Back emf of motor - Power Electronics

[elexhobby]

Hello friends,

I had a doubt while studying inverters & converters in Power Electronics. In most of the circuits, (semiconverter, full-wave bridge converter, midpoint converter) the load is assumed to be a RLE load (i.e resistance R, inductance L, & a battery E). Now, it is said that the battery E may be the generated counter EMF of a dc motor. This is what I don’t understand. Why do they need to show a separate battery when an inductor is already shown? The back emf occurs due to the inductance of the motor & hence is accounted for when an inductance is shown in the load circuit.

Kindly help me & let me know the flaws in my reasoning.
Thanks in advance.

Regards.

 

[Oznog]

Say we have a 3v DC battery. You hook an ohmmeter up to the motor and it reads 3 ohm. You hold the rotor so that it can't turn. If you hook the battery up to it it will pull 3 amps.

Take off the piece keeping the rotor from turning. Now it spins at 3000 rpm and only draws 50mA. How can we draw this and still meet the I=V/R? Inductance is not it, an inductor acts as nothing except its DC resistance when hooked up to DC.

Well, you'll find if you spin the motor at 3000 rpm with another motor driving the shaft, it will generate a couple of volts. This is a slightly different situation, but you get the idea. The current makes the motor spin, but as the speed picks up, it generates its own voltage which counters the battery voltage. Load the motor, make it slow down, and the current goes up. Makes sense.

 

[Roff]

EDIT: Never mind - I should have known to wait for Oznog.

 

[audioguru]

I should have known to wait for Oznog.

Yeah, he knows that 3V/3 ohms= 1A and that the power is 3W. He can correct his post himself.

 

[akg]

Say we have a 3v DC battery. You hook an ohmmeter up to the motor and it reads 3 ohm. You hold the rotor so that it can't turn. If you hook the battery up to it it will pull 3 amps.

whooaa.. cool.. :lol:

 

[elexhobby]

Hey Oznog,

Thanks a lot... Ur explanation makes things pretty clear...

 

[Styx]

Just to add some more bits of info for this

Yes the equiv cct for a brushed-DC (as well as a brushless) is a R-L-E load

The R is effectively static (although temp will play a part in it a bit).
The L isn't really it will have some variation across commutation (can be viewed as none-changing for Brushed-DC, for brushless this change can be for a well designed machine 20% change across commutation zones)
The E however, is a speed dependent potential

There are two constants with machines Kt and Ke (for an ideal machine they are equal, in practice not far off)
Kt is the torque constant T = Kt*i
Ke is the backEMF constant, E = Ke*w (w = speed of rotor in rad/s)

at zero-speed no potential is developed, with increase speed the machine acts like a generator and a potential is developed.

say you have this 3V battery for a stalled rotor and the terminal resistance is 3Ohms, then 1Amp will flow (not 3Amps :wink since it is DC the inductor will eventially not play a part.

Say we start spinning AND the rotor is loaded, if we spin at say 3000rpm with the same 3V battery and say the backEMF at thios speed is 1V, it means we only have 2V available to force current in (ie a maximum of 0.7Amps) THUS if we want to spin at 3000RPM we can ONLY do so IF the load is such that it doesn't require 0.7A


We can spin at 3000rpm un-loaded and say only 50mA is being drawn. Why so low? because of the inductance and commutation. A machine will only draw as much current as it need to meet the Torque-speed & power-speed of that operating point.

The inductance will limit the di/dt BUT more current will cause the rotor to spin faster, a faster spinning rotor will cross a commutation point sooner THUS the level that the current reaches is dependent on this


Start to load the machine and the rotor will slow down, this means current has a chance to flow (in that commutation zone) for longer giving it time to build to a higher level (to produce more torque) result is the speed catchs up to the 3000rpm BUT more current is being drawn

IF the load is too great for what the machine can deliver at that speed (part of the Torque-speed curve) the speed will collapse to meet that torque