Simple 555 PWM - disappointing performance

[Terry Given]

Pooh Bear wrote:

Herbert Blenner wrote:


Hence the transistor approximates a voltage switch.


Oh dear.

Pls explain the difference between a *voltage* switch and a *current* switch.
A switch is a switch is a switch....

Graham

Hi Graham,

Its entertaining, and if we do a decent job of explaining why its all
crap, then others (specifically Terry Pinnell) will learn some
analytical techniques, as well as another circuit to put in the "dont do
this" pile.

Cheers
Terry

 

[Terry Given]

Terry Pinnell wrote:

Tony Williams <tonyw@ledelec.demon.co.uk> wrote:


In article <qqgb915u62pm4ecjn0iamg9i4rd58uqavi@4ax.com>,
Terry Pinnell <terrypinDELETE@THESEdial.pipex.com> wrote:


Thanks a lot Tony, appreciate your going to the trouble. Very
timely, as I was pondering how to proceed with simulation in
parallel with my practical experiments, and didn't really know
where to start. Will study that circuit and simulation in detail.

The model is still under development and I'd welcome
any corretions or additions.


I'll try to take the key measurements for this motor and plug
them in. I already measured two key ones earlier. According to my
Atlas LCR, the motor has a DC resistance of 0.2 ohms and
inductance of only 14 uH, measured at 200 kHz. (The Atlas chooses
its own test frequency.) One odd point is that the DC resistance
initially measured about 1.0 ohm, but after running the motor
briefly that became 0.2! Same with two identical motors.

That would be a stall current of 12A when directly powered
off the original battery. Be warned though Terry. I burned
out my B&D battery screwdriver doing some similar expts
about a year ago. Run the thing off a current-limited
supply if possible.


Can't say I really see how some of those elements work (I1, I2,
R2 and C1), but I'll hold my queries until I've played with it.

I2 and R2 are trying to account for the gearbox losses, which
was the 1A my B&D drew with no external load. I1 is supposed
to simulate the external load. C1 is supposed to acquire a
voltage across it that represents the motor's generated back
emf.


Thanks for the follow-up. You didn't comment on that inductance? Can a
motor really have such a tiny value?

I managed to measure what I think is the stall torque. Motor (ex
HomeBase screwdriver) was allowed to pull a spring balance via a cord
fixed to a pulley of a little under 2 inch in diameter. The balance
registered about 7 lbs, and I was surprised to measure about 9.5 A. At
that point the voltage at the motor's terminals was a mere 1.0 V. At
the source, a set of 4 D-type Nicads, it had dropped from its unloaded
value of 5.0 V to 4.0V. Quite a lot of variation between successive
apparently identical measurements though. Had 10A at one point.

I'm I bit handicapped by not having a heavy duty variable voltage
power supply, but I'll see how much further I can get with batteries.
I can always get some variation with series diodes.

Well done Terry. I just love basic physics.

Cheers
Terry

 

[Tony Williams]

In article <sNAle.3344$U4.479905@news.xtra.co.nz>,
Terry Given <my_name@ieee.org> wrote:

As you've no doubt seen, since posting that question Tony has
kindly published his proposed simulation model, which looks
rather more complex.

I dont use LTSpice, alas.

I've been trying to model something that simulates
the back-emf that a motor generates whilst running.

--+-- 4V
|
\
/R1 (0.5R)
\
|
)
)L1 (3mH)
)
|
+----+-----+-------+
| | | |
_|_ | | _|_
| | \ |+ | |
I2| I | /R2 ===C1 | I |I1
|___| \ | |___|
\|/ | | \|/
| | | |
+----+-----+-------+
|
100R |/c
300Hz------/\/\---|Q1 = 2N3055
+12v to +0.4v |\e
|
-+-0v

I2 represents a fixed gearbox torque loss.... = 0.5A
R2 is the speed dependant torque loss........ = 1.9R
I1 is the external torque load... 0 to 1A so far.
C1 represents the energy storage of the rotor. The
value of C1 is roughly chosen to emulate the time
constant of the rotor. 2000uF as a first stab.

When chopping, the voltage that results across C1 should
represent the generated back-emf, which is also a measure
of the speed of the motor.

The LTspice simulation, at 300Hz 50:50, shows a pitiful
performance from the 'motor'. At an I1 of only 1A the
voltage across C1 is down to 0.16V, which suggests a
motor rpm of only 8% of the no-load speed at 2.4Vdc.

Suggestions to improve the model are welcomed.

--
Tony Williams.

 

[Ban]

Terry Given wrote:

Its entertaining, and if we do a decent job of explaining why its all
crap, then others (specifically Terry Pinnell) will learn some
analytical techniques, as well as another circuit to put in the "dont
do this" pile.

