Hi all,

Just a thought: Would it be possible to feed the stator field coil of an existing AC motor straight from pick-up shoe to mass, through a diode bridge + capacitor ? This would eliminate the need for a permanent magnet (at the cost of power consumption). What is the magnetic field strength that needs to be achieved

Gregor

Thanks Jorge,

Why should the current be cut when the train stops ? The magnet can be enabled all the time (like the real permanent magnet). It will have some power consumption, but that shouldn't be a problem.

Is there a way to measure the field strength of the permanent magnets ? This would enable me to create the same strenght using the old stator assy.

Best regards,
Gregor

If you just measure the DC resistance of the rotor and add a corresponding resistor, you will get a too low value (too high current). In analog, the motor is fed by AC, and the AC impedance is higher than the DC resistance. (Impedance and resistance isn't exactly the same thing, but for this purpose you can consider them equivalent.) Also, there is a backward EMF generated in the rotor when it rotates. This effectively reduces the voltage over the field coil, and thus the current.

With a given rotor voltage, the strength of the motor is proportional to the field strength of the field magnet, which is proportional to the field coil current up to a saturation point. When the iron is saturated, further increase of the field coil current will increase the field strength very little. The speed of the motor is inversely proportional to the field strength. So, higher field coil current will give a stronger but slower motor.

The fact that the speed is inversely proportional to the field strength may seem confusing at first. It is of course only valid at ideal conditions, but generally it is true that decreased field strength gives a faster motor, unless the load is very heavy.

So, why ? The rotor generates a backward EMF (Ve) which is proportional to speed (n) and field strength (f): Ve = k x n x f , where k is a constant specific to a certain motor. If the motor runs with low load, it accelerates until the EMF balances the applied voltage (V) on the rotor. Thus, if we ignore the losses in the motor, so that V = Ve, we get: n = V / (k x f), i.e. speed is inversely proportional to field strength.

I noted that I had made an error in my previous post: I wrote "With a given rotor current", it should be "With a given rotor voltage". I have edited the previous post to correct it.

A very good technical explanation perz!

Infact in Italy where locomotives are DC voltage fed (3000 V) to increase the speed the field strength is reduced as the running characteristics of such a motor has a hyperbolic behaviour (less speed=more torque; more speed=less torque). The field strength reduction is achieved reducing the current flowing through the field coil of the stator by means of shunt, i.e. short-circuiting some coils of the field coil. In this type of motor the field coil is connected in serie with the rotor winding (yes as in older Marklin AC motors).

One thing more: in AC fed circuits in addition to resistance and impedance also the so called inductance is present, due to the fact that the coils are made of many turns of isolated copper wire on a magnetic circuit. If you remember the Pitagora's law, you can consider the impedance as the longest side of a rectangular triangle, where the other sides are the resistance and the inductance (so 90 degrees turned one in respect to the other). A drawing would explain better what I said, but you can pick up a geometry book and refresh your school knowledge (I think this doesn't apply to perz).

Renato