# effect of BEMF ?



## gregc (Apr 25, 2015)

like a generator, a motor generates a EMF when turning: BEMF. 

When a voltage is applied to a non-spinning motor where bemf = 0, the current thru the motor is limited by the resistance of its windings. But as soon as the motor starts turning, bemf increases, opposing the voltage applied to the motor and the current is reduced -- (volt - bemf) / R.

let's define that a positive voltage causes the motor to turn CW and when the motor turns CW, it generates a negative EMF.

i believe when the voltage is reduced to zero, the motor continues to spin and generate bemf, but because there is not path through the driver circuit (or a very high resistance path), the bemf has little effect on the motor. In other words, it doesn't generate a current large enough to oppose the spinning motor.

however, if the driver circuit applies a negative voltage to slow down a motor turning CW and generating a negative EMF, is the effective voltage the sum of the negative drive voltage and negative EMF?

in other words, with the drive circuit generating a voltage not in opposition to the BEMF is their sum generating a combined current opposing the motion of the motor?


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## wvgca (Jan 21, 2013)

an electric motor has the current turned off for the period of time that it reads the emf .. the two are not active at the same time ..
at least this was the way i did it in my homemade decoders with MERG firmware


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## gunrunnerjohn (Nov 10, 2010)

The power has to be off to sample the back-EMF, think about it.  Typically, with a PWM drive, the back-EMF is sampled when the drive is off between drive pulses.


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## gregc (Apr 25, 2015)

yes. you can't measure the bemf if a voltage is applied across the motor terminals.

without a path for current to flow, it has no effect. It's like turning a generator w/o a load.

my question is does it have an effect when a voltage is applied of the same polarity to brake the motor.

with little or no friction or load, the bemf is approximately the same as the applied voltage. If the polarity the applied voltage is suddenly reversed, I believe the effective voltage across the armature almost doubles (opposing the motion).


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## wvgca (Jan 21, 2013)

there are , in most cases, only two variables for the emf [while on] ....
the first is the width of the null region , where it neither speeds up or down ..
and the second is the slope of the angle of aggressiveness, whether speed up or down


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## gunrunnerjohn (Nov 10, 2010)

You can certainly brake the motor by applying a reverse voltage! Even shorting the motor armature as I do in the Super-Chuffer brings it to a sudden halt. Without shorting the armature, the motor coasted for at least half a second with nothing but the inertia of the smoke impeller on it.


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## gregc (Apr 25, 2015)

so, a short provides a path for the bemf current.

and sounds like a voltage applied to brake the motor adds to the bemf (not opposes it) as well as providing a path for current due to their sum.

i'm not sure what the resistance in across the motor terminals if a drive circuit is generating zero voltage, compared to it generating even a small voltage (e.g. 0.1V) with the same polarity as the bemf. in other words, is 0.1V effectively a short?


WRT stopping at a specific encoder position -- how quickly can the motor be stopped when the target position is reached and there is significant load/friction (gear train, weight on bridge w/ and w/o a loco on it).


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## gunrunnerjohn (Nov 10, 2010)

A DC motor is a generator when it is turning without power. That's what back-EMF is, the power being generated. Since an electrical load increases the resistance of the motor armature to rotation, a direct short stops it pretty quickly. If you apply a reverse polarity, you can stop it even faster, however I get pretty fast stops with just a short. Electric cars use the same idea with dynamic braking charging the batteries when you apply the brakes.


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## wvgca (Jan 21, 2013)

the emf produced by a motor spinning without power applied is very , very small ..
it doesn't really have any effect when there is power applied .
it not even in the calculations for this stuation


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## gregc (Apr 25, 2015)

if bemf has no significant effect, what limits the speed of the motor?

i guess you disagree with the explanation in my original post, the the difference between applied voltage and bemf results in just enough voltage across the windings to apply a force to counter friction/load and maintain speed.


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## gunrunnerjohn (Nov 10, 2010)

The reason that motor current goes down as the motor gets to speed is the back-EMF is balancing out much of the voltage across the motor and thus limiting the current. When a motor is running at speed, there is only s fairly small difference between the applied voltage and the back-EMF voltage. As the load increases, the difference widens, which is why the motor draws more current under load.


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## wvgca (Jan 21, 2013)

gregc said:


> if bemf has no significant effect, what limits the speed of the motor?



while back electro motive force is in itself neglible, the measurement thereof allows the decoder to determine if a higher or lower voltage is required to maintain speed as set in the bemf angle


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## gregc (Apr 25, 2015)

gregc said:


> is 0.1V effectively a short?


? ? ?


