Back EMF - A General Question About Motors - Page 1
Forum section: electronics / beginners

author: NeuEnergy
author_rank: Newbie
timestamp: December 05, 2023, 03:45:02 pm
content: It's nice this site has a place for beginners to ask questionsI've been doing a series of tests on motor-generator sets for fun (let's notover the fact that these do not create "free energy"). Observing the data from my testing has left me with a general question about motors. I asked a friend of mine who's more knowledgeable on the subject and his response was "likely has to do with Back EMF" and recommended this forum. So here I am.Ok, so I connect my 12V power source to my 12V, 120W motor and we get 2500rpm's consuming 10A of current.Now since we can use a motor as a generator, anyone without a lot of knowledge on the subject (like myself) would assume "if I spin the motor at 2500rpm, it will give me 10A of current at 12V". Now I spin said motor (using it as a generator) up to 2500rpm's using a smaller motor, I have 12V coming out of my generator, but I connect a 5W device via an inverter and the motor cannot support it. The voltage drops and the inverter turns on/off trying to supply the 12V to my load.So the rate of spin, size of my coils and size of my magnets is not what is dictating current here or I'd have 10A and be able to supply up to 120W of power. So what is this extra force that dictates how many amps I'm able to pull from my motor that's at proper operational speed to produce the 12V I need? Why is the motor producing such low current despite having 2500rpm's of spin? Thanks in advance for enlightening me

author: CountChocula
author_rank: Unknown
timestamp: December 05, 2023, 04:19:45 pm
content: The reason the motor stalls is that whatever is driving is unable to provide a sufficient amount of power to overcome the physical resistance created by connecting the load.A generator doesn't “create” energy (otherwise, you enter the free energy crazy person realm), it justconvertsit from one form (rotational) to another (electrical). Therefore, in order to get a certain amount of current on the output, you need to apply a certain amount at the input. Since no generator is 100% efficient, the input amount will have to be greater than what is being drawn on the output.To make a somewhat contrived analogy, think about pushing an empty grocery cart at a fixed velocity. If someone drops a 20kg bag of flour inside the cart, your speed doesn't matter… you do, however, have to overcome the inertia created by the sudden increase in the mass of the cart.—CC

author: CatalinaWOW
author_rank: Super Contributor
timestamp: December 05, 2023, 04:25:10 pm
content: It has been decades since I have done motor/generator design, and don't have time to brush up at the moment. I think there are several parts to the answer, and I only have time to dredge up a couple here.1. When you are driving the motor, the power supplied does not all go into spinning the motor. Part is dissipated as heat in the windings due to their resistance. So in your example there is not 12V x 10A power represented in the rotational motion of the shaft. This probably accounts for something like 5-15 watts of the "missing" power.2. You are presumably needing 5-6 amps of current from the "generator" to supply the inverter. Depending on your connections and wire sizes this may result in very noticeable voltage drops between the generator and inverter.There are others, but I need to get on the road. Keep digging, it will all make sense eventually. No extra forces are involved.

author: IanB
author_rank: Super Contributor
timestamp: December 05, 2023, 04:39:55 pm
content: Quote from: NeuEnergy on December 05, 2023, 03:45:02 pmNo, not quite. Most likely it will consume a very small amount of current. The amount of current drawn by the motor depends on the size of load you put on the motor. There is a power balance, "power in (minus losses) = power out".QuoteOnce more, no, not quite. It may give you 10 A output, if, and only if, you put enough power into it (power balance again).QuoteMost likely, your smaller motor is not strong enough to turn the generator when you try to load it down.QuoteYou can only supply 120 W of power on the output if you provide at least 120 W of power on the input. That means the input to the generator needs a big, powerful motor of at least 120 W output.

author: radiolistener
author_rank: Super Contributor
timestamp: December 05, 2023, 04:49:42 pm
content: You're needs to apply more power to your motor than you want to obtain from generator connected to the motor. This is because part of power is radiated as electromagnetic waves and heat (due to friction in the generator and motor).As you know, any wire has some resistance and this resistance leads to power loss due to heating when you put current through this wire. The same any friction leads to power loss due to heating. This is why free energy is impossible, because you're needs to apply more power to the motor than you can get from generator connected to that motor. The rest power is radiated as a heat and electromagnetic waves.

