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 > >
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 W." wrote in message news: snipped-for-privacy@news.supernews.com... | In article , | snipped-for-privacy@bigpond.net.au says... | >
| >
| >"Bill W." wrote in message | >news: snipped-for-privacy@news.supernews.com... | >| | >| 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
The circuit again:
_________________________________ | | | | | armature | | > resistance >
| shunt < < | field > >
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.
"Bill W." wrote in message news: snipped-for-privacy@news.supernews.com... | In article , | snipped-for-privacy@bigpond.net.au 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
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:-)
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 / Rwhich 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.
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
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.
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
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.
----------- I would disagree with that- Please note that Vapplied-CEMF =0 in the ideal no-load situation (ignoring IR ) which implies that Vapplied is "in phase" with the applied voltage. Take any two equal sources which are in phase and connect them - nothing happens -which fits the "series opposing" situation. Unequal voltages- then there will be a current - In the case of a DC motor, the voltage at the motor terminals after disconnection from the supply will be of the same polarity (i.e. phase)as that of the supply.
Part of the problem is in how one looks at the situation or tries to apply Lenz's law. There is a tendency to use two different viewpoints at the same time. That doesn't work. Apply a voltage to a motor- there will be a speed voltage produced which "opposes" the applied voltage. This voltage "rise" is of the same polarity as the voltage "rise" across the supply. This is also true for a resistor. Simply put- the motor load appears as a resistance for DC and as a resistive impedance for AC. It is an energy sink as far as the supply is concerned. In analysis, use voltage rises throughout or use voltage drops throughout but don't mix the two viewpoints . Otherwise "conservation of energy" is apparently violated and we really don't want to try that. Whether the situation is a motor or a voice coil or...- it really doesn't matter.
-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
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 > >
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.
---------- This is true but still doesn't address my point that the applied voltage and the back emf or generated voltage of the motor are of the same polarity - I was ignoring the IR drop (taking a theoretical limit in the same way that internal resistance of a battery is often ignored) as in a good machine, at no load, there will be very little torque required to overcome friction and windage and the current and IR drop will be small- generally negligable compared to other factors. (i.e.for a 50HP, 230V motor the no load armature current will be about 7Aand the armature resistance about 0.05 ohms. Then the back emf at no load will be about 229.6V as opposed to my approximation of 230V or a difference of 0.2%).
--------
---------- See above. The point is that the voltages are of the same polarity or "phase" I was simply looking at a theoretical limit as an approximation. It does fit as "series opposing" is related to connection and polarity, not magnitude. Give the motor a mechanical boost and the "equal voltage" situation or the situation where the motor emf exceeds the supply voltage and the current reverses and we will still have "series opposing" All one has is two sources, one of which is speed dependent.
----------------snip of diagram of shunt motor-------->
--------- See above
-----------
Actually Lenz's Law reads (or should read) quite a bit differently than that. In the case of a source applied to the motor, there is no current generated by the CEMF or speed voltage in the armature-there is a reduction of currne due to the opposing voltage. If you say that the induced emf opposes the driving source and thus affects the current -fine.
The current that flows is determined by the load torque (including losses) in the motor so the motor slows under load reducing the speed dependent back emf (and increasing V-Ea where Ea is the armature induced speed voltage which you call CEMF and I call a generated voltage.) until sufficient current flows to handle the load's torque need. If there is no friction or windage, then there will be no need for torque and the current will be 0. In addition, magnetic fields do not generate currents. They generate voltages. Also note that the field produced by the armature current in a DC machine is in space quadrature with the exciting field. Ignoring non-linearities, the two fields do not interact. This is inherent in the design of the machine.
The generated/ induced speed voltage does, indeed, oppose the applied voltage. I have no objection to this. What you have is a case of two sources, the supply and the motor in which the supply voltage is higher than the motor voltage and power flow is from electrical to mechanical. If the generated voltage is higher - the flow is the other way.
--------- > >"opposes" the applied voltage. This voltage "rise" is of the same polarity
--------- . Yes E and Cemf are series opposing. I never denied this. However, the rest of the above makes no sense. The only armature load is the mechanical load. The armature resistance will always be in series with the armature speed voltage - both are internal to the machine so even the concept of having two voltage sources in parallel driving current through the armature resistance is a no go.
----------->
-------------- Except that I was considering conservation of energy as applying a bit further- i.e. in the conversion from electrical to mechanical. Remember that in any DC motor (steady state -i.e. no leakage inductance effects)the basic relationships boil down to 3 equations (1) V=IaR+Ea where
(2)Ea =K(phi)w Note that Ea and speed are intimately connected-current doesn't enter this relationship
V is the supply voltage, Ea is the motor generated voltage, phi is the flux , w is the speed and K is a bugger factor depending on the motor, Ia is the armature current (3) T =K(phi)Ia Torque and Ia are also intimately connected by the same K(phi) (MKS units)-speed or voltage doesn't enter this equation. Then: P in =VIa =I^2R +EaIa but EaIa =Tw = Mechanical developed output including mech losses. so conservation of energy leads to Pelectrical in =Plosses (electrical and mechanical) + Pmechanical out
The actual behaviour of the motor will depend on how phi is obtained - hence a series machine will not behave the same way that a shunt machine would. -- Don Kelly snipped-for-privacy@peeshaw.ca 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" .
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.
