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Confused with back emf,by DashIQ
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#1
Mar114, 01:37 AM

P: 98

I need help in understanding back emf, I'm a bit confused here.
Example(for understanding): Let's assume we have a 12 V DC motor, that draws 20Amps when it starts, as it reaches maximum speeds, current drops due to back emf, lets assume the resistance is 0.6 ohms, and b.emf = 10V. V_{tot} = 2V ,from ohms law I can find the current at maximum speed, it would now be I = V/R = 3.3Amps Now what about power? Initial power = 240W Power at max rpm = 6.6W? Will the motor be stable at 2V at maximum speeds? Or will it draw more voltage because of the b.emf and stay the same rate of power @ 240W with lower current? (here is where I'm struggling). If everything above correct? 


#2
Mar114, 02:50 AM

P: 1,072

A motor does not "draw" voltage. There is 12V across the motor at all times.
The 0.6 ohms represents 6.6W lost as heat in the motor windings. The motor is drawing 3.3A @ 12V which is 39.6 watts. The difference 39.66.6 = 33Watts of power delivered to the load (ignoring other motor losses). When it starts @ 20 amps it is producing high torque and delivering a lot of power to accelerate the load. You can do the numbers. 


#3
Mar114, 05:39 AM

P: 98

Please explain. 


#4
Mar114, 06:22 AM

HW Helper
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P: 5,129

Confused with back emf,



#5
Mar114, 06:59 AM

Engineering
Sci Advisor
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P: 6,921

At zero RPM, all the electrical power (I^{2}R) is converted into heat in the windings. 


#6
Mar114, 04:17 PM

P: 1,072

Sorry about that  at zero speed there is no power output. Just torque.
You can think of a dc motor as having a generator inside. When the motor is stalled the generator creates no voltage so the motor draws the maximum current as limited by its internal resistance. When the motor is running UNLOADED at maximum speed the internal generator creates maximum voltage which reduces the current the motor draws to the minimum. There is always 12V across the motor. Say you had a perfect motor with no load. It would draw an infintesimal current. But, how can you limit the current draw from 12V to be an infintesimal current? The internal generator creates 12V, so no current is draw from the supply. If the speed goes down, the generator voltage reduces, so more current is drawn. In your case the motor is drawing 12V x 3.3A = 39.6 Watts. 33W of that is doing work, the rest is dissipated in the internal resistance. If the motor is loaded down, it slows down a bit which causes the internal generator to produce less voltage which causes the current to increase to drive the load. It cannot get back to the same speed it was before the load was increased. Don't get hung up on thinking about the 2 volts. Think rather about an internal generator based on speed (a speed dependent voltage source) that controls the current drawn from the supply. More current means more power delivered to the load, and also less speed (because the motor is loaded down) The system will always settle at the point where the speed and load are balanced. 


#7
Mar114, 10:20 PM

P: 98

The example I used is random from my head, I have no specific model to study in detail.
I just wanted to understand back emf more. 


#8
Mar214, 12:37 AM

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P: 3,498

You might study this thread. Another PF'er was similarly confused, but he 'got it' by end of second page.
See if it helps. http://www.physicsforums.com/showthread.php?t=624785 


#9
Mar214, 12:59 AM

PF Gold
P: 1,420

In a simplified dc motor: and given your initial parameters: 12 vdcWe can see that we can't really solve the equation: F=ILB F being the force on the wire in newtonsBut one thing we can tell is that the motor is going to start rotating. There is another equation: emf = vBL sin θ (ref) which describes what happens when a conductor is pushed through a magnetic field. emf being the voltage generatedNow in the illustration that Hyperphysics has graciously provided, the angle θ changes constantly. But in real world dc motors, there are multiple loops of wire in the rotor, keeping the angle at relatively close to 90°. So the equation reduces to: emf = vBL If you were to make up some numbers to to fill in the missing variables, you can actually calculate how fast your motor will run under no load, ideal textbook conditions, which would be when your emf generated, equaled the voltage supplied. If cemf didn't exist, the implications of F=ILB, would be disastrous, for an unloaded motor. For F=ma, and if the current didn't reduce because of the cemf, the motor would accelerate, until it disintegrated. 


#10
Mar214, 10:52 AM

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P: 3,498

Nice Job OM !
One can empirically arrive at these two formulas for a DC machine, it's just lumping the constants together: Counter EMF = K X (flux) X RPM, where [K X (flux)] is usually written K[itex]\Phi[/itex], giving Counter EMF = K[itex]\Phi[/itex]RPM , [itex]\Phi[/itex] of course is flux, K encompasses pi and radius to get from RPM to velocity, and length L too. Drive the motor at some speed and measure the open circuit voltage it makes, you have K[itex]\Phi[/itex] Torque = same K[itex]\Phi[/itex] X Armature Current X 7.04 , gives footpounds As OM said  with no counter EMF , ie motor stalled, armature current goes sky high and torque is dramatic. Your automobile starter is a good example . It's a beautifully self regulating system. As RPM comes up so does counteremf, reducing current. Continuing to raise RPM (by driving the motor with something), counteremf overwhelms applied voltage so current reverses and you have a generator. 


#11
Mar214, 01:20 PM

PF Gold
P: 1,420

"Stalled", in this context, means the motor is not allowed to turn. We referred to this as "locked rotor", back in my day. On the submarine I sailed around on, they had what was called a "motorgenerator". 


#12
Mar214, 02:28 PM

P: 98




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