120V from both legs

hears it

Hello there. Here's a case for 'back voltage'

Picture a 120V 'supply' source connected to a 20 ohm resistor, which is then connected to ground. 6A flow.

Now add a variable voltage source to that circuit. The 120V source is still connected to the 20 ohm resistor, which is then series-connected to the new variable voltage source, then to ground. With the variable source at 0 volts, we have the same result as above, 6A flow from the source into the resistor, then the variable voltage source, and finally to ground.

Now set the variable voltage source at 20V such that it 'opposes' the

120V source. There are 100 V across the 20 ohm resistor. 5A flow from 'supply', through the resistor, through the variable voltage source, and to ground.

Set the variable voltage source at 40V. 4A flow.

Set the variable source at 60V. 3A flow.

Set the variable source at 80V. 2A flow.

Set the variable source at 100V. 1A flows.

Set the variable source at 120V. 0A flow.

Set the variable source at 140V. 1A flows from the variable source, INTO the 120V supply.

Set the variable source at 160V. 2A flows from the variable source, into the 120V supply.

I would liken back emf to the variable voltage source in the above simple example. Say the current into a dc machine goes from 6A at startup down to 1A at steady-state running speed. This would correspond - as far as the above example is concerned - with the back emf going from 0V at rest up to 100V in steady state. If the dc machine were driven by a prime mover, such that its speed increased further, the emf could rise to the point that no current flows into the machine. If the prime mover drives the armature even faster, the machine would enter generating mode and will drive current (supply power) into the 'supply'.

This is how I think of back emf. The back emf 'subtracts from' the supply voltage insofar as it decreases the votlage across the machine's resistance (the resistor in the above). In my example, there would always be 120V measured at the 'terminals' as you correctly pointed out would be required for a satisfactory model. The back emf voltage would be produced by the conductor (armature) 'cuting the flux' of the magnetic field. This should be a real, measurable voltage. It would be present even if the motor terminals were opened (no current flowing). In this way it seems to me that calling it a 'back' voltage is in keeping with the physical reality of the machine.

An induction machine is a little more complicated. But the term 'back emf' does get used used in that situation as well. I believe any emf induced in the rotor circuit will have current associated with it as the rotor is 'shorted' through the rotor resistance. The rotor voltages and currents will 'reflect' back to the stator through the transformer action that occurs. The rotor voltage reflected to the stator can be thought of as back emf.

A synchronous machine is often modelled very simply as an impedance in series with a voltage source, like the resistor and variable voltage source in my numerical example, except with an inductance rater than a resistance. The 'back emf' here is the excitation voltage caused by the rotating field of a revolving dc rotor inducing voltages in the stator.

However one can measure that current

I'm not certain what circuit configuration you would be suggesting. I note that a Norton equivalent of the circuit I described above would give accurate terminal quantities, and would have a current source to model a 'back current'. Maybe this is what you are proposing? Would your circuit consist of only E=120 feeding R=10 (connected then to ground) and a current source Ib parallel to R and injecting current into the circuit? If so, this is equivalent (at the terminals). An 'Ib' could be calculated at any instant of my numeraical example above as: Ib = (voltage of variable source) / (20 ohms).

j

-------- Any circuit model that we use must fit the observed behaviour of the device. The series Ra and Eg model does so better than the model that you propose. However, don't throw your model away yet- see below.

I will do the DC motor with separate (or permanent magnet) excitation fields for simplicity. First of all we can see that if the motor shaft is rotated from an external voltage source, there will be a voltage developed at its terminals.(re: Faraday) this voltage is proportional to speed and the field flux and is give by Eg=K(flux)w where w is the speed. This can easily be measured. The windings of the generator will have some small resistance Ra. If a load R is connected across the terminals then there will be a single loop circuit such that Eg=Ra*I +R*I =Ra*I +V Now, instead of a load resistor, consider a second voltage source V. If V is less than Eg the current will flow from Eg to V and the power flow will be from the mechanical side to the electrical side. The machine is still a generator. If V=Eg there will be no current and no power flow. If V>Eg then the current will reverse and the direction of power flow will also reverse. The machine is now called a motor. . In the motor case the " back emf" is simply the generated voltage.

