Stepper motor question

[thermo102]

I have a 6-wire motor which I plan to operate as a 5-wire motor. As such
I've found the two center tap wires and will treat them as a single wire
going to a supply voltage of 40.5 VDC. Measuring the resistance of the
coils gives me 2.2 ohms to each of the remaining wires.

The motor itself is rated for 2 A. Using i = V - (Rated A * motor R) /Rated
R yields a result of 18 ohms for a current limiting resistor.

The closest I can find (would like a power resistor) is 15 ohms which would
give 2.35 A to the motor. (17% over)

Will this overheat the motor?

The motor is on a desktop CNC machine run by a unipolar driver (???) and is
for the Z axis. Best guess is running short periods of time (30 seconds
max, but usually only a couple of seconds).

Or, can someone suggest a source for a 18 ohm resistor? (To replace an
Ohmite TCH35P7R0J resistor).

Thanks in advance,
Thermo

 

[John Fields]

On Fri, 26 Feb 2010 14:44:17 -0700, "thermo102" <thermo102@comcast.net>
wrote:

I have a 6-wire motor which I plan to operate as a 5-wire motor. As such
I've found the two center tap wires and will treat them as a single wire
going to a supply voltage of 40.5 VDC. Measuring the resistance of the
coils gives me 2.2 ohms to each of the remaining wires.

The motor itself is rated for 2 A. Using i = V - (Rated A * motor R) /Rated
R yields a result of 18 ohms for a current limiting resistor.

The closest I can find (would like a power resistor) is 15 ohms which would
give 2.35 A to the motor. (17% over)

Will this overheat the motor?

The motor is on a desktop CNC machine run by a unipolar driver (???) and is
for the Z axis. Best guess is running short periods of time (30 seconds
max, but usually only a couple of seconds).

Or, can someone suggest a source for a 18 ohm resistor? (To replace an
Ohmite TCH35P7R0J resistor).

---
That's a 7 ohm 35 watt resistor, so if it was part of the old circuit
there might be more going on than meets the eye and an 18 ohm resistor
might not be what you want.

What more can you tell us about how you'll be driving the stepper?

JF

 

[thermo102]

"John Fields" <jfields@austininstruments.com> wrote in message
news:f0tgo5ttla4mai2fbebu9ip1utn5i3fftu@4ax.com...

On Fri, 26 Feb 2010 14:44:17 -0700, "thermo102" <thermo102@comcast.net
wrote:

I have a 6-wire motor which I plan to operate as a 5-wire motor. As such
I've found the two center tap wires and will treat them as a single wire
going to a supply voltage of 40.5 VDC. Measuring the resistance of the
coils gives me 2.2 ohms to each of the remaining wires.

The motor itself is rated for 2 A. Using i = V - (Rated A * motor R)
/Rated
R yields a result of 18 ohms for a current limiting resistor.

The closest I can find (would like a power resistor) is 15 ohms which
would
give 2.35 A to the motor. (17% over)

Will this overheat the motor?

The motor is on a desktop CNC machine run by a unipolar driver (???) and
is
for the Z axis. Best guess is running short periods of time (30 seconds
max, but usually only a couple of seconds).

Or, can someone suggest a source for a 18 ohm resistor? (To replace an
Ohmite TCH35P7R0J resistor).

---
That's a 7 ohm 35 watt resistor, so if it was part of the old circuit
there might be more going on than meets the eye and an 18 ohm resistor
might not be what you want.

What more can you tell us about how you'll be driving the stepper?

JF

Guess I left out most important part. This motor is to replace another
motor which has 133 ohms arcross the coils. I'm pretty sure the 18 ohm value
is what I need for the new motor, but can't seem to find one of that value.

About 30 years ago, I was told by an engineer that most electrical stuff had
a fudge factor of about 10% or more built into them. Is this more or less
true today also? If so, then I can probably use the 15 ohm resistor without
burning up the motor.

