Re: Understanding PWM of Motor (current problem)


<makolber@xxxxxxxxx> wrote in message
news:2c9fed62-ee2e-4a36-b68a-30f50f16c85b@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Dec 16, 7:40 pm, "amdx" <a...@xxxxxxxxxxx> wrote:
Hi All,
I've been having a discusion about PWM of DC motors on an electric vehicle
forum.
I keep seeing the idea that battery current is different than motor
current
when using a
PWM controller. As quoted below.

http://www.4qdtec.com/pwm-01.html This site says this;
"You should see from the above that, if the drive MOSFET is on for a 50%
duty cycle, motor voltage is 50% of battery voltage and, because battery
current only flows when the MOSFET is on, battery current is only flowing
for 50% of the time so the average battery current is only 50% of the
motor
current! "

I understand that 50% duty cycle would cause a motor voltage equivalent to
50% of the B+.
And full current would flow for 50% of the time (ignoring inductance
caused
rise and fall time).

Quote;
"so the average battery current is only 50% of the motor current!"

But where are they getting the other 50% of the current?
If you say it is from the collapsing magnetic field of the field and
armature, then
I would say, but the battery had to build that field to begin with so
battery current during the
50% on time was had to be higher than the full cycle motor current.
Ok, I'm going to stop now, because I'm not sure I communicating!

This site has waveforms in Section #12 although my thought is he has the
500hz
and 20khz labels
reversed.http://www.picotech.com/applications/pwm_drivers/#chap12

Your help appreciated, Mike


i think that part you are missing is the CATCH DIODE.

No, I realize it's there and is passing the current caused by the
collapsing
magnetic field in the motor.

When the switch switches off, the current continues to flow through the
diode.

When you say "the current" I'm assuming you mean the current from the
collapsing field?

So with the right inducantce, and a 50% DF, the current in the load flows
all the time, the current through the battery flows 1/2 time so the
average output current is 2x the input current and the output voltage
is 1/2 the input voltage. Input power = output power. This is
simplifed and ignores losses of course.
Mark

What is DF?
Your saying the same thing as the EV group. Now I need to understand
where the (non battery supplied) current is coming from.
Mike


.

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