I have an extruder motor that has rating of 700HP 500vdcARM 300v/150vdc.field 8.9a/17.8 amp field @ 1750RPM. Currently under load we have 800amps on the armature @1750RPM. We are using 300vdc.field EXC Shunt.The manufacture say the motor is capable of 2400RPM. If we purchase a field weaking board. My first question is what will happen to arm.current at the same load go up or down and why?My 2nd question is where can I get some mathematical formulae on dcmotors because I measured the field resistance before installation and got 26ohms on the field.Using ohms law does not apply 300v/26ohms does not equal 8.9amps as rated on nameplate.
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FIELD WEAKENING 3
- Thread starter mag100
- Start date
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2
- #2
electricpete
Electrical
I'm not much on dc motors, but I think I can start to answer your questions. Reliance motor has some good theory on their web page.
My first question is what will happen to arm.current at the same load go up or down and why?
The equation for predicting effects on armature current is
Ia=(Vt-Eg)/Ra = (Vt-NKPhi)/Ra
where
Ia = armature current
Vt = Terminal voltage
Eg = generated voltage = NKPhi
N = speed
K= constant
Phi = flux.
Perhaps it depends what you mean by constant load.
If speed is kept constant (in spite of change in motor torque-speed characteristic with increasing field current), then the equation predicts that armature current will go down.
If horsepower is kept constant... let me think about that one (anyone else have answer?)
My 2nd question is where can I get some mathematical formulae on dcmotors because I measured the field resistance before installation and got 26ohms on the field.Using ohms law does not apply 300v/26ohms does not equal 8.9amps as rated on nameplate.
If you have measured the resistance at room temperature, then that is probably the cause of the difference.
Copper resistance increases approx 0.4% per increases in temperature in degrees C. If you increase the temp on the order of 60C, you get approx 23% increase in resistance, which I believe would account for the difference.
My first question is what will happen to arm.current at the same load go up or down and why?
The equation for predicting effects on armature current is
Ia=(Vt-Eg)/Ra = (Vt-NKPhi)/Ra
where
Ia = armature current
Vt = Terminal voltage
Eg = generated voltage = NKPhi
N = speed
K= constant
Phi = flux.
Perhaps it depends what you mean by constant load.
If speed is kept constant (in spite of change in motor torque-speed characteristic with increasing field current), then the equation predicts that armature current will go down.
If horsepower is kept constant... let me think about that one (anyone else have answer?)
My 2nd question is where can I get some mathematical formulae on dcmotors because I measured the field resistance before installation and got 26ohms on the field.Using ohms law does not apply 300v/26ohms does not equal 8.9amps as rated on nameplate.
If you have measured the resistance at room temperature, then that is probably the cause of the difference.
Copper resistance increases approx 0.4% per increases in temperature in degrees C. If you increase the temp on the order of 60C, you get approx 23% increase in resistance, which I believe would account for the difference.
rockcontrolman
Electrical
mag100:
The DC Variable Speed Drive will not just put out the specified voltage, but will actually act as a current source to supply the desired field current (the current is the important value here). Make sure and put this parameter (field current) in correctly; I have changed DC motors in the past and failed to make a presumably small change in the field current parameter for the new motor(less than 2 amps) and caused a motor to fail in 24 hours (nothing like learning the hard way).
Also note that if your drive is currently set up for "Voltage Feedback" instead of tach feedback, you will be unable to activate the field weakening feature in your drive; in field-weakened state, the armature voltage is no longer an accurate measure of the speed of the motor. You must have tach (or resolver) feedback to use this function. I'm assuming from your question that the motor is rated for this higher speed.
I believe I would also increase the frequency of brush/bearing maintenance on this motor at this increased speed.
Good Luck!
The DC Variable Speed Drive will not just put out the specified voltage, but will actually act as a current source to supply the desired field current (the current is the important value here). Make sure and put this parameter (field current) in correctly; I have changed DC motors in the past and failed to make a presumably small change in the field current parameter for the new motor(less than 2 amps) and caused a motor to fail in 24 hours (nothing like learning the hard way).
Also note that if your drive is currently set up for "Voltage Feedback" instead of tach feedback, you will be unable to activate the field weakening feature in your drive; in field-weakened state, the armature voltage is no longer an accurate measure of the speed of the motor. You must have tach (or resolver) feedback to use this function. I'm assuming from your question that the motor is rated for this higher speed.
I believe I would also increase the frequency of brush/bearing maintenance on this motor at this increased speed.
Good Luck!
For your first question, when you go past the base speed of the motor you will then run in constant horsepower mode but your torque will decrease as the speed increases from that point forward. With your 700HP motor you get 2100.8 lb-ft of torque up to 1750RPM ((HP*5252)/SPEED) and at 2400RPM you will have 1531.8 lb-ft of torque (Base Torque*(Base Speed/Run Speed - if > Base Speed)).