author: TimFox
author_rank: Super Contributor
timestamp: December 05, 2023, 04:51:38 pm
content: The major losses in a DC motor are the copper loss in the winding resistance and the bearing friction.The back emf in isolation is proportional to the product of the magnetic field and the RPM.The emf and loss apply to both motor and generator operation.The torque is proportional to the current, but the bearing friction is a waste of torque.If you stall the motor, the stall torque is determined by the stall current (no bearing friction or emf at 0 rpm), which is the applied voltage divided by the resistance.For a generator, at a given rpm, the terminal voltage is the emf minus the voltage drop across the copper loss, where the current is determined by the net torque (applied torque less bearing friction).From these basic considerations, you can work out the mechanical and electrical power into and out of the (ideal) device in both operating conditions, so long as you know the proportionality factors (emf)/(rpm) and (torque)/(current), and the loss terms (copper loss resistance and bearing friction torque).The no-load rpm will give you the emf proportionality factor, and either the stall torque or the drop in rpm with load will give you the torque proportionality factor.The torque loss due to bearing friction is not strictly constant, but can be estimated from the torque factor and the no-load current.

author: NeuEnergy
author_rank: Newbie
timestamp: December 05, 2023, 05:34:30 pm
content: Firstly - I want to personally thank each of you who has responded thus far for being kind and respectful.I do believe my first error here was mentioning the motor-generator part of this concept. Let's throw the motor out 100% because that's what I'll be doing on future projects utilizing solar, wind and hydro as forms of power. For now let's focus on hydro and wind which both turn a generator to create power.So we have our 120W generator spinning at 2500rpm, producing 12v and it's being spun by wind or water. I connect a 10W load to our generator and we see voltage drop and loss of power. What's the magic bullet I'm missing here? That whatever force is supplying the spin to our generator needs to have a specific amount of torque to produce X load? I do believe this is what you're saying TimFox, I simply want to confirm.For example (theoretically) it takes only 0.2lbs of torque to spin the magnets in our 120W generator, whereas it takes 2lbs of torque to sustain 10W of load from those magnets due to back EMF, 20lbs of torque to sustain 100W of load, etc...? And this is because applying a load creates current, which creates back EMF, which only increases the more current we have, which then cancels out a percentage of voltage we have (a form of magnetic resistance if you will) requiring more torque (push) to overcome it? Am I at least somewhere in the ballpark?

author: IanB
author_rank: Super Contributor
timestamp: December 05, 2023, 05:45:19 pm
content: Back EMF is confusing you. What TimFox wrote is very advanced theory and is far more than you need to consider for this problem.The situation is very basic. If you connect a 10 W load to your generator, it means you want 10 W of power to come out of the generator. The only way that can work is if you put more than 10 W of power into the generator. This could be 10 W of wind power, or 10 W of hydro power.This (conservation of energy, or power balances) is a fundamental consideration that always applies.After that, you have more detailed considerations, such as matching speed and torque using gearboxes or similar.For an analogy, consider a car with a manual transmission. If you try to go at 10 mph in 4th gear, the engine will bog down and stall. The engine has enough power, but the engine speed is not properly matched to the wheel speed. However, in first gear everything is fine.In a similar way, your wind turbine or hydro turbine speed has to be properly matched to the generator speed. If the match is not right, you will not get the expected power and the generator will bog down and lose output.

author: ejeffrey
author_rank: Super Contributor
timestamp: December 05, 2023, 05:45:48 pm
content: Quote from: NeuEnergy on December 05, 2023, 05:34:30 pmIt's not a loss of power. When the motor is spinning at 2500 RPM with no load attached, it's not supplying any current or power. The torque needed to spin it at that speed is just enough to overcome the internal losses. I'd ignore the wind or water, and just pretend you were turning the motor with a hand crank: you would find it very easy to do. While the generator might be rated at 120 W, it's currently producing zero.Now, connect the "10 W" load, which I will assume is designed to draw 10 W when supplied with 12V, i.e., it's like a ~14 ohm resistor. Now, the power from the generator increases from zero to ~10 watts. Your crank just got harder to turn, as the generator requires more mechanical input power to keep spinning. If you are strong enough, you can keep turning it at 2500 RPM, but if you aren't maybe the speed drops to 2000 RPM and the voltage drops. You are no longer supplying the full 10 watts electrical you expected, but still more than the zero watts when unloaded.