Thermo

 

[John Fields]

On Sat, 27 Feb 2010 09:55:03 -0700, "thermo102" <thermo102@comcast.net>
wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:f0tgo5ttla4mai2fbebu9ip1utn5i3fftu@4ax.com...
On Fri, 26 Feb 2010 14:44:17 -0700, "thermo102" <thermo102@comcast.net
wrote:

I have a 6-wire motor which I plan to operate as a 5-wire motor. As such
I've found the two center tap wires and will treat them as a single wire
going to a supply voltage of 40.5 VDC. Measuring the resistance of the
coils gives me 2.2 ohms to each of the remaining wires.

The motor itself is rated for 2 A. Using i = V - (Rated A * motor R)
/Rated
R yields a result of 18 ohms for a current limiting resistor.

The closest I can find (would like a power resistor) is 15 ohms which
would
give 2.35 A to the motor. (17% over)

Will this overheat the motor?

The motor is on a desktop CNC machine run by a unipolar driver (???) and
is
for the Z axis. Best guess is running short periods of time (30 seconds
max, but usually only a couple of seconds).

Or, can someone suggest a source for a 18 ohm resistor? (To replace an
Ohmite TCH35P7R0J resistor).

---
That's a 7 ohm 35 watt resistor, so if it was part of the old circuit
there might be more going on than meets the eye and an 18 ohm resistor
might not be what you want.

What more can you tell us about how you'll be driving the stepper?

JF

Guess I left out most important part. This motor is to replace another
motor which has 133 ohms arcross the coils. I'm pretty sure the 18 ohm value
is what I need for the new motor, but can't seem to find one of that value.

About 30 years ago, I was told by an engineer that most electrical stuff had
a fudge factor of about 10% or more built into them. Is this more or less
true today also? If so, then I can probably use the 15 ohm resistor without
burning up the motor.

---
Dunno.

Do you have a schematic and, again, what more can you tell us about how
you'll be driving the stepper?

http://www.mouser.com/Passive-Components/Resistors/Power-Resistors-Wire-Wound/_/N-2grd3?P=1z0yllmZ1z0x6z1Z1z0vl7s

JF

 

[thermo102]

"John Fields" <jfields@austininstruments.com> wrote in message
news:uodlo5ddejh1fsr6kugsctdchjcgb4odnv@4ax.com...

On Sat, 27 Feb 2010 09:55:03 -0700, "thermo102" <thermo102@comcast.net
wrote:


"John Fields" <jfields@austininstruments.com> wrote in message
news:f0tgo5ttla4mai2fbebu9ip1utn5i3fftu@4ax.com...
On Fri, 26 Feb 2010 14:44:17 -0700, "thermo102" <thermo102@comcast.net
wrote:

I have a 6-wire motor which I plan to operate as a 5-wire motor. As
such
I've found the two center tap wires and will treat them as a single wire
going to a supply voltage of 40.5 VDC. Measuring the resistance of the
coils gives me 2.2 ohms to each of the remaining wires.

The motor itself is rated for 2 A. Using i = V - (Rated A * motor R)
/Rated
R yields a result of 18 ohms for a current limiting resistor.

The closest I can find (would like a power resistor) is 15 ohms which
would
give 2.35 A to the motor. (17% over)

Will this overheat the motor?

The motor is on a desktop CNC machine run by a unipolar driver (???) and
is
for the Z axis. Best guess is running short periods of time (30 seconds
max, but usually only a couple of seconds).

Or, can someone suggest a source for a 18 ohm resistor? (To replace an
Ohmite TCH35P7R0J resistor).

---
That's a 7 ohm 35 watt resistor, so if it was part of the old circuit
there might be more going on than meets the eye and an 18 ohm resistor
might not be what you want.

What more can you tell us about how you'll be driving the stepper?

JF

Guess I left out most important part. This motor is to replace another
motor which has 133 ohms arcross the coils. I'm pretty sure the 18 ohm
value
is what I need for the new motor, but can't seem to find one of that
value.

