I am still thinking about the best way to provide power for servo
drives. As of now I went back to using 30A8 drives, which require 20
to 80VDC.

One alternative that I have is an autotransformer that was in the
Bridgeport control cabinet. This is a 3 kVa autotransformer that has
multiple taps. Its job was to take 220v and convert it to 380 volts
for the Bosch servo drives.

Being 3 kVa, it is a heavy item, weighing perhaps 50 lbs or even
more.

As I said, it has many taps, and among them specifically it has a 0v
tap, 460v tap and 180v tap.

If I connect 124 VAC that I get from the wall, to 0-460v taps, I would
then get 48 VAC on the 0-180v taps. When rectified and filtered, it
would be 68 volts DC. That's quite an acceptable voltage, providing a
bit of margin of safety.

However, it would not be isolated.

I will call Advanced Motion to confirm that 30A8 drives can be used
with non-isolated DC voltage.

I have two questions.

1) Is that somehow harmful to use non-isolated DC voltage as input for
those drives

2) What exactly does 3 kVa rating mean when applied to this
autotransformer, and how many amps can I get from the 0-180v tap in
the above described configuration.

Thanks
i

picture is here

http://igor.chudov.com/projects/Brid...ntrol-7669.jpg

On 2010-06-28, Ignoramus28517 wrote:
I am still thinking about the best way to provide power for servo
drives. As of now I went back to using 30A8 drives, which require 20
to 80VDC.

One alternative that I have is an autotransformer that was in the
Bridgeport control cabinet. This is a 3 kVa autotransformer that has
multiple taps. Its job was to take 220v and convert it to 380 volts
for the Bosch servo drives.

Being 3 kVa, it is a heavy item, weighing perhaps 50 lbs or even
more.

As I said, it has many taps, and among them specifically it has a 0v
tap, 460v tap and 180v tap.

If I connect 124 VAC that I get from the wall, to 0-460v taps, I would
then get 48 VAC on the 0-180v taps. When rectified and filtered, it
would be 68 volts DC. That's quite an acceptable voltage, providing a
bit of margin of safety.

However, it would not be isolated.

I will call Advanced Motion to confirm that 30A8 drives can be used
with non-isolated DC voltage.

I have two questions.

1) Is that somehow harmful to use non-isolated DC voltage as input for
those drives

2) What exactly does 3 kVa rating mean when applied to this
autotransformer, and how many amps can I get from the 0-180v tap in
the above described configuration.

Thanks
i

Ignoramus28517 fired this volley in
:

If I connect 124 VAC that I get from the wall, to 0-460v taps, I would
then get 48 VAC on the 0-180v taps. When rectified and filtered, it
would be 68 volts DC. That's quite an acceptable voltage, providing a
bit of margin of safety.


Iggy, providing a non-isolated supply to the drives might or might not be
harmful depending upon their local grounding. The schematics will tell
that.

However, I think you might not know that a transformer must receive a
certain minimum excitation voltage to efficiently operate. At roughly
1/4 the input at which it's designed to operate, there's a good chance
that core losses are going to go sky-high, resulting in overheating
(probably even without any load).

Because switching power supplies have become so inexpensive relative to
the cost of new iron, you'd be better off either finding a switcher rated
at the output you need, or thrash around on the surplus market for a
while to locate a transformer. Running a 460V winding on 125V most
likely won't work satisfactorily for you.

LLoyd

On Jun 28, 10:57*am, "Lloyd E. Sponenburgh"
lloydspinsidemindspring.com wrote:

However, I think you might not know that a transformer must receive a
certain minimum excitation voltage to efficiently operate. *At roughly
1/4 the input at which it's designed to operate, there's a good chance
that core losses are going to go sky-high, resulting in overheating
(probably even without any load).


LLoyd

I think you have this backwards. At 1/4 the input voltage the core
losses will be much less than at full voltage.


Dan

" fired this volley in news:ad72df97-
:

I think you have this backwards. At 1/4 the input voltage the core
losses will be much less than at full voltage.


Not below the minimum required excitation, Dan.

LLoyd

On Jun 28, 12:22*pm, "Lloyd E. Sponenburgh"
lloydspinsidemindspring.com wrote:
" fired this volley in news:ad72df97-
:

I think you have this backwards. *At 1/4 the input voltage the core
losses will be much less than at full voltage.