author: mikerj
author_rank: Super Contributor
timestamp: December 05, 2023, 05:52:12 pm
content: Quote from: NeuEnergy on December 05, 2023, 05:34:30 pmYes, whatever energy source is turning your generator has to be able to supply at least the amount of power your load is drawing, plus any conversion losses. Your brushed DC motor operating as a generator likely won't be very efficient at all, 50% is probably not an unreasonable assumption to start with. If you have a load that draws e.g. 100W that would mean you need a power source of at least 200W to drive the motor.Mechanical power is a product of torque and RPM, to a first order the voltage output from your generator will be proportional to it's RPM and the torque required to turn the generator will be proportional to the current it is supplying into the load. If your power source (wind or hydro turbine etc) cannot supply sufficient torque, then it's RPM will drop when connected to a load.

author: NeuEnergy
author_rank: Newbie
timestamp: December 05, 2023, 05:58:51 pm
content: Honestly you guys have been awesome thus far, thank you again. I do feel my confusion has been cleared up to a great extent. So the concept I was missing was torque. Thank you for not being condescending or treating me like an idiot.

author: TimFox
author_rank: Super Contributor
timestamp: December 05, 2023, 05:59:51 pm
content: Quote from: NeuEnergy on December 05, 2023, 05:34:30 pmWith no electrical load on your DC generator (open circuit), there is no current through the windings. Therefore, the required torque (in lb-ft or N-m) goes to overcome the bearing friction (at that RPM) and the terminal voltage is its maximum (emf constant times RPM).When you draw current (better variable than power for this calculation) from the terminals, while maintaining the same RPM, the emf is the same, but the terminal voltage falls due to the current through the motor and the copper loss (resistance). The torque required from the mechanical input increases due to the torque constant (N-m/A) times that current.All of this assumes the magnetic field is constant (as with a permanent-magnet motor).If you calculate the emf constant, torque constant, and resistance you can graph the terminal voltage and input torque as a function of load current (at a given RPM).From the load current and terminal voltage, you can calculate the output electrical power (at constant RPM).From the input torque and RPM, you can calculate the input mechanical power.It gets more interesting if the mechanical RPM from your source falls with increasing torque.Note that the (mechanical) input power is the productTx (2 pi) xf, whereTis the shaft torque andfis the rotation speed (in revolutions per second).Of course, the (electrical) output power is the productVterminalxI, which must be less than the input power.The difference is heat generated by bearing friction and copper-loss resistance at the operating RPM and current.

author: radiolistener
author_rank: Super Contributor
timestamp: December 05, 2023, 06:09:27 pm
content: Quote from: NeuEnergy on December 05, 2023, 05:34:30 pmThere is no magic bullet. Energy is not voltage but a power.The power is P = U * I. The same power can be carried with different voltage and current.For example 100 W can be transfered with 10 V x 10 A or with 100 V x 1 A or with 1 V x 100 Amps, etc.As you can see, if current (I) is low, there needs to be high voltage to keep the power value.When your generator output is open it has high impedance load and since I = U/R it has very low current.But once you connect some low impedance load, the current will rise up. And since there is too small power applied to generator (due to small and weak motor), the voltage needs to drop down to keep the same power.This is why voltage drops down when you connect low impedance load.In order to get higher voltage on the same load you're needs to increase rotation power applied to generator.Since it depends on the motor you're needs to increase power applied to the motor.You can do it by increasing voltage and current on the motor input.

author: NeuEnergy
author_rank: Newbie
timestamp: December 05, 2023, 06:51:41 pm
content: Quote from: TimFox on December 05, 2023, 05:59:51 pmDude you are a rockstar!This last part you mentioned above, "falls with increasing torque", do you mean like water over a waterfall where gravity increases the speed to add more energy to the weight of the falling water than said water would have if it were simply in a reservoir and the water pressure were pushing it through a hydroelectric generator?

