Motor torque and back emf

Does anyone have, or know where to find an explanation of the relationship between back emf and the mechanical power at the shaft of an AC electric motor other than I Eb. I'm
referring to the theory that the mechanical power at the shaft is due to the electrical power expended against the back emf or back current. TIA
Bill W.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
| | Does anyone have, or know where to find an explanation of | the relationship between back emf and the mechanical power | at the shaft of an AC electric motor other than I Eb. I'm | referring to the theory that the mechanical power at the | shaft is due to the electrical power expended against the | back emf or back current. TIA | | Bill W. |
Bill, I think you might have your wires crossed a little.
Let me have a go at explaining.
I will use a DC motor as an example because it is easier to get your head around, but the same thing applies to AC motors as well ( well kinda, at least )
An electric motor can only deliver power against a mechanical load. If you take away the mechanical load the motor will spin at it's unloaded speed, drawing very little power.
Counter Electro Motive Force ( CEMF ) is a voltage generated by the armature of a motor turning inside the magnetic flux of it's field. ( voltage is proportional to RPM and Flux density, along with a couple of constants)
Consider the simplist case of an unloaded DC motor, initially at standstill, powered by a battery, with a separately excited field (Shunt field ). At the instant the armature circuit is closed, the armature CEMF is zero because the motor is stationary. the only effective resistance in the circuit is the resistance of the armature. THe current that will flow in the armature circuit is simply the battery voltage divided by the resistance of the motor armature, I armature = V battery / R armature
this current will be realtively large and the resulting torque produced by the armature windings will start to accelerate the armature.
Once the armature starts to turn, you have a winding rotating in a magnetic field. This motion will start to produce a voltage, Which just happens to be of the opposite polarity as that of the battery, ( Thus COUNTER EMF ) Now our calculation of current has to take into account , this voltage subtracting from the battery voltage. Thus
I armature = (V battery - V cemf) / R armature.
considering the CEMF opposes the battery voltage, the net circuit voltage will be less then it was before, Thus the current will start to fall as the motor accelerates.
Once the motor reaches it's full no load speed, the generated counter EMF will almost match the battery voltage, thus the net voltage will be very small and the current flowing will be small. Basically a balance is reached where windage and friction of the unloaded motor slows the motor to a speed where the difference in voltage between the battery voltage and the CEMF is enough to have the right amount of current ( read force) to overcome these losses.
This is called the "No Load Current"
Lets assume we have the equipment to apply an infinitely variable load to this motor. If we apply some load to the motor shaft, The balance of forces would be upset because the total load would be greater then the force, the "No Load current" can provide. The result is the motor slows a little.
With this slowing , comes a reduction in the motor CEMF. The net result is the diffrernce between the battery voltage and the CEMF is greater, and thus the armature current will increase. This Increased current, results in increased torque ( read force) and at some stage a new balance is reached where again the torque produced by the motor armature current. matches the torque of the mechanical load. So effectively the motor has slowed. however the torque has increased to match the torque of the load. ( the amount the motor slows is referred to as the "droop" )
If we continue to increase the load on the motor , the speed will continue to fall, until such times as the "Full load current" is flowing.
If the applied Armature voltage is set at the "Motor rated voltage", and the voltage on the field is set to "motor rated field", The power delivered at this speed and torque is said to be "Rated power" of the motor, and the speed of the motor is said to be at "Rated Speed" .
These values are normally specified for a particular motor design and are stamped on the motor "Name Plate"
If the mechanical load was suddenly removed, the motor would quickly accelelrate back up to it's "No Load Speed" and sit there with " No Load curnent" flowing.
As long as we leave the terminal volts constant the behaviour under varying loads will be that the motor speed changes up or down until the current flowing produces enough torque to match what ever load is applied.
If the load is increased beyond motor rated, Nothing magical happens, The motor simply slows down even further and the current increases until the net motor torque matches the torque demanded by the mechanical load.
This is termed as "over loading the motor" , and if sustained for any length of time will result in the internal temperature rising to an excessive level. This excessive temperature will result in premature failure of the winding insulation, or in more common terms " Motor Burned out"
Now typically we employ control systems to keep the motor at some desired speed, independant of load, To do this, the "Battery voltage", so to speak, is increased as a function of load, instead of letting the motor slow, so that the desired armature current or torque is acheived without the armature speed (or CEMF) decreasing.
The Back EMF is a funtion of the motor speed and the field current ONLY. The actual speed the motor goes for any given load can vary depending on the control system employed.
Translating this to AC motor theory, is not so simple, but basically the field of an AC Induction motor is produced by current in the Squirrel cage winding, which itself is a function of the slip between the stator and the rotor. Counter EMF is still dependent on the motor speed ( which does not change a lot in the normal operating range) and the field flux, which is dependent on how much the motor slows for a given load.
Again the input voltage is fixed by virtue of the fixed line voltage. supply frequency is normally fixed so, as in the example above, the only thing the motor can do to respond to an increasing load is to slow down. As it slows, the slip between the stator and the rotor increases, causing the field (or excitation current) to increase.
The net result of the slowing motor speed and the increasing Field flux is a net reduction in Motor CEMF. Lower CEMF with fixed supply ( as above) will result in more Armature ( read LINE) current thus more torque.
Again A balance is reached where the torque produced by the combination of increased field, and increased line curent, matches the torque demanded by whatever the mechanical load is.
The counter EMF in this case is still the direct function of the motor speed and the motor excitation Flux. NOT directly related to mechanical load, but effected by it in any case.
Well That is my attempt, I know it is a little long but maybe it made sense
Good Luck Tom Grayson 10 Nov 2003
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
snipped-for-privacy@bigpond.net.au says...