The motor circuit will look like this

__________Ra_____________ + I-> + | V=R*I +Eg or I=(V-Eg)/Ra V Eg _________________________|

At start: the speed w is 0 and then Eg=K(flux)w =0. The current will be V/Ra and will be maximum. The torque is given by T=K(flux)*I (same K(flux) as above in MKS units and speed in radians/sec) The torque will accelerate the motor and as speed increases, so will Eg. Eventually the motor will reach a constant speed which would result in 0 current, 0 torque and Eg=V ----IF-there were no losses in the mechanical side. In practice there will be a no-load speed which will correspond to Eg being slightly less than V such that only enough current flows to balance the losses. Now a mechanical load is applied to the motor shaft. This will cause the motor to slow down a bit and the result is that Eg decreases a bit so the current will rise, producing a higher torque. The motor will settle down at a slightly lower speed such that the torque produced balances the torque required by the load. If the mechanical load decreases, the reverse occurs as the motor will speed up, reducing the current and torque. Looking at power: the input power will be Pin =V*I =(I^2)*R +Eg*I However Eg*I =Tdev*w =mech power developed =mech loss +mechanical power out.

There are variations on this depending on how the machine and these will affect the torque speed characteristic.

Now look at the AC analog of this motor/generator and substitute Za =Ra +jXa for the series resistance. Ignoring Ra which is relatively small, the power developed becomes (after looking at I=(V-Eg)/Za as a start) P=[(V*Eg)/Xa] sin (delta) where delta is the phase angle of Eg with respect to V. The effect of the magnitude of Eg is less than that of delta and Eg may be more or less than V for both motoring and generating. If Eg lags behind V then it is a motor, if Eg leads V, it is a generator. In detail it gets a bit messier but that's the basic gist of it.

The major motor in use is the induction motor: General analysis of this (as can be doen with the other motors) is on the basis of coupled current carrying windings. I won't go into detail but in this case, for steady state operation the model becomes that of a transformer where the load is a ficticious resistance dependent on the rotor resistance and the slip (difference in speed between the motor's rotating field and the rotor of the motor. At 0 slip, there is no mechanical power developed and the only current is the magnetising current. Now here is where your model comes in:

-----Rs+jXs------o-------Rr/s +jXr-----| Stator impedance Rs +jXs I---> | Ir---> | Rotor impedance Rr/s +jXr hwere s is the slip V Zm | Zm is the magnetising impedance branch ______________|________________| Torque proportional to (Ir^2)Rr/s Pmech =(120*pi*frequency/poles)* phases*ws* (Ir^2)Rr(1-s)/s

I do have a program that looks at the behaviour of DC motors of different kinds as well as that of synchronous and induction motors. It also shows the effects of non-linearity on magnetising currents and interactions between synchronous machines on a 2 machine system (including governor droop) If you want it, let me know. It is written in Turbo Basic and compiled to an exe DOS based executable. (Full screen under windows is best). However, it doesn't delve into the theory per se- just the behaviour of the machines in steady state operation.

Correcting my result above, by accounting for phase. Power factor would be R/Z = 0.2 (not a good motor..) Ib = E/R - E/Z PF = (120/10) - (120/50) * 0.2 = 1.92 amps

Excellent view of back emf you offer above. In your examples there must be a real back enf voltage. This would be neccessary for the variable voltage (back emf) to transfer energy to the source when its voltage is greater than that of the source (work as a generator), which is the reality of what actually happens. Therefore the same would apply with back emf less than the voltage source. So I would suggest there must be a source of real back emf in any motor model.

I believe one needs to account for phase (add power factor to your equation).

Northstar

Thank you. Not to complain, but the ASCII schematics however are jumbled and unreadable, and some terms not defined.

Northstar

I hope I gave a suitable description. It was definitely a long ways short of being an analysis. Power factor is, as you say, important for any of the ac machines.

j

Actually, your description was very good. There is a real back emf- the voltage produced by windings moving in a magnetic field. In an AC machine, this voltage may or may not be in phase with the applied voltage (and generally is not in phase). However, the "power factor" of a motor only relates to the voltage and current at the terminals- not the relative phase of two voltages. The magnetising current of an induction motor is the main culprit there.

Very suitable. It appears to illistrates that back emf is a real voltage.

Northstar

Much jumble snipped.

Again thank you, but unreadable. I believe you said basically

________ Ra______ | | | | Vin Eg | | |_________________|

Vin = voltage applied Ra = armature resistance Eg = back emf

Then Vin = I Ra + Eg

Northstar

Damn- I am having problems- equation is correct for a DC machine as you have written it. Also it appears that I got carried away with too much detail at once- sorry.

