Motor RPM to Voltage Ratio

[wvw]

Hello

I am currently working on a project where I want to use a DC motor (like the ones bought at electronic shops) and apply mechanical power to the shaft. This will in turn provide a voltage across the motor's terminals from where I can connect a load.

The questions I have are as follows:
1.) What is the ratio between the shaft revs per minute to the Voltage output?
I understand that the faster the shaft is turned the higher the Volts output, but I
would like to know if there is a known ratio or formula to determine this?
2.) How does the motor's torque affect the ratio from question 1.)?
Can this be added to a formula to calculate (and ultimately plot a graph) the
Voltage output?

Your help is much appreciated.

wvw

 

[waross]

If this is a permanent magnet motor the volts per RPM ratio is pretty linear. The greater the load, the more deviation from linearity.
The torque required to turn the motor will depend on the power being drawn and the speed. (plus losses).
The ratio will depend on the motor. Buy it and try it!

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

 

[cswilson]

The ratio of terminal voltage to motor speed is usually called the "back EMF constant" (Ke). As Bill says, for a permanent-magnet motor, it really is essentially a constant ratio (as long as currents do not push the iron into magnetic saturation). Many, if not most, of these motors will show this number in the data sheet.

These motors also have a "torque constant" (Kt) - the ratio of motor torque to current. Again, for this kind of motor, the ratio really is constant over a wide range of operating conditions. These two constants are really the same thing - two sides of the same coin (by conservation of energy). In consistent units (like SI), they are the same number. If you change the motor design to increase one, the other increases by the same percentage as well.

With no load, the voltage you measure across the terminals should be directly proportional to the speed. As you apply an electrical load, you will start drawing current. The measured voltage across the terminals will drop by the product of this current times the resistance of the motor winding. The current drawn will also apply a load torque to the motor, proportional to the torque constant.

Curt Wilson
Delta Tau Data Systems

 

[mikekilroy]

with the excellent descriptions above out of the way, your equations are:

Vterminal=Ke * N + I * R
Torque = Kt * I

with these you can plot it all

 

[wvw]

Hi Everyone

Thank you so very much for all the awesome responses!

@Bill - I fully understand what you mean with Buy it and try it I always enjoy putting things to the test, however it can sometimes turn out to be a bit time consuming that's why I thought of asking in order to cut my workload in half.

@Curt - What happens when the current pushes the iron into magnetic saturation?

I would also like to ask my next question. Apart from buying a motor off the shelf, I was thinking about hand winding my own motor (ie construct the motor myself). With the end result being about the same physical size than the "off-the-shelf" motors. The reason for this is that I am looking for max Volts/Current output with minimum rotation at the shaft end.

What I have been thinking about is the following:

If I hand-build the entire motor I can pick the best materials to be used (read data sheets of the different components) and from there build my own motor which should end with the result I want? Or has this already been covered by motor manufacturers?

 

[waross]

The rotor (armature) windings generate AC and the rotor steel is subject to an AC magnetic field. AC magnetic fields produce eddy current loses in iron cores. For that reason a small DC motor will run cooler and with less losses with a laminated rotor. A large DC motor with a solid rotor will be quite inefficient due to eddy current losses in the solid core.
Stamping, insulating and assembling the rotor laminations is probably best left to a manufacturer.
Saturation: When the iron is saturated an increasing current will no longer result in a increased magnetic field strength. All other factors being equal, the voltage can no longer rise in response to an increasing excitation current. The volts per rpm is no longer linear.
Also, in an AC circuit, the inductive reactance does not limit that part of the current above the current required below saturation. in some situations rapid over-heating and failure can result from operation in saturation.
Saturation affects more than just motors, many AC devices such as transformers, relays and other devices may be subject to saturation.
The wound field of a DC motor motor is subject to saturation but the resistance is the current limit so the current does not become disproportionately high.
Generally saturation is the result of over voltage and results in heat and non-linearity.

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