About 30 years ago, I was told by an engineer that most electrical stuff
had
a fudge factor of about 10% or more built into them. Is this more or less
true today also? If so, then I can probably use the 15 ohm resistor
without
burning up the motor.

---
Dunno.

Do you have a schematic and, again, what more can you tell us about how
you'll be driving the stepper?

http://www.mouser.com/Passive-Components/Resistors/Power-Resistors-Wire-Wound/_/N-2grd3?P=1z0yllmZ1z0x6z1Z1z0vl7s

JF

Since my last post, I downloaded mouser's catalogue. Took my time to find
the page with the ohmite resistors and printed it. Then I found an on-line
calculator for resisters in parallel. Trial and error gave me the
following:

A 24 ohm and a 75 ohm in parallel will give me 18.18 effective ohms. That's
close enough for me. I was hoping for a simple change out, but this will
work. I plan on sandwiching the two resistors back to back on a homemade
heatsink and letting the leads of each one share the spot where the single
resistor was previously. (Which is making them parallel, right?)

To answer your question, I don't have a schematic, and the stepper(s) are
being driven via a PC parallel port.

I'll order the resistors tomorrow and see what happens. Wish me luck!

Thanks,
Thermo

 

[John Fields]

On Sun, 28 Feb 2010 12:54:48 -0700, "thermo102" <thermo102@comcast.net>
wrote:

Since my last post, I downloaded mouser's catalogue. Took my time to find
the page with the ohmite resistors and printed it. Then I found an on-line
calculator for resisters in parallel. Trial and error gave me the
following:

A 24 ohm and a 75 ohm in parallel will give me 18.18 effective ohms. That's
close enough for me. I was hoping for a simple change out, but this will
work. I plan on sandwiching the two resistors back to back on a homemade
heatsink and letting the leads of each one share the spot where the single
resistor was previously. (Which is making them parallel, right?)

---
Right.
---

To answer your question, I don't have a schematic, and the stepper(s) are
being driven via a PC parallel port.

I'll order the resistors tomorrow and see what happens. Wish me luck!

---
Good luck!

JF

 

[JeffM]

thermo102 wrote:

A 24 ohm and a 75 ohm in parallel will give me 18.18 effective ohms.

18 ohms is a standard value.

36 ohms is a standard value
(and will divide the power dissipation equally).

....and it's not at all necessary to blockquote 4-deep.
While *some* context is advisable,
trimming **much** of the previous text is advised.

 

[Jasen Betts]

On 2010-02-28, thermo102 <thermo102@comcast.net> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
On Sat, 27 Feb 2010 09:55:03 -0700, "thermo102" <thermo102@comcast.net
"John Fields" <jfields@austininstruments.com> wrote in message
On Fri, 26 Feb 2010 14:44:17 -0700, "thermo102" <thermo102@comcast.net
I have a 6-wire motor which I plan to operate as a 5-wire motor. As

What more can you tell us about how you'll be driving the stepper?
Guess I left out most important part. This motor is to replace another
Do you have a schematic and, again, what more can you tell us about how
you'll be driving the stepper?
To answer your question, I don't have a schematic, and the stepper(s) are
being driven via a PC parallel port.

pc parallel port can't produce enough drive to turn your stepper motor.
what's between ground and the legs of the motor?

Darlington transistors will drop about 0.7v

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[Jasen Betts]

On 2010-02-28, thermo102 <thermo102@comcast.net> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:uodlo5ddejh1fsr6kugsctdchjcgb4odnv@4ax.com...
On Sat, 27 Feb 2010 09:55:03 -0700, "thermo102" <thermo102@comcast.net
wrote:

The closest I can find (would like a power resistor) is 15 ohms which
would
give 2.35 A to the motor. (17% over)

Will this overheat the motor?

how close will this 82W heater be to the motor?

it realy depends on how hot the ambient temperature is.

have you considered PWM, or using a lower supply voltage?

