.e.Nm.

Fundamental unit of rotational speed is radians per second.

1920rpm = 1920*2*pi/60 = 201rad/sec
Power = speed * torque = 201 *0.16 = 32W
Power = Voltage * Current so
Current = power/voltage
so you can get an idea of torque by measuring current.

All assuming 100% efficiency (some hopes!)

The torque IS the load on the shaft, it is the twisting force. The current the motor takes will be proportional to torque but some torque will be used inside the motor to keep it turning against friction.
 
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