To learn something needs thorough understanding of the proper function. The
same is valid for measuring, especially on a motor. A motor has an
inductance, which can be measured at stall, but when turning an additional
voltage source comes in play(Back EMF), so the only way to determine the
force is to measure the current through the motor, which Terry so far has
not managed to do. Exept a few static values. So we do not know if the
repetition rate of the PWM is too high. He also seems to insist to use the
"quarter-bridge" he has found somewhere on the web together with the 3055.
I do not see any insight here, so whatever anybody suggests takes a long
time to be evaluated and the final result will be accordingly.

--
ciao Ban
Bordighera, Italy

 

[John Popelish]

Terry Given wrote:
(snip)

I did a simple simulation in Simetrix, using an MJD2955 and a 1mH
inductor (no winding R), chopped at 100Hz with 15V square-wave base
drive. Peak current is controlled *entirely* by Rbase (ergo by Hfe, so
all over the show with time, temperature and type of transistor). To get
2.5A, I needed Rb = 50 Ohms, and Vcesat actually pulls right up to the
supply rail! The turn-off current commutation behaviour is as described.

In conclusion, this is a *terrible* idea, and will not work.

Using a superconducting inductor as a model for a motor is a poor way
to come to a valid conclusion about a motor load, since its current is
unlimited and will force an average voltage (through a switching
cycle) of exactly zero volts. Obviously some other component (like
the transistor) must set the current limit, in that case.

 

[John Popelish]

Terry Pinnell wrote:

Did you see my query about the value of 14 uH measured by my Peak
'Atlas LCR' instrument? And that odd inconsistency of R?

Your measured brush resistance riding on the "chocolate" oxide coating
on copper commutator is quite normal.

 

[Mike Monett]

Tony Williams wrote:

In article <sNAle.3344$U4.479905@news.xtra.co.nz>,
Terry Given <my_name@ieee.org> wrote:


As you've no doubt seen, since posting that question Tony has
kindly published his proposed simulation model, which looks
rather more complex.

I dont use LTSpice, alas.

I've been trying to model something that simulates
the back-emf that a motor generates whilst running.

--+-- 4V
|
\
/R1 (0.5R)
\
|
)
)L1 (3mH)
)
|
+----+-----+-------+
| | | |
_|_ | | _|_
| | \ |+ | |
I2| I | /R2 ===C1 | I |I1
|___| \ | |___|
\|/ | | \|/
| | | |
+----+-----+-------+
|
100R |/c
300Hz------/\/\---|Q1 = 2N3055
+12v to +0.4v |\e
|
-+-0v

I2 represents a fixed gearbox torque loss.... = 0.5A
R2 is the speed dependant torque loss........ = 1.9R
I1 is the external torque load... 0 to 1A so far.
C1 represents the energy storage of the rotor. The
value of C1 is roughly chosen to emulate the time
constant of the rotor. 2000uF as a first stab.

When chopping, the voltage that results across C1 should
represent the generated back-emf, which is also a measure
of the speed of the motor.

The LTspice simulation, at 300Hz 50:50, shows a pitiful
performance from the 'motor'. At an I1 of only 1A the
voltage across C1 is down to 0.16V, which suggests a
motor rpm of only 8% of the no-load speed at 2.4Vdc.

Suggestions to improve the model are welcomed.

--
Tony Williams.

Tony, I just popped in briefly and probably missed a lot of important
stuff. But I came across this SPICE model that may have some bearing on
the problem:

http://www.ecircuitcenter.com/Circuits/dc_motor_model/DCmotor_model.htm

Mike Monett

 

[Terry Pinnell]

John Popelish <jpopelish@rica.net> wrote:

Terry Pinnell wrote:

Did you see my query about the value of 14 uH measured by my Peak
'Atlas LCR' instrument? And that odd inconsistency of R?

Your measured brush resistance riding on the "chocolate" oxide coating
on copper commutator is quite normal.

Thanks. Surprisingly consistent, to 1 dp across two motors!

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Terry Given]

John Popelish wrote:

Terry Given wrote:
(snip)

I did a simple simulation in Simetrix, using an MJD2955 and a 1mH
inductor (no winding R), chopped at 100Hz with 15V square-wave base
drive. Peak current is controlled *entirely* by Rbase (ergo by Hfe, so
all over the show with time, temperature and type of transistor). To
get 2.5A, I needed Rb = 50 Ohms, and Vcesat actually pulls right up to
the supply rail! The turn-off current commutation behaviour is as
described.

In conclusion, this is a *terrible* idea, and will not work.


Using a superconducting inductor as a model for a motor is a poor way to
come to a valid conclusion about a motor load, since its current is
unlimited and will force an average voltage (through a switching cycle)
of exactly zero volts. Obviously some other component (like the
transistor) must set the current limit, in that case.

this is true, but I spice'd it to prove the disappearing-base-drive
"feature" of Herberts suggested "improvement."

