Victors Non-linear!!!

[gc02]

...inductance doesn't appear to have that effect...

Alright looks like I'm going to have to break out the big guns here, graphs! I wrote up a program to graph inductor waveforms and calculate the RMS curent, here's the result:

RMS current of swithced inductor

[Joe Ross]

I don't have a lot of time, as I'm once again posting from an internet cafè in Mexico.

I seem to remember a series of posts from a while ago that stated that there was a band at the outer limits that produced a full voltage output. Perhaps someone could do some research on that.

[Al Skierkiewicz]

Originally Posted by gc02

Alright looks like I'm going to have to break out the big guns here, graphs! I wrote up a program to graph inductor waveforms and calculate the RMS curent, here's the result:

RMS current of swithced inductor

Great set of graphs, you should all take a look.

Joe, I think you are referring to the fact that the controllers should put out pure DC voltage at the extremes of the PWM scale. (i.e. at 0 and at 255) This statement is correct only if the controllers are calibrated to your control system/joysticks. If you do not calibrate it is possible to hit DC before the end of travel on the joystick or you may not be able to achieve DC with the joystick fully up or down.


You are not supposed to make the rest of us feel bad by telling us your in Mexico. I, for one, am back here in the cold, bleak winter, getting ready for kickoff. Brrr, it's cold up here! I am going to go put on another pair of socks and a sweater. (Am I making you feel bad yet?)

[Joe Ross]

[quote=Al Skierkiewicz]Joe, I think you are referring to the fact that the controllers should put out pure DC voltage at the extremes of the PWM scale. (i.e. at 0 and at 255) This statement is correct only if the controllers are calibrated to your control system/joysticks. If you do not calibrate it is possible to hit DC before the end of travel on the joystick or you may not be able to achieve DC with the joystick fully up or down.
[\QUOTE]

Al, what I think I'm refering to is that the controllers would put out 12v in a range, for example 240-255 (much like the dead spot around 127). Like I said before I'm not sure, though.

If nobody else can find it, I'll look in a few days.

[Joe Ross]

Here is what I was refering to:

http://www.chiefdelphi.com/forums/sh...12&postcount=4
http://www.chiefdelphi.com/forums/sh...69&postcount=2

[Caleb Fulton]

Originally Posted by MikeDubreuil

Exactly,
It's general knowledge that the Victors pulse 12V to the motors to control their speed. Except at the extremes when the FET's give the full 12V. It would be very difficult to measure this on a general purpose multimeter.

I am extremely skeptical of the voltages you are getting. A better test would be to see if the frequency is linear. I plan on checking that when I get back to the lab after the holidays.

I'll do the same, and try to remember to post my results as well.

In response to Andy's post:

I can't tell how familiar you are with circuit analysis in the time domain, but it seems like you have a pretty good grasp of it qualitatively.

Here's a "formal" way of going about the series inductor-resistor's response to an impulse.

The impulse can be modeled by the use of the unit step function:

Vin = Vm(u(t) - u(t-Tw)),

where Vm is the voltage magnitude of the pulse and Tw is the length of the pulse. The unit step function u(tau) is simply 1 for tau >= 0 and 0 otherwise.

The "physics" description of how the inductor responds to voltages and currents is good for qualitative analysis, but I often get confused with the signs corresponding to the use of the term EMF instead of voltage drops. Here's the way I visualize it (this should be fun in ascii):

 Current, i(t)
   |   |
   |   |
   |  \|/
   |   v
+  |
    ) <-this "comes out at you"
  (   <-this "goes into the background"
    )
  (
    )
-  |
   |

emf = -L * di(t)/dt

An increasing i(t) over some period of t will cause the magnetic flux cutting the coils to increase in the downward direction. This causes an emf to be generated in the opposite direction, which is in the upward direction for the orientation of the coils above. Here's where my memory gets fuzzy, but I *think* it's right: The emf going upward, having the same units as Voltage, causes a voltage to develop across the terminals as marked on the figure. THe reason for this orientation, as I think of it, is that the emf is the result of a circular electric field that causes positive charges to go up in the figure. Thus, if a "load" existed in parallel with the inductor, it would act like a "battery" with a "voltage" that caused the current to go in the direction of the emf.

Thus, we can write a nice voltage/current relationship for the inductor and eliminate the negative sign and the emf term:

v(t) = L * di(t)/dt,

where the voltage and current orientation are as in the figure above. This corresponds to the passive sign convention, commonly used in linear circuit analysis, and the sign convention it follows is also used for resistors and capacitors: the current enters the positive node of any two-terminal element.