Thanks for the excellent explanation, Tom.
The only question I would have concerns the phase of the CEMF. Intuitively, it seems out of phase. Looking at a shunt DC motor with the field essentially independent of changing other parameters
_________________________________ | | | | | armature | | > resistance > | shunt < < | field > > + < < constant > | DC power | | source | | _ | + ___ | | - | | CEMF ___ | | _ - | | | |______________|__________________|
Viewing this as source and CEMF in series then the source and CEMF are in series-opposing, or out of phase. Viewed as the two in parallel, negatives are common, thusly the two are in phase. Even if CEMF polarity is noted as incorrect, the same theory holds. I've seen it stated in textbooks as in phase, in other textbooks as out of phase. Generally the phase is not noted... rather a statement such as "A CEMF is generated that opposes the drive voltage." I tried the following with permanent magnet loudspeakers, regarding both the magnitude and phase of CEMF:
Back emf or CEMF
A tandem system was built with two 8 inch drivers having identical motors, one mounted behind the other. Most of the cone was removed from the rearward driver, in order to lower its mass and not restrict air flow. The voice coils were coupled together with a lightweight rod going through the front drivers back plate vent hole, such that driving the rear unit caused the undriven front unit to move in unison. The tandem speaker was then installed in a one cubic foot box with no stuffing. With the rear driver at open circuit (no connection), the front driver was driven at fundamental resonance (58.6 Hz) through a current meter with 1.41 volts across the speaker terminals, and a microphone placed 1/4 inch from the cone, noting the current and exact SPL on digital meters.
Calculated back emf was
Eb = E - (I Re) = 1.127
where (I Re) is the net or effective voltage across the coil-armature. which may be referred to as armature voltage.
The rear driver was then driven at a level giving the same exact SPL reading, with only a meter connected to the front driver to monitor its generated voltage. At the same SPL (read excursion), the generator action of the motor should generate the calculated back voltage. The generated voltage was noted, and the entire process was repeated at higher and lower input levels for 5 trials. The generated voltage values matched the calculated values for back emf within 0.21 percent average, for excellent agreement between theory and actual operation. That the tandem driver was not too different from average in function is shown by
Qmc = 5.72 Qec = 1.43 Qtc = 1.14
Back emf Phase
The rear motor of the tandem speaker was driven at resonance, with channel A of a Tektronix oscilloscope monitoring the waveform at the speaker input terminals. The open circuit terminals of the front motor were connected to channel B of the scope, to monitor the generated voltage. All driver and scope grounds were common to each other, of course. The two traces were in perfect phase, even allowing perfect overlap, when amplitude was matched on the scope gain controls, showing back emf to be in phase with the applied voltage in this evaluation mode.
I would appreciate your (and others) view on the premise of the evaluation mode, and overall view as well.
Now.. perhaps my wording as to intent was lacking in the original post above. If I state "The energy expended against the CEMF appears as the torque on the motor shaft", then is this accurate? In other words, is the current or power that's driven against the CEMF, what generates the actual force that moves the load?
Bill W.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
| >| | >| Does anyone have, or know where to find an explanation of | >| the relationship between back emf and the mechanical power | >| at the shaft of an AC electric motor other than I Eb. I'm | >| referring to the theory that the mechanical power at the | >| shaft is due to the electrical power expended against the | >| back emf or back current. TIA | >| | >| Bill W. | >| | > | > | >Bill, I think you might have your wires crossed a little. | > | >Let me have a go at explaining. | > | >I will use a DC motor as an example because it is easier to get your head | >around, but the same thing applies to AC motors as well ( well kinda, at | >least ) | > | >An electric motor can only deliver power against a mechanical load. If you | >take away the mechanical load the motor will spin at it's unloaded speed, | >drawing very little power. | > | >Counter Electro Motive Force ( CEMF ) is a voltage generated by the | >armature of a motor turning inside the magnetic flux of it's field. ( | >voltage is proportional to RPM and Flux density, along with a couple of | >constants) | > | >Consider the simplist case of an unloaded DC motor, initially at standstill, | >powered by a battery, with a separately excited field (Shunt field ). At the | >instant the armature circuit is closed, the armature CEMF is zero because | >the motor is stationary. the only effective resistance in the circuit is the | >resistance of the armature. THe current that will flow in the armature | >circuit is simply the battery voltage divided by the resistance of the motor | >armature, I armature = V battery / R armature | > | >this current will be realtively large and the resulting torque produced by | >the armature windings will start to accelerate the armature. | > | >Once the armature starts to turn, you have a winding rotating in a magnetic | >field. This motion will start to produce a voltage, Which just happens to | >be of the opposite polarity as that of the battery, ( Thus COUNTER EMF ) | >Now our calculation of current has to take into account , this voltage | >subtracting from the battery voltage. Thus | > | >I armature = (V battery - V cemf) / R armature. | > | >considering the CEMF opposes the battery voltage, the net circuit voltage | >will be less then it was