Best guess is running short periods of time (30 seconds
max, but usually only a couple of seconds).

the motor is probably on even when stopped.


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[John Fields]

On 6 Mar 2010 02:44:56 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

have you considered PWM,

---
Have you read the subject line?


JF

 

[Jasen Betts]

On 2010-03-06, John Fields <jfields@austininstruments.com> wrote:

On 6 Mar 2010 02:44:56 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:


have you considered PWM,

---
Have you read the subject line?

are you saying that PWM can't be used to limit the current in stepper
motors?

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[John Fields]

On 8 Mar 2010 08:32:59 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:

On 2010-03-06, John Fields <jfields@austininstruments.com> wrote:
On 6 Mar 2010 02:44:56 GMT, Jasen Betts <jasen@xnet.co.nz> wrote:


have you considered PWM,

---
Have you read the subject line?


are you saying that PWM can't be used to limit the current in stepper
motors?

---
Try it for yourself and see how it works...

JF

 

[thermo102]

Well, I tried! The power resistors I installed are acting as bleed off's I
guess. A waste of money, but they don't seem to be harming anything, so I
plan to leave them there.

The new motor did run, but with exaxtly the same current draw as the old
one.

Hopefully the following is meaningful, and will help answer my question:
What do I have to do to supply the new motor with approximately 2.0 amps
operating current?

Pins 2,3,4,&5 of the parallel port go to a DM7 400N Logic NAND gate.

A trace goes from the NAND gate to another transistor which I cannot
identify, and then on to a large condensor. There is a trace branching off
(through a 1K resistor) to the 'base' of TIP 120. 5 volts from the parallel
port here?

Am I correct that this 1K resistor is the actual current limiting resistor?

I 'think' there is a 42 Volt DC applied to the 'emitter' of TIP 120. Then
the 'collector' goes to the motor.

Of course I have four TIP 120's with the same configuation. (one for each
'coil' of the motor).

The OLD motor had 133 ohms resistance across the coils, and the above gave
approximately 0.29 operationg amps.

The NEW motor I'm working with has 2.2 ohms resistance across the coils.
What value resistor should be used to replace the existing 1K's to achieve
approximately 2.0 operating amps?

Thanks in advance,
Thermo

 

[Jasen Betts]

On 2010-03-13, thermo102 <thermo102@comcast.net> wrote:

Well, I tried! The power resistors I installed are acting as bleed off's I
guess. A waste of money, but they don't seem to be harming anything, so I
plan to leave them there.

The new motor did run, but with exaxtly the same current draw as the old
one.

Hopefully the following is meaningful, and will help answer my question:
What do I have to do to supply the new motor with approximately 2.0 amps
operating current?

Pins 2,3,4,&5 of the parallel port go to a DM7 400N Logic NAND gate.

A trace goes from the NAND gate to another transistor which I cannot
identify, and then on to a large condensor. There is a trace branching off
(through a 1K resistor) to the 'base' of TIP 120. 5 volts from the parallel
port here?

Am I correct that this 1K resistor is the actual current limiting resistor?

I 'think' there is a 42 Volt DC applied to the 'emitter' of TIP 120. Then
the 'collector' goes to the motor.

Of course I have four TIP 120's with the same configuation. (one for each
'coil' of the motor).

The OLD motor had 133 ohms resistance across the coils, and the above gave
approximately 0.29 operationg amps.

The NEW motor I'm working with has 2.2 ohms resistance across the coils.
What value resistor should be used to replace the existing 1K's to achieve
approximately 2.0 operating amps?

Is your 42V supply capable of producing the extra 70W needed?
(140W if you need to energise two coils at the same time)

The 2.2 ohm motor is designed to run on a 5V supply with a darlington
transistor (like TIP120) to ground,

If your 5V supply can spare another 10W run the motor from that.

