1. Lecture 18. Electric Motors simple motor equations and their application 1

2. I will deal with DC motors that have either permanent magnets or separately-excited field coils This means that all I have to think about is the armature circuit. We apply a current to the armature. The current cuts magnetic field lines creating a local force This results in a global torque, making the armature accelerate The motion of the armature means that a conductor is cutting magnetic field lines which generates a voltage that opposes the motion, the so-called back emf All of this is governed by just three equations. 2

3. The torque is proportional to the armature current The back emf is proportional to the rotation rate These are connected by the voltage-current relation For most control applications the time scales are slow enough that we can neglect the inductance Ohm’s law is good enough 3

4. 4 The voltage in Ohm’s law is the sum of the input voltage and the back emf The two proportionality constants are generally more or less equal Combine all this to get

5. 5 The maximum rotation rate is at zero torque: the no load speed The maximum torque is at zero rotation: the starting torque   Power is torque times speed, so its maximum is at half the no load speed.

6. 6 You can find K and R from the starting torque and no load speed at whatever nominal voltage is given for the motor There are other things you can do. This is discussed in the text.

7. 7 Take a look at some simple dynamics One degree of freedom system — the red part moves

8. 8 You can easily verify that the ode is which we can rearrange We can solve this for

9. 9 To visualize this let K, R, C = 1 and suppose the input voltage to be sinusoidal

10. 10 Let’s look at a simple control problem: find e i to move  from 0 to π Define a state Write the state equations, which are linear Define an error vector

11. 11 The error equations are The x d term does not actually appear in the equation because the first column of A is empty

12. 12 So we have Controllability which is obviously of full rank (square with nonzero determinant)

13. 13 We can define a gain matrix, here a 1 x 2 and write The controlled (closed loop) equations Characteristic polynomial

14. 14 In this case we don’t actually need a g 2 to stabilize the system — the motor can do that for us — but we can use it to place the eigenvalues Substituting gives us a formula for the input voltage

15. 15 Combining these two equations leads us to the same input voltage We can look at this from the nonlinear perspective, even though it it a linear problem

16. 16 What happens if we replace the torque as the controlling element for the robot with voltage?

17. 17 Each torque is given in terms of its motor constants, input voltage and shaft speed I’ll suppose the motors to be identical for convenience’s sake

18. 18 We need to figure out the  s and size the motor The motor has to have enough torque to hold the arms out straight Speed is not a big issue

19. 19 The maximum torque required is that needed to hold the arms out straight I can set the motor starting torque equal to twice this, from which Choose K = 1 and e MAX = 100 volts The no load speed is 100 rad/sec = 955 rpm

20. 20 We can see how all this goes in Mathematica