Motor and ESC Theory - Current Draw 2

[CWAnthony]

Hi guys,

I'm trying to understand (DC brushed) motor behaviour in certain situations when coupled with a PWM ESC (powered by 24V battery). The following situation is not realistic but will help me to understand, I'd appreciate any replies.

Imagine my motor is in an application (a wheeled vehicle) where drag force of the vehicle is constant, and unrelated to vehicle speed. Again, not necessarily true in the real world, but play along

Let's say, we are trundling along at 40mph, 100% PW, with the full 24V from my battery. Drag force (at all speeds, remember) is 40N, giving 40A current draw from the motor.

Now let's say I halve the duty cycle of the PWM, 50%, giving only an effective 12V to motor. This in turn decreases speed (20mph). Drag force though, stays the same, at 40N, so the motor, technically, still wants to draw 40A. What current does the motor actually get? Does the PWM affect the current the motor receives? If so, how?

Thanks in advance,

Chris

 

[waross]

40 Amps at 24 Volts is twice the work as 40 Amps at 12 Volts. Constant load, less speed, less work done.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

If you wanted to make a simplistic statement to describe the motor operation, you could say,
- current and torque are proportional
- voltage and speed are proportional

 

[CWAnthony]

Chaps, I appreciate the replies, but I'm afraid neither tells me what I would like to know. Waross, I understand the equations of energy and power in their simplest form. Lionelhutz, I understand the basic rules of DC motor operation too. My confusion stems from the DC motor's interaction with an ESC using PWM. Let me explain my query in a bit more detail.

I know that the ESC will change the effective voltage from the batteries. This is how motor speed is changed, because "voltage and speed are proportional". So on a 50% PWM duty cycle, the battery's 24V is "averaged out" to 12V as far as the motor is concerned, hence speed is halved.

I am wondering if the ESC does the same thing to the current flow. Surely the current supply also varies according to the PWM duty cycles. So with this in mind, I can picture two possible outcomes to my hypothetical situation. These I will list below, after copying in my original questions as a refresher:

"Now let's say I halve the duty cycle of the PWM, 50%, giving only an effective 12V to motor. This in turn decreases speed (20mph). Drag force though, stays the same, at 40N, so the motor, technically, still wants to draw 40A. What current does the motor actually get?"

Outcome 1 - Current to the motor stays at an average of 40A, due to PWM peaks of 80A (troughs of 0).
Outcome 2 - PWM peaks of 40A are seen, but due to the 50% duty cycle (and troughs of 0), an average current draw of only 20A is seen at the motor.

Can anyone tell me which outcome will occur? Or will another outcome that I have not foreseen occur?

Thanks,

Chris

 

[Brad1979]

It's going to do neither of your 2 options. It has to pull 40 Amps because your torque stays the same. And the current isn't going swing from 80 Amps to 0 Amps because the motor's inductance is going to make it resist fast changes to the current.

 

[CWAnthony]

Thanks for the reply Brad. So the motor gets the same amount of current, that's good. You'll have to forgive me, I'm pretty inexperienced with electrical theory and don't know much about inductance. But that is at the motor end, I'll accept what you said and learn exactly what it means later!

What about current flow through the speed controller then. When we plot voltage versus time through the ESC, you see some square-shaped peaks and troughs, right? What would a similar plot for current look like?

 

[Brad1979]

Well, it's going to depend on the switching frequency of the PWM, the motor speed, the motor inductance and the motor resistance. But in general, during steady state operation, the current will "oscillate" around the 40 Amps. By "oscillate" I mean that the current will rise exponentially during the ON portion of the PWM and fall exponentially during the OFF portion of the PWM. This is called current ripple and it is usually small compared to the average current. See here for a decent explanation.

 

[waross]

The difference between a DC motor and a DC generator is just a few RPM or a few Volts.
Imagine your motor is hooked up to a dyno which has the ability to over drive the motor.
The motor is generating almost as much voltage as is applied to it. When the motor is running on the dyno at no load the current will be small. If the motor is then driven faster, the back EMF (the voltage generated by the motor) will rise and the current will drop.
You will soon reach a speed where the current is zero. If the speed is increased further the motor becomes a generator and has the ability to drive current through the source. (Whether it does depends on the source.
Now start to load the motor. As the load increases the speed drops and the back EMF drops. When the back EMF drops relative to the applied voltage, the current increases. With 12 Volts applied the difference between 12 volts and the back EMF is almost the same as the difference between 24 Volts and the back EMF at the same torque load.
The relationship between applied torque load and speed drop is almost constant across a very broad speed range. It is the speed difference between no load and full load which will determine the current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[cswilson]

It's best to look at this through equations. The motor voltage equation is:

Vm = Ke*w + I*R

The motor torque equation for generated torque T is:

T = Kt*I

In your first case, you are supplying 48 volts to the motor. Some of this is to "overcome" the back EMF of the motor when the vehicle is moving at 40mph (Ke*w). The rest is to push current through the winding (I*R).

Now you reduce the supply voltage to 24 volts. It will no longer be able to push current through the winding at this speed, so less torque will be generated. Since the load torque is constant, the car will decelerate, reducing the back EMF, until the back EMF is low enough that you have enough voltage headroom to push enough current through the motor to generate torque matching the load torque. This speed will be LESS than half of the steady state speed from 48 volts.

To throw some semi-plausible numbers at this scenario:

40 mph = 17.78 m/sec

Using an effective drive radius of 0.01m (through gearing), the motor is turning at 283 rad/sec (2700 rpm) at this speed.

Using a back EMF constant Ke of 0.15V/(rad/sec), the back EMF at this speed is 42.45V.

Kt must be 0.15 N*m/A. To produce a force F of 40N with a drive radius of 0.01m, the torque must be 0.04 N*m, and the current must be 0.04/0.15 = 0.266A.

