Will these two brushed DC motors reach the same RPM

  • Thread starter William123
  • Start date
Hi
I'm working on a project in school (or at home) about brushed DC motors. I have recently built a prototype myself but instead of using permanent magnets around the armature I'm using an electromagnet from another motor that died. I intend to build another motor using the same armature but this time with permanent (neodymium) magnets and I'm going to measure the RPM of the two motors when they draw the same amount of current. What are your predictions? I read somewhere on the internet once that voltage determines the speed (this is probably not the whole truth) and according to this statement the first prototype should run much faster because the resistance is higher in that motor, which means higher voltage to reach the same amp draw. This doesn't really make sense to me though, I feel like they should reach about the same RPM.

I also noticed that the back emf generated by this first prototype wasn't all that much. It was noticeable but I'd like to see a bigger difference in the current draw when the motor is running vs when it is stalled. At 40-45V the amp draw was 1.8 when stalled and 1.6 when running maybe (i can double check this later). Does this have anything to do with the fact that I'm not using permanent magnets or is it all based on the friction inside the motor?

Thanks in advance :)
 

jim hardy

Science Advisor
Gold Member
2018 Award
Dearly Missed
9,813
4,866

davenn

Science Advisor
Gold Member
8,770
5,820
I intend to build another motor using the same armature but this time with permanent (neodymium) magnets and I'm going to measure the RPM of the two motors when they draw the same amount of current. What are your predictions?
There's a lot of variables, the main one being the variations in the magnetic field strength

I read somewhere on the internet once that voltage determines the speed (this is probably not the whole truth) and according to this statement the first prototype should run much faster because the resistance is higher in that motor,
but if it's the same armature, then the resistance should be the same
 
That's a good platform for experimenting.

Flux speed and current determine back emf.
Flux and current determine torque.
With a separate field you can vary flux.

Look up "generator saturation curve"
for starters try http://www.electrical4u.com/magnetization-curve-of-dc-generator/
Thank you!

There's a lot of variables, the main one being the variations in the magnetic field strength



but if it's the same armature, then the resistance should be the same
I don't think I mentioned it but the resistance of the stator alone is about 15 ohms maybe and it's wired in series with the armature, so when I replace that with permanent magnets the resistance will be only 4 ohms (the resistance of the armature). I'm going to build version 2 as soon as I can and see which version reaches the highest RPM (but really this will be a race between the electromagnetic stator and Nd magnets).
 

jim hardy

Science Advisor
Gold Member
2018 Award
Dearly Missed
9,813
4,866
There are two equations that define external characteristics of a dynamo.
Flux is fundamental. With permanent magnets for a field it's constant, with wound field it's proportional to field current.
Symbol for flux is Φ .

1. Counter EMF = K X Φ X RPM , removing the X's gives KΦRPM . You determine constant K by open circuit test , that's your saturation curve as in the link i gave you.
With permanent magnets you'll measure KΦ as a single constant.
With wound field you control Φ by field amps.

2. Torque = same KΦ X 7.04 X Iarmature . That gives torque in foot pounds , if you use Newton-meters it's an easy units conversion.

Observe this about behavior when when field is wired in series with armature .
Apply voltage and large current flows because there's no counter EMF.
Large current hrough field makes high flux. So there's a lot of torque.
Machine accelerates. Counter EMF appears reducing current. So torque decreases,
but so does field current hence flux, meaning more RPM is required to maintain counter EMF.
So machine speeds up some more.
If there's no external mechanical load the only thing limiting speed is friction of bearings and air.
It is possible to overspeed a series motor so be careful.
When that happens the wires fly out of the armature by centrifugal force and it wrecks the machine.

old jim
 
There are two equations that define external characteristics of a dynamo.
Flux is fundamental. With permanent magnets for a field it's constant, with wound field it's proportional to field current.
Symbol for flux is Φ .

1. Counter EMF = K X Φ X RPM , removing the X's gives KΦRPM . You determine constant K by open circuit test , that's your saturation curve as in the link i gave you.
With permanent magnets you'll measure KΦ as a single constant.
With wound field you control Φ by field amps.

2. Torque = same KΦ X 7.04 X Iarmature . That gives torque in foot pounds , if you use Newton-meters it's an easy units conversion.

