Re: [CAD_CAM_EDM_DRO] Re: Mariss' power supply circuit

Folks,

I've been reading these comments on braking techniques for servos/steppers,
and though I don't think any of the conversations to date have reached
sufficient current levels, one needs to watch the peak current level in
permanent magnet motors so as to not exceed the demaginazation current
level. This is the point at which the magnetic field in the motor becomes so
high that it can destroy/demagnatize even rare earth magnets. This problem
is most common in emergency stop situations where a resistor or short is
used to brake a motor.

Also note that the fastest way to stop a motor in an e-stop situation is to
use a resistance equal to the winding inpedance. watch out for note above.


John H. Berg
Berg Engineering, LLC
1133 Kensington
Grosse Pointe Park, Michigan 48230

Phone: (313) 999-1227
Fax: (313) 881-6421
E-Mail: berg1133@...

----- Original Message -----
From: "mariss92705" <mariss92705@...>
To: <CAD_CAM_EDM_DRO@yahoogroups.com>
Sent: Tuesday, June 18, 2002 5:14 PM
Subject: [CAD_CAM_EDM_DRO] Re: Mariss' power supply circuit


> Les,
>
> I think there are 3 energy sources that need to be dissipated. The
> kinetic energy you say is 25J and the capacitor energy is 72J.
> Guessing that your motor inductance is 10 mH, at 30A that would
> contribute an additional 4.5J of inductive energy.
>
> I am not sure your 25J mechanical component is a factor regarding
> capacitor voltage assuming you have a brush DC motor. It can be
> modeled as a voltage source in series with armature inductance and
> resistance. The voltage would be proportional to motor speed and can
> never exceed your power supply voltage.
>
> Adding it all up you have about 102J to get rid of. Making things
> easier, the kinetic energy will decay exponentially so it can be
> treated as an equivalent capacitor. The "capacitor" would be 102 / 72
> times 20,000 uF or 28,333 uF.
>
> Now assume you want the energy discharged in 1 second. If 3 time
> constants constitutes a discharged supply, calculate a value for R7
> from T / C or 0.333 / 0.028333, or 11.75 ohms. Call it 12 ohms as the
> closest standard value. Peak current would be 85 / 12 or 7.23A.
>
> 102 joules (or watt-seconds) have to be dissipated. A 10W wire-wound
> can tolerate 100W for 5 seconds (500 watt-seconds), so it should do
> nicely. Obviously a 6 ohm resistor would dump the energy in 0.5
> seconds at a peak current of 14.5A. That would be 50% of the IRF540
> max rated Ids of 27A. Interestingly 10W would still be OK because the
> energy is the same.
>
> Mariss
>
> --- In CAD_CAM_EDM_DRO@y..., "Les Watts" <leswatts@r...> wrote:
> > I think I see. C1 charges up through R2 linearly thus forming the
> sawtooth
> > wave portion of the modulator.
> >
> > I am building a new servo supply for my machine. The current one is
> not
> > stiff enough and causes following error estops with the high
> acceleration
> > high speed moves I want to do. I will incorporate this circuit in
> the supply
> > I think.
> >
> > I need about 30 amps cont at 85v and the 50 or so pounds of
> transformer
> > iron laying on the floor says I will get it. I am using stepdown
> isolation
> > transformers at half voltage so that iron won't be at the full flux
> density
> > but there is still plenty of room. I will nominally go from 120v to
> 60
> > so I should get 60*1.414= 84.8vdc. My amps turn off at 92v and have
> > 100 wv caps in them.
> > Since my line voltage is notoriously high I may put a few bucking
> turns
> > on the transformers. I read why power companies like higher voltage-
> > revenue goes up as the square of that voltage! Too bad they can't
> control
> > it.
> >
> > As for sizing R7 in your circuit- well lets see.... my 25 joule of
> max
> > regeneration
> > energy could attempt to pump up the 20,000uf supply cap from its
> normal
> > 72 joule @ 85v to 72+25=97 joule. That would be 98.5v. It would be
> some less
> > from losses but that is a maximum. With your circuit at 100% duty
> cycle
> > dumping
> > 25 joule in say .05 sec would require 500 watts as a linear
> approximation.
> > So the value would be about 20 ohms. I use Dale 10 ohm 25w
> wirewounds
> > to short the servos in my estop circuit and they don't even get
> warm
> > so 25w should be fine. I could use a lower value.. if I chose the
> resistor
> > current
> > to be the ps rated current it would be about 4 ohms but if I go too
> low
> > I think I will estop on every power flicker.
> >
> > As a check for power failure 3*R7*C= 1.2 seconds ; long enough. My
> estop
> > will short the servos in a sustained failure anyway.
> >
> > Les
> >
> >
> > Leslie Watts
> > L M Watts Furniture
> > Tiger, Georgia USA
> > http://www.rabun.net/~leswatts/wattsfurniturewp.html
> > engineering page:
> > http://www.rabun.net/~leswatts/shop.html
> > ----- Original Message -----
> > From: "mariss92705" <mariss92705@y...>
> > To: <CAD_CAM_EDM_DRO@y...>
> > Sent: Monday, June 17, 2002 8:09 PM
> > Subject: [CAD_CAM_EDM_DRO] Re: Mariss' power supply circuit
> >
> >
> > > Les,
> > >
> > > I appreciate you taking the time to understand the circuit. The
> PWM
