Another DC motor question

Hi Group,

Being that I have a 120VDC motor and I am full-wave recifing the mains
and then using a MOSFET to chop the sine.

I generally turn on the FET at the zero cross and then off sometime
later to control the motor average current, thus speed of the motor.
The only problem is that by reducting the "power" it also reduces the
stalling torque of the motor so if driving a variable load, the system
will slow and speed up.

Another method I would like to try is to allow the voltage to reach the
peak and then turn on the FET for a small duty cycle. Say 1mS or so.

Since the motor is now getting the 166V peak with say a small kick of
voltage, what if any, will be the reaction to the motor rpm or motor
current while driving a load? Will this method get me anything in
terms of drive speed?

Thanks!

PDRUNEN

 

[Ken Smith]

In article <1109740337.522548.257910@g14g2000cwa.googlegroups.com>,
<pdrunen@aol.com> wrote:

Hi Group,

Being that I have a 120VDC motor and I am full-wave recifing the mains
and then using a MOSFET to chop the sine.

I generally turn on the FET at the zero cross and then off sometime
later to control the motor average current, thus speed of the motor.
The only problem is that by reducting the "power" it also reduces the
stalling torque of the motor so if driving a variable load, the system
will slow and speed up.

Is this a PM or universal motor?

On PM motors, the "back EMF" will remain when the switch is off. You can
use this fact to get a crude speed regulation out of it. Linear
technology makes a "isolated flyback regulator" chip. If your motor is PM
I suggest you look at their data sheet and App-notes. You can treat the
back EMF like the flyback voltage.

There is a trick you can use on some universal motors. You can put a
large rectifier across the field winding. (Be very careful of polarity;
you want it off when the transistor is on) What this does is allow the
current in the field coil to continue briefly when the transistor shuts
off.

If you are lucky, you will see the drain of the MOSFET do this:


.......*.................
.......**................
.......*.*..*************
.......*.*.*.............
.......*.**..............
.......*.................
*******.................
^
!
This bit is new and indicates the speed (sort of)

Another method I would like to try is to allow the voltage to reach the
peak and then turn on the FET for a small duty cycle. Say 1mS or so.

You would really be better off not feeding high frequencies to the iron in
the motor. If you power it through an inductor, delaying the turn on may
not be such a bad idea.


--
--
kensmith@rahul.net forging knowledge

 

[Ken Smith]

In article <112bsfgmd6da03c@corp.supernews.com>,
Tim Wescott <tim@wescottnospamdesign.com> wrote:
[...]

Building a real switching amplifier that presents a low-impedance load
to the motor may help a lot, but to really make the motor go a constant
speed you need to take the suggestion already given and wrap the motor
with a closed-loop controller.

You can make the output impedance negitive and do even better. Run the
motor unloaded at some speed and measure the voltage and current. Attach
the load and adjust the voltage for the same speed and measure voltage and
current. You now know that you need to increase the voltage by about X
volts when it draws Y current. You don't want to increase it by any more
than X so you may want to fall a little short of the voltage you measured.


--
--
kensmith@rahul.net forging knowledge

 

[Larry Brasfield]

"Ken Smith" <kensmith@green.rahul.net> wrote in message
news:d05u25$60b$2@blue.rahul.net...

In article <112bsfgmd6da03c@corp.supernews.com>,
Tim Wescott <tim@wescottnospamdesign.com> wrote:
[...]
Building a real switching amplifier that presents a low-impedance load
to the motor may help a lot, but to really make the motor go a constant
speed you need to take the suggestion already given and wrap the motor
with a closed-loop controller.

You can make the output impedance negitive and do even better. Run the
motor unloaded at some speed and measure the voltage and current. Attach
the load and adjust the voltage for the same speed and measure voltage and
current. You now know that you need to increase the voltage by about X
volts when it draws Y current. You don't want to increase it by any more
than X so you may want to fall a little short of the voltage you measured.

A fairly close approximation of the right source
impedance to effect speed regulation is to set
it at the additive inverse of the coil resistance.
Regardless of whether the stationary field is
setup by permanent magnets or a winding,
this drive arrangement can be thought of as a
flux forcing function, holding the derivative
w.r.t. time proportional to the control voltage.
This occurs only at a particular speed, at least
on a timescale where the inductance can be
ignored. The output impedance plus the coil
resistance should stay in the left half-plane in
order to avoid instability.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Tony Williams]

In article <1109740337.522548.257910@g14g2000cwa.googlegroups.com>,
<pdrunen@aol.com> wrote:

Being that I have a 120VDC motor and I am full-wave recifing the
mains and then using a MOSFET to chop the sine.
[snip]
Since the motor is now getting the 166V peak with say a small
kick of voltage, what if any, will be the reaction to the motor
rpm or motor current while driving a load? Will this method get
me anything in terms of drive speed?

Do you have LTSpice from Linear Technology? It can
be used to 'experiment' with ideas like that, and
see if they are worth trying for real. A rough
model of a PWM'd dc motor is shown below. It is my
rough guess, which is still under construction.


Vsupply+---+---------------+
| __________|___________
| | | |
| | \ MOTOR |
| | R / |
| | \ |
| | | |
| | )|| |
| | L )|| |
| | )|| |____
| | | |
| | +----+--+--+----+ <-- |
| | | | | | | |
| | [I1] [C1] [I2] [C2] |V1 |
| | | | | | | |
| | +----+--+--+----+ <-- |
_|_ |__________|________________|
D1 /_\ |
| |
+---------------+
|
_ _ +
_| |_| |_---> / Switch
+
|
0v+----------------------------+

R and L represent the motor's resistance and inductance.
I1 is a constant-current sink representing the no-load
current (torque), and C1 represents the motor's own
moment of inertia. Similarly I2 and C2 simulate an
external torque load and moment of inertia.

The (resultant) voltage V1 is the generating back emf
of the motor, which also indicates the rpm achieved for
various duty-cycles and loads.

R, L, and I1 are usually given by the mfr. C1 can
be estimated by assuming a no-load rundown time of
T seconds and using C*Vsupply = I1*T to calc C1.
Umm..... I think.

Full load current is also given by the mfr, so this
I2 can be set for various values, representing loads
up to full load. C2 is a bit of a guesstimate,
depending on the type of load.

With LTSpice various PWM frequencies and duty cycles
can be run, and the resultant V1 and I-motor looked at.

Of particular interest is I-motor. It must never be
allowed to drop to Zero during the PWM cycle. If it
does the torque-speed curve goes soggy, and losses
increase. This is a matter of making sure that the
PWM Off-time is always small compared to the L/R of
the motor.

I suspect that your 100Hz PWM does allow I-motor to
drop to Zero, and this may be why the torque-speeed
is not what it should be.

A simulation may be useful.

--
Tony Williams.