Connect the motor common wires to 5V and
the branch wires to the collectors

Ahe TIP120 transistors may need heatsinking as they'll dissipate 1.5W
of heat (and all through one of them) when the motor is stopped.

The TIP120 has a gain of about 1000 so it needs atleast 2mA input to
turn on enough to conduct 2A for switching applications (like this) it's
recommended to exceed that by a factor of ten (so 20mA)

The parallel port won't be able to produce enough 5V curren to turn the
chips on.

you mention 'another transistor' is that a branch off the path to the
TIP120 or in the path to the TIP120

Because if it's in the path you probably just need to reduce the 1K
resistor to 180 ohms.

Else you'll need to replace the DM7400N with someething stronger;
the DM7400N chip can only produce 0.4 ma so it's not going to work,
A SN74AHC00N chip can procude 8mA on the outputs which is probably
close enough to 20mA for it to work, replace the 1K resistors with
wire links.

either way you'll probably need a 5V supply, this can be taken from
a keyboard socket, game controller port, or usb socket. (if you're not
intending to run the motor from this 5V supply, if you are you could
try using one of the hard drive connectors if your PC can spare the power)


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[thermo102]

Else you'll need to replace the DM7400N with someething stronger;
the DM7400N chip can only produce 0.4 ma so it's not going to work,
A SN74AHC00N chip can procude 8mA on the outputs which is probably
close enough to 20mA for it to work, replace the 1K resistors with
wire links.

Thanks for your help. My power transformer only had a rating of 1 Amp, so I
picked up another one today with a rating of 4 Amps. While at the
electronics store, I also picked up a couple of the SN74AHC00N chips you
mentioned. Is the logic the same as for the DM7400N?

As soon as I get the new transformer mounted and wired in, I'll remove the
1K resistors and replace with some of smaller value to see what happens.

Thanks again,
Thermo

 

[Jasen Betts]

On 2010-03-16, thermo102 <thermo102@comcast.net> wrote:

Else you'll need to replace the DM7400N with someething stronger;
the DM7400N chip can only produce 0.4 ma so it's not going to work,
A SN74AHC00N chip can procude 8mA on the outputs which is probably
close enough to 20mA for it to work, replace the 1K resistors with
wire links.


Thanks for your help. My power transformer only had a rating of 1 Amp, so I
picked up another one today with a rating of 4 Amps. While at the
electronics store, I also picked up a couple of the SN74AHC00N chips you
mentioned. Is the logic the same as for the DM7400N?

As soon as I get the new transformer mounted and wired in, I'll remove the
1K resistors and replace with some of smaller value to see what happens.

The "74" series of logic chips have been reworked several times in
different variations of power ratings and speed, they all have the
same pinout for the same base part number. both those parts are
"7400" variants, I don't recall explicitly checking the pinpouts, but
I did compare the written descriptions and did glance at the pinout
diagram, but did not make a detailed comparison.
www.alldatasheet.com will find you the datasheets for both parts.
(or just google "SN74AHC00N datasheet")

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[John Fields]

On Mon, 15 Mar 2010 19:24:24 -0700, "thermo102" <thermo102@comcast.net>
wrote:

Else you'll need to replace the DM7400N with someething stronger;
the DM7400N chip can only produce 0.4 ma so it's not going to work,
A SN74AHC00N chip can procude 8mA on the outputs which is probably
close enough to 20mA for it to work, replace the 1K resistors with
wire links.


Thanks for your help. My power transformer only had a rating of 1 Amp, so I
picked up another one today with a rating of 4 Amps. While at the
electronics store, I also picked up a couple of the SN74AHC00N chips you
mentioned. Is the logic the same as for the DM7400N?

---
Yes, and the pinouts are identical.

JF

 

[thermo102]

---
Yes, and the pinouts are identical.

JF

Thanks for the info! I think I fried tne DM7400N yesterday. At least I have
a replacement chip on hand.