So let's say that we have a voltage source, given by Vin = Vm(u(t) - u(t-Tw)), in series with our inductor, L, and reisistor, R.

current
 -->
 ______
|      |
|      R
+      |
Vin    L
-      |
|______|

Kirchoff's voltage law gives the following equation:

Vin(t) = R * i(t) + L * di(t)/dt

or:

di(t)/dt + R / L * i(t) = Vin(t) / L

Using the integrating factor method (you can look this up if you haven't seen it before):

Let the integrating factor be exp(R / L * t) (where exp(x) = e^x)

then:

exp(R / L * t) * (di(t)/dt + R / L * i(t)) = d/dt((exp(R / L * t) * i(t)) = exp(R / L * t) * Vin(t) / L

Changing the variable of integration to tau and integrating both sides from 0 to t (Vin is 0 for tau < 0):

int(d/dt(exp(R / L * tau) * i(tau)), tau, 0, t) = int(exp(R / L * tau) * Vin(tau) / L, tau, 0, t)

For 0 <= t < Tw, Vin(t) = Vm, so evaluating the integrals gives:

exp(R / L * t) * i(t) - i(0) = Vm / L * ( L / R) * exp(R / L * t) - Vm / L * ( L / R) = Vm / R * exp(R / L * t) - Vm / R

We can say, for this case, that i(0), the initial current through the inductor, is 0, so:

exp(R / L * t) * i(t) = Vm / R * exp(R / L * t) - Vm / R

i1(t) = Vm / R (1 - exp(-R / L * t))

I call this i1(t) because it is only valid for the first part of the Vin excitation to the system.

From here, one can define t' = t - Tw such that t' = 0 corresponds to t = Tw

We can now solve for i2(t'), which is valid for all t' > 0 (or t > Tw) using the same method. It will come down to this equation again:

di(t')/dt' + R / L * i(t') = Vin(t') / L = 0 for t' > 0

Separating variables:

di(t')/i(t') = -R / L * dt'
or
di(tau')/i(tau') = -R / L * dtau'

Integrating from 0 to t', since we have taken into account tau' < 0 in the previous section of the problem:

ln(|i(t')|) - ln(|i(0)|) = -R / L * t'

exp(ln(|i(t')|) - ln(|i(0)|)) = exp(-R / L * t')

i2(t') = i2(0) exp(-R / L * t') where i2(0) = i1(Tw), a constant

In terms of t, i(t) for t > Tw is:

i1(Tw) exp(-R / L * (t - Tw))

Thus, the current over all time is:

i(t) = {0, t < 0 
       {Vm / R (1 - exp(-R / L * t)), 0 <= t < Tw
       {i1(Tw) exp(-R / L * (t - Tw)), t >= Tw

So this is what the waveform for the current looks like when there is a resistor in series with an inductor as a response to a pulse.

[Manoel]

Originally Posted by Joe Ross

Al, what I think I'm refering to is that the controllers would put out 12v in a range, for example 240-255 (much like the dead spot around 127). Like I said before I'm not sure, though.

If nobody else can find it, I'll look in a few days.

Actually, I think you are confusing the fact that the Victor will output a green light (full forward) in the 240-255 range, but its output won't necessarily be the supply voltage.

[Al Skierkiewicz]

Originally Posted by Caleb Fulton

i(t) = {0, t < 0
{Vm / R (1 - exp(-R / L * t)), 0 <= t < Tw
{i1(Tw) exp(-R / L * (t - Tw)), t >= Tw

[/code]

So this is what the waveform for the current looks like when there is a resistor in series with an inductor as a response to a pulse.

Caleb,
Nice work on the math derivation. The equation should simplify to

I(t)= (V*(1-e^(-(t*R)/L)))/R
Where t is the time the voltage has been applied
L is the inductance of the coil
R is the series resistance of the coil + all other resistive series losses
e is the natural log
V is the applied voltage
If you look closely you will see Ohm's Law (I=V/R) In the case of a motor at speed, the true applied voltage is the supply voltage minus the counter EMF and the inductance is reduced to the average inductance of the windings as they turn.
For all as rules of thumb, the time constant of an L,R circuit is approx. L/R and the current will be about 66% of maximum in one time constant. Current will reach very near maximum between 5 and 6 time constants. So for Greg's example motor above, L/R=40 exp -6/.0859=.465mSec. A 10% duty cycle at 120 Hz is .833mSec so the current at this duty cycle is not quite at 100% but greater than 66% of maximum. Backing into the math you can see that 6 time constants is a little over 2.4mSec or about 30% duty cycle. The graphs that Greg has referenced show this.

Originally Posted by Manoel

Actually, I think you are confusing the fact that the Victor will output a green light (full forward) in the 240-255 range, but its output won't necessarily be the supply voltage.

Can you elaborate? At full up the Victor should output DC at power supply input minus the series "on" resistance of the FETs.

[Camerzn]

James,
At my lab, we have one really nice Hameg oscilloscope whith neat features like the ability to freezeframe. We could use that to test these guys' crazy theories at some point, if you want.