High Rmotor indeed limits the current eg 4GOhms gives 1nA (reductio ad
absurdum). If its a grunty little motor (out of an electric screwdriver)
then I would expect Rmotor to be relatively small (say 10%) thus
initially, ignoring it isnt *that* bad.

ignoring back-EMF, OTOH, is a far more grievous sin

Cheers
Terry

 

[Terry Given]

Herbert Blenner wrote:

Replace Q1 with a PNP transistor in common collection configuration.
Now this circuit can take advantage of the sizable current sink
capability of the 555 timer. Further since you are now switching
voltage across the motor, D1 is not needed.


Thanks, Herbert, but according to the LM555 datasheet, a 555 "output
can source or sink 200 mA."

and besides, he is completely wrong about the "dont need the diode",
which of course you still do. The diode is placed to ensure the motor
current has somewhere to commutate to when the switch turns off. for a
pnp switch, the pnp collector goes to the motor and the cathode of the
diode; the other end of the motor and the diode anode are at zero volts.

advice like that you can live without.


Connecting the emitter to the motor places the pnp transistor in a
common collector configuration. Hence the transistor approximates a
voltage switch. The current however will slew between on and off due to
the inductance of the motor.

thats a fairly piss-poor description, indicative of a deep lack of
understanding.


---+-- +4V
|
[motor]
|
|
|
|/e
555--[R]-----| pnp
|\c
|
---+-- 0V


Dont forget the 555 runs from +15V.

When the 555 o/p pulls low, the motor turns on, but where does the
substantial base current required come from?


If my memory is correct, the output stage of the 555 has an open
collector and grounded emitter. In this case a second resistor from
the 4 volt supply to the base of the pnp transistor would ensure prompt
turn off.
Alternately if the output stage of the 555 has active source and sink
stages then you must reconnect the timer to a lower voltage and omit
the second resistor.

Incorrect, its always been a totem-pole output stage (which is why 555's
draw 300mA current spikes on each switching edge, as both totem-pole
transistors momentarily conduct, hence the need for a hefty bypass cap
right across the pins of the device). The totem-pole output can sink
much more than it can source, for much the same disappearing Ibase
reason as this terrible idea.


Since the output sinks more than it sources, the turnon base current in
a pnp transistor would be higher than with a npn unit.

*NO* not with the circuit as drawn, because the voltage across which the
base current is developed, GETS SMALLER as the transistor turns on. This
is because when the transistor turns on, Vce tends to zero (ish).
Therefore Vbe tends to zero, and so does the voltage across Rbase
(because the 555 output is at zero-ish)

I am not discussing turn off here at all, merely turn on. As the pnp
turns on, the emitter voltage falls, therefore so does the base,
therefore so does the voltage across the 555-to-base resistor.


Adjust the resistor to compensate for the change in voltage. The bottom
line is to get the base current up to the proper value.

this gets tricky, as the voltage across Rb changes a *lot* - from 4V or
so at the onset of turn-on to a few hundred mV when the transistor is on
(or at least as on as it gets)

my simulation shows this clearly, and I use an ideal voltage source (ie
one that can sink and source many trillions of amps)


As the emitter pulls down,
the voltage across Rbase drops, so Ibase drops, so PNP turns off.
Besides, even if we magically required almost no base current (eg
darlington, Ibase = 3A/10,000 = 300uA) the emitter will *always* be Vbe
above the base, losing a large chunk of the available 4V supply. The
reducing base drive "feature" of this POS *forces* the transistor to
dissipate far more power than it would if switched properly.


You forget that bipolar transistors are not the only kind.

*you* said "Replace Q1 with a PNP transistor"

please explain to me how I can have a pnp transistor that is *not* bipolar?


After you objected to the higher Vce of the pnp transistor, I reminded
you that not all transistors are bipolar.

Herbert, the *reason* Vce is high is twofold:

1) Ve is *always* Vbe *above* the base voltage, which (if you want any
base current) is *always* above the 555 output "zero". This is due to
the circuit topology, rather than the bjt itself. If you want low
Vcesat, dont do it this way. Likewise, dont use a Darlington, for much
the same reason. the original NPN arrangement has a much lower Vcesat,
yet is a BJT, because Vbe doesnt get in the way.

2) the disappearing base drive causes real problems here.

When the 555 o/p goes to +15V, where does the motor current flow to? by
Lenz' law, the emitter voltage will rise, until the base-emitter diode
is forward biased (IOW about 15.7V) at which point the pnp will turn on,
shunting the motor current to 0V. Of course Vce = 15.7V, with 3A across
it, ramping down to zero in time t = Lmotor*Ipeak/15.7V. dissipation
acros the pnp is thus 23.5W for however long it takes. The motor current
ramps down a lot faster than it ramps up, as it has almost 3x the
voltage across it.


