Re: What passes as Pulse Width Modulation in DC Motor Control?

amdx wrote:
"Dan Coby" <adcoby@xxxxxxxxxxxxx> wrote in message news:N5adnf0AQc5yiB3UnZ2dnUVZ_hadnZ2d@xxxxxxxxxxxxxxxx
amdx wrote:
"Jamie" <jamie_ka1lpa_not_valid_after_ka1lpa_@xxxxxxxxxxx> wrote in message news:heNfl.27335$2o3.14283@xxxxxxxxxxxxxxx
Rich wrote:

Can anyone please tell me what passes as PWM in Motor Control?

Now that it's all sorted out.
Assuming a PWM 50% duty cycle,
If I put an ampmeter in the positive lead from the battery
and a second ampmeter in series with the motor,
Will the two ampmeters read the same?
No. That may seem to be a little strange but it is true.

Lets take an example. We have an ideal motor (i.e. no resistance losses,
etc.) which is being driven by an ideal PWM (i.e. a PWM with no losses).
Also let us assume that the PWM is being powered by 20 volt DC supply and
the PWM is switching the motor drive between 20 volts and 0 volts. And
we are driving the motor with a 10% duty cycle and that the average motor
current is 3 amps. We will also assume that the PWM frequency is high
enough that there no significant changes in the motor speed and current
during a cycle. (This is usually a reasonable assumption. PWMs are
usually run at a high enough frequency to make this true. See below.)

The average voltage at the motor is 2 volts (20v times 10%). The power
in the motor is 3 amps times 2 volts = 6 watts. Since there are no losses
with our ideal motor and ideal PWM, then the power from the 20 volts power
supply must also be 6 watts. The power from the supply is only provided
while the PWM is turned on (i.e. during the 10% on part of the cycle).
During the on part of the cycle we have to be providing 60 watts of power
(6 watts / 10% = 60 watts). This means that we are getting 3 amps (60 watts
divided by 20 volts during the 'on' cycle. (So in this sense the
currents are equal since we are putting 3 amps into the motor and we are
drawing 3 amps from the power supply during the 'on' cycle.) HOWEVER the
ammeter connected to the power supply will indicate the average current
from the supply. Since we are supply 3 amps for 10% of the cycle and
0 amps for 90% of the cycle, we will an average current of 0.3 amps.

So the average motor current will be 3 amps and the average power supply
current will be 0.3 amps.

This leaves a few questions:

1) What is supplying the 3 amps of motor current during the 90% part of
the duty cycle when the PWM circuit is 'off'?

The PWM is switching the drive between 20 volts and 0 volts. During the
'off' part of the cycle, the 3 amps is being pulled from the 0 volt side
of the PWM (i.e. the motor is pulling current from the circuit ground).


2) Is the motor's current and speed really constant the entire PWM cycle?

In theory, no. There is always a small amount of speed and current ripple
but the amount of ripple can be made very small by increasing the PWM
frequency. Since we are assuming an ideal motor (i.e. no resistance losses)
then our motor looks like it has a back EMF and a series inductance.
the back EMF of the motor will be equal to the 2 volt average voltage at
the motor. (A real motor with a non-zero winding resistance would have
a lower back EMF since there would be some voltage drop across the winding
resistance.)

During the 'on' part of the PWM cycle, there is an 18 volt difference (20 volt
power supply versus a 2 volt back EMF) between the motor's back EMF and the
power supply. This voltage difference will try to increase the current flowing
into the motor. The inductance of the motor will limit the increase in current.
But there is a small (usually very small) increase in current during the 'on'
part of the cycle. This increased current will try to accelerate the motor.
However the inertia of the motor and its connected load will oppose the
acceleration but there will be a small change in speed.

During the 'off' part of the cycle, there is -2 volt difference between the
motor's back EMF and the drive voltage from the PWM circuit. This -2 volts
will try to decrease the motor current. The decrease in current will decrease
the motor's torque and motor will start to slow down. Once again, the motor's
inductance will oppose the change in the motor's current. The inertia of the
motor and its connected load will oppose the change in speed. The rate of
decrease in current speed will be 1/9 (18 volts version 2 volts) of the rate
of increase in speed and current during the on part of the cycle but it also
lasts 9 times as long. Thus the total increases and decreases will balance.

So there is some small changes in motor current and speed during the PWM's
period. However if the PWM frequency is 100 kHz and the motor and its connected
load weigh 10 pounds, the changes in speed will be very very small.


A couple of final notes:

Please note that PWM circuits are not magical, even though we are only
drawing an average of 0.3 amps from the power supply, we are drawing
6 watts. This is the same 6 watts that is being provided by the motor
to its load. We are not getting something for nothing.

In much of the previous analysis, it is very important that we are looking
at a motor being driven by a PWM. If, instead of motor, we are driving a
simple resistor, then the current through the resistor and the current
from the power supply would be exactly the same. The resistor does not
store any energy. However the motor stores energy in both its inductance
and the inertia of the motor and its connected load.


Dan

Hi Dan,
I new the answer, I just wanted to see the responses to the inquiry.
It took me a while to get my head around it. When someone finally
pointed out Power in = Power out. Duh! So if we have 20v battery pack
flowing .3 amps that's 6 watts. If we have a 50% duty cycle
the average voltage is 10 volts so the current must be .6amps.
You described it well, especially where that current comes from.

There are a few more complications.

Several things happen with the power into a motor. A real motor has several
types of losses. There is power lost into heating the resistance of the motor
windings. There is some power lost in circulating currents in the magnetic
core and other magnetic losses There are friction losses in the bearings in
the motor and its attached load. And finally there is the actual power which
is delivered to the load, i.e. the real usable power from the motor.

With an ideal motor we can ignore all of the losses so we consider the power
into the motor being equal to the power out. (That is not a very good assumption
but it makes the analysis much easier.)

In the earlier discussion, I mentioned the 'back EMF' of the motor without
describing any of its details. The term EMF is short for 'electromotive force'.
It is basically a voltage that is created as the motor turns. It is usually
called 'back EMF' since it opposes the current flow into the motor. The faster
the motor turns, the higher the back EMF. For ideal motors, the back EMF is
proportional to speed. With an ideal motor, the power out of the motor is
equal to the back EMF times the motor current. (With non ideal motors, you
have to subtract off the losses in the magnetics, friction, etc.)

In the earlier discussion, you should note that with our ideal motor, the back
EMF was equal to the average voltage from the PWM circuit. If you increase
the duty cycle of the PWM to 50% then the average voltage will increase to
10 volts (as you said). The motor will respond to this increase in voltage
by increasing its speed until the back EMF is also 10 volts (or a little less
with a non ideal motor). This means that the motor speed increases by a factor
of 5.

For many loads that are connected to motors, the increase in speed will also
increase the power required to drive the load at the higher speed. This
higher power required for the load will mean that the power delivered to the
motor must also increase. How the power requirements of the load vary with
speed depends upon the nature of the load. If the power required to drive the
load at a 5 times faster speed also increases by a factor of 5 then will need
to deliver 30 watts (6 watts times 5) to the load. This means that the motor
current stay the same (3 amps). The current from the power supply will increase
to 1.5 amps (30 watts divided by 20 volts).

For many types of loads, the power requirements will increase with speed at a
rate higher higher than simple proportionality. For instance if the motor is
driving an electric car, the power requirement increase rapidly with speed
since a higher speed implies both a larger distance per unit time and higher
drag forces (and power is force times distance per unit time).

The final result is that increasing the duty cycle will increase the motor
speed but probably not reduce the power requirements.


Dan

.

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