Not below the minimum required excitation, Dan.

LLoyd

Okay, if the core losses are lower at full voltage, and I assume zero
at no voltage in, then at what voltage is the core losses a maximum?

Dan

On Mon, 28 Jun 2010 11:22:46 -0500, "Lloyd E. Sponenburgh"
lloydspinsidemindspring.com wrote:

" fired this volley in news:ad72df97-
:

I think you have this backwards. At 1/4 the input voltage the core
losses will be much less than at full voltage.


Not below the minimum required excitation, Dan.

LLoyd


You might want to have another look at some hysteresis curves!


Regards
Mark Rand
RTFM

On Jun 28, 12:22*pm, "Lloyd E. Sponenburgh"
lloydspinsidemindspring.com wrote:
" fired this volley in news:ad72df97-

I think you have this backwards. *At 1/4 the input voltage the core
losses will be much less than at full voltage.

Not below the minimum required excitation, Dan.

LLoyd

Lloyd, have you ever tested and measured transformers?

jsw

Lloyd E. Sponenburgh wrote:
" fired this volley in news:ad72df97-
:


I think you have this backwards. At 1/4 the input voltage the core
losses will be much less than at full voltage.



Not below the minimum required excitation, Dan.

Can you define this "minimum required excitation"? I've never heard of
such a thing.
Induction MOTORS definitely do have such a thing, the excitation must be
enough to develop the field in the rotor, or it llos like a transformer
with a dead short on the output.

But, with transformers, the core flux is proportional to the applied
voltage X time, and opposes the magnetizing current in the primary windings.
This may not apply to substation-class transformers, but should be
perfectly true of a 3 KVA autotransformer.

Jon

Lloyd E. Sponenburgh wrote:

However, I think you might not know that a transformer must receive a
certain minimum excitation voltage to efficiently operate. At roughly
1/4 the input at which it's designed to operate, there's a good chance
that core losses are going to go sky-high, resulting in overheating
(probably even without any load).


Lloyd is obviously not an electrical engineer. Core losses are almost
perfectly proportional to excitation X frequency.
So, there is NO WAY underexciting a transformer will overheat it. Try
it if you have any doubts, but he's just wrong.

You WILL lose power rating proportional to the voltage reduction, so a
480 V 3 KVA transformer will only provide
about 775 VA output on 124 V.

Jon

That's huge. Almost exactly what I use in my Hardinge CHNC. Don't need
isolation, just separate circuit and keep wiring as far as possilbe from
rest of control.

here's a tutorial on DC supplies:

http://www.kpsec.freeuk.com/powersup.htm

Karl

On 2010-06-28, Karl Townsend wrote:
That's huge. Almost exactly what I use in my Hardinge CHNC. Don't need
isolation, just separate circuit and keep wiring as far as possilbe from
rest of control.

You mean the transformer is huge?

And you think that it will work for the intended application?

here's a tutorial on DC supplies:

http://www.kpsec.freeuk.com/powersup.htm

I will read that right now.

Thanks

i

"Ignoramus28517" wrote in message
...
On 2010-06-28, Karl Townsend wrote:
That's huge. Almost exactly what I use in my Hardinge CHNC. Don't need
isolation, just separate circuit and keep wiring as far as possilbe from
rest of control.

You mean the transformer is huge?

And you think that it will work for the intended application?

here's a tutorial on DC supplies:

http://www.kpsec.freeuk.com/powersup.htm

I will read that right now.

Thanks

i

I just buy 2:1 transformers and step 110 down by 1/2 for the A8 series of
AMC drives, then a bridge recticifier and an elctrolytic capacitor. My mill
uses a 1KVA, that's 1000 watts. I used an old 5 KVA on the hardinge. Divide
by voltage to get amps and allow for some loss. I'm sure a EE type will
chime in and give the exact formuals.

These things are real simple to build. Even I can do it.