Observe this about behavior when when field is wired in series with armature .
Apply voltage and large current flows because there's no counter EMF.
Large current hrough field makes high flux. So there's a lot of torque.
Machine accelerates. Counter EMF appears reducing current. So torque decreases,
but so does field current hence flux, meaning more RPM is required to maintain counter EMF.
So machine speeds up some more.
If there's no external mechanical load the only thing limiting speed is friction of bearings and air.
It is possible to overspeed a series motor so be careful.
When that happens the wires fly out of the armature by centrifugal force and it wrecks the machine.

old jim
This information will be valuable later when I start writing the report so thanks again.

I discovered that the rotor turns faster when I put 2 neodymium magnets on the stator. I think the next step is to improve the commutator and brushes to reduce the sparking and friction because it's running surprisingly well right now. A bit hot maybe but it's all good.
 

CWatters

Science Advisor
Homework Helper
Gold Member
10,528
2,291
I discovered that the rotor turns faster when I put 2 neodymium magnets on the stator.
In general a DC permanent magnet motor will accelerate until the back EMF equals the applied voltage. The back emf is proportional to the flux/strength of the magnetic field so a motor with weak magnets will tend to spin faster than one with stronger magnets. That may seem counter intuitive but it doesn't mean a motor with weak magnets is "better". It just means the motor constant is different. Typically the efficiency will be worse. The peak power may also be lower.

By attaching these two magnets you probably altered the field in the motor and made it weaker. If you want the motor to go faster on a particular voltage then removing turns from the rotor should give better results (lower winding losses).
 
In general a DC permanent magnet motor will accelerate until the back EMF equals the applied voltage. The back emf is proportional to the flux/strength of the magnetic field so a motor with weak magnets will tend to spin faster than one with stronger magnets. That may seem counter intuitive but it doesn't mean a motor with weak magnets is "better". It just means the motor constant is different. Typically the efficiency will be worse. The peak power may also be lower.

By attaching these two magnets you probably altered the field in the motor and made it weaker. If you want the motor to go faster on a particular voltage then removing turns from the rotor should give better results (lower winding losses).
I had no idea about this. If I can find a good source on this I can include this in my report aswell. You can basically choose between torque/efficiency and higher RPM by altering the strength of the field then? (depending on what you're going to use the motor for)

And thank you for the answer, I appreciate it.
 

CWatters

Science Advisor
Homework Helper
Gold Member
10,528
2,291
Yes. Usually motors are designed to produce a specified rpm when operating from a specified voltage. The motor constant k = rpm/voltage. So a 9v 9000 rpm motor has a motor constant of 1000 rpm per volt. You can achieve that using ferrite magnets and a lot of turns or stronger rare earth magnets and fewer turns. A motor with fewer turns has lower winding resistance and hence lower losses due to I^2R. So rare earth magnets motors can be said to have lower "copper losses" compared to ferrite magnet motors.
 

jim hardy

Science Advisor
Gold Member
2018 Award
Dearly Missed
9,813
4,866
I don't think I mentioned it but the resistance of the stator alone is about 15 ohms maybe and it's wired in series with the armature, s

figure out how it's wired.

Then play with those two equations.
Terminal voltage is the sum of counter-emf and IXR product of armature current and resistance .

High armature resistance means you can't make much torque. IXR hogs the available voltage leaving not much to do useful work.
 

CWatters

Science Advisor
Homework Helper
Gold Member
10,528
2,291
You can basically choose between torque/efficiency and higher RPM by altering the strength of the field then? (depending on what you're going to use the motor for)
Yes, although there are a lot of factors that effect the decision on what type of magnets to use. If there are few constraints then Ferrite Magnets are normally used because they cost less to make than rare earth magnets. Rare Earth magnets can also loose their strength more easily if overheated.

I've been struggling to find good references that cover the subject at the right level but perhaps see..

Page 5..
http://www.gearseds.com/files/Lesson3_Mathematical Models of Motors.pdf

Theoretically the motor armature should continue to accelerate to a higher and higher speed unless there is a force that works in opposition to the battery voltage. Clearly the motor armature does not continue to accelerate but instead reaches a finite top speed.

