From: Doug McKean ([email protected])
Date: Thu Nov 09 2000 - 17:00:08 PST
"SEOW,ERWIN-SP (HP-Singapore,ex1)" wrote:
>
> Hi all,
>
> I have a question about DC motors.
>
> Lets say I have a DC souce connected to a DC motor. The motor is running for
> a while, what happens when I disconnect the switch at time t=0? Please see
> attached figure.
>
> Which is the dominating factor of current flow? The armature inductance or
> the back EMF?? If I neglect back EMF, the circuit will be easily explained.
> Do back EMF come into this picture?
>
> Any comments??
The diode will take the brunt of the current at t0.
Basically, define what the voltage will be across
the motor plus define what the DC winding resistance
will be. Transform the motor into a voltage source
with an internal resistance equal to the winding
resistance. Now, put the diode across the terminals
forward biased. That's what the diode will have to
endure when you open that switch.
Back emf is the opposable field in the armature due
to applied current. It's proportional to the speed
of the motor. So, current is a max at start up and
decreases when the motor is at speed. It's why car
batteries are so big. The batteries initially "see"
only the starter winding resistance which is extremely
small. Thus, the initial current through your battery
when you start your car is in the several hundred
ampere range.
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