Just let us say that the resistance and the back emf are in series. (Both DC and AC synchronous machines) The polarity of the back emf Eg is the same as that of the source voltage.

Also, to get down to basics- Back emf is the voltage expressed by Faraday's Law and the direction given by Lenz' law. These apply to coils, transformers or motors- wherever there is a changing flux linking a winding.

There is no equivalent in terms of current.

The commonly used term "induced current" is, in fact wrong and misleading. Induced voltages do exist and these can cause current to flow or modify existing current.

Of interest is the relation of the load voltages (I Ra + eg) to the electrical power factor (cosine angle between current and input voltage). Any comment?

Northstar

------------ Yep: a) DC machine -phase angle is not of consequence and pf is not a useful concept (always 1). The only reason that Vs and Eg can be different is due to the IRa drop. Note that the "load" voltage is actually Eg -not IR+Eg as Eg is the internal speed voltage. The voltage that you have given is the applied voltage at the terminals.

b) in an AC machine the leakage reactance is much larger than the resistance so the supply voltage is Vs =IZa +Eg as phasors. Again, the only reason Vs and Eg can be different is due to the IZa drop.

c) Rewrite: I =(Vs-Eg)/Za so the phase angle of the current depends on both the relative phase of Vs and Eg as well as the angle associated with Za. This applies to the synchronous machine. In the case of a synchronous machine, the magnitude of Eg is usually controllable (by controlling the DC field current) and the power factor can be swung from lagging to leading by doing so. A synchronous motor which is overexcited (Eg>Vs) will be capacitive. If underexcited (Eg

Thank you. So what is the electrical power factor in the form of R/Z for an ac motor? TIA

Northstar

---------- The question makes no sense. An AC motor is not an impedance load. R/Z isn't meaningful. As the load changes, the power factor changes. Power factor can be expressed in terms of R/X for a passive resistance/reactance device but stick to the original definition in terms of current and voltage for motors.

Suppose that at a given load, a 240V motor draws 10A at 0.8 pf lag. Under these conditions, and ONLY under these conditions, it may as if it has an R/Z of 0.8 and the motor will to have a Z =24 ohms at angle

36.87 degrees or a resistance of 19.2 ohms in series with an inductance of 14.4 ohms. In fact the motor may have an actual R =0.1 ohms and an actual reactance of 1 ohm in series with a voltage source. Knowing this doesn't give any useful information as, when load or voltage changes, the calculated from the current and voltage will change. In other words if I know that at 240V,10A 0.8 pf the =24 ohms, that knowledge is absolutely useless for finding the current at 230V. Ohms law (E=IZ for Z constant) isn't worth a $3 bill. Sure it would be possible to calculate the apparent impedance for different conditions but this would require calculation or measurement of the V and I under a wide range of conditions, followed by a lot of mathematical analysis. Why bother- it serves no useful purpose? Motors, generators and batteries are active devices, involving energy conversion and cannot be properly modelled by an impedance.

I have replied to this under a new subject heading: motor models

What is the original definition?

Sorry but I disagree. What about shaker tables and speaker motors?

Northstar

---------- Look at DC where power =V*I Now look at AC where it became obvious (about 100 years ago) that power is not V*I . The term V*I was then called apparent power and the real power was realised to be V*I *cos(angle of I with respect to V). Essentially then, power factor was defined as cos (angle of I with respect to V) or real power (watts)/apparent power(volt-amps). Reactive (vars) then was defined as V*I*(angle...). In the case of a passive load Z, where one can say Z=V/I the angle associated with Z is the inverse of the angle between current and voltage. V=100V @ 0 degrees. I =10A at -30 degrees. S(apparent power) =1000Volt amps. P(real power) =1000*cos (-30) =866Watts pf=0.866 lag( I lags V) If the load is passive then one can say Z=100/10 =10 ohms and R =8.66 ohms, X=10*sin(30) =5 ohms.

-----------

angle

as,

the

Good question: The difference is in how the motor is loaded and what the operating conditions are. Speakers and industrial (or washing machine) motors operate in different regimes with different performance criteria.