You should never allow Vbe of a pnp transistor to go positive. This is
especially true for power transistors, whose base-emitter junctions
have low reverse breakdown voltages.


read it again. I said "until the base-emitter diode is forward biased"
which of course means Ve > Vb for a pnp. This is implicit in the "about
15.7V" statement immediately thereafter.


Driving the base resistor from a 15 Volt output of a 555 reverse biases
the base-emitter junction of the pnp transistor and causes reverse
breakdown.

only for a very short period of time. The 555 output has a finite,
non-zero slew rate. As Vb rises above 4V, the transistor turns OFF, and
the inductive current has to go somewhere. It wants to keep flowing in
the same direction, so charges up whatever C is lying around, and makes
the pnp emitter voltage *rise* (again, finite non-zero slewing). It
keeps rising until its at 15.7V or so, at which point the pnp B-E
junction becomes forward-biased again, and the motor current flows thru
the emitter to 0V - except for 1/Hfe, which flows out the base and
*into* the 555 (where else can it go?)

this is exactly analagous to the npn switch case - when the npn turns
off, Vce rises, forced by the inductive load attempting to commutate
somewhere (anywhere). If no commutation path is found, Vce rises to +4V
+ LdI/dt (actually more like 4+IZo but thats a whole 'nother thang).
Normally this is a very high value. Consider 1A, 1mH and 1us turn-off -
Vce rises to about 4V + 1mH*1A/1us = 1004V, which will kill the npn
switch (thereby giving the current somewhere to commutate to...). Hence
the so-called freewheeling diode.

although it is possible, depending on stray capacitances and the 555
dV/dt, to apply a momentarily high reverse voltage across the
base-emitter junction. yet another reason not to do it this way.


If the output stage of the 555 has active source and sink stages then
connecting the timer supply to four or five volts ensures proper
biasing of the transistor.

I should mention that, unless the 555 has output clamp diodes (or a FET
output stage) the pnp wont work at all, as all base current has to flow
*into* the 555 output when it is high.


When the output of the 555 is high, the pnp transistor is off and no
base current flows.

if the pnp is OFF, where does the motor current go to? its an
inductor.....if when the switch is on, current is flowing from +4V into
the inductor, its going to want to keep flowing that way when the switch
turns off. And it does, as both a simple explanation and SPICE show.

An LMC555 will do that just fine
(FETs dont care which way the current goes) but I cant recall OTTOMH
what the vanilla 555 output stage looks like. They probably do have
clamp diodes, and its probably *not* a good idea to bung more than a few
mA thru them, lest you trigger the internal SCR which sits across the
device supply rails (thereby shorting out the 15V supply thru the 555,
and cooking it quick-smart).

changing the supply voltage to 4V does not change the circuit behaviour,
only the voltage at which the pnp turns on again (Ve = 4.7V or so), and
of course the rate-of-change of motor current. Its still a ratshit
circuit, which explains why *nobody* does this.

the magical disappearing base current alone is enough to consign this
POS to the wastepaper basket, where it truly belongs




If D1 is left connected across the motor, this turn-off behaviour does
not happen, as motor current commutates thru D1, keeping about 0.7V
across the motor (so current downslope is about 5-6 times *slower* than
the up-slope)

I did a simple simulation in Simetrix, using an MJD2955 and a 1mH
inductor (no winding R), chopped at 100Hz with 15V square-wave base
drive. Peak current is controlled *entirely* by Rbase (ergo by Hfe, so
all over the show with time, temperature and type of transistor). To get
2.5A, I needed Rb = 50 Ohms, and Vcesat actually pulls right up to the
supply rail! The turn-off current commutation behaviour is as described.

In conclusion, this is a *terrible* idea, and will not work.


I suggest doing the circuit analysis. You will find that switching
current in an inductance produces voltage spikes. By contrast switching
voltage across the inductance gives a continuous change in current with
slope discontinuity. Voltage switching in inductive circuits is not an
idea, instead it is standard practice in electromechanical controls
systems.

Herbert

I'd love to see your circuit analysis of this. Admittedly I dont know
much about pissant little motors (definition: anything a human can lift
unaided), but my calculus is OK. yes, V=dLambda/dt = LdI/dt (if and only
if L is constant, otherwise add in an IdL/dt term) so dI/dt = V/L.


The response, i(t), of a series connection of R and L to a voltage
source V u(t) is V / R ( 1 - e ^ ( - R / L t) ).
^^^^^^^^^

no its not, but I assume its just a typo on your part. e^(-Rt/L)

Now a current source, I u(t), driving the series connection of R and L
develops a voltage v(t) = I R u(t) + L I du(t) / dt = a step plus an
impulse.

yep. that of course has nothing to do with you appalling circuit.

all that happens in this case (apart from the appallingly low
efficiency, due to piss-poor base drive and linear operation) is that
the pnp transistor functions as the clamp diode, but in a much more
lossy fashion. For Vcc(555) = 4V, 0.7V is dropped across the motor
(exactly the same as with D1) *but* the C-E voltage of the pnp
transistor is 4.7V, so it dissipates 4.7/0.7 = 6.7 times more power than
D1 would in the same situation.


