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Measuring field and armature resistance on a dc shunt motor(3)

Have a small DC shunt wound gearmotor. Nameplate shows 115vdc; armature current .33A. If I measure the leads coming from the motor, I get 145 ohms for the armature, and 1268 ohms for the field. Knowing that the field and armature are internally connected in parallel, I am wondering what the measurements from the leads represent? From the np data of .33A at 115v, that would indicate the armature is V=IR or 349 ohms. I want to build a field voltage supply using a bridge rectifier and smoothing capacitor. With 120 volts ac in, I think the bridge rectifier circuit yields 1.414X120 or about 168 volts dc. Is that right? Then would need to drop the voltage about 50 volts with a rheostat in series with the field. I'm trying to size the resistance range of the rheostat. . . and the field resistance is needed. Can anyone help? I don't have experience in electonics. 

You say you have 145 ohms for the armature and 1268 ohms for the field, and then you ask what the armature and field resistances are. Something doesn't make any sense there. Your .33A at 115v does not equate to an armature resistance of 349 ohms because you have not included the back emf. You would have about 67 volts of back emf under the condition of .33A and 115V. 

waross (Electrical) 
24 Mar 06 14:12 
#1 you won't get an accurate resistance measurement of the armature if you measure through the brushes. Measure the resistance between comutator bars. It will be much lower. #2 The resistance of the armature limits the starting currrent. The starting current is very high. The running current is limited by the back EMF. #3 You have correctly calculated the peak voltage. Unfortunately, your power supply won't operate at peak voltage unless it's unrealistically oversized. OK for electronics, expensive for motors. #4 If the field and armature are connected internally, the leads usually won't be brought out individually. #5 You haven't mentioned what you intend to use for a current source for the armature. This will be a bigger challenge than the field supply. Start with a resistor about 25% or 30% of the resistance of the field (when you are sure that you are actually measuring the field resistance) and adjust it for optimum performance. Field voltage too low, motor runs fast. Field voltage too high, motor runs slow. The very old text books describe a technique called "Field weakening" that was used to increase the speed of a DC motor over a moderate range. yours 

DavidBeach
David . . . I didn't know about the emf in the equation. If I subtract out the 67volt emf, which leaves 48v and apply V=Ir . . . the resistance I would get matches what I measured across the external armature leads. Does that confirm that armature resistance is 146 ohms . . . just like measurement? Is it ok to use the measured lead field resistance of 1265 ohms or is there more to that? I am just trying to size a dropping resistor or rheostat to get the voltage from the bridge rectifier down to 115v. Your reply was very appreciated. . . and as you can see, I am inexperienced with this. 

Waross Thank you for your reply. After I find out whether the 1265 ohms I measured is correct for the field, then I will put a rheostat in series with it and experiment with field weakoning. I have a 0 to 90 vdc controller that is meant for pm and will just use it and set it at 90 for the armature. I would build a bridge circuit for that but I heard that the armature likes pure dc while you can used a "smoothed dc" and for the field. Are you saying that what I measured with the leads is probably what it is? Richard 

UKpete (Electrical) 
24 Mar 06 17:12 
Richard
Yes, assuming you have actually found the field connections properly (are they marked?) then the measured value of 1268ohms is the value you will use.
If I were you I would first set the field current to the original value i.e. 115/1268 = 90mA, then start with the armature voltage at zero. As you raise the armature voltage the motor will accelerate (not quite in proportion). If you instantly apply 90V you will get a big surge in armature current, maybe too big, and a sharp acceleration.
The equation for the armature circuit is as follows:
V applied = EMF + I*Ra + Vb
where EMF is the back emf (proportional to the product of speed and field flux) I is the armature current Ra is the armature resistance Vb is the brush volt drop (may be only about 2V so could be ignored).
The 67V figure quoted above assumes that your armature resistance is 145ohms and has been calculated using the above equation. But as Waross says, your actual armature resistance will be less than this as it should be measured directly on the commutator at the brush positions. Actually, I don't think you need to know the armature resistance to operate this machine, so don't worry about it.
Field weakening will result in an increase in speed, although the converse is not really true. If you keep on increasing the field current it will saturate the field and it will have no further affect on the machine (except that the field windings will get hotter).


aolalde (Electrical) 
24 Mar 06 17:16 
richard29e5
The field is a passive circuit, then the resistance you measure is what limits the current. The lower ripple you apply the best DC field and performance for the motor.
The armature circuit is an active circuit, the EMF = k*phi*rpm, yes it is dynamic and changes with field exitation and speed. EMF=0 whith the rotor standstill.
The magnetic flux, phi = k2*Ifield
For the armature circuit:
[b] V = EMF + Ia*Ra or Ia=(VEMF)/Ra 

So in putting this together, if field resistance is 1265 ohms and voltage coming out of bridge rectifier circuit is 168 volts and only 115 v dc is needed or a difference of 53 volts . . . the the resistance needed in series with the field is v=ir or 53=I r andextimating I in field at .1A or 530 ohms ball park. Is that about right?
All I want from this shunt motor is to run at normal fl speed and at somewhere rated load. I don't need to change field volts or armature volts. Could this be hooked up to a common bus (one supply) at 100vdc for both field and armature . . . or is that a no no. Motor will never overheat with the load that will be on it. 