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## gregc (Apr 25, 2015)

wvgca said:


> while back electro motive force is in itself neglible, the measurement thereof allows the decoder to determine if a higher or lower voltage is required to maintain speed as set in the bemf angle


i understand that bemf is proportional to speed.

how much voltage have you seen resulting from bemf when measured during the short periods when voltage is not applied to the motor?

it's voltage will be dependent on the impedance of the measurement circuit, Rme, and than of the motor, Rmo. In other words, Rme / (Rme + Rmo).





looks like a short can be created using an H-bridge (top right)


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## wvgca (Jan 21, 2013)

an easier way to explain it is a 220v single phase motor driving a 440v 3 phase motor .. in which case a bemf generated 440v is accomplished ..although at a much lower current ..but not through a VFD or similar, which in some cases can drive a 3phase motor from a single phase supply

which can be roughly calculated if you know the power factors of both the motors ..


a dead short [or close] will slow a DC motor more quickly than no short, but as far as i know turntables in HO made by Atlas do not make use of this, final speed of the reveloution is not going to be affected by motor speed


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## gregc (Apr 25, 2015)

wvgca said:


> an easier way to explain it is a 220v single phase motor driving a 440v 3 phase motor .. in which case a bemf generated 440v is accomplished ..although at a much lower current ..but not through a VFD or similar, which in some cases can drive a 3phase motor from a single phase supply


didn't follow this at first. I think you point is ...

connect the shafts of two identical(!) motors, drive one with a voltage and measure the the voltage being generated by the other. The measured voltage should be closed to the bemf in the motor being driven.


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## wvgca (Jan 21, 2013)

gregc said:


> connect the shafts of two identical(!) motors, drive one with a voltage and measure the the voltage being generated by the other. The measured voltage should be closed to the bemf in the motor being driven.



actually the motors are NOT identical, that's the reason for doing so ..it's commonly used in machine shop equipment to drive 3phase motors with single phase [which we already have] driver motors ..you lose the power factor of both motors in the process, but it's usually cheaper and more available ..
the voltage generated by the driven motor IS the EMF, or bemf as it's sometimes called ..
but generated voltage AND run voltage DO NOT happen at the SAME TIME, in the same motor, in other words, as long as a run voltage is applied, NO bemf is generated, and cannot be measured ... the time spent measuring the bemf is usually pretty neglible compared to the time in the run mode, in model train decoders, otherwise it would affect the power factor too much ..the idea of dynamic brakes in locomotives is based on this principle, they are just very large resistors cooled by some fans


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## gunrunnerjohn (Nov 10, 2010)

The trend nowadays it so use that dynamic braking energy to charge batteries instead of wasting it as heat. )


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## gregc (Apr 25, 2015)

ok, i understand why described a 3-phase generator.

but i disagree that a turning motor is not generating an EMF that opposes the voltage applied to its terminals -- as already described

what limits the speed of a motor that has no mechanical load?



yes, connecting an electrical load to a generator causes a mechanical load on the motor as the armature current creates a magnetic field in opposition to the permanent magnet field


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## wvgca (Jan 21, 2013)

gregc said:


> o
> but i disagree that a turning motor is not generating an EMF that opposes the voltage applied to its terminals -- as already described
> 
> what limits the speed of a motor that has no mechanical load?
> ...


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## gregc (Apr 25, 2015)

the current thru the armature creates a magnetic field that pushes against the permanent magnets to push the armature. The current is simply the voltage across the armature windings divided by the armature resistance -- volt / R 

That current generates a torque that accelerates the motor very quickly from being stopped. Without any reduction in that torque, the acceleration would remain constant

why does the acceleration that is so high when the motor is stopped drop to zero (constant speed) when the same voltage is present when the motor is both stopped and at speed.



as already described, any armature turning in a magnetic field generates a voltage (the mechanical motion generates an electric potential, an emf).

It doesn't matter if the shaft is turned mechanically or magnetically due to an external voltage causing a current to pass thru the armature.

... but, as the armature starts turning, the polarity of the emf caused by that motion opposes the voltage applied at the terminals of the armature. it's a back emf because it opposes the voltage causing the motion.

as the speed increases, the bemf increases and the armature current decreases -- (volt - bemf) / R, acceleration decreases, ...

equilibrium is reached when the residual current results in a force sufficient to just overcome friction. as john said, under higher loads, the motor slows, reducing the bemf resulting in a higher current and a higher force.



there's no way to measure the bemf when applying a voltage to the terminals of the motor. One way, as you described, is to momentarily set the voltage applied to the motor to zero and measure the voltage at the terminals being generated by the motor.


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## ncrc5315 (Jan 25, 2014)

The speed of a AC Synchronous, is a function of the frequency, of the AC sine wave. The formula is: (120*Frequency)/#of Motor Poles. This formula, does not take in to account any slip.


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## gunrunnerjohn (Nov 10, 2010)

gregc said:


> there's no way to measure the bemf when applying a voltage to the terminals of the motor. One way, as you described, is to momentarily set the voltage applied to the motor to zero and measure the voltage at the terminals being generated by the motor.


AFAIK, that's the only way. One important point, it's not "setting the voltage to zero", it's really opening the circuit to have the motor drive in a high impedance state, and then measuring the voltage generated by the spinning motor.


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## wvgca (Jan 21, 2013)

ncrc5315 said:


> The speed of a AC Synchronous, is a function of the frequency, of the AC sine wave. The formula is: (120*Frequency)/#of Motor Poles. This formula, does not take in to account any slip.





thank you!