When Vce is 4.7 V the pnp transistor is off and the current is zero.
During turn on Vce is about 1 volt, which entails three times the power
dissipation of a saturated common emitter transistor. However this
additional dissipation is merely cause to fet.

Herbert




BTW, the point Ban made is quite correct. The motor has some inductance,
Lmotor. This is what "forces" the current to keep flowing when the
2n3055 switches off. When the switch is on, current ramps up, current
slope = dI/dt = Vbattery/Lmotor. Eventually this current gets so high
the voltage drop across the winding resistance gets in the way (eg in a
relay, Rwinding sets Irelay). When the transistor switches off, the
voltage on the winding reverses (Lenz' law) forward biasing the diode
to, say, 1V. The inductor current now ramps down, dI/dt = Vdiode/Lmotor


In common emitter configuration the 2N3055 approximates a current
switch. Now switching the current produces voltage spikes and
necessitates the diode.

Herbert




Because Vdiode is small (1V cf 15V) the current ramps down slowly. This
will limit the performance you get from the machine. A simple fix for a
low power single-ended circuit is to add a zener in series with the
diode (or a few more diodes), but beware the losses!

Cheers
Terry

 

[John Popelish]

Terry Given wrote:

John Popelish wrote:

Using a superconducting inductor as a model for a motor is a poor way
to come to a valid conclusion about a motor load, since its current is
unlimited and will force an average voltage (through a switching
cycle) of exactly zero volts. Obviously some other component (like
the transistor) must set the current limit, in that case.


this is true, but I spice'd it to prove the disappearing-base-drive
"feature" of Herberts suggested "improvement."

High Rmotor indeed limits the current eg 4GOhms gives 1nA (reductio ad
absurdum). If its a grunty little motor (out of an electric screwdriver)
then I would expect Rmotor to be relatively small (say 10%) thus
initially, ignoring it isnt *that* bad.

ignoring back-EMF, OTOH, is a far more grievous sin

Not if you are modeling the stalled motor condition.

 

[Tony Williams]

In article <429739F1.199B@spam.com>,
Mike Monett <no@spam.com> wrote:

Tony, I just popped in briefly and probably missed a lot of
important stuff. But I came across this SPICE model that may
have some bearing on the problem:

http://www.ecircuitcenter.com/Circuits/dc_motor_model/DCmotor_model.htm

Nice one Mike, thank you. That looks like it will
enable a far better modelling of the generated Vbemf.

--
Tony Williams.

 

[Fred Bloggs]

Mike Monett wrote:

Tony Williams wrote:

In article <sNAle.3344$U4.479905@news.xtra.co.nz>,
Terry Given <my_name@ieee.org> wrote:



As you've no doubt seen, since posting that question Tony has
kindly published his proposed simulation model, which looks
rather more complex.

I dont use LTSpice, alas.

I've been trying to model something that simulates
the back-emf that a motor generates whilst running.

--+-- 4V
|
\
/R1 (0.5R)
\
|
)
)L1 (3mH)
)
|
+----+-----+-------+
| | | |
_|_ | | _|_
| | \ |+ | |
I2| I | /R2 ===C1 | I |I1
|___| \ | |___|
\|/ | | \|/
| | | |
+----+-----+-------+
|
100R |/c
300Hz------/\/\---|Q1 = 2N3055
+12v to +0.4v |\e
|
-+-0v

I2 represents a fixed gearbox torque loss.... = 0.5A
R2 is the speed dependant torque loss........ = 1.9R
I1 is the external torque load... 0 to 1A so far.
C1 represents the energy storage of the rotor. The
value of C1 is roughly chosen to emulate the time
constant of the rotor. 2000uF as a first stab.

When chopping, the voltage that results across C1 should
represent the generated back-emf, which is also a measure
of the speed of the motor.

The LTspice simulation, at 300Hz 50:50, shows a pitiful
performance from the 'motor'. At an I1 of only 1A the
voltage across C1 is down to 0.16V, which suggests a
motor rpm of only 8% of the no-load speed at 2.4Vdc.

Suggestions to improve the model are welcomed.

--
Tony Williams.


Tony, I just popped in briefly and probably missed a lot of important
stuff. But I came across this SPICE model that may have some bearing on
the problem:

http://www.ecircuitcenter.com/Circuits/dc_motor_model/DCmotor_model.htm

Mike Monett

Excellent link-

 

[Terry Given]

John Popelish wrote:

Terry Given wrote:

John Popelish wrote:


Using a superconducting inductor as a model for a motor is a poor way
to come to a valid conclusion about a motor load, since its current
is unlimited and will force an average voltage (through a switching
cycle) of exactly zero volts. Obviously some other component (like
the transistor) must set the current limit, in that case.



this is true, but I spice'd it to prove the disappearing-base-drive
"feature" of Herberts suggested "improvement."