Karl

On Jun 28, 11:10*am, "Karl Townsend"
wrote:
That's huge. Almost exactly what I use in my Hardinge CHNC. Don't need
isolation, just separate circuit and keep wiring as far as possilbe from
rest of control.

here's a tutorial on DC supplies:

http://www.kpsec.freeuk.com/powersup.htm

Karl

Industrial controls are a little different from standard electronics
practice, they have added problems with long noisy wire runs, ground
loops and some idiot spearing the control box with a forklift, then
stepping off into a puddle of coolant.

jsw

On 2010-06-28, Jim Wilkins wrote:
On Jun 28, 11:10?am, "Karl Townsend"
wrote:
That's huge. Almost exactly what I use in my Hardinge CHNC. Don't need
isolation, just separate circuit and keep wiring as far as possilbe from
rest of control.

here's a tutorial on DC supplies:

http://www.kpsec.freeuk.com/powersup.htm

Karl

Industrial controls are a little different from standard electronics
practice, they have added problems with long noisy wire runs, ground
loops and some idiot spearing the control box with a forklift, then
stepping off into a puddle of coolant.

Well, at home I may have an idiot, but no forklift.

i

On Jun 28, 11:39*am, Ignoramus28517 ignoramus28...@NOSPAM.
28517.invalid wrote:
...

Industrial controls are a little different from standard electronics
practice, they have added problems with long noisy wire runs, ground
loops and some idiot spearing the control box with a forklift, then
stepping off into a puddle of coolant.

Well, at home I may have an idiot, but no forklift.

i

I meant that as an example of the different rationales for the design
rules.

BTW I got to keep the damaged Variacs from a similar forklift
incident, and shuffled their parts into a few good ones which I built
into power supplies. The cores roll off too fast with frequency to
make subwoofer crossovers.

jsw

I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

i

Ignoramus28517 fired this volley in
:

The cost is $168 and I am sure that it will be money well
spent.


Indeed. Good on ya'.

LLoyd

On 06/28/2010 09:50 AM, Ignoramus28517 wrote:
I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

To make 20-80VDC for your drive? That should be good. What current do
the drives pull? You need to size your caps so that the instantaneous
voltage to the drive never falls below 20V -- I calculate from the peak
voltage to 20V in 1/120th of a second, assuming maximum line sag
(usually 100V if it's a 120V line). You really have more than 120th of
a second, but that's an easy number to remember and it's conservative.

A surplus place may have something for less money; since it's a one-off
you won't care if it's not a production part. I like Surplus Sales of
Nebraska (http://www.surplussales.com) and Herbach & Rademan
(http://www.herbach.com). I've never had a problem ordering from them.
I took a quick look expecting to find joy and they may not have
exactly what you want -- but it's not a bad idea to look.

On 2010-06-28, Tim Wescott wrote:
On 06/28/2010 09:50 AM, Ignoramus28517 wrote:
I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

To make 20-80VDC for your drive? That should be good.

Yes, 20A at 80 VDC. Drives are limited to 20-80 VDC. (they refuse to
work if voltage is outside this spec)

What current do the drives pull?

15A per drive continuous, 30A per drive "peak".

You need to size your caps so that the instantaneous voltage to the
drive never falls below 20V -- I calculate from the peak voltage to
20V in 1/120th of a second, assuming maximum line sag (usually 100V
if it's a 120V line). You really have more than 120th of a second,
but that's an easy number to remember and it's conservative.

I will have 20,000 uF.

A surplus place may have something for less money; since it's a one-off
you won't care if it's not a production part. I like Surplus Sales of
Nebraska (http://www.surplussales.com) and Herbach & Rademan
(http://www.herbach.com). I've never had a problem ordering from them.
I took a quick look expecting to find joy and they may not have
exactly what you want -- but it's not a bad idea to look.

It does not look so.

A lot of used toroidals (and transformers in general) are sold with
bundles of wire stocking out, with no clue as to which are primary,
secondary etc.

I will just get one at DigiKey and will get on with my life. Hopefully
by the end of this week, I will have a working power supply and will
not need to worry about it ever again.

i

On 06/28/2010 10:45 AM, Ignoramus28517 wrote:
On 2010-06-28, Tim wrote:
On 06/28/2010 09:50 AM, Ignoramus28517 wrote:
I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

To make 20-80VDC for your drive? That should be good.