I removed some heatsinks, etc so I could ID the unknown chip. It's a 7805 1
Amp positive voltage regulator with an output of 5 volts and max 1 Amp. The
input to this chip is approximately 32 VDC.

Am I correct in my thinking that I can replace the 7805 with a L78S05 2 Amp
positive voltage regulator to get approximately the 2 amps for my new motor.
I assume the other two motors would not be affected by this change as long
as the output voltage of the L78S05 is 5 volts???

The Darlington chips are configured as:

Emitter: Ground
Base: receives 5 volts from the 7805 (across a 1K ohm resistor)
This same 'feed' goes to the DM7400N
Collector: goes to a coil winding of a motor

Any suggestions on what I should do next?

Thermo

 

[Jasen Betts]

On 2010-03-19, thermo102 <thermo102@comcast.net> wrote:

---
Yes, and the pinouts are identical.

JF

Thanks for the info! I think I fried tne DM7400N yesterday. At least I have
a replacement chip on hand.

I removed some heatsinks, etc so I could ID the unknown chip. It's a 7805 1
Amp positive voltage regulator with an output of 5 volts and max 1 Amp. The
input to this chip is approximately 32 VDC.

Am I correct in my thinking that I can replace the 7805 with a L78S05 2 Amp
positive voltage regulator to get approximately the 2 amps for my new motor.
I assume the other two motors would not be affected by this change as long
as the output voltage of the L78S05 is 5 volts???

you're going to need a serious heatsink if you go that route.
you'll be burning 54 watts or energy up as heat in the regulator to supply
2A to the motor,

if you only need a few tens of milliamps to run the nand gate and the bases of
the darlintons then the 7805 was a good solution

the 78S05 from STmicro has a 3 degrees per watt thermal resistance
(TO220 case) so even if you weld it to a perfect heatsink in an
ice-water bath at 54W dissipation the junction would be at 3X54=162C
which is 12 degrees too high.

thses figures can be improved by putting a 50W 12 ohm resistor in
series with the input to the regulator, then mos of the heat will go
into the resistor and the regulator will only get about 8W of heat.

your first idea with the series resistors on the motor was better.

psaically P=VxA is your enemy when you want to run a low voltage
device from a higher voltage supply.


I'd consider a 2A 5V powersupply like this one:
http://jaycar.com/productView.asp?ID=MP3316
only get one with the right plug for the power outlets at your place,
although I guess you could use that one inside the machine and connect
it using insulated quick-connect terminals

The Darlington chips are configured as:

Emitter: Ground
Base: receives 5 volts from the 7805 (across a 1K ohm resistor)
This same 'feed' goes to the DM7400N
Collector: goes to a coil winding of a motor

you can boost the current output of the 7400 by using NPN transistors (eg
PN2222) as emitter followers.

collector to 5V
base to the 7400 output
emitter to a 100 ohm 0.5W
resistor with the other end to the TIP120 base.


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[thermo102]

thses figures can be improved by putting a 50W 12 ohm resistor in
series with the input to the regulator, then mos of the heat will go
into the resistor and the regulator will only get about 8W of heat.

your first idea with the series resistors on the motor was better.

psaically P=VxA is your enemy when you want to run a low voltage
device from a higher voltage supply.


After reading your post, I studied the sketch I'm making up of my board

layout and it looked like I had a configuration similar. Then a google
search turned up the concept of a voltage divider circuit. Which I guess I
have.

Measuring the actual voltage at the input side of the 7805, I was surprised.
A really low value, somewhere around 2 volts, and the output was crazy. So
I started trial and error resistor values untill I had an actual 22 volt
input and a 5 volt output. (replaced a 500 ohm '2 watt' resistor with a 100
ohm '2 watt' )

Did not see any ill effects by operating 2 motors simultaneously for a short
period.

My 3rd motor still doesn't run, so I guess I blew the transistors also. Is
there a good way to check them with a multimeter without removing them from
the circuit?

Again, thanks for all your help,
Thermo