before, Thus the current will start to fall as the | >motor accelerates. | > | >Once the motor reaches it's full no load speed, the generated counter EMF | >will almost match the battery voltage, thus the net voltage will be very | >small and the current flowing will be small. Basically a balance is reached | >where windage and friction of the unloaded motor slows the motor to a speed | >where the difference in voltage between the battery voltage and the CEMF is | >enough to have the right amount of current ( read force) to overcome these | >losses. | > | >This is called the "No Load Current" | > | >Lets assume we have the equipment to apply an infinitely variable load to | >this motor. If we apply some load to the motor shaft, The balance of forces | >would be upset because the total load would be greater then the force, the | >"No Load current" can provide. The result is the motor slows a little. | > | >With this slowing , comes a reduction in the motor CEMF. The net result is | >the diffrernce between the battery voltage and the CEMF is greater, and | >thus the armature current will increase. | >This Increased current, results in increased torque ( read force) and at | >some stage a new balance is reached where again the torque produced by the | >motor armature current. matches the torque of the mechanical load. So | >effectively the motor has slowed. however the torque has increased to match | >the torque of the load. ( the amount the motor slows is referred to as the | >"droop" ) | > | >If we continue to increase the load on the motor , the speed will continue | >to fall, until such times as the "Full load current" is flowing. | > | >If the applied Armature voltage is set at the "Motor rated voltage", and the | >voltage on the field is set to "motor rated field", The power delivered at | >this speed and torque is said to be "Rated power" of the motor, and the | >speed of the motor is said to be at "Rated Speed" . | > | >These values are normally specified for a particular motor design and are | >stamped on the motor "Name Plate" | > | >If the mechanical load was suddenly removed, the motor would quickly | >accelelrate back up to it's "No Load Speed" and sit there with " No Load | >curnent" flowing. | > | > | >As long as we leave the terminal volts constant the behaviour under varying | >loads will be that the motor speed changes up or down until the current | >flowing produces enough torque to match what ever load is applied. | > | >If the load is increased beyond motor rated, Nothing magical happens, The | >motor simply slows down even further and the current increases until the net | >motor torque matches the torque demanded by the mechanical load. | > | >This is termed as "over loading the motor" , and if sustained for any length | >of time will result in the internal temperature rising to an excessive | >level. This excessive temperature will result in premature failure of the | >winding insulation, or in more common terms " Motor Burned out" | > | >Now typically we employ control systems to keep the motor at some desired | >speed, independant of load, To do this, the "Battery voltage", so to speak, | >is increased as a function of load, instead of letting the motor slow, so | >that the desired armature current or torque is acheived without the | >armature speed (or CEMF) decreasing. | > | >The Back EMF is a funtion of the motor speed and the field current ONLY. | >The actual speed the motor goes for any given load can vary depending on the | >control system employed. | > | >Translating this to AC motor theory, is not so simple, but basically the | >field of an AC Induction motor is produced by current in the Squirrel cage | >winding, which itself is a function of the slip between the stator and the | >rotor. Counter EMF is still dependent on the motor speed ( which does not | >change a lot in the normal operating range) and the field flux, which is | >dependent on how much the motor slows for a given load. | > | >Again the input voltage is fixed by virtue of the fixed line voltage. supply | >frequency is normally fixed so, as in the example above, the only thing | >the motor can do to respond to an increasing load is to slow down. | >As it slows, the slip between the stator and the rotor increases, causing | >the field (or excitation current) to increase. | > | >The net result of the slowing motor speed and the increasing Field flux is a | >net reduction in Motor CEMF. Lower CEMF with fixed supply ( as above) will | >result in more Armature ( read LINE) current thus more torque. | > | >Again A balance is reached where the torque produced by the combination of | >increased field, and increased line curent, matches the torque demanded by | >whatever the mechanical load is. | > | >The counter EMF in this case is still the direct function of the motor speed | >and the motor excitation Flux. NOT directly related to mechanical load, but | >effected by it in any case. | > | >Well That is my attempt, | >I know it is a little long but maybe it made sense | > | >Good Luck | >Tom Grayson | >10 Nov 2003 | | | Thanks for the excellent explanation, Tom. | | | The only question I would have concerns the phase of the CEMF. | Intuitively, it seems out of phase. Looking at a shunt DC | motor with the field essentially independent of changing other | parameters | | _________________________________ | | | | | | | armature | | | > resistance > | | shunt < < | | field > > | + < < | constant > | | DC power | | | source | | | _ | + ___ | | | - | | | CEMF ___ | | | _ - | | | | | |______________|__________________| | | | Viewing this as source and CEMF in series then the source | and CEMF are in series-opposing, or out of phase. Viewed | as the two in parallel, negatives are common, thusly the | two are in phase. Even if CEMF polarity is noted as | incorrect, the same theory holds. I've seen it stated | in textbooks as in phase, in other textbooks as out of | phase. Generally the