Yes, 20A at 80 VDC. Drives are limited to 20-80 VDC. (they refuse to
work if voltage is outside this spec)

What current do the drives pull?

15A per drive continuous, 30A per drive "peak".

You need to size your caps so that the instantaneous voltage to the
drive never falls below 20V -- I calculate from the peak voltage to
20V in 1/120th of a second, assuming maximum line sag (usually 100V
if it's a 120V line). You really have more than 120th of a second,
but that's an easy number to remember and it's conservative.

I will have 20,000 uF.

A surplus place may have something for less money; since it's a one-off
you won't care if it's not a production part. I like Surplus Sales of
Nebraska (http://www.surplussales.com) and Herbach& Rademan
(http://www.herbach.com). I've never had a problem ordering from them.
I took a quick look expecting to find joy and they may not have
exactly what you want -- but it's not a bad idea to look.

It does not look so.

A lot of used toroidals (and transformers in general) are sold with
bundles of wire stocking out, with no clue as to which are primary,
secondary etc.

I will just get one at DigiKey and will get on with my life. Hopefully
by the end of this week, I will have a working power supply and will
not need to worry about it ever again.

i

Since this actually touches on things I do for work and it's specific, I
feel compelled to add a disclaimer: burn down your shop, destroy your
neighborhood, turn you and your family into crispy critters and it's
_not my fault!_. (I.e., this is free advice and worth every penny of it).

The voltage will ramp down proportional to current and inversely
proportional to capacitance. At 30A, that's (30A)/(20000uF) =
1500V/sec. In 1/120th of a second, the voltage will drop by 12.5V.

With 100V into the transformer, you'll get (50V)(100V)/(120V) =
41.6VRMS, or 58.3V peak to the rectifier. Assuming a 1.4V drop (full
bridge) that's 57V peak (with rounding up) to the cap. Drop that by
12.5 and you're down to 44.5V at the trough of the wave.

Check the current capabilities of the caps -- 30A out means lots more
than 30A in when the rectifiers are actually conducting. Ditto the
rectifier maximum instantaneous current -- it'll be much higher than 30A.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Mon, 28 Jun 2010 11:50:44 -0500, Ignoramus28517
wrote:

I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

i


Thats pretty much what an OmniTurn CNC lathe uses.

Check on the Left side of the control
http://picasaweb.google.com/gunneras...84689516471026

Gunner



One could not be a successful Leftwinger without realizing that,
in contrast to the popular conception supported by newspapers
and mothers of Leftwingers, a goodly number of Leftwingers are
not only narrow-minded and dull, but also just stupid.
Gunner Asch

On 2010-06-28, Gunner Asch wrote:
On Mon, 28 Jun 2010 11:50:44 -0500, Ignoramus28517
wrote:

I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

i


Thats pretty much what an OmniTurn CNC lathe uses.

Check on the Left side of the control
http://picasaweb.google.com/gunneras...84689516471026


Looks good.

i

On Mon, 28 Jun 2010 11:50:44 -0500, Ignoramus28517
wrote:

I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

i

If you full-wave rectify and filter it with just a capacitor, it will
give you more like 12A at 60-70VDC without overheating.

You could also consider something like this (probably overkill)
http://cgi.ebay.com/Kikusui-55-20L-D...item33576042e4

On 2010-06-28, Spehro Pefhany wrote:
On Mon, 28 Jun 2010 11:50:44 -0500, Ignoramus28517
wrote:

I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

i

If you full-wave rectify and filter it with just a capacitor, it will
give you more like 12A at 60-70VDC without overheating.

Why would it not give me 20A?

i

You could also consider something like this (probably overkill)
http://cgi.ebay.com/Kikusui-55-20L-D...item33576042e4

On Mon, 28 Jun 2010 13:37:53 -0500, Ignoramus28517
wrote:

On 2010-06-28, Spehro Pefhany wrote:
On Mon, 28 Jun 2010 11:50:44 -0500, Ignoramus28517
wrote:

I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

i

If you full-wave rectify and filter it with just a capacitor, it will
give you more like 12A at 60-70VDC without overheating.

Why would it not give me 20A?

i

Because the current it draws from the supply is not sinusoidal- it
will be drawn in pulses at the peak of the sine wave, so the I^2*R
heating of the windings is higher than for a sinusoid.