High Rmotor indeed limits the current eg 4GOhms gives 1nA (reductio ad
absurdum). If its a grunty little motor (out of an electric
screwdriver) then I would expect Rmotor to be relatively small (say
10%) thus initially, ignoring it isnt *that* bad.

ignoring back-EMF, OTOH, is a far more grievous sin


Not if you are modeling the stalled motor condition.

Touche.

Cheers
Terry

PS nice work on the TANH circuit.

 

[John Popelish]

Terry Given wrote:

PS nice work on the TANH circuit.
Thanks. I'm still having lots of fun with it. This is the first time

I have dealt with amplifier nonlinearity as a good thing.

 

[Terry Pinnell]

"Ban" <bansuri@web.de> wrote:

Terry Given wrote:

Its entertaining, and if we do a decent job of explaining why its all
crap, then others (specifically Terry Pinnell) will learn some
analytical techniques, as well as another circuit to put in the "dont
do this" pile.


To learn something needs thorough understanding of the proper function. The
same is valid for measuring, especially on a motor. A motor has an
inductance, which can be measured at stall, but when turning an additional
voltage source comes in play(Back EMF), so the only way to determine the
force is to measure the current through the motor, which Terry so far has
not managed to do. Exept a few static values. So we do not know if the
repetition rate of the PWM is too high.

You're right that I've not so far managed to do that. I'm still
working at it though, as it interests me. What practical further
advice do you have on measuring this motor's dynamic torque?

He also seems to insist to use the
"quarter-bridge" he has found somewhere on the web together with the 3055.
I do not see any insight here, so whatever anybody suggests takes a long
time to be evaluated and the final result will be accordingly.

You really do have a bee in your bonnet about this, don't you, Ban!
Are you on some sort of worldwide commission scheme for promoting the
bridge approach? I explained this at least once up-thread. One more
time: the project was finished a YEAR ago. Past tense. So, with due
allowance for your english, it makes no sense to say 'he also seems to
insist to use the "quarter-bridge"...'

If you want to criticise a decision I took a year ago, fine, but
that's OT from this thread. FWIW, I was fully aware then of the
relative advantages and disadvantages of different approaches,
including the half and full bridges. Ever think there might have been
reasons for going the relay-based route? Ever think there might be
practical reasons for not now scrapping a successfully working project
and starting from scratch, to achieve a minor improvement?

And what's this all about: "he has found somewhere on the web together
with the 3055."? Have you got something against 'the web', BTW? And
what makes you think I 'found' that trivially simple 555/3055 PWM
circuit anywhere? I've been building circuits with 555s and/or 2N3055s
for over 25 years, so why would I need to 'find' it?

I do not see any insight here, so whatever anybody suggests takes a long
time to be evaluated and the final result will be accordingly.

English may not be your first language, but you're not bad on the
personal insults! Lighten up, Ban, you're a grandad now, remember? And
maybe read the threads a little more carefully, and consider the
background, before taking such a patronising and sanctimonious tone.

BTW, I'm curious what 'the final result' is supposed to mean? Do you
mean the improvement of my curtain controller - i.e. slowing it down
while maintaining adequate torque, the stated objective? Or my
ultimate personal destiny as an electronics hobbyist? <g>

Have a nice Sunday!

--
Terry Pinnell
Hobbyist, West Sussex, UK
Sun 29 May 2005, 05:56 UK time

 

[Terry Pinnell]

No obvious place in the thread to post these, the latest results of my
static tests with the motor under discussion.

MISC NOTES:
No PWM involved. Tests were performed in the reverse order to the more
logical sequence shown. IOW, I started with the four-battery set, so
battery source voltage declined as nicads were discharged heavily.
Used a set of D-type NiCads, wired with taps. Current measured with a
vintage 0-15A analog meter I've been trying to use for a decade or so.
Voltages with a couple of DMMs. Forward and reverse polarity made no
significant difference. Stalling was with motor spindle locked and
held for a second or so (less on the potentially damaging 4-battery
test). By 'Motor unloaded' I mean with the motor running freely, but
driving its built-in gear train. I've rounded amps to 1dp.