phase is not noted... rather a statement | such as "A CEMF is generated that opposes the drive voltage." | I tried the following with permanent magnet loudspeakers, | regarding both the magnitude and phase of CEMF: | | | Back emf or CEMF | | A tandem system was built with two 8 inch drivers having identical | motors, one mounted behind the other. Most of the cone was removed | from the rearward driver, in order to lower its mass and not restrict | air flow. The voice coils were coupled together with a lightweight | rod going through the front drivers back plate vent hole, such | that driving the rear unit caused the undriven front unit to | move in unison. The tandem speaker was then installed in a one | cubic foot box with no stuffing. With the rear driver at open | circuit (no connection), the front driver was driven at fundamental | resonance (58.6 Hz) through a current meter with 1.41 volts across | the speaker terminals, and a microphone placed 1/4 inch from the | cone, noting the current and exact SPL on digital meters. | | Calculated back emf was | | Eb = E - (I Re) = 1.127 | | where (I Re) is the net or effective voltage across the coil-armature. | which may be referred to as armature voltage. | | The rear driver was then driven at a level giving the same exact SPL | reading, with only a meter connected to the front driver to monitor | its generated voltage. At the same SPL (read excursion), the generator | action of the motor should generate the calculated back voltage. | The generated voltage was noted, and the entire process was repeated | at higher and lower input levels for 5 trials. The generated voltage | values matched the calculated values for back emf within 0.21 percent | average, for excellent agreement between theory and actual operation. | That the tandem driver was not too different from average in | function is shown by | | Qmc = 5.72 | Qec = 1.43 | Qtc = 1.14 | | | | Back emf Phase | | The rear motor of the tandem speaker was driven at resonance, | with channel A of a Tektronix oscilloscope monitoring the waveform | at the speaker input terminals. The open circuit terminals of the | front motor were connected to channel B of the scope, to monitor | the generated voltage. All driver and scope grounds were common | to each other, of course. The two traces were in perfect phase, | even allowing perfect overlap, when amplitude was matched on the | scope gain controls, showing back emf to be in phase with the | applied voltage in this evaluation mode. | | | I would appreciate your (and others) view on the premise of the | evaluation mode, and overall view as well. | | | Now.. perhaps my wording as to intent was lacking in the | original post above. If I state "The energy expended against | the CEMF appears as the torque on the motor shaft", then is | this accurate? In other words, is the current or power that's | driven against the CEMF, what generates the actual force | that moves the load? | | Bill W. |
Bill, With regards to your comment "................................I've seen it stated | in textbooks as in phase, in other textbooks as out of | phase. Generally the phase is not noted... rather a statement | such as "A CEMF is generated that opposes the drive voltage."
I deliberately avoided using any referene to phasing for this reason.
remember "Flemmings Rules" Hold your Thumb, First finger and second finger at right angles to represent the three axes of a coordinate system.
Left hand for motors , First finger for Field or flux direction , Thumb for the direction of Motion, Second finger for the Direction of current flow.
Right hand for a generator First finger for Field or flux direction , Thumb for the direction of Motion Second finger for the Direction of the generated EMF
Hold both hands out with the, first fingers pointing away from you, ( same field direction) Thumbs up, ( same direction of motion) and note that the two second fingers are pointing towards each other.
For your Left hand (motor) the motion or force direction is up because of the interaction of the field (away from you) and the the current flowing to the Right.
For your Right hand, (Generator) The direction of the generated EMF is to the Left because of the interaction of the moving conductor upwards, and the field direction away from you.
So inherently in a motor, ( or generator) the direction of the generated Counter EMF is always in opposition to the current flowing in the motor windings that is producing the motion in the first place. :o)
After this, what you call "In Phase" or "Out of Phase" is really just a function of definition.
Remember that, in theory, a motor and generator are technically the same, where the operation switches between motoring and generating seamlessely, depending on the dynamic situation at any given instant.
following up your speaker experiment, Which , by the way , sounds well thought out.
Sounds like you have an interesting problem. I am not an authority on Speakers, but the only thing that I can suggest at first guess is your reference to driving the rear speaker at "its resonant frequency". Is this frequency the published Resonant frequency of the drivers or is it the "Experimentally measured Resonant frequency" of the whole box / Speaker assembly.
This is just a "WAG" but perhaps the resonance itself has something to do with shifting phase. I would find this highly unlikely , however in the past I have had stranger things happen to me.
Seeing you do not need to measure sound preasutre level for this second part of your experiment, I would try the experiment again, with the drivers out of the box, in "Free air" so to speak. Perhaps even working at different frequencies as well, Thou it seems highly likely that you would have already tried this.
Another thought. Do the two drivers have the same polarity on the fields?
Happy hunting Tom
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
snipped-for-privacy@bigpond.net.au says...