Here is a decent application note on the matter-- I use 1.7 as a rule
of thumb, the document says 1.6~1.8 so we are in complete agreement-
see the bottomline in the table on the last page.

http://www.audiofaidate.org/it/mater...terRatings.pdf

Note that the output _power_ will be only 20-30% less than 1000W,
because the filtered voltage is higher than the unfiltered.


You could also consider something like this (probably overkill)
http://cgi.ebay.com/Kikusui-55-20L-D...5V-0-20A-1100W


Actually, now that I see your actual requirements of 30A peak per
axis, a power supply of this type would definitely NOT be a good idea-
it will current-limit at the rated current and your drive will stop
working, whereas a xfmr + filter will only be limited by the circuit
breaker you put in the primary.

BTW, toroidal transformers draw absolutely wicked primary surges on
power up sometimes (depends on their magnetization state when they
were shut off and where in the cycle it's turned on), so you may want
to put a slow circuit breaker on the primary and a fuse on the
secondary after the filter cap.

On 06/28/2010 11:37 AM, Ignoramus28517 wrote:
On 2010-06-28, Spehro wrote:
On Mon, 28 Jun 2010 11:50:44 -0500, Ignoramus28517
wrote:

I have decided that I have enough problems already and do not want to
create any more. I will buy a toroidal 1000 VA 50v transformer at
Digikey. It converts 115 VAC into 50 volts, or 123 VAC (my typical
voltage) to 57 VAC, which is basically perfect for me. It has
isolation and I can wire it to share the same ground as signals,
etc. The cost is $168 and I am sure that it will be money well
spent.

It produces 20 amps at 50 VAC.

i

If you full-wave rectify and filter it with just a capacitor, it will
give you more like 12A at 60-70VDC without overheating.

Why would it not give me 20A?

Because 20A out of the DC supply is way more than 20A RMS out of the
transformer. You're pulling current out of the transformer for maybe
1/400th of a second at a time (whatever a 37 degree conduction angle is
at 60Hz), which means that to hit an average of 20A you'd have to be
pulling much more than 20ARMS.

The transformer current rating is there to keep the coils from
dissipating too much power, and virtually all of that is I^2*R losses,
which is RMS current.

Decrease the cap values and it'll be better (at the cost of more
ripple), but you'll never get it down to 1:1.

Cadillac supplies of this type use a choke-input filter; i.e. the
rectifiers feed a honking big coil that then feeds a cap. They have far
better voltage regulation as long as you're pulling some minimum current
at the cost of buying that honking big coil and lugging it around, and
at the cost of the output voltage being about 90% of the transformer's
output RMS voltage, less the rectifier drop.

Use a choke-input supply, and the RMS transformer current will be almost
exactly 100% of the DC output current, instead of way more.


choke
___ DC out
o----. ,------o---|-----o----UUU-----o-------o
AC in )|( | | |
)|( .--)---|-----' |
o----' '---o | ---
| '---|--. --- cap
| | |
'------|--o |
| |
| |
=== ===
GND GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

On 06/28/2010 07:28 AM, Ignoramus28517 wrote:
I am still thinking about the best way to provide power for servo
drives. As of now I went back to using 30A8 drives, which require 20
to 80VDC.

One alternative that I have is an autotransformer that was in the
Bridgeport control cabinet. This is a 3 kVa autotransformer that has
multiple taps. Its job was to take 220v and convert it to 380 volts
for the Bosch servo drives.

Being 3 kVa, it is a heavy item, weighing perhaps 50 lbs or even
more.
(snip)

2) What exactly does 3 kVa rating mean when applied to this
autotransformer, and how many amps can I get from the 0-180v tap in
the above described configuration.

You've already bagged the idea, but to answer your question:

The kVA rating of a piece of AC gear is the maximum design RMS voltage
(i.e. 220V) times the maximum design RMS current (i.e. 14A), when it's
hooked up normally.

The reason that it's a "kVA" rating and not "watts" is because when you
are driving reactive loads (i.e. inductive or capacitive) you can have
lots of current without any real power being consumed -- yet in an
electrical machine, much of the loss (and hence heat build up) is caused
not by the power going through it, but by the applied voltage causing
core losses, and the current through the coils causing I^2*R losses.

