Extremely well written.

All I was going to say was that the PWM method of driving a low voltage
motor at high voltages allows significantly better torque than normal
voltages. Sometimes that can be a real plus in the application, such as
when driving a motor who is designed to vibrate with an eccentic cam; they
are easier to startup.

--Bob

At 11:29 AM 6/14/2003 -0400, you wrote:
>It isn't very relevant to running a little tiny 10mA motor but IMHO, it is
>a mistake to assume that 400 to 2000Hz is a good typical PWM frequency.
>
>(WARNING: here comes a long explanation :-)
>
>There are two regimes or modes that PWM can operate in: low frequency and
>high frequency.
>
>Which one you are operating in depends on whether your PWM period is
>significantly higher or lower than the motor's inductive-resistive (RL)
>time constant, which is L/R (so it is determined by whether your frequency
>is much larger or smaller than R/L). If your switching element and wiring
>has significant resistance, you need to include this in the R here.
>
>If the low frequency region, you usually connect the motor to the power
>supply during the "ON" time and leave it open circuited (with a flyback
>diode) during the "OFF" period. The current waveform in the motor then
>looks like a constant current during the ON time and the current very
>quickly decays to zero at the beginning of the OFF period and remains zero
>for the majority of the OFF time.
>
>This results in torque pulses being applied to the motor (since torque is
>proportional to current). You rely on the mechanical time constant of the
>system to smooth out these torque pulses into steady motion.
>
>In this region of operation, your main concern is that you do not want so
>low a PWM frequency that you cause mechanical vibration and you don't want
>so high a frequency that it becomes very audible and annoying.
>
>The average motor current equals the average battery current and the
>applied torque (at a fixed motor RPM) is a linear function of the PWM duty
>percent.
>
>The motor output power is average current times back EMF (motor open
>circuit voltage as it spins and acts like a generator). System input
>power is average current times battery voltage. Roughly, the motor speed
>and therefore EMF will go up linearly with duty percent, so you can have a
>very inefficient situation in this PWM region. Consider a situation where
>the motor EMF is only 1 volt and your supply voltage is 10 volts (PWM duty
>of about 10%). This means that your system will only be 10% efficient. The
>extra power is lost in I^2*R losses in the motor winding resistance, which
>is NOT Iavg^2*R but is actually the average(I^2)*R, which is much higher
>since your current consists of very large pulses of short duration.
>
>The other mode of PWM operation is high speed mode, where your PWM period
>is so short that the RL time constant tends to keep the motor current
>constant during each PWM cycle.  Here, you connect the motor to the supply
>during the ON time but SHORT it during the OFF time.
>
>This rubs a lot of people the wrong way because they realize that normally,
>shorting a motor causes breaking (the back EMF acts to put current through
>in the direction that causes torque opposite the direction of rotation).
>However, what they don't realize is that when you suddenly short a motor,
>you actually get positive torque for a very short period of time (due to
>the stored energy in the inductance being used very efficiently to drive
>the motor) followed by breaking. If you are using a high enough PWM
>frequency, then the breaking never happens (unless you are at a lower PWM
>duty than you should be for the current speed and load torque).
>
>In the high frequency mode, the motor and PWM switch are being used like a
>little buck SMPS (switching power supply) to efficiently convert from the
>high supply voltage down to the lower back EMF of the motor. This happens
>because the inductor forces a constant current, and so has a high voltage
>across it when the supply is connected to the motor. This high voltage
>bucks the supply and prevents the current from increasing. Since voltage
>times current equals power, this represents power going into the inductor
>during the on time, Then, during the off time, the inductor acts like a
>little power supply most of that power back to the motor.
>
>The net result is that there is very little current ripple in the motor so
>that things are efficient (Iavg^2 is about equal to average(I^2)) and also
>the applied torque is very uniform.
>
>For typical motors, we will be taking about somewhere around 10kHz as a
>typical high frequency PWM, although plenty of motors need 30 or even 60kHz
