Motor torque and back emf



Make that the electromechanical resistance of the motor = (BL)^2/R. Bill W.
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My reply this time around is all here between the + lines.
You stated: "Given a mechanical resistance = velocity/force"
Should this not be mechanical impedance = force/velocity ?

How do you define e=(Bl)v electrical (back emf)? Is it just BL times velocity, or are you converting (BL)v, and if so how?

f =(Bl)i could be taken two ways per earlier discussion. Please be specific.
Your diagrams below are scrambled, and not readable with certainty of content. In any event, no point in delving deeper until the above is cleared up.
Bill W.
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?
To Daestrom and Don Kelly
Looking back through the thread, looks like you two are the nice guys. My apology if I stepped on a toe. The few websites I've visited have these hostile participants who bite newcomers, and I sensed a tad of that here, and was on the defense...
I sincerely appreciate the help from both of you, and in particular to Daestrom for the enlightment on the o-scope polarity bit. I had scratched my head on that more than once. :-)
Nice place here, maybe I'll hang around. Bill W.
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says...

I've worked with a reversible type of motor-generator used for converting AC to DC to charge batteries, and then in emergency convert the DC from the batteries to AC.
Now, an interesting thing about this setup is that by slightly varying the shunt field of the DC machine, the CEMF of the DC motor can be raised to the point where it is exactly equal to the battery. Raising the shunt field current a little more, and the 'CEMF' is now higher than the battery, current reverses and we start charging the battery. The 'EMF' induced into the armature of the DC machine by the speed of rotation and field flux could be higher or lower than the battery ('E applied') simply by adjusting the field current. (This is when the AC side was locked in-sync with another, larger AC generator).
So, depending on how much energy is dissipated through mechanical friction, the CEMF can be very close to E applied. If another mechanical energy source is attached to supply those losses, it *can* be exactly equal, or even higher.

polarity
Now this is where I still disagree with your terminology. If we draw a 'flowpath' around the circuit from battery, through armature resistance, through CEMF source and back to battery, then label the first terminal we reach for each component '1' and the terminal where the current leaves each component '2', then we have a standard set of markings.
The voltage on terminal '2' of the battery is positive with respect to terminal '1'. Now we measure the voltage of terminal '2' of the resistor with respect to terminal '1'. It is negative. Similarly, the voltage of terminal '2' of the CEMF source is negative with respect to terminal '1'. Adding all the voltages (paying attention to assigned signs) we get zero. Kirchoff's voltage law. And only by applying this systematic markings (with polarity), can you get the 'right' answer when measuring all the voltages in the loop.
When you use your o-scope setup and you see a positive 'spike' from the battery as the speaker cone moves forward, you measured the voltage and polarity of the battery. But when you disconnected the battery and manually moved the speaker forward, you *didn't* have the scope connected properly. If you didn't reverse the leads, you had them on 'wrong'. Because your leads are on backwards, you saw a 'positive' spike when you should have seen a 'negative' and you seem to think they two sources are thus 'in-phase'.
For example, say that the scope ground is on negative of battery and scope lead is on positive terminal of battery. So 'current' is flowing around the circuit, entering the battery terminal where the ground lead is, and leaving the terminal where the scope lead is. You now want to measure the voltage across the speaker coil. You should now connect the ground lead of the scope where the current enters the coil and the scope lead where the current exited the coil. Notice when you do this, you have reversed the leads.
If you aren't seeing this, put a resistor in each leg of the circuit between battery and speaker. Figure out how you would measure the voltage across each component so that Kirchoff's law works out. Notice how the leads/polarity when measuring across the speaker is different than on the battery?
When you measure CEMF in this manner, it is 'negative' while the supply is 'positive'. I.e. it *opposes* the battery supply (just as you feel it should).
But the term 'in-phase' is misleading since it could mean 'aids the battery in the production of current flow', or 'opposes the current flow produced by the battery'. I think we all agree it opposes the current flow produced by the battery.
Just a matter of interpreting what 'in-phase' is supposed to mean. Don and I seem to feel 'in-phase' to mean be additive with the battery and 'aid the battery in production of current flow', so we don't agree with your use of it here.
To support our position, when doing AC vector analysis, the term 'in phase' is used to describe two vectors that are at the same angle and when adding the two vectors together, the result is a vector of the same angle as the originals, but twice the magnitude. 'in phase' vectors aid each other, not oppose.
'In phase' for two parallel sources is interpreted differently than for two series components. Parallel sources, the voltage is taken as the reference, and the currents are usually discussed. The current in your battery and the current in your speaker are 'out of phase' if you are thinking of them as two parallel sources. (flows out positive terminal of one, and into the positive terminal of the other)
But in two series components, the current is taken as the reference. With that as your reference, the battery and speaker voltages are 'out of phase'. (one opposes the other and only the difference is available to drive current) (Don and my viewpoint)
A simple, two component circuit can be considered either, but the results are the same. Add more components in series, it becomes clearer (at least to me ;-).

