# Permanent Magnet DC Motor Transfer Function

D

#### Damian

I am working on a PMDC motor for which I have found the transfer function to be as follows:

[Kt]/[(sL+R)(sJ+C)+(KtKe)]

Could you please help by showing me the steps on how to convert this to a tranfer function with respect to the electrical and mechnical time constants?

Derivation of the electrical, mechincal time constants and the final simplified transfer function would be of great help... THANKS

T

#### Tom Bullock

My partner, George Younkin, wrote a paper on this. Send me your e-mail address and I'll forward 2 papers on the subject.

Tom

Thomas B. Bullock, President
Bull's Eye Research, Inc.
Fond du Lac, WI 54935-9507
Ph: 920: 929-6544
Fax: 920: 929-9344
E-mail: [email protected]
www.bullseyenet.com

S

#### sreid

If you post a FAX number I will send it to you.

S

#### sreid

Tom,

I'd like a copy for the paper. My email address is

[email protected]

Steven Reid

C

#### Curt Wilson

First, remember that the concepts of mechanical and electrical time constants are approximations here -- that they are not truly independent in this context.

To see where the mechanical time constant concept comes from, assume that the inductance L is so small that it can be ignored (set L to 0). This is assuming that the current response to a voltage step is so quick that it is effectively a step as well. Your denominator reduces to:

s + (Ke*Kt)/(J*R)

which is s + 1/Tm. (This assumes that C -- which I think you mean is a mechanical damping term -- is also 0.)

Next, assume you had a locked rotor so the motor could not move (no back EMF). In this case, your current would react to a voltage step with an electrical time constant of L/R.

Finally, assume your transfer function is composed of independent electrical and mechanical time constants. You would have a transfer function of:

1 / [Ke * (Te*s + 1) * (Tm*s + 1)]

Process the denominator until you get it in the form:

s^2 + As + B

The B term will be

R/L + (Ke*Kt)/(JL)

In the real transfer function, this term is just R/L. But if R/L is much bigger than the other term, which it usually is, you have a reasonable approximation.

Curt Wilson
Delta Tau Data Systems

C

#### Curt Wilson

Sorry -- In the above post, I meant to say:

The B term will be

(Ke*Kt)/(JL)

the same as in the actual transfer function.

The A term will be

R/L + (Ke*Kt)/(JR)

In the real transfer function, this term is just R/L...

Curt Wilson
Delta Tau Data Systems