author: IanB
author_rank: Super Contributor
timestamp: December 05, 2023, 07:02:33 pm
content: "falls with increasing torque" simply means "slows down if you put some resistance on it".For instance, if you have an electric screwdriver and you grip it with your hand, you might slow it down and stop it turning. In this case, gripping it with your hand applies torque to the shaft, and this results in a decrease in rpm.

author: radiolistener
author_rank: Super Contributor
timestamp: December 05, 2023, 07:12:39 pm
content: Quote from: NeuEnergy on December 05, 2023, 06:51:41 pmVery strange and confusing analogy.Things are pretty easy. Power = Voltage * Current. Power doesn't appears from nothing, so you can get less power from generator than is applied to it's rotor. The output power less than input because some amount of power will be lost due to friction and wire heating.In short when input power is constant the output power will be constant and voltage will depends on the load resistance.

author: TimFox
author_rank: Super Contributor
timestamp: December 05, 2023, 08:22:15 pm
content: Note that powerP=VxI=Fxv(mechanically), but "power" is the time rate of "energy", not the energy itself.

author: NeuEnergy
author_rank: Newbie
timestamp: December 05, 2023, 09:41:15 pm
content: Thank you everyone! My confusion has been resolved per rotations and current. When I have my next question I know where to post it. I learned a lot in a short period of time and it was much simpler than watching a dozen videos (which I tried) or google searches. Thank you for your time and willingness to share.

author: TimFox
author_rank: Super Contributor
timestamp: December 05, 2023, 09:46:50 pm
content: Continuing the discussion of motors and generators:In my replies above, I required that the magnetic field in the motor be constant.If we vary that field value, we find some interesting results.For simplicity, assume that there are no losses in an ideal motor: the winding resistance and bearing friction are both zero.Of course, without loss, the mechanical and electrical power in operation will be equal (100% efficiency): each is the product of eitherVxIorSxT(speed x torque).However, consider the ratios of speed/torque (mechanical) and voltage/current (electrical) at the two ends of the machine.We have two important constants when we factor out the magnetic fieldB:The voltage (emf) is proportional to the product of field and speed:V=kVxSxB, whereSis the rotational speed (rpm or Hz or rad/sec, per choice).Similarly, the torque is proportional to the product of field and current:T=kIxIxB.Both voltage and torque are proportional to the field.The two "k"s depend on the geometry of the magnets and windings, but are independent of field.Therefore, the ratio(electrical) (V/I) =kIkVB2x (S/T) (mechanical)which can be varied by changing the magnetic field (for example, if it is from another winding).A colleague of mine needed to make a non-ferromagnetic electric motor to adjust a capacitor inside an MRI machine, where he could make use of the large magnetic field from the electromagnet, but not distort the field when the motor was turned off.He tried his contraption (the rotor coil was wound on a non-magnetic core) in two machines: one at 5,000 Gauss, and the other at 15,000 Gauss.Consistent with my simplified discussion above, he found that the speed of the motor was slower at the higher field, but the torque available was higher at the higher field.

author: IanB
author_rank: Super Contributor
timestamp: December 05, 2023, 10:03:03 pm
content: Quote from: TimFox on December 05, 2023, 09:46:50 pmA very practical application of this was in the DC traction motors of old trains on the London Underground, before AC drive electronics took over.The DC trains, when they wanted to reach maximum speed, had a "weak field" switch. Given that the line voltage is constant, weakening the field makes the speed increase, so the trains could go faster. Only longer distance trains that needed to go very fast had this setting. Subway or metro trains that were stopping and starting at every station a mile or so apart had no need of it.

author: NeuEnergy
author_rank: Newbie
timestamp: December 05, 2023, 10:26:12 pm
content: Quote from: TimFox on December 05, 2023, 09:46:50 pmDo you happen to have the equation to calculate the back EMF?I'm gonna have to analyze your last post a few times before it'll sink in. When you have V = kV, both "V"s in this equation are voltage correct? Now how is K calculated?If I remove the kVand kI(as we don't currently have a K variable to work with) it makes what you're saying easy to understand. Increase speed and your field is smaller. Increase the field and speed is less. Similarly torque will increase with a larger field despite having less current. Smaller field, higher current.Quote from: IanB on December 05, 2023, 10:03:03 pmFor curiosity's sake, how do you weaken the field?