But Kirchoff states for a series circuit....
0 = E1 + E2 +E3
Noting the polarity of the battery is opposite the polarity of IR and CEMF, we get...
0 = -E + (IR) + (CEMF)

And playing the 'sign game' some more from Kirchoff's voltage theory and that all currents in a simple series circuit are equal...
0 = (-E)*I + I^2R + I*CEMF

True. But the term 'in-phase' depends on how you connect your meter/scope. And how you interpret the term ;-)
daestrom
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says...

battery
by
by
we do ------------

and
the
---------- I seem to have a different view here. I think of there being two voltage sources as shown 1 s 2 |-------/ -------| + + V1 V2 |____________|
If V1 and V2 are of equal magnitude and phase (i.e. V1 and V2 phasors are identical) then the voltage across the switch is V1-V2 =0 I am looking at both V1 and V2 as the voltage drops from the top or + end to the bottom or reference node. It doesn't matter whether the voltages are produced by sources or are drops across impedances. In the motor case the voltages are both sources. If they are 180 degrees out of phase, then the voltage across the switch will be 2V1 (shades of a 120/240V system). Think of it in the case of parallelling two generators -do you want them in phase or out of phase at the time you throw the switch? However, from what you say below, it is evident that this is what you mean rather than what you have above .

phase'
not
---------- However, in this case , as you go around the loop taking voltage drops throughout you will have (-V1)+V2 =0 (taking both as "drops" in the direction of the loop for KVL ) so V1=V2 ( both as drops from the + terminals to the bottom common terminal which I consider as "in phase" and closing the switch will result in 0 current. ------------------

two
reference,
the
phase'.
---------- OK -see what you mean - but if I take voltage V1 as reference and refer both V1 and I to that reference then V1 and V2 are "in phase" It is not always convenient nor useful to take current as the reference. Suppose you want to calculate the voltage at the sending end of a short line, given receiving end conditions. You can use any reference you want but I for one would use voltage at the receiving end. The resulting voltage at the sending end is typically nearly in phase with the reference. Using current as the reference better give the same result.
We agree but just have a difference in how we express it. --------snip--------

meter/scope.
--
There's the rub :>{
--
Don Kelly
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daestrom@NO_SPAM_HEREtwcny.rr.com says...

Again all well and good, but please note that my diagram described a motor with a *constant* voltage supply source, with which back emf cannot equal or exceed supply voltage.

I believe I stated if viewed in parallel, but noted this was not the case in operation, i.e. that in parallel and in phase did not apply in operation. In any event, I had not figured out the dynamics in technical detail such as you have described in a clear and concise manner. Thanks for the clarification. I appreciate it.

I believe we have become miscommunicated and/or misconbobulated.. :-)
Again (I haven't checked back) I belive I stated in phase if viewed in parallel, but not in operation. Again, thanks for the clarification.

Hey... I knew the retarding side is negative, i.e. E + IR + CEMF = 0 I was wondering when this would come up. :-)

Agreed. Thanks yet again. The phase of this CEMF stuff is tricky. Perhaps best to stay with the polarity terminology.
Bill W.
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Would you please give your results in the following three equations? Let me know if you need values that are not noted below. System compliance Cmt at fc = 0.000292 (stiffness k = 3,425)

-------------------------------------------------------------
Yes, you are correct, the reference is a mass and spring, not forced oscillation. For the 12 drivers, the 2 equations a = wv and a = F/m the average difference in magnitude of acceleration at 227.4 Hz is 2.38%, so that F/m gives a pretty close approximation. The downside is that results vary from 0.2% to 10.2%. When an average difference is greater than 1% from what I expect, I look with askance, but I did not check this at the outset here. Sorry.
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--------
(and you affirm it above), you correctly treat the "R's" as

---------
Zmec at fc = Rmt = 11.12. 23 is Rec
-------- This should be of help:
Electrical impedance:
Z = E/I = Re + Rec @ fc = 28.78 @ RP = 6.21
Rec = Eb/I = Z- Re @ fc = 23.0 @ RP = 0.43
where: RP = "resistive point" = low Z point above resonance Z = electrical impedance, as above E = applied voltage = 1.41 I = current @ fc = 0.049 @ RP = 0.227 RE = DCR of coil = 5.78 Rec = resistance of electrical impedance other than Re as noted above. I may have noted as Res earlier.. Eb = back emf = E-(I Re) @ fc = 1.127 @ RP = 0.098
-----------
Mechanical impedance: Noted as ratio of force to velocity Kinsler eq. 1.30
_ _ Zmec = F/v @ fc = 11.12 @ RP = 35.50
otherwise 1 Zmec = sqrt [Rmt^2 + (wm - -------)^2] @ fc = 11.12 w Cmt @ RP = 35.50
where: Zmec = mechanical impedance = as above _ F = average applied force = E (Bl)/Re = 1.75 _ v = average velocity @ fc = 0.157 @ RP = 0.0493 Rmt = total mechanical resistance = Rms + (Bl)^2/Re = 11.12 Rms = system mechanical resistance = 2.23 (Bl)^2/Re = Rme = mechanical resistance due to back emf = 8.89 Bl = force factor = 7.17 wm = mass reactance = 36.15 Cmt = system compliance = 0.000292 = 1/4 pi^2 fc^2 m
Xmec at RP agrees with the 33.7 figure you note above:
Xmec = sqrt (Zmec^2 - Rmt^2) = 33.7
Xmec at fc = 0
Bill W.
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Yes.