[Manoel]

Originally Posted by Al Skierkiewicz

Can you elaborate? At full up the Victor should output DC at power supply input minus the series "on" resistance of the FETs.

Yes, that's what I meant... With an Rds of 0.012 Ω per FET, I just disconsidered it in the context of what I was writing. Of course, with high currents, you do get a noticeable voltage drop on the Victors (and battery wire, and bad crimped connectors, oh, the losses! )

[Al Skierkiewicz]

Originally Posted by Manoel

Yes, that's what I meant... With an Rds of 0.012 Ω per FET

Don't forget that there are three FETs in parallel, .012/3=.004 or about the resistance of four feet of #10.

[KVermilion]

Hey guys. During the winter break I was working on the current sensors... I came across this post and decided to check it out for my self... I wired up the current sensors and graphed some points (all of them.. twice) and the results are astonishing.

I'm writing it up now, and will publish it via our site (www.saratogarobotics.com) I'll post back with a link, when I'm done.

Here you go.

Wrote this tonight too... interesting, no?

[Manoel]

Originally Posted by Al Skierkiewicz

Don't forget that there are three FETs in parallel, .012/3=.004 or about the resistance of four feet of #10.

Wouldn't it be three FETs in parallel sourcing current plus three FETs in parallel drawing current, and hence 0.008 Ω? It's an H-Bridge after all, isn't it?

Originally Posted by KVermilion

Hey guys. During the winter break I was working on the current sensors... I came across this post and decided to check it out for my self... I wired up the current sensors and graphed some points (all of them.. twice) and the results are astonishing.

I'm writing it up now, and will publish it via our site (www.saratogarobotics.com) I'll post back with a link, when I'm done.

Here you go.

Wrote this tonight too... interesting, no?

Interesting data, just be aware when comparing your results with Rickertsen2's that you used a resistor and he used a motor (resistor + voltage source + inductor model). Of course, your tests are better, because this resistor setup will not interfere with the Victor's output waveform.
With your data, we can say for sure that the Victor's are linear (something the IFI guys already told us, for the matter), I'm just a little puzzled about the current "crop" at the graphic's extremes, were you using a battery to feed the Victor or maybe a different power supply with a maximum output current?

[andy]

%$#& forum kicked me off mid post! (no offence Brandon!)

Cool stuff!

Very good work Caleb I learned alot! However, Al's simplification does not make a whole lot of sense to me...

It started when I wanted to solve for VOLTAGE not CURRENT to compare Jame's work and the expected model. So, if you examine the simplified equation,

I(t)= (V*(1-e^(-(t*R)/L)))/R

and use the multiplication rule we can take out V and R

I(t)= (1-e^(-(t*R)/L))*(V/R)

If V = IR then V/R = I

Soo... now

I(t)= (1-e^(-(t*R)/L))*I

and the I s cancel? Something is screwy....

Some interesting links while I'm at it!
inductors

Inductors

Step functions

PS, Al is it ok if we ignore the 0.004 ohms resistance in the FETs?

[Rickertsen2]

Originally Posted by Manoel

Wouldn't it be three FETs in parallel sourcing current plus three FETs in parallel drawing current, and hence 0.008 Ω? It's an H-Bridge after all, isn't it?



Interesting data, just be aware when comparing your results with Rickertsen2's that you used a resistor and he used a motor (resistor + voltage source + inductor model). Of course, your tests are better, because this resistor setup will not interfere with the Victor's output waveform.
With your data, we can say for sure that the Victor's are linear (something the IFI guys already told us, for the matter), I'm just a little puzzled about the current "crop" at the graphic's extremes, were you using a battery to feed the Victor or maybe a different power supply with a maximum output current?

.


His data is not "better". Neither is mine for that matter. Put together they reinforce the inductance theory. My graphs were corect. What was not was my assumption that my curve was caused by a nonlinearity of the victors. Given your data we can see this is not the case.I think at this point we can conclude (to IFIs credit), that the victors are linear. This does not mean that my data was wrong. BOTH sets of results are correct. The tests were different. KVermilion measured current across a RESISTOR ONLY with a constant resistance and varying PWM input to the speed controller. This would be nice if we were powering heaters with the victors but we are powering motors.

A motor's torque output is porportional to the current through it regardless of why that amount of current is going through it. It has been shown that while the victors output a linear voltage, the load is not purely resistive and this does not lead to a linear current. It has an inductive Component. It may still be useful to compensate for the the pwm vs. current curve that we now know is caused by inductance. Even though I mistaked the cause of the curve, it is still there. There is still a possibility that my voltage test was wrong due to my meter not being able to properly read a square wave however the pwm vs. speed curve is still viable. After all, we are trying to control speed here, so that is the ultimate test



Btw. Andy, it happens to me about every other post. I have learned to copy my post before submittting. Sessions either always time our for me or arn't working properly.