The motor shaft will not continue to rotate faster because the spinning of the armature coils within the permanent magnetic field of the motor generates a voltage or a back emf that opposes the applied voltage.
https://en.wikipedia.org/wiki/Motor_constants

Motor velocity constant, back EMF constant
snip
The field flux may also be integrated into the formula:[9]

Eb = Kωϕω
where Eb is back EMF, Kωis the [motor] constant, ϕ is the flux, and ω is the angular speed
snip
By Lenz's law, a running motor generates a back-EMF proportional to the RPM. Once the motor's rotational velocity is such that the back-EMF is equal to the battery voltage (also called DC line voltage), the motor reaches its limit speed
So the stronger the magnets, the greater the Back EMF at any given rpm and the lower rpm at which the Back EMF equals the supply voltage.
 
figure out how it's wired.

Then play with those two equations.
Terminal voltage is the sum of counter-emf and IXR product of armature current and resistance .

High armature resistance means you can't make much torque. IXR hogs the available voltage leaving not much to do useful work.
So this is why you would want a big counter-emf, right? I ordered a non-contact tachometer this week for the purpose of testing this. I have a question about that second formula though. If Torque = KΦ X 7.04 X I and I'm using an electromagnet instead of permanent magnets, the torque doesn't increase linearly, does it? Because as I increase the current, Φ grows too, doesn't it? If I would use permanent magnets on the other hand, Φ would stay the same all the time and the current through the armature would be the only thing changing. Is this correct?

So the stronger the magnets, the greater the Back EMF at any given rpm and the lower rpm at which the Back EMF equals the supply voltage.
This makes sense. I also realised that maybe I don't need a source to tell me that:
a motor with weak magnets will tend to spin faster than one with stronger magnets.
I could prove it myself mathematically and through testing. Thank you for the gearseds source though, it will definitely come in handy! :)

Edit: And thank you for telling me about Lenz's law. Did you mean to add I*R to the back-emf when you said "Once the motor's rotational velocity is such that the back-EMF is equal to the battery voltage", because as Jim said, terminal voltage is equal to I*R + back-emf.


Edit #2: I found another source confirming that Vt = I*RArmature + Eb
 
Last edited:

jim hardy

Science Advisor
Gold Member
2018 Award
Dearly Missed
9,813
4,866
So this is why you would want a big counter-emf, right?
Yes !
I have a question about that second formula though. If Torque = KΦ X 7.04 X I and I'm using an electromagnet instead of permanent magnets, the torque doesn't increase linearly, does it? Because as I increase the current, Φ grows too, doesn't it?
If it's series wound, you are correct. Armature and field amps are the same so torque equation has amps in it twice,
once in Iarmature and once in Φ .
That's why automobile starters are series wound they make immense torque to overcome static friction in the engine and to push those pistons up against the compressed gas in the cylinders . Listen to yours and you'll hear it labor to do that as each cylinder makes its compression stroke. When you've trained your ear you can tell by that sound when your battery is nearing end of life and you have time to look for a sale.

If I would use permanent magnets on the other hand, Φ would stay the same all the time and the current through the armature would be the only thing changing. Is this correct?
Yes.

It is very heartening to see somebody figuring things out and improving his understanding.
Apply your new insights to everyday experience and you'll grow more of them quickly.

Thanks for sharing. It helps us old guys feel maybe we're helping out.

DC machinery was one of my favorite courses . We're surrounded by them in everyday life so we owe it to ourselves to understand them.
Payback is you're not helpless when an electric window on your car quits working.
We live in fascinating times. Have fun with the technology.

old jim
 

CWatters

Science Advisor
Homework Helper
Gold Member
10,528
2,291
Do you have a way to apply a variable load and measure the torque? If you do then you can calculate the output power and efficiency. It's relatively easy to make a motor that delivers the required power at some specified rpm but the ultimate test of how good it is would be to compare the efficiency under load.
 

Want to reply to this thread?

"Will these two brushed DC motors reach the same RPM" You must log in or register to reply here.

Related Threads for: Will these two brushed DC motors reach the same RPM

  • Posted
Replies
24
Views
1K
  • Posted
Replies
9
Views
2K
  • Posted
Replies
5
Views
3K
  • Posted
Replies
1
Views
8K
Replies
5
Views
839

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top