Only my term Rms fits your D, i.e. the mechanical resistance of the suspensions Rms = 2.23.
Such that at fc Zmec = Rmt = Rms + (Bl)^2/Re = 11.12
where Rmt is the total mechanical resistance.
But how does this relate to a result for your above eq. ?
f =mdv/dt +Dv _K(integral of v.dt)
It would be very helpful if you would include the resultant value of your equations. TIA.

No, only one driver of each type (model) is used, IOW no two drivers were the same type. However as I noted, all are 8 inch.

Above resonance and with increasing frequency, the impedance has a depression (decreases to a low value, then increases again). At the low point the impedance appears as resistive, in this case at 227.4 Hz. I define RP as the frequency of this low point in the impedance, i.e. 227.4 Hz.

Exactly, as I note above. We have discussed this before but without a defining term, now being RP = 227.4 Hz.

But our concern here is at RP and below.

No. Square the mechanical reactance term, wm-1/wCmt
Zmec = sqrt [Rmt^2 + (wm -1/wCmt)^2]
otherwise as
Zmec = sqrt [Rmt^2 + (wm - k/w)^2]

The total mechanical resistance of the motor at resonance, as noted above. Rmt = Rms + (Bl)^2/Re = 11.12 as noted above
Likely we are on the same page. Looking in the mechanical domain, since in effect wm and k/w cancel at resonance, this leaves resistance to be the load. Rms is the resistance due to suspensions, so this leaves (Bl)^2/Re to be resistance encountered in overcoming the back emf.

Good. Regarding power factor, I have not dropped it by the wayside, but as you suggested, am reviewing. This has been a sticky issue for me for some time, as was the back emf polarity bit. I have values for electrical power factor, i.e. Re/Z as well as a term for mechanical power factor, that although may not be definable as power factor in this case, are *very* useful and accurate as related to a contradiction in back emf terms. Regarding this would you please:
1. Define power factor as you view it in this case.
2. State if you view electrical input power and mechanical output power as scalars or vectors in this case.
3. State if you view power factor in this case as: PF = 1 means 100% of input energy becomes useful output PF = .9 means 90% of input energy becomes useful output PF = .5 means 50% of input energy becomes useful output
etc. TIA
Bill W.
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Continuing re back emf:
Measured values
E = 1.41 volts = applied sinusoidal voltage Re = 5.78 ohms = DCR of armature coil I @ fc = 0.049 amps = current at resonance I @ RP = 0.227 amps = current at 227.4 Hz fc = 58.6 Hz = resonant frequency RP = 227.4 Hz = 227.4 Hz "resistive point"
The motor formula is
E-Eb I = ------- R
where R in this case = Re = DCR of coil/armature I = current E = applied sinusoidal voltage Eb = back emf
so that
Eb = E-(I Re)
Eb @ fc = 1.127 Eb @ 227.4 Hz = 0.098
-------- With measured values again of
Bl = 7.17 _ v = 0.0493
the other basic equation for generated back emf gives Eb = Blv
Eb @ fc = 1.126 Eb @ 227.4 Hz = 0.353
Note the two equations agree close enough at fc, i.e. at resonance, but not above resonance. Measured generated _ back emf voltage on the dual motor driver agrees with BLv, so apparently Blv is the generated back emf and E-(I Re) is the net or effective back emf. This because one must use E-(I Re) = 0.098 to obtain the correct mechanical power at 227.4 Hz of I Eb = 0.0222 watt, which agrees with power in = power out
IE = I^2 Re + I Eb = 0.320 watt
where I^2 Re = 0.298 is the power lost as heat in the armature coil (copper loss), and I Eb = 0.0222.
My reasoning for the lack of agreement is that at resonance, the reactance due to acceleration of the mass is out of the picture, being in effect canceled by the reactance of suspension stiffness. However, I can not come up with _ a simple or intuitive expression to relate E-(I Re) to Blv.
Bill W.
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that is:
_ v = 0.0493 at 227.4 Hz _ v = 0.157 at 58.6 Hz

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says...

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Make that net power.

Make that Pmec net = I Eb = 0.227*0.98 = 0.022
Sorry. I'll proof-read twice next time...
Bill W.
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What's wrong with public discussion? I prefer such, as the input (and knowledge exchanged thereby) is not limited to two people. As example, I recall the useful input by Daestrom re back emf polarity, which was greatly appreciated by me. Much of your input has also been appreciated as well.
Bill W.
My address is easy to decode. I have a

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