> > > modulation is actually a linear function of the over-voltage seen
> by
> > > the supply. The amplitude of the 120 Hz sawtooth (R2,C1) is set to
> > > about 2V peak-to-peak, this being the "regulation" range of the
> > > circuit. The "trough" of the full-wave rectified reference clamps
> the
> > > negative excursion of the sawtooth to the difference voltage.
> > >
> > > The IRF540 has an Rds of about .07 ohms. It always operates as a
> > > switch, the current being limited by the 10 to 20W shunt resistor.
> > > The resistor must be wire-wound because wire-wound resistors have
> a
> > > superb overload capability, typically 10 times their rated wattage
> > > for 5 seconds, 50 times rated for 200mS.
> > >
> > > In an undervoltage situation the circuit would attempt to adjust
> > > itself to the new, lower voltage. It would be abley assisted by
> the
> > > load in this endevor.
> > >
> > > Like all silicon, MOSFETs invariably fail shorted if abused. The
> > > consequence would be the shunt resistor would rapidly overheat and
> > > fuse open, clearing the circuit. Everything else would fail
> > > harmlessly and would leave the basic supply uneffected.
> > >
> > > Another thing about the circuit is the component value choice for
> the
> > > +12VDC supply. The size of C3 is chosen so that the circuit
> remains
> > > functional for .5 second after AC power removal to fully discharge
> > > the supply.
> > >
> > > Thanks again for your comments; my pleasure is designing
> minimalist
> > > circuits the way others might enjoy working crossword puzzles.
> > >
> > > Mariss
> > >
> > > --- In CAD_CAM_EDM_DRO@y..., "Les Watts" <leswatts@r...> wrote:
> > > > I finally found the .pdf file and had a look.
> > > > As I understand it the circuit will perform shunt regulation
> > > > using PWM at 120 hz. Duty cycle will vary from 0-100% over a
> > > > couple volts diff between the filter cap and transformer
> secondary
> > > > peak. I think it will modulate with a raised cosine law. Right?
> > > >
> > > > Question: what is the on resistance of the Fet you selected?
> Just
> > > > want to calculate heat sink thermal resistance. Also, what
> happens
> > > > in a line power undervoltage scenario?
> > > >
> > > > And as a little FMEA thing, what would be the most likely part
> to
> > > fail?
> > > > Would it be the fet if say the primary mains went very high?
> Would
> > > it
> > > > fail open or shorted? (I currently have a floating neutral
> argument
> > > with
> > > > my Georgia Power supplier and their electrical feed)
> > > >
> > > > I want to publicly thank you for publishing this as open source.
> > > > I appreciate your weekend efforts.
> > > > These things cost about $300 from other suppliers.
> > > >
> > > > Les
> > > >
> > > > Leslie Watts
> > > > L M Watts Furniture
> > > > Tiger, Georgia USA
> > > > http://www.rabun.net/~leswatts/wattsfurniturewp.html
> > > > engineering page:
> > > > http://www.rabun.net/~leswatts/shop.html
> > > > ----- Original Message -----
> > > > From: "mariss92705" <mariss92705@y...>
> > > > To: <CAD_CAM_EDM_DRO@y...>
> > > > Sent: Sunday, June 16, 2002 3:54 PM
> > > > Subject: [CAD_CAM_EDM_DRO] Re: Power Supply Circuit
> > > >
> > > >
> > > > > Hi,
> > > > >
> > > > > OK, here's my Saturday's project. I posted a "Power supply
> > > > > circuit.pdf" in the files section of this list.
> > > > >
> > > > > It wound up being different than what I started with. Sensing
> > > current
> > > > > was too crude. Instead this circuit compares the power supply
> > > voltage
> > > > > against the xformer secondary voltage. It pulse width
> modulates
> > > the
> > > > > dump resistor and has a very nice proportional response. The
> PWM
> > > goes
> > > > > from 0% to 100% as the power supply voltage climbs to 2V above
> > > the
> > > > > nominal voltage. It has been tested and works very well.
> > > > >
> > > > > Mariss
> > >
> > >
> > > Addresses:
> > > FAQ: http://www.ktmarketing.com/faq.html
> > > FILES: http://groups.yahoo.com/group/CAD_CAM_EDM_DRO/files/
> > >
> > > OFF Topic POSTS: General Machining
> > > If you wish to post on unlimited OT subjects goto:
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> reach it if
> > you have trouble.
> > > http://www.metalworking.com/news_servers.html
> > >
> > > http://groups.yahoo.com/group/jobshophomeshop I consider this
> as a
> > sister site to the CCED group, as many of the same members are
> there, for OT
> > subjects, that are not allowed on the CCED list.
> > >
> > >
> > >
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>
>
> Addresses:
> FAQ: http://www.ktmarketing.com/faq.html
> FILES: http://groups.yahoo.com/group/CAD_CAM_EDM_DRO/files/
>
> OFF Topic POSTS: General Machining
> If you wish to post on unlimited OT subjects goto:
aol://5863:126/rec.crafts.metalworking or go thru Google.com to reach it if
you have trouble.
> http://www.metalworking.com/news_servers.html
>
> http://groups.yahoo.com/group/jobshophomeshop I consider this as a
sister site to the CCED group, as many of the same members are there, for OT
subjects, that are not allowed on the CCED list.
>
>
>
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