DC motor starter

What horse power is the motor? Small motors aren't much of a problem, as relays rated for switching DC currents up to around 20 amps are readily available. Likewise, DC ouput, zero-crossing, solid state relays (SSR's) can switch rectified, un-filtered, AC power, which is typically what DC shunt-wound motors are rated for.

the problem statement is :

Draw a neat diagram of a 4-point dc shunt motor starter. Write one paragraph explaining the performance of this starter.

Are you the same "Guest" poster who wrote the original post?

A general description of a DC Shunt Motor is that the field windings have two leads, and the brushes have two leads. When field and armature (brush leads) are connected to a DC source with the same polarity, the motor turns in direction A. When either the field or armature leads are reversed in polarity, the direction of the motor reverses, which we'll call direction B.

Now, given field winding direction or magnetic field polarity, and armature winding direction or magnetic field polarity,following any acceptable convention, the direction A or B of the armature for a given DC polarity may be determined apriori.

See http://www.distel.co.uk/DC_MOT_CON1.htm or http://www.mitedu.freeserve.co.uk/Circuits/Switching/reverse.htm for diagrams, but note that each shows a diagram for a permanent magnet field DC motor. The only benefit to using a wound-field DC motor is that the field current may be reversed, which is usually much smaller than the armature current and therfore easier/cheaper to switch with less robust components. Speed control is also possible via field current control (usually much smaller than armature current as stated previously), which is directly related to magnetic field strength: As the armature turns in a constant magnetic field, it will reach a speed proportional to the magnetic field (back EMF). If the field strength varies, the speed will vary.

yah that's me,

i really appreciate your help.

what about the 4 point starter ?

Well, the 4 point starter is merely two contacts for the armature and two for the field, which usually operate on the same maximum voltage.

Either the armature or the field may be operated at other than maximum voltages, each giving speed control. Please note that field current is proportional to applied field voltage and magnetic field strength. By reducing the field voltage (and therefore the field current) the armature will speed up. The reverse is true for varying the armature voltage: The higher the armature voltage, the higher the speed.

OK, going back a few years, a four point starter for a DC machine was a four stage across the line starter. This was done for fractional through hundreds of HP sized motors. Technique in a nutshell was to use series resistance to limit the current to the motor by adding external resistance. As the motor comes up to speed the external resistance is decreased in stages.

Here is an example of a three stage DC motor starter.

Say you have a 500VDC line, motor with Name Plate 100amps full load, with a 50V IR drop from no load to full load (motor R = 50/100=0.5ohms) . So if the motor is stopped and you drop it across the 500V you will draw 1000 Amps and the motor blows up. So, add series resistance.

Stage 1) allow maximum of 200% current on stopped motor (Eg = k*rpm = 0). Imax = 200% *100 = 200 Amps, Rtotal=500V/200amps=2.5ohms, therefore Radded = Rtotal-Rmotor = 2.0ohms.

stage 2) now the motor will accelerate, creating a counter emf, and this will reduce the current draw. Say when the current drops to 100%, you short out some of that resistance. At 100 Amps and Rtotal=2.0ohms you develop 200volts with resistance, and the cemf is now 300 volts. So we now want 200amps again, so Rtotal becomes = 200/200 = 1.0ohms, so external resistance drops to .5 ohms, so you just short out 2.0-.5 = 1.5 ohms.

stage 3) At 100 amps and Rtotal = 1.0 ohms, IR drop is 100 volts and cemf is 400 volts. 200 amps again, so Rtotal = 100v/200amps = .5ohms, Radded = 0ohms, so you short out the remaining external resistance.

So you can see that the number of stages is dependent on the max allowed motor current, rated IR drop of the motor, and switching points of current (say 110% instead of 100%).

The wattage sizing of the resistors becomes very dependent on the inertia of the started load. It was quite common to size the resistors on their watt second capacity to keep them as small as possible. For high inertia loads the resistors become 100% rated for continuous dissipation. It was costly in that the resistors were large, and you needed contactors rated for making and breaking the DC currents.

With some playing around with the math you should be able to determine the number of stages and the resistance value required. I just can't put my hand on it at the moment. Good question for an undergrad.

Hope this has caught the "thread" of your question.

Ahhh . . . I wasn't thinking about large motors! I asked what HP he was concerned with, but as an ME student of unknown class rank, I figured it was an elementary course.

A diagram for large HP motors is available at http://www.angelfire.com/nt/ramson/Motorstarter.htm and a diagram for small motors is available at http://zone.ni.com/devzone/cda/ph/p/id/54 complete with reversing circuit.

It's always interesting seeing different solutions coming from dofferent applications.

My application was from the paper industry where the motor came up to line speed on a common generator supply for the machine. It allowed starting and stopping of individual sections and the total machine speed was controled by the main generator supply. (Individual sections speed trimmed by field.)

Your sample drawings drew me back to the controls for electric street cars in Toronto Canada, before SCR controls!

The starting procedure for a shunt connected motor is similar to that of the series motor. In both types a faceplate starter is used to add resistance in to the circuit during the starting process.

When starting the shunt connected motor the resistances are connected into the armuture circuit only as this is where the high starting currents will be. When the armuture is rotating at full speed and the maximum back e.m.f is being produced, the resistors are shorted out. Overcurrent and no voltage protection are similar to thoses on the series type motor

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Thank you and good night

Hi my name is Omar, I'm also a student at KFUPM. I came across this thread from a google search, while searching for a way to design a certain starter, that keeps the current at 150% of the load, I just wanted to thank you guys for the help, the thread really helped me a lot and I will start directing my colleagues here.