The circuit again:
_________________________________ | | | | | armature | | > resistance > | shunt < < | field > > + < < constant > | DC power | | source | | _ | + ___ | | - | | CEMF ___ | | _ - | | | |______________|__________________|
Take a speaker, connect a scope across the input terminals with ground to ground, positive to positive, apply a DC pulse of ~ 1.5 volts (D cell battery) with positive from the battery to positive on the speaker. If the speaker is polarized correctly, the cone will move forward (out of the frame) and the scope trace will jump upward, indicating the applied pulse begins as positive. Now lay the battery aside, and give the cone a push forward. You will see the scope trace jump upward again. To me this shows the phase of applied voltage and that of the CEMF are in-phase relative to common ground. However, in operation they function in series opposing, just as two barreries in series opposing, i.e. non-aiding.

The latter.

CEMF = Blv at resonance only, otherwise acceleration of the system mass is in the picture.

Yes - been there, done that.

Yes, a positive pulse moves the coil forward (away from the magnet on both).

However.. the original question remains: Is the work done by the power source against the CEMF what moves the load?
Bill W.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
says... | > | > | > Bill, | > With regards to your comment | > "................................I've seen it stated | >| in textbooks as in phase, in other textbooks as out of | >| phase. Generally the phase is not noted... rather a statement | >| such as "A CEMF is generated that opposes the drive voltage." | > | >I deliberately avoided using any referene to phasing for this reason. | | | The circuit again: | | _________________________________ | | | | | | | armature | | | > resistance > | | shunt < < | | field > > | + < < | constant > | | DC power | | | source | | | _ | + ___ | | | - | | | CEMF ___ | | | _ - | | | | | |______________|__________________| | | | Take a speaker, connect a scope across the input terminals | with ground to ground, positive to positive, apply a DC | pulse of ~ 1.5 volts (D cell battery) with positive from | the battery to positive on the speaker. If the speaker is | polarized correctly, the cone will move forward (out of the | frame) and the scope trace will jump upward, indicating the | applied pulse begins as positive. Now lay the battery aside, | and give the cone a push forward. You will see the scope | trace jump upward again. To me this shows the phase of | applied voltage and that of the CEMF are in-phase relative | to common ground. However, in operation they function in | series opposing, just as two barreries in series opposing, | i.e. non-aiding. | | | >remember "Flemmings Rules" | >Hold your Thumb, First finger and second finger at right angles | >to represent the three axes of a coordinate system. | > | > | >Left hand for motors , | > First finger for Field or flux direction , | > Thumb for the direction of Motion, | > Second finger for the Direction of current flow. | > | >Right hand for a generator | > First finger for Field or flux direction , | > Thumb for the direction of Motion | > Second finger for the Direction of the generated EMF | > | >Hold both hands out with the, | >first fingers pointing away from you, ( same field direction) | >Thumbs up, ( same direction of motion) | >and note that the two second fingers are pointing towards each other. | > | > For your Left hand (motor) the motion or force direction is up because | >of the interaction of the field (away from you) and the the current flowing | >to the Right. | > | >For your Right hand, (Generator) The direction of the generated EMF is to | >the Left | >because of the interaction of the moving conductor upwards, and the field | >direction away from you. | > | >So inherently in a motor, ( or generator) the direction of the generated | >Counter EMF | >is always in opposition to the current flowing in the motor windings that is | >producing the motion in the first place. :o) | > | >After this, what you call "In Phase" or "Out of Phase" is really just a | >function of definition. | > | >Remember that, in theory, a motor and generator are technically the same, | >where the operation switches between motoring and generating seamlessely, | >depending on the dynamic situation at any given