Designer69
Senior Member
Folks, if I can recall correctly I believe I heard once that DC motors do not have current inrush.
Is this truth or pure nonsense?
Thanks!
Is this truth or pure nonsense?
Thanks!
Inrush on DC Motors
- Thread starter Designer69
- Start date
- Status
If you directly apply rated voltage, the instantaneous current is very large.
And the DC inrush current should be the DC resistance divided into the applied voltage. Once the motor is spinning it generates a back emf which reduces the current draw.
There is some inductance there also. Would it be enough to limit the current until the motor gets moving?
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110222-1817 EST
Designer69:
You need a fundamental understanding on how a DC motor works, then you will be able to answer this question and a number of others.
So maybe the correct question to ask is how does a DC motor work?
.
Designer69:
You need a fundamental understanding on how a DC motor works, then you will be able to answer this question and a number of others.
So maybe the correct question to ask is how does a DC motor work?
.
Designer69
Senior Member
The inductance will slow the rate of rise of current - a little.
We make DC chopper drives. Typically on a 600Vdc system, we add an external inductance of about 1 mH to reduce ripple current. If you applied the full DC supply to the motor you would get a rate of rise of current in the region of several hundred amps per ms.
DC Starters
DC Starters
Mechanical DC starters ( don't know about solid state ones) had resistors in series with the armature. The resistors limited current to the armature circuit. As the motor increased speed the resistors were shorted out by contactors in parallel with them. The resistors were usually shorted out by timing relays.
DC Starters
Mechanical DC starters ( don't know about solid state ones) had resistors in series with the armature. The resistors limited current to the armature circuit. As the motor increased speed the resistors were shorted out by contactors in parallel with them. The resistors were usually shorted out by timing relays.Designer69
Senior Member
yes that is exactly what we have and I was kind of trying to see if we are choking soo much current with our resistors that the armature can't start spinning.
however delving into it a bit, I found the resistor is 8 ohm wire wound with center tap (dropping out 4 Ohms at a time). With 240VDC that still allows 30A to the motor which is 3HP 11FLA. So it should be plenty.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110223-1028 EST
Designer69:
The book I had for DC Machinery was "Direct Current Machinery" by Hempstead S. Bull, 1939, John Wiley & Sons. I had Tarboux as the teacher of this course with this book. But Tarboux's teaching approach was quite different than Bull's method in the book. Quite a difference between these two professors in their teaching methods. Unlikely you will find this book.
In a shunt wound DC motor for fixed field excitation the output torque is proportional to armature current. Thus, all you need to start the motor is at least enough current to supply the load torque. Load torque in this case includes motor friction. Note: Permanent magnet motors are essentially the same as a fixed excitation shunt wound motor.
The time of acceleration to full speed will be a function of the input current. Thus, once you supply enough current to overcome friction losses, then the amount of additional current supplied will determine how fast the motor and load accelerate.
If you had a true constant current source up to the point of full speed, and switched to constant voltage as the source when speed was reached, then the results of acceleration would be quite different than just using series resistance. Note: if series resistance is used to start the motor, then as speed increases the voltage difference between the counteremf and the applied voltage gets smaller and less current can pass thru the fixed resistance. Thus the reason for stepping down the resistance as the speed increases.
If you have a shunt wound motor, then you want full excitation to the field for starting, and only put the current limiting resistance in series with the armature. This provides the greatest torque. Once up to base speed if you want higher speed, then the field excitation current can be reduced, but torque drops.
.
Designer69:
The book I had for DC Machinery was "Direct Current Machinery" by Hempstead S. Bull, 1939, John Wiley & Sons. I had Tarboux as the teacher of this course with this book. But Tarboux's teaching approach was quite different than Bull's method in the book. Quite a difference between these two professors in their teaching methods. Unlikely you will find this book.
In a shunt wound DC motor for fixed field excitation the output torque is proportional to armature current. Thus, all you need to start the motor is at least enough current to supply the load torque. Load torque in this case includes motor friction. Note: Permanent magnet motors are essentially the same as a fixed excitation shunt wound motor.
The time of acceleration to full speed will be a function of the input current. Thus, once you supply enough current to overcome friction losses, then the amount of additional current supplied will determine how fast the motor and load accelerate.
If you had a true constant current source up to the point of full speed, and switched to constant voltage as the source when speed was reached, then the results of acceleration would be quite different than just using series resistance. Note: if series resistance is used to start the motor, then as speed increases the voltage difference between the counteremf and the applied voltage gets smaller and less current can pass thru the fixed resistance. Thus the reason for stepping down the resistance as the speed increases.
If you have a shunt wound motor, then you want full excitation to the field for starting, and only put the current limiting resistance in series with the armature. This provides the greatest torque. Once up to base speed if you want higher speed, then the field excitation current can be reduced, but torque drops.
.
Last edited:
I think you're right. There is an R L time constant I didn't consider.
Memo to self: Think, then wait, then think again, then post. . .
Hameedulla-Ekhlas
Senior Member
- Location
- AFG
In DC motors
Voltage equation is = V = E - IaRa
where
V = terminal voltage
E = Back emf
Ia = Armature current
Ra = Armature Resistance
In the beginning E = 0 and V = IaRa
where Ra is always very less about = 0.01 ohm
So, if you calculate for Ia = V/Ra, you will get large current.
- Status