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 +E3Noting 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
we do
------------
---------- I seem to have a different view here. I think of there being two voltage sources as shown
1 s 2 |-------/ -------|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 .
---------- 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.
------------------
---------- 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.
??? Where is the misunderstanding here? I have not said otherwise, in fact as my diagram shows (which was snipped again), and as I have noted, the negative side of applied voltage and that of back emf are common to ground. This means they are of the same polarity.
Please see above.
I was simply looking at a theoretical limit as an approximation. It
One can find literaly dozens of interpretations of Lenz' law, take your pick. One professor stated "There is no back emf, it's just a communist plot." :-) If one tapped the armature coil at several points and measured voltages from supply through the taps to ground there would be no evidence of back emf, just the appropriate voltage drops. Also there is only one current. I view back emf and back current as calculated or effective values.
From a force point of view (which I prefer), then from standstill, and with sufficient applied voltage, velocity increases until the retarding force of the load (which increases along with velocity) equals that of the applied voltage which is E BL/R. During this time, back emf increases (decreasing current) until a balance of force between driving force and retarding force is reached, i.e. until force in equals force out. achieving constant velocity. Of coures, the reverse applies with a decrease of load.
Just as I thought.. the wife is a generator...
---------
Assuming you refer to the above version of Lenz's law, see above. The above version was clipped from my archives, I can find several more. Since I didn't write it, here is my version as applied to a motor: As a conductor moves through a magnetic field, a force is generated that opposes the force causing the motion.
?The only armature load is the mechanical load.
??? again. This implys that I said "the concept of having two voltage sources in parallel driving current through the armature resistance" was a valid concept. This is absolutely not true.
All good hi-tech stuff I suppose, but the bottom line is this: Current theory states that energy can be changed in form, but not created or destroyed, and the power equation conforms to this. One can manipulate the power equation I gave
2 IE = I R + I CEMFto describe energy or work.
--
Yes, I try not to counter the wife...
Bill W.
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.
OK- it appears we are both misinterpreting each other's statements to some extent but are agreeing on this.
-------
approximation
--------------------- True - all are wordy and often definitions do not apply to the case under consideration. As a point of interest, of the 5 machines texts that I have copies of, 4 don't even mention Lenz's law and the oldest gives a definition but notes that using voltage drops gets rid of the sign problem and never refers to it again. That is an indication of its real usefullness or lack thereof.
One doesn't need to tap the coil as the voltage measured at the terminals is a voltage drop. This drop is partly due to IR and partly due to the generated voltage and one cannot separate the two by such a measurement. The separation does exist in the circuit model but this is simply a model which gives the correct terminal behaviour and the internal representation can be anything that does this, including a little old man with a crank and a variable resistor.
The so-called back emf is the speed voltage that is generated- it exists but the only way to measure it is to run the machine as a generator ands measure the open circuit voltage. That is actually what is done (open circuit magnetisation curve givning the voltage as a function of the exciting current- done at one speed and scaled linearly proportional to speed.)
------------
--------------- I don't think we are disagreeing here even though you have oversimplified a bit and I have a quibble (below). I was not addressing the nature of the load's transient or steady state torque speed characteristic vs the motor's. I can if you wish. If they don't intersect then there is a real problem. Given a steady state situation then if there is a change in load, the motor torque will not then match the load torque and must rise or fall to match if a new operating point is to be reached. If there is a decrease in load torque then the excess motor torque accelerates the load and this in turn increases the back emf and decreases the current - reducing the torque difference- eventually a new balance. The opposite in the case of a load increase. Giving a transient condition such as starting, then the load dynamics and the motor dynamics are related by E=K(phi)w (or BL(velocity) and Torque =K(phi)i -just include factors such as inertia and inductance,etc.
My quibble is that I do have a problem with the expression you give for force unless by E you mean the difference in voltage between the supply and the motor generated voltage. You haven't defined your terms or the basis for this expression (Bli, Blv ?). Please clarify.
------------
--------------- A conductor moving through a field doesn't necessarily have a counter force generated. I am assuming you are talking about a mechanical force. There will be a voltage generated (if by "force" you mean electromotive force, say so), and the direction of this voltage is such that it will to produce a current which in turn will produce an opposing force. No current- no opposing force.
Simply looking at power flow avoids problems with Lenz's law. If generating, delivering energy to the electrical system, there is a counter force (mechanical) against the prime mover. If motoring, or delivering energy to the mechanical side, the voltage (electromotive force) acts to reduce the motor current and associated mechanical torque or force.
----------
True. You brought it up the concept (derisively) and did not imply that it was a valid concept. I was wondering why it was even mentioned as it was so far out. Hence my comment.
----------
--------- And what I said agrees (P=VIa =Ia^2R +EaIa is the same equation with the terms defined) with this, but I simply went a bit further to include the mechanical output. After all a motor is simply a way to convert electrical energy to mechanical energy. The circuit model only represents a breakdown of the electrical side of the power balance (again, as a model its validity is there only at the terminals). I should have said that "conservation of energy is satisfied", rather than "conservation of energy leads to...." The equations that I gave are not hi tech by any means but just very simplified equations for a DC machine in steady state (I've run and tested enough of these to know).
Note also that it is possible to model a motor by a kind of transformer model (one side electrical, other mechanical and ideal transfromer in between), and represent the mechanical parameters by equivalent R, L and C. There is little benefit in the case of a DC motor but there is a benefit in using the transformer approach for a speaker.
-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer
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