author: CaptDon
author_rank: Super Contributor
timestamp: December 05, 2023, 10:27:00 pm
content: As stated previously, you put 120 watts in to the 'motor' and probably would get only 30 to 50 watts of energy at the shaft. Your 120 watts 'in', was that with no load on the motor? Then you had 120 watts in and nothing out? Now as for your motor being a generator, they suck unless specifically designed for the task. As a motor the brushes 'lead' the pole pieces in order to overcome current lagging behind voltage in the inductive windings of your motor. Now as a generator the brushes lag the pole pieces and efficiency is shot! Kind of like taking a race engine with 20 degrees of total timing advance and retarding the timing to 20 degrees retard!! That would make a powerful V8 run like a lawnmower engine if it even started at all! Super well built motor-generator sets probably don't even reach 40% efficiency. There is a more efficient device called a dynamotor where the field poles are shared by the motor and generator and the motor windings and generator windings share space on the same armature generally with a commutator on each end of the rotating assembly with brush lead/lag optomized at the design speed and current. So bottom line a D.C. brush motor is a poor generator and likewise, back in the day when they used a Delco Remy 12 volt generator (exactly like the ones on cars) as a motor to start 8 to 12 horsepower garden tractor engines they sucked as motors, mostly due to being optomized as a generator and not as a motor. Cheers mate!!

author: TimFox
author_rank: Super Contributor
timestamp: December 05, 2023, 10:57:37 pm
content: Varying the field:Some DC generators use an electromagnet to make the magnetic field, rather than a permanent magnet, so changing the current in the "field" coil changes the field.The power dissipated in this field coil should be quite a bit less than the output current of the generator.(A permanent magnet participates in a PM motor without losing energy to the output.)Another application is the "Ward Leonard" control system, used to regulate the DC output of a motor-generator.https://en.wikipedia.org/wiki/Ward_Leonard_controlThat article includes more complicated equations including inertial effects in the mechanics.The calculation of the constantskVandkIrequires engineering calculations of the coil geometry.The basis is that the back emf is the time derivative of the flux linked by the rotor windings as it rotates through the field, and that the force on the rotor windings (that gives you the torque) comes from the product of the current through the coil wires and the field at the rotor coil.For details, one should consult textbooks on electric machinery.In general, you have two electrical variables (VandI), two mechanical variables (TandS), and the magnetic fieldB.The two constantskVandkIare determined by the construction geometry and dimensions.You can't control all five variables at once: however, you can specify, e.g.,V(applied voltage) andT(load torque), for a givenB, and see what speedSand currentIvalues result.

author: IanB
author_rank: Super Contributor
timestamp: December 05, 2023, 11:01:03 pm
content: Quote from: NeuEnergy on December 05, 2023, 10:26:12 pmThe motor has two sets of coils, field coils and armature coils. These coils and their associated motors can be connected in various arrangements of series and parallel according to speed and load, and rheostats might also be used to reduce the current. When the motor has reached maximum speed and needs to go faster, the field current can be reduced by switching in a bypass shunt resistor to divert some of the current around the winding and weaken the magnetic field. In some cases this will be done manually, and in others it is done automatically.DC traction motors on trains often feel like they have "gears" due to the jolt that may be felt as each switch in the motor circuits happens.

author: TimFox
author_rank: Super Contributor
timestamp: December 05, 2023, 11:12:03 pm
content: The original DC traction motors on the CTA 'L' system had discrete control settings for the motor coils:Parallel (fastest),Series (slower), andResistor (slowest, with series resistance added)and the speed control was not very smooth.(The applied voltage is the (constant) third-rail 600 V DC.)The newest rolling stock has gone to VFD-controlled induction motors from the third-rail.