Let's look at speakers: We have an electrical network coupled to a mechanical network and the main variable is frequency . Assuming linearity (as analysis does in this case) the relationships are- for steady state at any given frequency: E=ZcI+Eg where Zc is the coil impedance R+jwL) at radian frequency w=2*pi*f Eg=CV where C is essentially flux of the magnet times the effective coil length and Vis the coil velocity. The mechanical force is F=C*I The mechanical load consists of damping D, mass M, and spring compliance K - all constant- independent of frequency force or velocity (again linearity assumed) Then, mechanically the load consists of F=Dv +j(M*w-1/K*w)V =Zmech*V From the electrical side Eg = C*V=C*F/Zmech =C*C*I/Zmech Then E={Zc +(C^2)/Zmech]* I and the apparent impedance is Zc +(C^2)/Zmech (You may refer to Beranek- "Acoustics" -if you bow 3 times beforehand.)

After all this - what is the difference?

The apparent impedance of the speaker is frequency dependent but not dependent on voltage or current - given a speaker in a particular situation (i.e. enclosure or room), it's mechanical load is strictly (within limits) dependent on the force (current) and velocity (back emf) as well as the frequency. With the supply voltage magnitude fixed, the current is dependent on frequency only. It looks like a somewhat buggered up transformer as does the coupling between the speaker cone and the acoustic load. Since the frequency is variable, a plot of Z vs frequency is meaningful (in terms of the load on the source) and considerations of resonances are important. Impedance matching may be a concern where it isn't for an industrial motor.

The speaker load may be represented by an impedance in normal operating conditions. This is rarely a useful approach for a general motor load.

The objective, in any situation is to use a model which reflects, as well as possible, the reality of the situation. The basic relationships are essentially the same but the mechanical load calls the shots- The motor , in either case, must meet the load requirements.

In the case of the speaker, the mechanical load looks, within limits, like a constant R,L, C circuit under variable frequency conditions. In the case of a fan or pump, the mechanical load doesn't look like a constant R,L,C circuit and is usually under constant frequency conditions (variable frequency complicates things somewhat).

Example: A speaker with an input of 10V may have an output of 10 watts. At 5 V the current is also halved so the output is roughly 2.5 watts. This is consistent with an impedance model. A motor at 10V may have an output of 10 watts but at 5 V the output may be 5 watts or 2.5watts or 1 watt, depending on the load's speed torque characteristic.The impedance model doesn't represent this correctly unless one goes to a non-linear model which multiplies the analytical effort by a hell of a lot.

(Example: 100=10*I, find I vs 100=(0.8*I +0.2*I^3 ) - find I- which is more work?)

It is possible to use an impedance model- I don't deny that. However, in most motor applications, it doesn't save work or improve understanding. In the case of an induction motor, the load may be modelled by a slip dependent resistance load on a transformer. It works. However determining the operating point when the motor is driving a given load becomes a graphical analysis or successive approximations- where do the motor and load torque speed curves cross?

Part of the difference, outside of this, is that "Load" in utility or industrial terms is consdered to be power (including volt-amps as apparent power etc) where in the electronic industry, it is an impedance. This leads to a lot of confusion.

Thank you for the excellent analysis. Very interesting. The only thing I see needed to complete the analysis is the relationship or relationships for mechanical power. Again, thanks.

Northstar

-----------much snipped.------------

Take the DC case for simplicity: Pin =V*I =R*(I^2)+Eg*I R*(I^2) is loss in the winding resistance. Eg*I is the power transferred from the electrical side to the mechanical side. As Eg=K*(flux)*w and Torque T=K*(flux)*I where K is a bugger factor of the individual machine and flux is the magnetic flux per pole from a field winding or a permanent magnet. w is the mechanical speed ( radians/ sec that makes the K the same in the MKS system) Then Eg*I =Tw =the mechanical power developed (watts) Pout is then Tw-mechanical losses.

AC Synchronous motor: Essentially the same relationship but phasor relationships must be included. The power converted will be Tw = Eg*I* cos(angle between Eg and I) w will be the same at all loads and is ws (see below).

Induction motor: In the usual steady state model, the mechanical load is represented by a resistance Rr/s where Rr is the rotor resistance and s is (ws-wm)/ws where ws is the synchronous speed determined by the frequency (same as in synchronous motor) -say 1200rpm for a 4 pole 60 Hz motor and wm is the actual speed -say 1150 rpm. Then s =50/1200=0.0417 or 4.17% The torque is then (poles*phases/(2*pi*frequency)*(Ir^2)Rr/s The speed is ws(1-s) and from these the Twm can be found. (mechanical losses can be deducted from this) The current Ir can be found from the circuit model.