instant. | > | >following up your speaker experiment, Which , by the way , sounds well | >thought out. | > | >Sounds like you have an interesting problem. | >I am not an authority on Speakers, but the only thing that I can suggest at | >first guess is your reference to driving the rear speaker at "its resonant | >frequency". Is this frequency the published Resonant frequency of the | >drivers or is it the "Experimentally measured Resonant frequency" of the | >whole box / Speaker assembly. | | | The latter. | | | > This is just a "WAG" but perhaps the resonance itself has something to do | >with shifting phase. I would find this highly unlikely , however in the | >past I have had stranger things happen to me. | | | CEMF = Blv at resonance only, otherwise acceleration of | the system mass is in the picture. | | | >Seeing you do not need to measure sound preasutre level for this second part | >of your experiment, I would try the experiment again, with the drivers out | >of the box, in "Free air" so to speak. Perhaps even working at different | >frequencies as well, Thou it seems highly likely that you would have already | >tried this. | | | Yes - been there, done that. | | | >Another thought. Do the two drivers have the same polarity on the fields? | | | Yes, a positive pulse moves the coil forward (away from the magnet | on both). | | | >Happy hunting | >Tom | | | However.. the original question remains: Is the work done | by the power source against the CEMF what moves the load? | | Bill W. | | Ths short answer is NO
What moves the load is the force produced on the driver coil by the interaction of the current in the Driver coil, with the Field flux in the permanant magnet.
My humble opinion is that
All the counter EMF does is produce a voltage in opposition to the supply voltage, which limits the current in the windings, more then the simple impedence of the coil would.
Keep in mind that if you are actually experimenting and looking at this , then you are already way in front of me. I tend to agree with you that these drivers should be behaving in a similar way as a motor because the the driving power results in windings moving in magnetic fields. If the load was simply an inductor with a little resistance, I think the only thing limiting the current would be the impedance. (much like a stalled motor, I would imagine)
Another point that you might find relivant is that I have been building a "BAND PASS" subwoofer. The books say that this design does not need a cross over because
"Quote...............It uses a unique enclosure which eliminates the need for a crossover, All one sees on the outside is a port from which the noise emanates. .................... "Bandpass is used in it's name because the enclosure itself acts like a crossover so the subwoofer passes only a limited range of frequencies to the outside air.................... End Quote"
my thought here is that, because the driver and the box will not allow the driver to respond to the frequencies above this band, that quite a lot of amplifier power in the higher frequencies will be effectively feeding into a short. I am still not sure what is correct here.
Maybe the results of what you are doing here will help me here. My plan is to put an active crossover before the amplifier so that it only responds to frequencies the Subwoofer can use.
:o)
TOm
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

cross
noise
a
to
Tom, I have no desired to get flamed again for writing about speaker systems, but FWIW I totally agree with your thinking as posted above.
Using an active crossover before the amplifier is standard procedure on all large concert audio setups and makes tuning far easier than using passive crossovers in the speaker cabinet.
I would think that, for low power systems, your book's comment is quite correct. They are likely assuming that the power handling capacity of the speaker itself will be far greater than the normal listening level (ie. high dynamic range with little chance of being driven into distortion) assisted by the response of the speaker + enclosure being naturally poor outside the design frequency range. A lot of high-frequency energy into a low-frequency sub certainly *will* damage the coil over time.
Just my A$0.02..
Cameron:-)
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
snipped-for-privacy@bigpond.net.au says...