author: CaptDon
author_rank: Super Contributor
timestamp: December 06, 2023, 12:12:51 am
content: The very early E.M.D. 'F7' locomotives came with a manual 'transition' control lever which would do the different combinations of series / parallel connections of the traction motors by way of huge relays. Later editions of the F7 and nearly everything after it had automatic transition control. Those early locomotives had a maximum 600 V.D.C. generator output. The last locomotives I worked on had a D.C. link voltage up to 1400 and had 12 huge capacitors (1700uf @ 2000VDC) 2 per inverter with one inverter per traction motor. In dynamic braking the motors could pump the link voltage up to 1200vdc into a grid resistance of about .5 ohms for 1200 volts X 2400 amps of braking effort. B.T.W. all 12 of those capacitors were essentially in parallel and take the stored energy at 1200vdc plus the electrical output of over 4000 H.P. from the alternator and 3 phase rectifier and think about the consequences when one of the capacitors dead shorts internally when overheated due to high E.S.R., yes, they are oil filled!!!! I have seen 'Auxilliary Cab' doors blown clean off the locomotive along with the bonus 'oil fire' afterwards. Ever think about the consequence when over 40,000 pounds of rotating mass (engine + alternator 900RPM) suddenly sees a dead shorted 3 phase rectifier blowout!!!! Prior to the safety 'demagnetization' circuit to kill the alternator field locomotives were virtually 'flipped' on their side causing derailment!

author: IanB
author_rank: Super Contributor
timestamp: December 06, 2023, 12:53:16 am
content: Quote from: CaptDon on December 05, 2023, 10:27:00 pmI usually think of a motor-generator set as being a motor coupled to a generator on the same shaft. One example I have seen was to convert 240 V, 60 Hz to 240 V, 50 Hz. This was installed in the USA to provide power to equipment ultimately destined for a 50 Hz destination. I'm not sure how that was done, exactly, but there might have been a gearbox in between motor and generator to convert between 3600 rpm and 3000 rpm. It was certainly noisy when installed on a shop floor.Another example was on some of the old London Underground trains, such as the 1938 stock. These had a motor-generator set with a 630 V DC motor on the traction current, driving a 120 V AC generator to provide train lighting and other services. It made a nice interesting hum under the floor of the train.

author: eKretz
author_rank: Frequent Contributor
timestamp: December 06, 2023, 04:15:42 am
content: We had a 100 HP DC motor-generator set on a 72" swing lathe where I served my machinist apprenticeship. While starting, it sounded about like a turbine engine firing up. I remember there was a two way knife switch for starting it. One way for starting, then flip to the other side once it had accelerated up to speed. I don't remember exactly how it was described to me then, but looking back, I think it may have been a manual switch to do with pole switching.

author: NeuEnergy
author_rank: Newbie
timestamp: December 06, 2023, 02:38:29 pm
content: Quote from: CaptDon on December 05, 2023, 10:27:00 pmThank you for the comparison to vehicle timing, as I've built and worked on race vehicles, rebuilt their motors, did their timing, advanced timing, etc... so it definitely clicked for me. The 120W I was talking about was more or less theoretical (max watts of motor). I do know these motors I'm using are not very efficient, perhaps in the 55%-65% range. I'm using these motors because 1) they're cheap, 2) all the kids with their "PE" videos online use the crappiest motors they can find from old electronic toys with horrible efficiency. 3) It's not easy finding small generators in the 20-100W range. I'm "recreating" if you will, the "working models" from PE youtube videos to show all the reasons why their working models don't actually work, and you guys are helping me learn part of the why that has eluded me (aka mechanical torque if we leave efficiency aside). I am familiar with dynamotors, but as you said they're more or less a motor-gen in one and the "experiment" requires them to be separate. Do you happen to know where I can find a decent 50W-100W brushless generator? If not I may just make one. I think I still have a decent amount of neodymium magnets lying around somehwere.

author: Kleinstein
author_rank: Super Contributor
timestamp: December 06, 2023, 04:39:57 pm
content: Brushless motors and generators are not that much different. For the power rating the limiting factor is often the current that the winding can withstand and thus the torque it can produce. Quite often one can vary the speed over quite some range (e.g. limited by the bearings) and this way also the power rating will change. So the same more may be 100 W at 2500 RPM or 400 W at 10000 RPM.The generators for motorcycles could be an option. They may be in the right power range, though I don't know if there are brushless ones - some I are knows to use bushes.For the efficiency this tends to get better the larger the motor / generator. This is kind of a general trend with electromagnets, transformers and the like. This is a reason why small DC motors usually use permananet magnets, while really large one tend to use electromagnets, as the permanent magnets scale well also down to small size and the larger the more attractive an electromagnet becomes (this was especiall an issue in the old days before Nd magents). With small ones a higher frequency / RPM can help a little - so small ones tend to run faster.