Voltage (DC V)
Test description At source At motor Current (A)
---------------- --------- -------- -----------
1 x NiCad:
- No load 1.28
- Motor unloaded 1.18 0.96 1.8
- Stalled 0.98 0.34 4.4

2 x NiCads:
- No load 2.60
- Motor unloaded 2.35 2.12 2.1
- Stalled 1.67 0.67 8.5

3 x NiCads
- No load 3.95
- Motor unloaded 3.64 3.41 2.3
- Stalled 2.19 0.95 11.5

4 x Nicads
- No load 5.35
- Motor unloaded 4.97 4.74 2.5
- Stalled 2.77 1.26 14.1

Finished these last night, so have yet to attempt any conclusions on
how they help me model the motor. Any observations would be welcomed
please.

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Bob Monsen]

Terry Pinnell wrote:

No obvious place in the thread to post these, the latest results of my
static tests with the motor under discussion.

MISC NOTES:
No PWM involved. Tests were performed in the reverse order to the more
logical sequence shown. IOW, I started with the four-battery set, so
battery source voltage declined as nicads were discharged heavily.
Used a set of D-type NiCads, wired with taps. Current measured with a
vintage 0-15A analog meter I've been trying to use for a decade or so.
Voltages with a couple of DMMs. Forward and reverse polarity made no
significant difference. Stalling was with motor spindle locked and
held for a second or so (less on the potentially damaging 4-battery
test). By 'Motor unloaded' I mean with the motor running freely, but
driving its built-in gear train. I've rounded amps to 1dp.

Voltage (DC V)
Test description At source At motor Current (A)
---------------- --------- -------- -----------
1 x NiCad:
- No load 1.28
- Motor unloaded 1.18 0.96 1.8
- Stalled 0.98 0.34 4.4

2 x NiCads:
- No load 2.60
- Motor unloaded 2.35 2.12 2.1
- Stalled 1.67 0.67 8.5

3 x NiCads
- No load 3.95
- Motor unloaded 3.64 3.41 2.3
- Stalled 2.19 0.95 11.5

4 x Nicads
- No load 5.35
- Motor unloaded 4.97 4.74 2.5
- Stalled 2.77 1.26 14.1

Finished these last night, so have yet to attempt any conclusions on
how they help me model the motor. Any observations would be welcomed
please.

I've been playing with the motor model that was suggested by someone
upthread. However, the parameters required for the model are things like
motor inductance and resistance, a 'back emf' constant, and torque at a
particular current. I tried varying the particular values to match your
dataset, but was unsuccessful. You may want to give it a try, though. It
was here:

http://www.ecircuitcenter.com/Circuits/dc_motor_model/DCmotor_model.htm

Here is an LTSpice implementation of their model:

Version 4
SHEET 1 928 820
WIRE -384 64 -384 32
WIRE -384 176 -384 144
WIRE -224 32 -384 32
WIRE -80 32 -224 32
WIRE -80 96 -80 32
WIRE -80 208 -80 176
WIRE -80 320 -80 288
WIRE -80 432 -80 400
WIRE -80 544 -80 512
WIRE 128 80 128 32
WIRE 128 176 128 160
WIRE 208 32 128 32
WIRE 352 32 208 32
WIRE 352 64 352 32
WIRE 352 176 352 144
WIRE 352 288 352 256
WIRE 352 400 352 368
WIRE 512 64 512 48
WIRE 512 160 512 144
WIRE 512 176 512 160
WIRE 608 48 512 48
WIRE 688 48 608 48
WIRE 688 64 688 48
WIRE 688 160 512 160
WIRE 688 160 688 128
WIRE 816 48 688 48
WIRE 816 64 816 48
WIRE 816 160 688 160
WIRE 816 160 816 144
FLAG -80 544 0
FLAG -384 176 0
FLAG 352 400 0
FLAG 128 176 0
FLAG 512 176 0
FLAG 608 48 position_in_radians
FLAG -224 32 Motor_Voltage
FLAG 208 32 Torque
SYMBOL voltage -384 48 R0
WINDOW 3 -365 194 Left 0
WINDOW 123 -361 160 Left 0
WINDOW 39 24 130 Left 0
SYMATTR Value PWL(0MS 0V 1MS 10V 1000MS 10V 1010MS 0V 2000MS 0V)
SYMATTR Value2 AC 1
SYMATTR InstName V_AMP
SYMBOL ind -96 80 R0
SYMATTR InstName LM
SYMATTR Value {La}
SYMBOL res -96 192 R0
SYMATTR InstName RM
SYMATTR Value {Ra}
SYMBOL bv -80 304 R0
SYMATTR InstName B_EMF
SYMATTR Value V=I(VSENSE2)*Kemf
SYMBOL voltage -80 416 R0
WINDOW 0 43 48 Left 0
SYMATTR InstName VSENSE1
SYMATTR Value 0V
SYMBOL bv 128 64 R0
SYMATTR InstName B_TORQ
SYMATTR Value V=I(VSENSE1)*Kt
SYMBOL ind 336 48 R0
SYMATTR InstName LJ
SYMATTR Value {Jm}
SYMBOL res 336 160 R0
SYMATTR InstName RB
SYMATTR Value {Bm}
SYMBOL voltage 352 272 R0
SYMATTR InstName VSENSE2
SYMATTR Value 0V
SYMBOL bi2 512 64 R0
SYMATTR InstName B1
SYMATTR Value I=I(VSENSE2)
SYMBOL cap 672 64 R0
SYMATTR InstName CPOS
SYMATTR Value 1
SYMBOL res 800 48 R0
SYMATTR InstName RPOS
SYMATTR Value 1MEG
TEXT 24 -24 Left 0 !.param Kemf 25m
TEXT 24 -184 Left 0 !.param Ra=0.5
TEXT 24 -152 Left 0 !.param La=1.5mH
TEXT 24 -120 Left 0 !.param Kt=50m
TEXT 24 -88 Left 0 !.param Jm=250u
TEXT 24 -56 Left 0 !.param Bm=100u
TEXT 24 -224 Left 0 !.tran 10ms 2000ms
TEXT 528 16 Left 0 !.ic v(position_in_radians)=0
TEXT -296 -280 Left 0
;http://www.ecircuitcenter.com/Circuits/dc_motor_model/DCmotor_model.htm
TEXT 224 -24 Left 0 ;ratio of back EMF to angular speed of the motor
TEXT 224 -184 Left 0 ;Internal resistance of the motor
TEXT 224 -152 Left 0 ;Internal inductance of the motor
TEXT 224 -120 Left 0 ;Ratio of torque to motor current
TEXT 224 -88 Left 0 ;Inertia of motor + load
TEXT 224 -56 Left 0 ;Coulomb Friction of motor and load
TEXT 112 248 Left 0 ;current here is \nangular speed of motor