>drive to get well into this region (you usually want less than 30% current
>ripple for high efficiency). This is especially true of very high
>performance motors for electric flight, for example, which have low
>inductance.
>
>Another plus of this method is that you get the option of breaking as a
>free byproduct. If you set the PWM duty to zero, you get full breaking and
>anything lower than the equilibrium point that sustains your speed will be
>breaking to some extent. This is different than the low-speed method where
>you need a separate switching element to do breaking (or at least a
>separate control signal). Setting the PWM duty to a lower value than the
>equilibrium in low speed operation just causes the motor to slow down due
>to the attached load, which might be very light and cause a slow speed
>change. This can be a problem for feedback control systems because your
>ability to apply torque is highly asymmetrical (you can apply high positive
>torque but almost no negative torque).
>
>The down-side of the high frequency method is that you need to design your
>switching circuit very well so that it can switch efficiently at these high
>frequencies. As Herbert said, you need to minimize the time spent in
>transition. You also still need a flyback diode because you need to prevent
>voltage spikes during each transition (voltage spikes are due to the
>inductor trying to enforce constant current through a circuit with high
>resistance, such as when both switching elements are open).
>
>Average supply current is now duty percent times motor winding current. In
>both PWM methods, if your battery or supply has a significant internal
>resistance, you often want to bypass it with large capacitors (that can
>handle the ripple current) so that your supply voltage doesn't go up and
>down with each current pulse. Even though the high frequency method causes
>constant current in the motor, it still draws pulses from the supply.
>
>Another minor downside is that the equations become slightly more
>complicated. Current is now a nonlinear function of duty percent and you
>can get situations where peak motor torque is achieved at a PWM duty
>percent below 100% and actually drops off above that. In extreme cases
>(which only occur when your supply resistance is very high), you might
>actually need to take this into account to achieve maximum torque during
>acceleration. In other words, a higher PWM duty will always result in a
>higher final speed, but just pegging it at 100% and then dropping it down
>to the equilibrium PWM duty for your new speed, as you would assume would
>give the fastest response, might actually be slower than, say, going to
>80%, sliding up to 90% during the acceleration, and then bringing it down
>to the new duty.
>
>When you bypass your power supply with capacitors such that it sees a
>constant current, then you get:
>
>Imotor = d*Vb/(d^2*Rw+Rm)
>Ibatt=d*Imotor
>
>where Imotor is motor winding current, Vb is battery (or supply) open
>circuit voltage, Rw is the resistance of the battery (or supply) and all
>the wiring connecting it to the circuit. Rm is the resistance of the motor
>plus switching element plus any wire between them.
>
>Note that this becomes the simple linear formula when Rw is much smaller
>than Rm (as might be the case with NiMH or NiCd batteries).
>
>I just went through all of this with a high performance brushless motor
>controller design and I managed to get somewhere around 90% efficiency from
>battery to load even at 50% duty. It sure took a while to figure it all
>out, though! It is also amazing how few people seem to know this, although
>I did find one site that explained it well.
>
>Sean
>
>
>
>
>
>
>At 07:11 AM 6/14/2003 -0400, you wrote:
>> > >     Well, it all depends one what you want to do. You could
>> > simply drive the
>> > > motor with a 50% duty cycle waveform and probably get an average pretty
>> > > close to 1.5V. The one great thing about motors is that they have a very
>> > > high inductance, which means they filter incoming pulsed waveforms quite
>> > > well. The frequency of operation isn't too critical, I'd say
>> > anywhere from
>> > > 400Hz to maybe 2000Hz would PROBABLY be OK.
>> >
>
>--
>http://www.piclist.com#nomail Going offline? Don't AutoReply us!
>email listserv@mitvma.mit.edu with SET PICList DIGEST in the body

---------------
NOTICE

1. This account can accept email & attachments up to 10M in size.
2. Federal Monitors: At request of client, some attachments are encrypted.
Please DO NOT delay traffic; please reply with credentials for password.
--------------

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email listserv@mitvma.mit.edu with SET PICList DIGEST in the body