At 48V supply with a back EMF of 42.45V, the remaining voltage is 5.55V to produce this 0.266A, giving a resistance of 20.8 ohms (oh well...)

At a supply voltage of 24V, when producing enough torque to match the 40N load force, there is still an I*R drop of 5.55V. So the back EMF will be 24-5.55=18.45V, and the motor speed will be 18.45/0.15 = 123 rad/sec. This corresponds to a vehicle speed of 17.4mph, a little less than half of the 40mph at twice the supply voltage.

(Feel free to check my math for errors. I did this quickly.)

Curt Wilson
Delta Tau Data Systems

 

[waross]

The difference between a DC motor and a DC generator is just a few RPM or a few Volts.
Imagine your motor is hooked up to a dyno which has the ability to over drive the motor.
The motor is generating almost as much voltage as is applied to it. When the motor is running on the dyno at no load the current will be small. If the motor is then driven faster, the back EMF (the voltage generated by the motor) will rise and the current will drop.
You will soon reach a speed where the current is zero. If the speed is increased further the motor becomes a generator and has the ability to drive current through the source. (Whether it does depends on the source.
Now start to load the motor. As the load increases the speed drops and the back EMF drops. When the back EMF drops relative to the applied voltage, the current increases. With 12 Volts applied the difference between 12 volts and the back EMF is almost the same as the difference between 24 Volts and the back EMF at the same torque load.
The relationship between applied torque load and speed drop is almost constant across a very broad speed range. It is the speed difference between no load and full load which will determine the current.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

Using the simple rule of thumb that the current is proportional to the torque tells you the current stays the same (40A).

Explaining how/why a H bridge or PWM circuit can reduce the voltage while the current stays the same is beyond the scope of an "Engineering Tips" web site.

 

[CWAnthony]

It is not the site that is the limiting factor though Lionel, it is the knowledge of the users of the site, and their willingness to share that knowledge. You might think explaining the answer is beyond the scope of this site/yourself, fair enough. But so far I've had a lot of seemingly relevant information presented to me by others which makes a good start at explaining what I wanted to know.

The first part of my question was "what happens". I didn't even know that the current would stay constant, even when a PWM is involved (I thought that might change things w.r.t the DC motor rules). "How does it happen" was the second part of my question. I'm sorry, and with all respect, but your reply didn't help clearly answer either. Please don't be offended by me saying that.

To everyone else, thanks for descriptions and links. I'm still reading through most of them. I'll respond (probably with more questions) once I'm up to speed!

 

[waross]

Bye

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[davidbeach]

Bye bye.

 

[CWAnthony]

lol. what?

 

[LionelHutz]

If you want to fully understand what is happening then you need to learn both the motor theory and the PWM controller theory. Chapters of books have been written on both so it's rather hard to explain in a few short paragraphs in a response box. Have you spent any time with Google looking for answers to get a start on understanding this?

 

[CWAnthony]

Yeah I started with google, I've spoken to a few people in person who I regard to be up on this kind of thing as well. I had some conflicting answers between my sources, hence why I thought I'd get some opinions in here. But yes you are right, it is clear that more knowledge of the subject is what I need. I'll go and have a read some more, the above material has given me another good starting point. Thanks for your reply, your understanding and your maturity. Chris.

 

[mikekilroy]

I was going to write just what Curt did above, but my very similar version got dumped by mistake 3/4 of the way thru. CsWilson's reply tells it all. Study that. read and reread his perfect explanation, and you should be able to 100% understand all the other partial answer posts and all should make perfect sense.

Once you understand the motor terminal voltage equation and play what ifs with it enough times, the pwm width question should be a no brainer and you probably will ask yourself why you even asked it to begin with, since it has nothing to do with your motor understanding.

Good luck and happy learning!

http://www.KilroyWasHere<dot>com

 

[electricpete]

In my first read through Curt's reply, I got a little confused by the numerical example. Some people (like me) follow symbolic equations more easily and plug in the numerical values only at the last possible minute. So, fwiw, here is a rewrite in a more symbolic, less numerical form

Curt's equations for shunt dc motor:
Vm = Ke*w + I*Ra
T = Kt*I

Solve each of above equations for I, and equate the two resulting expressions:
T/Kt = (Vm-Ke*w)/Ra

Solve for w:
w = (Vm - Ra* T/Kt)/Ke

This equation could represent a speed-vs-torque curve for constant voltage Vm, or a speed-vs-Vm curve for constant torque T. We're looking at the constant T case for op.

What is the behavior of the equation w = (Vm - Ra* T/Kt)/Ke ?
For Ra* T/Kt<<Vm, then w~Vm. This is the approximation that Lionel mentioned. Drawing w vs Vm would be straight line through the origin with positive constant slope.
OTOH, considering the effects of non-neglible Ra* T/Kt, thhe relationship w vs Vm would be another line a fixed distance below the first line drawn above. Since that fixed distance is a higher percentage at low values, if you double (from 12 to 24 volts), you more than double the result.

Under our assumption of constant torque, then what happens I?
I = T/Kt.
Nothing happens to I (it depends on torque, not affected by speed if torque held constant). That's the other thing Lionel said. And this one is exact (to within the assumptions of the model).

Summary of the two things Lionel mentioned that we proved:
I ~T is exact (within the model)
w~V is approximate (assumes Ra* T/Kt<<Vm)


=====================================
(2B)+(2B)' ?

 

[electricpete]

I ~T is exact (within the model)
w~V is approximate (assumes Ra* T/Kt<<Vm)

I used "~" to mean "proportional to". It was easier than trying to find a way to type the fish-looking proportional symbol.

=====================================
(2B)+(2B)' ?