Yes, the basic premise re the mechanical side starts with the fact that force is responsible for doing work on the load. That force works against the system mechanical impedance, which is the resistance and reactance, i.e. the square root of the sum of the squares of resistance and reactance. The force equation then is
force applied = velocity x mechanicel impedance
However it gets a little sticky, since part of the mechanical resistance is the the motor electromechanical resistance,
2 Bl / R
which depends on the magnitude of CEMF.
-------
Regarding the subwoofer, I would look at what you describe with askance if either the front or rear of the woofer radiate directly into the air outside the enclosure. Even if not, I am unclear as to how anything other than a very slow rolloff is obtained and that at a rather high frequency.
Bill W.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

This is perhaps where you are making a mistake. When you apply the battery pulse, your o-scope showed a 'positive' pulse and the cone moved 'forward'. When you disconnect the battery and manually push the cone forward, you again see a 'positive' pulse.
But you're thinking this CEMF is 'in-phase'. It is not so easy to say this. Just because the pulse is 'positive' across the same terminals, don't jump to the conclusion that it is 'in-phase'. When the cone/load generates a 'positive' pulse, it opposes the voltage you applied before with the battery. Only the *difference* between the CEMF generated by the cone moving forward, and the applied battery is available to create current flow. The reason we use the *difference* (i.e. subtract the two voltage readings) is because the CEMF is 180 from the input. If you closed the external circuit, the current from the second experiment would flow in the opposite direction than that from the first experiment. A voltage that was 'in-phase' with the first pulse would cause current to flow in the same direction and create a 'negative' pulse on the o-scope (as *crazy* as that s ounds ;-)

In *simple*, *ideal* DC motor theory...
(I'm using '==' here for proportionality, can't write a symbol easily in ASCII)
Power == Torque X Speed (or force X speed for a linear mechanism) Torque == current X Field CEMF == Speed X Field ==> Speed == CEMF/Field
Ergo...
Power == current X CEMF And as you mentioned before... current = (Vsupply - CEMF)/R.
So for a fixed-field strength, you can work out quite a few interesting relationships. Consider that large DC machinery have a total armature resistance below 0.1 ohm, one can also see why DC motors have very high starting currents (no CEMF developed yet) and how even a modest increase in R (by inserting a 'starting resistor') can reduce this starting current to acceptable levels.
This would be 'developed power'. Part of that power has to supply mechanical losses (windage, bearing friction, magnetic drag on the iron caused by eddy currents). The remainder will accelerate/power the connected load.
In an AC system, there are also leakage reactances to overcome as well and another set of eddy currents to consider.
Now, in a vibrating system, there is a formula for the acceleration a mass undergoes as it oscillates back and forth. This acceleration is function of the amount of displacement (how many mils does the object move from one extreme to the other) and how fast it moves back and forth (the frequency of vibration). This acceleration is caused by a force acting on the mass. Whether the force is in phase with the movement is another issue. As the frequency is changed through the resonant frequency(s) it will be 'in' and 'out' at different frequencies.
This 'force X distance' is work done on the mass, not necessarily the air adjacent to the cone.
daestrom
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
daestrom@NO_SPAM_HEREtwcny.rr.com says...

No mistake. :-) Perhaps not phrased clearly.. Note I said in phase with respect to ground, meaning the negatives of CEMF and the voltage applied are common. I noted also that *in operation* CEMF and applied voltage are in series-opposing, which would certainly not be termed as in phase.

_ CEMF Yes v = ------- BL

Net work done on the air is acoustic power times time
po 2 W = -------- (w U) t 2 pi c
All well and good, but are you saying the that work done by the power source against the CEMF is what moves the mass?
Bill W.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
says...