D.C. motors can basically perform two functions according to the name plate and I can only comment from experience with Siemens machines. The motor is designed to operate on full torque delivery from 0 to full armature volts with constant full voltage field and this is what dc motors are even today best equipped for in the low rpm range with tacho feedback. At this full speed point is where the fun begins as the shunt field is weakened the speed increases and we are in the constant power range but not constant torque range as torque falls off in proportion to speed to a speed and current usually stated this is classic of spindle motors on machine tools so you will hear motor will not drive load at speeds much greater than base speed and that's because it's not supposed to, really a fine line between speed and current.
Mars
Mars
D.C. motors can basically perform two functions according to the name plate and I can only comment from experience with Siemens machines. The motor is designed to operate on full torque delivery from 0 to full armature volts with constant full voltage field and this is what dc motors are even today best equipped for in the low rpm range with tacho feedback. At this full speed point is where the fun begins as the shunt field is weakened the speed increases and we are in the constant power range but not constant torque range as torque falls off in proportion to speed to a speed and current usually stated this is classic of spindle motors on machine tools so you will hear motor will not drive load at speeds much greater than base speed and that's because it's not supposed to, really a fine line between speed and current.
Mars
Mars
cbarn24050
Industrial
Hi, your fisrt question has no real meaning because the field will still be at nominal at that speed.If you were to reduce the field current the motor spped would rise and therefore the load would rise and therefore the armature current would rise.
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1
- #8
electrageek
Electrical
I only started with Eng-Tips recently so this may be too late to help.
I use the following equations to help me understand the characteristic of a DC motor.
Terminal Voltage Equation:
Vt = Vcemf + Ia*Ra + BD
Where Vt = terminal volts (what you can measure)
Vcemf = volts counter EMF (used by motor)
Ia = armature current
Ra = armature resistance
Ia*Ra = internal motor voltage loss
BD = brush drop (consider 2volts and ignored)
Speed equation
n = Vcemf/(N*If)*K1
Where n = speed
N = Number of turns in field winding
If = field current
N*If = field flux
K1 = constant
Torque Equation
T = Ia * (N*If) * K2
Where T = Torque
K2 = Constant
Horsepower Equation
HP = (T * n)/746
Where HP = horsepower
746 = constant KW to HP conversion
From the equations you can see that speed is directly proportional to Vcemf and inversly proportional to If. Double the armature voltage and you double the speed. Cut the field current in half and you also double the speed. However you notice that the Torque equation is not related to speed or voltage. As you weaken the field in a constant torque application the speed will go up but the armature current will have to also increase to keep the torque at the same value. So you can only weaken the field until you either reach top speed or the armature current reaches 100% then your out of Torque and HP.
I use the following equations to help me understand the characteristic of a DC motor.
Terminal Voltage Equation:
Vt = Vcemf + Ia*Ra + BD
Where Vt = terminal volts (what you can measure)
Vcemf = volts counter EMF (used by motor)
Ia = armature current
Ra = armature resistance
Ia*Ra = internal motor voltage loss
BD = brush drop (consider 2volts and ignored)
Speed equation
n = Vcemf/(N*If)*K1
Where n = speed
N = Number of turns in field winding
If = field current
N*If = field flux
K1 = constant
Torque Equation
T = Ia * (N*If) * K2
Where T = Torque
K2 = Constant
Horsepower Equation
HP = (T * n)/746
Where HP = horsepower
746 = constant KW to HP conversion
From the equations you can see that speed is directly proportional to Vcemf and inversly proportional to If. Double the armature voltage and you double the speed. Cut the field current in half and you also double the speed. However you notice that the Torque equation is not related to speed or voltage. As you weaken the field in a constant torque application the speed will go up but the armature current will have to also increase to keep the torque at the same value. So you can only weaken the field until you either reach top speed or the armature current reaches 100% then your out of Torque and HP.
Guest
set field controller for current regulation and dial in the desired current. don't wory about the field resistance. you'll need tach feedback on the dc drive. set the minimum field current you can tolerate. and yes you will need more armature current. the armature current will rise quickly. its a trade off. if your running at full load now then there's nothing more you can do. you can't squeeze blood out of a turnip.
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: Reference:
1. Gordon R. Slemon "Magnetoelectric Devices, Transducers, Transformers and Machines," John Wiley and Sons, Inc., 1966
Section 4.6.2 Armature and Field Resistance Control
Figure 4.38 Shunt-connected motor
Please, notice that a variable resistor Re in the shunt branch sets the flux of the machine, which in turn controls machine. The armature starting current can be controlled by a resistor Rd in series in the armature circuit.
The DC shunt motor equations are often found in DC motor textbooks and on web sites. Visit
for more info
1. Gordon R. Slemon "Magnetoelectric Devices, Transducers, Transformers and Machines," John Wiley and Sons, Inc., 1966
Section 4.6.2 Armature and Field Resistance Control
Figure 4.38 Shunt-connected motor
Please, notice that a variable resistor Re in the shunt branch sets the flux of the machine, which in turn controls machine. The armature starting current can be controlled by a resistor Rd in series in the armature circuit.
The DC shunt motor equations are often found in DC motor textbooks and on web sites. Visit
for more info
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