---
Regards,
Bob Monsen

 

[Terry Pinnell]

Bob Monsen <rcsurname@comcast.net> wrote:

I've been playing with the motor model that was suggested by someone
upthread. However, the parameters required for the model are things like
motor inductance and resistance, a 'back emf' constant, and torque at a
particular current. I tried varying the particular values to match your
dataset, but was unsuccessful. You may want to give it a try, though. It
was here:

http://www.ecircuitcenter.com/Circuits/dc_motor_model/DCmotor_model.htm

Here is an LTSpice implementation of their model:

<snipped very useful LTSpice circuit>

That's great, thanks. I had seen the model and article (linked by Mike
Monett) but not had time to study it. Your LTSpice circuit will make
that easier, although I'm not very skilled at working with LTSpice.
Have you now switched from Circuitmaker, BTW?

Did you also see Tony's earlier LTSpice model up-thread?

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Tony Williams]

In article <8eji91h3197k3a837hgcmuq7r63rcqovif@4ax.com>,
Terry Pinnell <terrypinDELETE@THESEdial.pipex.com> wrote:

The motor looks like this;

I Rmotor Lmotor
V-Atmotor +->----/\/\----))))---[Vback-emf]---+0v

Under steady state (dc) conditions the motor equation is.

V-Atmotor = I*Rmotor + Vback-emf.

Voltage (DC V)
Test description At source At motor Current (A)
---------------- --------- -------- -----------
- Stalled 0.98 0.34 4.4
- Stalled 1.67 0.67 8.5
- Stalled 2.19 0.95 11.5
- Stalled 2.77 1.26 14.1

When stalled, Vback-emf = Zero.

So V-Atmotor = I*Rmotor, or Rmotor = V-Atmotor/I.

So Rmotor = .34/4.4 or .67/8.5 or 0.95/11.5, or 1.26/14.1
which nicely averages out to Rmotor= 0.08 ohms.
~~~~~~~~~~~~~~~~~

- Motor unloaded 1.18 0.96 1.8 (0.816)
- Motor unloaded 2.35 2.12 2.1 (1.952)
- Motor unloaded 3.64 3.41 2.3 (3.226)
- Motor unloaded 4.97 4.74 2.5 (4.54 )

When running, the equation is V-Atmotor = I*Rmotor + Vback-emf.

eg, 0.96 = 1.8A*0.08ohms + (Vbemf at that speed, = 0.816V).

The added things in brackets are the calc'd internal Vbemfs
that the motor generates at the various rpm's.

That gearbox-only load will have a fixed current representing
the fixed friction, in parallel with a speed-dependant load,
ie, proportional to the Vbemf.

It roughly looks like Imotor = 1.7A + Vbemf/5.3ohms in the
unloaded condition. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I 0.08R Lmotor
V-Atmotor +->----/\/\----))))-+-[Vback-emf]-+-+0v
| |
Fixed loss--> +--[ 1.7A ]---+
| |
Speed-dependant loss--> +----/\/\/----+
| 5.3R |
| |
. .
. .
+---[I-Ext]---+

An external torque load will be another current in parallel
with Vbemf.

--
Tony Williams.