No, in my 'not so clear way', I was trying to say that the energy of CEMF x I doesn't all go into the acoustic power of the speaker. Some will 'go' into the speaker's magnet as heat and the dampening losses in the cone and structure.
With a speaker, the current is actually AC so there are eddy current losses in the magnet's body and the frame supporting the coil/magnet. I don't know how much is done in speaker design to reduce these.
A vibrating mass has a certain amount of energy in it due to the vibration (combination of kinetic and potential energy). For a speaker, this energy came from the electric signal. Once set in motion, the speaker's moving mass is in a sort of dynamic equilibrium, the energy dissipated as heat in the eddy currents, bending of the cone material and moving the air in front/behind the cone are balanced by the incoming electric signal. When the signal is cut off, the vibrating mass's energy continues to drive the speaker until it dissipates the mass's energy. If the circuit is setup for it, I suppose the moving mass could convert some of the mass's energy back to electrical form and dissipate it in the circuit. This might 'sharpen' the cutoff of acoustic energy when the electric signal is abruptly stopped.
Of course, I suppose the speaker's moving mass is kept very low so that this vibration energy storage/release is kept small. If the electric signal is constantly changing frequency and amplitude, then the kinetic/potential energy of the vibrating mass is constantly changing as well. So the larger the 'moving mass' of the system, the more distortion between the electric signal and the acoustic energy output.
daestrom
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
daestrom@NO_SPAM_HEREtwcny.rr.com says...

Simplicity of mechanical structure allows the ability to provide close tolerance, such that these losses are considered negligible.

power in = power lost as heat + mechanical power
2 IE = I R + I CEMF
2 I R is the power lost as heat in the armature coil. Heating the motor coil produces no mechanical power, leaving I CEMF to be the mechanical power, which means the current working against the CEMF (or work done against the CEMF) is what moves the mass, i.e. does work on the mass loading.

Yes, acceleration = force / mass, so the lower the mass for a given force, the higher the acceleration, and the more accurately the mechanical movement can follow the varying electrical signal. Hope the above was of interest, and thanks for your input.
Bill W.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
remove the urine to answer
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

But in operation one cannot ignore IR, therefore CEMF must always be less than applied voltage E
E - CEMF = >0

Sorry, but it doesn't fit. Again, CEMF cannot be equal to applied voltage E, it *must* be less. Here's the circuit from my original post:
_________________________________ | | | | | armature | | > resistance > | shunt < < | field > > + < < constant > | DC power | | source | | _ | + ___ | | - | | CEMF ___ | | _ - | | | |______________|__________________|
Note that in operation with current flowing, there must be a voltage drop across the armature resistance, such that
CEMF = E - (IR)
so that CEMF cannot equal E.

Lenz's law states that an induced electromagnetic force generates a current that induces a counter magnetic field that opposes the magnetic field generating the current. Accordingly, induced CEMF opposes E in the series-opposing configuration as drawn above.

Yes, per the drawing above, and viewed as two voltage sources in parallel, the negatives are common, making CEMF and E "in phase." However, one does not connect the positives together and then place the armature load between that connection and back to ground, such that CEMF and E are in parallel and driving current through the armature resistance. E and CEMF clearly are in series-opposing.

No conflict as I see it, since conservation of energy holds with voltage in equaling voltage out
E = IR + CEMF
then multiplying by current to obtain power, power in equals power out
2 IE = I R + I CEMF

A voice coil is the armature of a loudspeaker motor.
Bill W.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
remove the urine to answer
P.S. I hate the term CEMF and prefer to simply call it a generated voltage as that is what it is It is the same speed voltage that is produced when running the machine as a generator with rotation in the same direction and it is not a "counter anything" .
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Yes, I try not to counter the wife...
Bill W.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
remove the urine to answer
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
remove the urine to answer
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Does that mean I shouldn't worry about those two bare wires sticking out of the wall with 120 volts between them, since the average voltage is zero? :-)
The zero average for voltage and current reflects the characteristic of a sine function (or cosine), but that doesn't mean energy is not being transferred. _ _ P = I^2R
Even though voltage and current change sign each half-cycle, the square of the current is positive.

My Fluke and Simpson meters read RMS.
The result of the above equation E BL/R gives applied mechanical force, and and is most useful in the mechanical domain. This applied force then is numerically equal to - (velocity times mechanical impedance), where the mechanical impedance includes the total of the mechanical resistance plus the mechanical reactance. I should have noted that coil inductance is negligible below ~ 250 Hz, and thusly is generally ignored, as it is above. Sorry.

Back emf becomes part of the mechanical resistance, i.e. the mechanical resistance of the motor = (BL)^2/R.

Electrical to mechanical efficiency is simply CEMF/E, derived from power out/power in = I CEMF/IE. Yes, efficiency is low, generally under 10%. Mass is under constant acceleration or decelleration, taking a toll on the supplied energy. Note we are not working with constant velocity, and typical motor efficiency.

Note that for a given applied voltage, we have constant applied force that is not frequency dependent.

The mechanical load is included in the equation.

Considering the above, I believe not an approximation.

Sorry, miscommunication.
Bill W.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.