I'm trying to determine the transfer function of a simple DC motor connected to a resistive load, the added difficulty is in measuring the temperature of the resistor as it heats up (with a thermistor).

The equation dealing with temperature changes are Q = m*c*(dT/dt), where the energy put into a material is equal to the mass of the material multiplied by the specific heat capacity of the material and the change in temperature of the material. In this case, it is a wire-wound resistor.

The motor's shaft is being rotated and this produces energy which will heat up the resistor. So basically the transfer function is of the form: Temperature/Speed (Output/Input). When I do the resulting working out (see pictures), I am left with Temperature/Speed^2 and this is confusing me, what have I done wrong?

Note: s = Laplace
J = moment of inertia
kT = torque constant of motor
w = speed of motor

I got i = (s*J*w)/kT by equating:

kT*i = Jdw/dt, solving for i, and converting to Laplace domain

The integral of Torque*w (Power) is the energy of rotational movement = 0.5*J*w^2, and the integral of the power dissipated by an inductor is 0.5*L*i^2

 

Hi,

Are you saying that you are driving a resistor with a motor using the motor as a generator?

Also, are you taking into account the cooling effect on the resistor due to the surrounding air? Other wise you have to mount it in an insulated sleeve of some kind and run the motor for a limited amount of time.

If you know the power delivered by the motor and the material of the resistor, you should be able to calculate the temperature rise. If there is heat loss due to convection it will be harder to figure out so you might just end up estimating it.

 

That's right, motor is acting like a generator. How do I take into account the cooling effect in the transfer function?