Feeding more current to a motor using a Step down transformer?

[Hol]

A motor under no load uses 10 Amps, however at max load draws 120 Amps at 12 Volts.
Usually, I would apply more voltage to the motor(without load) to increase the current. If the total input power is 1440W, and the power source is 1500W, why can't I use a step down transformer to force the voltage to become 12V(or near it) and have that maximum current without any load?

 

[jraef]

Your question points to a basic misunderstanding of what motors are and how they work. What is it you are attempting to accomplish here?

"Will work for (the memory of) salami"

 

[OperaHouse]

"Sometimes you can't tell the AC's from the DC's." Firesign Theater

What would be the intended purpose of driving this motor at no load with 120A? The likely outcome would be it would fly apart.

 

[squeeky]

Lots of helpful guys here Hol but weneedto understand the question or better still the desired outcome with the limitations. At12v we assume it is a DC motor. 120A at 12v is 1440w input power so your calculations correct. At no load it draws 10A just to make it turn. You don't give the no load voltage but I have assumed it to be constant. Now why do you want to draw max current? If you increase the voltage at no load, the current will drop. On a universalDC circuit W = VxI so therefore I = W/V. V up, W constant, I goes down. If you reduce the voltage, the current will rise. If you keep reducing the voltage the motor will eventually stall as the torque comes from the voltage.

 

[OperaHouse]

Sure the motor looks DC, but where does the transformer come in?

 

[Hol]

Sorry for my terrible misunderstanding, I'm not an expert in this field at all. Hence why I come here for help.
This is a lift motor(i.e winch) that can carry objects over 1000 lb(load). This particular DC motor, has the potential to generate 4448 of force on objects that weight 1000 lb. Can that mechanical output(the force) be applied on smaller lighter objects that weight less than 1000 lb, maybe 100 lb or 50 lb?

I understand that voltage increases RPM of the motor, however current increase the torque?

 

[jraef]

When a motor rating is provided, it is almost always giving you the MAXIMUM capability, and even then it means the maximum capability without serious harm to itself.

The work (torque and speed) a motor performs is only what the load needs from it, up to and including that maximum capacity. Anything less than that just means the motor is not having any trouble at all. You even said it in your statement;

...has the potential to generate 4448 of force...

, but the flaw in your thinking is what you tagged onto that;

... on objects that weight 1000 lb.

The motor is capable of 4448 lb. ft. (assumed) of torque, regardless of the weight of the material connected to the shaft. The 1000lbs. would be the lift mfrs relation to what that much torque means to their lift, including the mechanical strength of other components, gearing etc. Anything less than that is no problem.

And yes, torque and current go hand in hand. But artificially lowering the voltage to increase the current does NOT increase the torque.

"Will work for (the memory of) salami"

 

[mikekilroy]

Usually, I would apply more voltage to the motor(without load) to increase the current.

Really? What planet are you on? If you are suggesting your dc motor pulls more than a small increase of current by applying a large increase in voltage, you must not be on earth. Our physics just does not work that way; here, that would just make it go proportionately faster!

http://www.KilroyWasHere<dot>com

 

[mikekilroy]

Now back to your original post, without enough info to make any really good guesses, I can make a bad guess that you are making really good sense... Does that make sense?

Let me guess this: You want 1440watts of power, wth the voltage being 12vdc; hence you want the current to be 120amps dc. Since on earth a Dc motor has a rating called Kt in units of torque/amp, then say you want 4448#-ft of torque, and your motor has a Kt=37#-ft/amp.... then if you put in 120 amps, you get out 4448#-ft. it is a pure simple linear relationship.

NOw to your power source. Let's guess it is 150volts dc. That means it has up to 10 amps max available..... hmmmm.... so you can only get 37#-ft/amp * 10amp =370#-ft of torque out of the poor motor.... but we want 4,448 #-ft..... we are what they call SOL or up the creek without a paddle....

what if.... we could magically transform the 150vdc down to 12v dc at the same 1500 watts power capacity? VIOLA! We got it!

But alas, dc voltage cannot be changed with just a transformer.... so more guessig how to get where you want to go.... I am out of guessing at this point. Need more info....

http://www.KilroyWasHere<dot>com

 

[waross]

If the lift can lift 1000 lbs. it can easily lift 100 lbs or 50 lbs. No need to try to mess with the voltage or current. It will draw as much current as it needs to lift the load.

Motors and voltage:
At first look it seems that double voltage on a DC shunt motor would double the field strength so that the back EMF per unit of RPM would be doubled and the armature would run at half speed, but the armature voltage is doubled also, so the speed remains the same. The field resistance is the same so the field current would double. The armature current would reflect the work done so at the same speed, with the same load, the kW would be IE. E has been doubled so I will be one half. Yes lower current on some planet.
Here on earth, motors are not generally over-built. At around 115% or 120% of rated voltage the field will probably saturate. From then on we will be increasing the armature voltage with no increase in field strength. The motor will try to turn faster and will probably draw more current at voltages above the field saturation point.
I hope that this is not a series motor or a PM type motor. (My 6 liter truck engine is started by a PM motor, so 120 Amps at 12 Volts is well within the range of PM motors.)


Bill
--------------------
"Why not the best?"
Jimmy Carter