# Motor Voltage, Amps and Torque Relationship

There are a few motors I am interested in working with a project although they only show the 6V values.
The website said that reducing the voltage to 3 would double the current to maintain the same torque values.

I understand with Ohm's law that voltage and current are proportional. How does DC motor torque play into this relationship?

If I cut the rated voltage in half, would the rated torque be cut in half as well?

If I cut the rated voltage in half, and wanted to keep the same torque would the amps double?

Which states that because motors are not directly proportional devices, ohms law can not apply.

My primary question still would be to find out the effects on torque and amperage if you lower the voltage. Are the proportional? Or is there an equation to find them?

The overall goal is to be able to look at 6V motor data sheets and calculate their 3V values respectively.

This only applies to permanent magnet dc motors.

1. Speed is proportional to voltage
2. Torque is proportional to current
3. Power is the product of speed times torque.

The three values above must fall within the ratings listed in the motor datasheet. In general, to get the same power out of a motor at half the voltage, the current must double to get double the torque to ultimately provide the same power.

It is important to understand that the load determines the torque. If you do not have enough current to meet the torque demand, the speed falls off. If the torque demanded by the load exceeds the capability of the motor, you can usually overload the motor for brief periods, within the design limits of the motor and the power supply.

Hopefully this will help: Applying PMDC motors | Machine Design

You need to be sure that "torque" is what you mean, it's a much misunderstood and misused term.

But if you halve the voltage and maintain the same current you will have the original torque (turning force) but at only half the speed.

To maintain the same power (not torque) then halving the voltage would need you to double the current and that would increase the torque.

Because the basic relationship between Kv (rpm per volt) and Kt (torque per ampere) is fixed most motor specifications only give you Kv and leave you to calculate the other if you need it.

Steve

"a twisting force that tends to cause rotation"

Sure, we can make it more complicated than the above but is that really necessary?

But if you halve the voltage and maintain the same current you will have the original torque (turning force) but at only half the speed.

This can only happen with a constant torque load. A variable torque or constant power load will not demonstrate this behavior.

The motors mechanical load determines the torque requirement which ultimately determines the current. You never "maintain the same current" unless you are running a motor in torque limit at which point the speed is no longer proportional to the applied voltage.

avr_fred:

"a twisting force that tends to cause rotation"

Sure, we can make it more complicated than the above but is that really necessary?

You need to be precise, not compicated. Torque is the change in energy per unit angle. Just as force
is the change in energy per unit distance.

I suggest the conceptually clear way to think about torque is in joules/radian, but in practice we use
newton-metres.
Newtons are joules/metre, and hence newton-metres are really just joules (ie the idea of
angle has dropped out of our units) - thinking in terms of joules/radian avoids that.

Often mechanics calculations are simpler if done in terms of energy.

For our electric motors we can consider the electrical energy change is approximately the mechanical
energy change (for a quality DC motor at least!), so we can equate voltage x charge = mechanical energy.

Thus voltage x charge = torque x angle
thus voltage x current = torque x angular velocity
(for an ideal motor). Also in an ideal motor the torque is proportional to current, so we get

torque = k x current, which implies voltage = k x angular velocity. Its the same k, the motor constant (a property of the motor and its windings).

Remember all units are SI, so angle in radians, torque in Nm or J/rad, angular velocity in rad/s

The goal is to be able to transport a fixed weight, at a fixed speed. So the load value will not change.

avr_fred:

1. Speed is proportional to voltage
2. Torque is proportional to current
3. Power is the product of speed times torque.

Based on the above , I believe the answer I was looking for is ;

Find a motor with the correct torque, appropriate current, but 2x the speed at 6V I can just feed it 3V to bring it to my speed needed and maintain the same current and torque values.
Is this the correct approach ?
(assuming the motor is rated for 3V)

I am trying to accomplish this using a fixed 3.7V battery system. It appears most 6V motors can move this amount at the speed I need.

MarkT:
in an ideal motor the torque is proportional to current

Based on this, it sounds like the websites suggestion of "half the voltage, double the current to maintain the same values" was in reference to maintaining the same speed.

But if I half the voltage, doubling the current should only increase torque , not speed correct ? Theretore the website was incorrect in which values are affected by the current.

As they said the torque will stay and speed will increase but that's impossible if voltage is proportional to speed
And current is proportional to torque.

If you half the voltage the speed will reduce by AT LEAST a factor of 2 and both current and torque will also decrease. Think about it : Motor stall current at voltage V = A amps, so if supply is 0.5V then logically it follows that stall current is 0.5A since the only circuit affecting current is armature resistance. Mr Ohm at play.

It therefore follows that reducing a motors supply voltage must correspondingly reduce its current demand and hence reduce its torque output.