Had you hooked it up backwards as you had planned, you would have had to
assume that the wire up to the 380V terminal was lighter than the wire
up to the 220V terminal, and derated that 3kVA rating. The safe rating
would have been whatever the current would have been out of that 380V
terminal -- a PE with experience with that sort of hacking might be able
to figure out a higher current rating than that, but would have to take
care that the transformer wouldn't experience spot over heating.

Ignoramus28517 wrote:

I am still thinking about the best way to provide power for servo
drives. As of now I went back to using 30A8 drives, which require 20
to 80VDC.

I'm putting my $5 that you will be scrapping the underpowered drives and
power supply and replacing them with items that will power the servos at
the same levels the Bridgeport engineers spec'd them for immediately
after you attempt your first project requiring reasonably heavy cuts.

On 2010-06-28, Pete C. wrote:

Ignoramus28517 wrote:

I am still thinking about the best way to provide power for servo
drives. As of now I went back to using 30A8 drives, which require 20
to 80VDC.

I'm putting my $5 that you will be scrapping the underpowered drives and
power supply and replacing them with items that will power the servos at
the same levels the Bridgeport engineers spec'd them for immediately
after you attempt your first project requiring reasonably heavy cuts.

I will be the first to let you know that.

The motors are rated at 30 amps peak, 15 amps continuous, 145 VDC. At
that current they develop stall torque 3.2 n-m. or 2.4
ft-lbs. Converted to force 2.4*2*12*10*0.8 = 460 lbs of force.

The speed will be about 2x less.

These drives do not lose out on the torque/force front much and the speed
would be about 55% of the max speed derived from motor nameplate.

It is also not a given that the original Bosch drives were capable of
delivering nameplate torque and speed to the motors.

i

Pete C. wrote:
Ignoramus28517 wrote:

I am still thinking about the best way to provide power for servo
drives. As of now I went back to using 30A8 drives, which require 20
to 80VDC.


I'm putting my $5 that you will be scrapping the underpowered drives and
power supply and replacing them with items that will power the servos at
the same levels the Bridgeport engineers spec'd them for immediately
after you attempt your first project requiring reasonably heavy cuts.

Well, maybe. I have very underpowered 1/8 Hp continuous motors on my
Series I Bridgeport.
Yes, it rounds off corners a bit when trying to make a sharp corner at
higher speed. I really don't mind that, it rarely can even be detected.
I calculated that neglecting losses, I can deliver about 750 Lbs of
linear force to the table. It is able to snap off a 3/8" end mill
without the servo amps faulting or getting a following error. (That's
why I now do setup at lower speeds - the machine is faster than my eyes
can detect interference.)

Iggy mentioned 500 - 600 Lbs force, that might be iffy on a Series II.

Jon

On 2010-06-29, Jon Elson wrote:
Pete C. wrote:
Ignoramus28517 wrote:

I am still thinking about the best way to provide power for servo
drives. As of now I went back to using 30A8 drives, which require 20
to 80VDC.


I'm putting my $5 that you will be scrapping the underpowered drives and
power supply and replacing them with items that will power the servos at
the same levels the Bridgeport engineers spec'd them for immediately
after you attempt your first project requiring reasonably heavy cuts.

Well, maybe. I have very underpowered 1/8 Hp continuous motors on my
Series I Bridgeport.
Yes, it rounds off corners a bit when trying to make a sharp corner at
higher speed. I really don't mind that, it rarely can even be detected.
I calculated that neglecting losses, I can deliver about 750 Lbs of
linear force to the table. It is able to snap off a 3/8" end mill
without the servo amps faulting or getting a following error. (That's
why I now do setup at lower speeds - the machine is faster than my eyes
can detect interference.)

Iggy mentioned 500 - 600 Lbs force, that might be iffy on a Series II.

I may be mistaken in my calculation. Let me redo. Stall torque of the
motor is 3.2 N-m = 2.5 ft-lbs. 2:1 reduction gives me 5 ft-lbs on the
ball screw.

T=F*l/(2pi*nu)

T is torque, L is lead (ratio) of the screw, I think 0.1, nu is efficiency. Say nu
is 0.8.