UKpete (Electrical) 
24 Mar 06 18:29 
Richard, if it is a shunt wound motor then it was designed to run with 115V on both the armature and the field. So why not connect the armature and field circuits in parallel and apply your rectified voltage to both (if you get one of the polarities wrong, the motor will simply run in reverse). As you say, the rectified voltage is too high; you can use a single resistor in series with the rectifier to drop the voltage down.
To calculate the resistance: the total motor current will be 0.33 + 0.09 = 0.42A approx. If you assume 168V rectifier output (although it will be a bit lower than this, as waross states) but you need 115V, then required resistance is: (168115)/0.42 = 126ohms and the power dissipated in the resistance is 22W. This is approximate because the voltage will be lower than 168V so the required resistance will be a bit lower otherwise your motor will run a bit slow.
Ignore my previous comment about first applying the field current then slowly ramping up the armature voltage, this applies to larger motors than yours really. 

Hi: If the aramture and field are connected internally, leads should normally be marked, if not, it can be a bit tricky getting correct values and you just have to keep trying diffent resistors until motor runs as you want it. Unfortunately, I made the mistake of measuring R through the armature brushes, which resulted in quite a delay in getting the job done correctly. If you don't have time to experiment, make sure your values are correct before starting anything. Good Luck!
Mikael 

waross (Electrical) 
24 Mar 06 21:02 
A couple of suggestions. Try it on a 12 volt car battery. That will verify that your connections and rotation are correct.
Energise your field with a slightly oversized resistor in series with your homebrew field voltage supply. This will hopefully avoid overheating if the voltage is too high. Measure the actual voltage and calculate the proper resistance. Doing it this way will factor in the voltage drop that your supply exhibits under load. It won't be exact because as you change the resistance, you change the loading somewhat. It should be close enough for practical purposes. It will probably be the simplest way to determine the resistance required. After you adjust the resistor value and remeasure the field voltage you should be able to hit it right on the money with one more adjustment.
I would then connect the 090 volt controller and ramp it up from zero volts. Why ramp it up? My very old textbook gives a general value for armature IR of 5% at full load. What is IR? Well if the applied voltage is 100 volts and the back EMF is 95 volts, the armature IR is 5 volts. This means that the resistance of the armature is determined from the IR voltage, not the applied voltage. In your case, 5% of 115 volts is 5.75 volts. 5.75 volts/.33 amps = 17.4 ohms. 115 volts/17.4 ohms = 6.6 amps starting current. With your small motor, the actual IR may be 3 or 5 times as much, but this is still a lot more accurate than trying to measure through the brushes. This also means that the current at locked rotor may be 20 times running current. With such a small motor it probably won't matter, but why take a chance. Once you get it running well you can go to across the line starting. It will probably work. If it doesn't you will know why. After you repair your power supply you can go back to ramping up the starting voltage. It pretty much depends on the ratings and internal protection of your controller. If you have a 5 amp controller it will probably survive. If you have a 0.5 amp controller, then hope it has internal protection. Keep us posted on your progress.
For those who want definite values, the percent speed regulation of a DC motor closely equals the percent IR. If you can measure the speed drop between no load and full load the percentage speed drop will equal the percentage IR. Because of windage it is sometimes difficult to achieve no load conditions. Go from 10% load or 50% load to full load or whatever loading is doable and extrapolate. Measure the load with an ammeter. Keep the voltage fairly constant. respectfully 

DickDV (Electrical) 
25 Mar 06 9:08 
The original post says that the field and armature are in parallel but then lists resistances for each separately. This seems inconsistent.
If the motor leads are labelled F1, F2, A1, and A2, then there is no internal connection and things get much easier.
Proceeding under that assumption, you can measure the field resistance with an ohmmeter but that will give you cold resistance. The warm running resistance will be somewhat more. The best way to measure warm resistance is to bring the field leads up to rated current for 10 minutes and then measure voltage across the field. Ohm's Law will give you warm resistance.
Armature resistance should be measured from the commutator as mentioned above to keep brush resistance out of the measurement. Armature resistance will be the cold resistance also but the difference between cold and hot resistance in an armature is too small to be significant. Measuring cold resistance should be good enough. 

Thanks for all of the help. I am going to use the circuit that UKPETE suggested. . . which was to connect both field and armature in parallel to one common supply bus from the bridge rectifier and use a dropping resistor in series with that connection. Although the wave form to the armature won't be that great, it will probably suffice with the small load on the motor. Thanks again 

waross (Electrical) 
25 Mar 06 16:31 
Should work. Let us know how you make out. Often simplest is best. respectfully 