To maintain constant torque when you decrease voltage by a factor of 2 you need to add an extra stage of gearing which will decrease speed by yet another factor of 2. So in effect the shaft speed will decrease by at least a factor of 4.

ericfragola:
There are a few motors I am interested in working with a project although they only show the 6V values.
The website said that reducing the voltage to 3 would double the current to maintain the same torque values.

Assuming you want the motor to run at the same speed with either voltage then what is missing from that is the role of voltage in "forcing" the current through the motor.

If you have a specific motor and if you reduce the voltage from 6v to 3v then you CANNOT get the same current int he motor, never mind getting double the current.

On the other hand if you get a different motor that can provide the same torque with 3v rather than 6v you will probably find that it needs more current because it will probably have fewer turns in the motor coils.

The torque of a motor depends (approximately) on the current multiplied by the number of turns in the coils. If you increase the number of turns it is more difficult for the current to flow - so you need a higher voltage to get the same current.

The maximum current flows when the motor is stationary (stalled) and the only obstacle to current flow is the very low resistance of the wire in the coils. When the motor starts turning the motion generates a voltage that opposes the battery voltage and at maximum speed the opposing voltage will have risen so much that the difference between it and the battery voltage only allows sufficient current to flow in the coils to offset the friction in the system.

...R

MarkT:
Just as force is the change in energy per unit distance.

I don't understand that. It is possible to have a force without having any movement to provide the "distance" factor.

Similarly one can have torque without motion.

...R

But if I half the voltage, doubling the current should only increase torque, not speed correct?

That is a correct statement. Dropping the voltage will reduce the speed, not the available torque. In order to maintain the same power output at half speed one must double the torque. This may or may not be possible due to 1) the load, 2) the motor and 3) the power source. Those three issues, one at a time:

The load type connected to the shaft will determine if the torque required will change with speed. I don't know what the OP's load is so here are the three basic load types.

• Constant torque A load where the torque required is independent of the speed applied. A material conveyor or screw feeder are examples. The power required is linearly proportional to speed since torque remains constant.

• Variable torque Sometimes referred to as Quadratic torque. The most common example would be a fan or propeller and most pumps. The torque required to turn a variable torque load goes up at a rate of the speed squared. Since power is speed times torque, the power required to turn such a load goes up at a rate of the speed cubed. This has a huge effect on the torque required and good example is this: For a given system at 100% power and 100% speed, reducing the speed by 50% would reduce the power by 87.5%
With a variable torque load, reducing speed will always result in a significant reduction in actual torque, you cannot increase the torque required when decreasing speed.

• Constant power The required torque is inversely proportional to the speed. An example would be a grinding wheel.
The problem with small hobby motors is that the specifications generally do not provide a maximum power rating. Voltage and torque ratings are given but good luck finding useful data like torque per amp/milliamp, you have to extrapolate that from the stall torque and current. Without a power rating, it is difficult to predict what the motor life will be for a given set of conditions.

Small hobby PM motors are high speed with corresponding high frictional losses - which cause most users to misinterpret what they see with regard to the three basic physical relationships that exist (stated in post #2) in a PM DC motor. Ultimately, is the motor selected capable of running at the required torque? How far away from the stall torque is that rating? I'd venture a guess and say anything more than 50% of stall may lead to shortened motor life. I'm not knowledgable there so I leave that to others to comment.

Lastly, there is the issue of available current. Can your power supply provide what the motor requires to produce the desired torque? If not, the speed falls off since the torque is being limited by way of the supply current. That's the easiest of the lot to solve As for some of the other material posted here, it is misleading at best. For example:

What is missing from that is the role of voltage in "forcing" the current through the motor. ...snip...

It is missing from the discussion because it does not exist. The voltage applied has zero effect on the available torque. A motor is not a resistor, amps are in no way proportional to applied volts.

If you have a specific motor and if you reduce the voltage from 6v to 3v then you CANNOT get the same current in the motor, never mind getting double the current.

Of course you can. Your point is only valid when discussing stall currents. A one amp motor running at 6V and 200ma is certainly capably of running at 3V and 400ma. It's a question of load behavior, not applied voltage.

avr_fred:
As for some of the other material posted here, it is misleading at best. For example:

What is missing from that is the role of voltage in "forcing" the current through the motor. ...snip...

It is missing from the discussion because it does not exist. The voltage applied has zero effect on the available torque. A motor is not a resistor, amps are in no way proportional to applied volts.

If you have a specific motor and if you reduce the voltage from 6v to 3v then you CANNOT get the same current in the motor, never mind getting double the current.

Of course you can. Your point is only valid when discussing stall currents. A one amp motor running at 6V and 200ma is certainly capably of running at 3V and 400ma. It's a question of load behavior, not applied voltage.

Thank you for these comments. I see that I forgot to say that my comments referred to the motor operating at the same RPM. I had assumed that that is what the OP would want - but I should have stated it explicitly. I will amend my Reply #8.

...R