F = T*2*pi*nu/l

If ballscrew is 1 inches in diameter, we get

F = 5*2*3.14*0.8/0.1 = 251 lbs.

Now, I contend that 300 lbs is a lot of force when it comes to milling.
It is far more than I am able to exert on my manual Bridgeport.

I may have an error in my calculation. I just did an experiment. I ran
my X table at 30 volts. It moved at appx. 1 inch per second at 1.45
amps.

I pushed against the table fairly hard (being somewhat space
constrained). Maybe I exerted 70 lbs force. The table did not slow
down at all (of course) and the amps rose from 1.45 to something like
1.67.

i

On Jun 29, 1:14*am, Ignoramus28517 ignoramus28...@NOSPAM.
28517.invalid wrote:
....
I pushed against the table fairly hard (being somewhat space
constrained). Maybe I exerted 70 lbs force. The table did not slow
down at all (of course) and the amps rose from 1.45 to something like
1.67.
i

At the maximum handle force I would apply to a 1/2" end mill, a spring
scale between the spindle and table reads 60 lbs.

jsw

On 2010-06-29, Jim Wilkins wrote:
On Jun 29, 1:14?am, Ignoramus28517 ignoramus28...@NOSPAM.
28517.invalid wrote:
....
I pushed against the table fairly hard (being somewhat space
constrained). Maybe I exerted 70 lbs force. The table did not slow
down at all (of course) and the amps rose from 1.45 to something like
1.67.
i

At the maximum handle force I would apply to a 1/2" end mill, a spring
scale between the spindle and table reads 60 lbs.

Makes perfect sense to me.

i

Ignoramus28517 wrote:

I may be mistaken in my calculation. Let me redo. Stall torque of the
motor is 3.2 N-m = 2.5 ft-lbs. 2:1 reduction gives me 5 ft-lbs on the
ball screw.

T=F*l/(2pi*nu)

T is torque, L is lead (ratio) of the screw, I think 0.1, nu is efficiency. Say nu
is 0.8.

F = T*2*pi*nu/l

If ballscrew is 1 inches in diameter, we get

F = 5*2*3.14*0.8/0.1 = 251 lbs.

These calculations based on slope of the screw thread make my head hurt.
A leadscrew can be made equvalent to a drum with a rope around it.
For a .200" lead (5 TPI) the equivalent drum is 0.0318" radius. (This
assumes the rope is infinitely thin.) Then you can convert torque to
linear force very easily. So, 2.5 Ft-Lbs from the motor is 2.5 * 16 =
40 In-Lbs. With a 2:1 belt reduction, that is 80 In_Lbs at the
leadscrew. 80/0.0318 = 2513 linear pounds force on the table.
I **think** I have done this calculation correctly, I have been using
this scheme for a decade.

Jon

On 2010-06-29, Jon Elson wrote:
Ignoramus28517 wrote:

I may be mistaken in my calculation. Let me redo. Stall torque of the
motor is 3.2 N-m = 2.5 ft-lbs. 2:1 reduction gives me 5 ft-lbs on the
ball screw.

T=F*l/(2pi*nu)

T is torque, L is lead (ratio) of the screw, I think 0.1, nu is efficiency. Say nu
is 0.8.

F = T*2*pi*nu/l

If ballscrew is 1 inches in diameter, we get

F = 5*2*3.14*0.8/0.1 = 251 lbs.

These calculations based on slope of the screw thread make my head hurt.
A leadscrew can be made equvalent to a drum with a rope around it.
For a .200" lead (5 TPI) the equivalent drum is 0.0318" radius. (This
assumes the rope is infinitely thin.) Then you can convert torque to
linear force very easily. So, 2.5 Ft-Lbs from the motor is 2.5 * 16 =
40 In-Lbs. With a 2:1 belt reduction, that is 80 In_Lbs at the
leadscrew. 80/0.0318 = 2513 linear pounds force on the table.
I **think** I have done this calculation correctly, I have been using
this scheme for a decade.

Jon, your calculation makes some intuitive sense. Yesterday at night,
I realized that I forgot 2*pi in my calculation, which would then
yield a number similar to yours.

i