Magnetic circuit of PMDC motor
Magnetic circuit of PMDC motor is composed of permanent magnet (as magnetic source), soft magnetic material (as magnetic circuit) and air gap (as energy exchange medium).
Figure 8 is a typical magnetic circuit distribution drawing of a 3P DC motor. Let us get to know the essential features of magnetic lines first.
●They tend to choose the shortest path.
●They tend to choose the magnetic path with the smallest resistance.
●They close the loop by themselves (from n-pole back to s-pole)
![Magnetic circuit of PMDC motor Magnetic circuit of PMDC motor]()
For easier analysis, we ignore the magnetic flux leakage. The magnetic flux Φ (magnetic line) goes out of n-pole and get into one tooth of the rotor through the air gap. It is divided into 2 parts there and reaches the other 2 teeth through the rotor yoke. They are united together there, go through the air gap at the other side and get into the s-pole. After that, it splits into 2 parts and goes back to n-pole through the yoke of the stator.
Figures 9,10 are the equivalent magnetic circuits for a 3P DC motor.
![Magnetic circuit of PMDC motor Magnetic circuit of PMDC motor]()
From figure 10, we can get the magnetic flux Φ
Φ=FFe/(2Rmδ+1½Rmt+½Rmy1+½Rmy2) (5)
This formula only works for the magnetic circuit of a 3P motor. 5P, 7P and even-slot motor have different structure. Flowing direction of the magnetic flux is slightly different. Besides, the position of the rotor also influences the flowing direction of the magnetic flux.
There are a lot of nonlinear soft magnet materials such as silicon steel and galvanized steel used in the magnetic circuit of the motor. Their magnetic resistance is not a fixed value. It varies along with its magnetic flux. So, it is unpractical to calculate the magnetic flux by formula (5) in engineering practice.
In actual motor design, usually we set a series of Φ1、Φ2…… Φn values first. Then we calculate the corresponding magnetic flux density B1, B2……Bn according to area of each section of magnetic circuit. After that, we get the magnetic field strength of each section H1、H2……Hn by the formula H=B/μ. We will get the curve Φ=f(H). We can get the true magnetic flux together with the demagnetization curve ΦFe=f(HFe). From formula E=KE•Φ•n(KE =2W/60 ) we can get the turns of the coil W. Further on we can get the wire diameter by the coil space factor and linear load. We will not discuss further on these in this article.
Finally, there is one thing to be emphasized: Any problem in design work of any product can be concluded as known/request binary problem.