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## gregc (Apr 25, 2015)

don't typical AC motors use squirrel cage rotors (armatures) that have no external electrical connections?

the AC current creates an alternating field in the fixed winding in the shell of the motor that induces a current in the rotor, which in turn creates a magnetic field and torque causing the rotor to turn.


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## gregc (Apr 25, 2015)

coupled 2 similar motors together (drilled a dowel and press-fit both motor shafts into it)

applied 12V to 1st motor
measured 7.8V generated by 2nd motor (fluke 21 meter)

measured 33 ohm resistance in 1st motor
measured 130 ma in 1st motor when 12V applied

130 ma thru 33 ohm corresponds to 4.3 V,
consistent that if 12V is applied to the motor, there is ~7.7V (bemf) opposing the 12V and why the current is not 363 ma



wvgca said:


> i don't follow ?? how can a motor be driven AND a driver at the SAME time ??


answered ?


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## wvgca (Jan 21, 2013)

i don't know on that, i'm not an electrical engineer ..
what i do know is that my MERG supplied firmware on my homemade decoders shuts off DC power to the motor to check bemf, on THAT situation, and sets it to an open circuit [if it was set to zero volts and NOT open, then it would in effect be a brake]


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## gunrunnerjohn (Nov 10, 2010)

I guess I'm not clear on the intent of this whole thread. Exactly what information are you trying to obtain? What are the real questions?


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## gregc (Apr 25, 2015)

so you're not convinced that there a bemf present when applying a voltage to the motor?

in my experiment with the coupled motors, I would briefly measure the same voltage (bemf) on both motors if i disconnected the 12V from the motor. This is what your MERG decoder does intermittently

it's like connecting a 12V in series with a resistor and 9V battery where the positive of both batteries are connected to either side of the resistor. There's only 3V across the resistor.


can you explain (see?) why the same voltage that accelerates the motor up to speed so quickly when stopped reaches a point where the motor no longer accelerates?


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## gunrunnerjohn (Nov 10, 2010)

gregc said:


> so you're not convinced that there a bemf present when applying a voltage to the motor?


And exactly where did you read that, certainly not anything I said! A voltage is induced into the armature windings anytime they cut through the magnetic field of the stator magnets.

When the back-EMF and friction of the motor rotation balances the energy in, the motor will stop accelerating in speed. As the motor increases in speed, the apparent impedance tends to go up which limits the current. That and the back-EMF determine the maximum speed. When you apply higher voltage, you get more current flow at a given RPM, so the motor speed increases until a new balance is achieved. 

Give it enough voltage and current and the magic smoke escapes and you have a pile of junk.


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## wvgca (Jan 21, 2013)

gregc said:


> so you're not convinced that there a bemf present when applying a voltage to the motor?



actually i don't, but i have nothing to base it on ...
nope, nothing at all ....
i can [usually] make something work, sometimes in an unusual manner, but i'm not real good with numbers on paper, it's more of a hunch or some thing like that ..


if it makes any difference, i'm a 65 year old semi retarded farmer, lol ... usually work on stuff a whole lot bigger, and nowheres near 'leading edge' ...
sorry


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## gregc (Apr 25, 2015)

gunrunnerjohn said:


> The power has to be off to sample the back-EMF, think about it.  Typically, with a PWM drive, the back-EMF is sampled when the drive is off between drive pulses.


this was never a question about how to measure bemf

it became a question of whether it exists when an external voltage is applied to a motor



gregc said:


> in other words, with the drive circuit generating a voltage not in opposition to the BEMF is their sum generating a combined current opposing the motion of the motor?


i believe answer to my question is of course, yes

if a positive external voltage causes a motor to turn CW and generates a negative bemf (opposing it) a
negative external voltage applied to a motor turning CW will combine with the same polarity to oppose the motion of the motor.

this is true if you believe a non-zero bemf is always present when a motor is turning regardless of whether an external voltage is applied.


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## gregc (Apr 25, 2015)

gunrunnerjohn said:


> And exactly where did you read that, certainly not anything I said!


i was responding to Warren's post preceding yours.


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## gunrunnerjohn (Nov 10, 2010)

The generated back-EMF is the primary limiting factor in the free-running speed. That and friction is what determines the terminal speed at a given voltage.


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## gregc (Apr 25, 2015)

wvgca said:


> gregc said:
> 
> 
> > so you're not convinced that there a bemf present when applying a voltage to the motor?
> ...


i don't think you're alone.

i've read explanations that describe the build up and collapse of magnetic fields like it's a supernatural force, ignoring the ohms law aspect.

(how can there only be 130 ma when 12V is applied across 33 ohms)?


these types of discussion help me better understand things ... clearing the fog of understanding.


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## gunrunnerjohn (Nov 10, 2010)

gregc said:


> (how can there only be 130 ma when 12V is applied across 33 ohms)?


Well, that's really simple. If the back-EMF is 8 volts, you really have 10 volts bucking 12 volts and you have 4 volts across the 33 ohms, roughly 130 ma.


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## gregc (Apr 25, 2015)

john
i realize you understand this and appreciate you confirmation. i've been trying to help Warren understand using Socratic method.


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