Stepper Motor 5V Pulse

Andreik · Jun 26, 2012

Good day,

I have a little problem with my Stepper Motor and PLC. From the attached image, The Pulse signal for Stepper Motor is 5VDC. However, My "COMMON" in PLC Output is -24VDC which covers Pulse and ordinary Output signals.

According to the attached excerpt from Keyence Manual, 24VDC Power Supply can be used but a current limiting Resistor should be implemented.

I'd like to know if it is possible to give +5VDC to CW+/CCW+ and -24VDC to CW-/CCW- as I indicated on my diagram.

Thank you very much,
Andrei K.

shooter · Jun 26, 2012

yes your schema will work, however the PLC is not fast enough to have the motor running at a decent speed.
have a look at arduino for this.
cheap and easy.

rpoet · Jun 26, 2012

shooter,
It depends on the PLC. Some have the ability to configure several of the outputs as high-speed pulse outputs. I have a DL-06 PLC from AutomationDirect running a stepper motor using its built-in HSIO features. The DL-06 can output at up to 10khz, which is fine for my application. The OP may need a HSIO card, depending on the type of PLC he's planning on using.

-rpoet

Andreik · Jun 27, 2012

shooter said:
yes your schema will work, however the PLC is not fast enough to have the motor running at a decent speed.
have a look at arduino for this.
cheap and easy.

Oh, so, will it work even if the "-" of 5VDC is like floating? I'm wondering what Voltage would I get if I measure between +5VDC and "-" of 24VDC.

keithkyll · Jun 27, 2012

It will not work with the -5 floating. If you want to use the 5 volt supply, connect the -5 to the wire that connects the PLC output common and -24. This assumes you have NPN outputs. Without the Keyence model number, I'm forced to guess.
The Keyence has a unique feature. It has 1.6K Ohm resistors built in. The attached drawing is wrong. The resistor outputs are R500, R501, and R502.

Use two of the three outputs 500-502 with the 5 volt supply, or R500-R502 with the 24 volt supply.
I also need the model number of the stepper driver to verify that it has 5 volt inputs, and the 1.6K resistor will drop the other 19 volts.

The Keyence spec sheet mentions 1 axis output, capable of 50KHz. Assume this PLC is fast enough.

Andreik · Jun 27, 2012

keithkyll said:
It will not work with the -5 floating. If you want to use the 5 volt supply, connect the -5 to the wire that connects the PLC output common and -24. This assumes you have NPN outputs. Without the Keyence model number, I'm forced to guess.
The Keyence has a unique feature. It has 1.6K Ohm resistors built in. The attached drawing is wrong. The resistor outputs are R500, R501, and R502.

Use two of the three outputs 500-502 with the 5 volt supply, or R500-R502 with the 24 volt supply.
I also need the model number of the stepper driver to verify that it has 5 volt inputs, and the 1.6K resistor will drop the other 19 volts.

The Keyence spec sheet mentions 1 axis output, capable of 50KHz. Assume this PLC is fast enough.

Hi,

So what you're saying is I should "short" the "-" of 24VDC and 5VDC Power Supply with the COMMON? Yes, I used NPN. The PLC is KV24-DT. The Step Driver is supplied with 24VDC and the Pulse signal is 5VDC as show in the attached document.

Another thing, I'm very much wary about the "0503" Output address of the Keyence PLC. Reading through the documents, "0503" is used for CW/CCW direction selection. What's bugging me is that, off course I'm not going to connect the Pulse signal from "0503" since it's not a Pulse Output unlike 0500, 0501 and 0502. Reading sample Programs, I can't find 0500, 0501 and 0502 so I don't even know which of these should I connect to CW- and CCW-.

Thank you very much,
Andrei K.

keithkyll · Jun 27, 2012

Andreik said:
So what you're saying is I should "short" the "-" of 24VDC and 5VDC Power Supply with the COMMON?

That's what I said, but let me revise that. Disconnect the line to -24 volts. That's not needed.
One single connection from -5 to PLC Output Com.

Andreik said:
"0503" is used for CW/CCW direction selection. What's bugging me is that, of course I'm not going to connect the Pulse signal from "0503" since it's not a Pulse Output unlike 0500, 0501 and 0502. Reading sample Programs, I can't find 0500, 0501 and 0502 so I don't even know which of these should I connect to CW- and CCW-.

Direction (CW/CCW) doesn't need to be high speed. Connection to 0503 is fine.
You can connect 0500, 0501, or 0502 to Pulse. If you can't find it in the program, run it and look for the LED that's pulsing.
If they used 0503 for Direction, my first guess is 0502 for Pulse. Is "Stepper Wire.jpg" the drawing associated with the program example? If so, then 0502 for Pulse is correct.

Do not use the R50x outputs with 5 volts, as shown in your first diagram, "Stepper Wire.jpg". The drawing should show connections to "0502" and "0503". "R0502" and "R0503" are typos. There isn't a "R0503", a clue that it's a typo. The rest of the diagram is correct.

Andreik · Jun 27, 2012

keithkyll said:
That's what I said, but let me revise that. Disconnect the line to -24 volts. That's not needed.
One single connection from -5 to PLC Output Com.

Direction (CW/CCW) doesn't need to be high speed. Connection to 0503 is fine.
You can connect 0500, 0501, or 0502 to Pulse. If you can't find it in the program, run it and look for the LED that's pulsing.
If they used 0503 for Direction, my first guess is 0502 for Pulse. Is "Stepper Wire.jpg" the drawing associated with the program example? If so, then 0502 for Pulse is correct.

Do not use the R50x outputs with 5 volts, as shown in your first diagram, "Stepper Wire.jpg". The drawing should show connections to "0502" and "0503". "R0502" and "R0503" are typos. There isn't a "R0503", a clue that it's a typo. The rest of the diagram is correct.

That's the problem now. Our Output devices are supplied with 24VDC. Well, I guess the solution now is to use 5VDC Relays. Off course my COM should be "-"5VDC.

You mean, there's no method for the programmer to choose from 500, 501 and 502, meaning random? For example I use 500 for CW, and 502 for CCW. Is it still 503 for the direction?

keithkyll · Jun 27, 2012

It will work with 24VDC. You don't need relays. With 24VDC, you must use the "R" outputs, R0500 - R0502.
Relays won't work with high speed pulses.

Sorry for the confusion. Most stepper drive inputs are Pulse and Direction. This one is CW and CCW.

The programmer (you) chooses the output. Not random.

Connect -24 to PLC Com.
Connect +24 to +CW and +CCW.
Connect R050x to -CW.
Connect R050x to -CCW.

Andreik · Jun 27, 2012

keithkyll said:
It will work with 24VDC. You don't need relays. With 24VDC, you must use the "R" outputs, R0500 - R0502.
Relays won't work with high speed pulses.

Sorry for the confusion. Most stepper drive inputs are Pulse and Direction. This one is CW and CCW.

The programmer (you) chooses the output. Not random.

Connect -24 to PLC Com.
Connect +24 to +CW and +CCW.
Connect R050x to -CW.
Connect R050x to -CCW.

Will it be alright if I supply +24 to +CW and +CCW that are +5V in specification?

Sorry, what I mean about the +24V Output devices are our normal Solenoid Valves and Lamps. The 5VDC is dedicated only for the Pulse signal going to the Stepper Motor.

keithkyll · Jun 27, 2012

Andreik said:
Will it be alright if I supply +24 to +CW and +CCW that are +5V in specification?

Yes. There will be 5 volts at the Stepper Controller. The other 19 volts will be dropped across the "Current Limiting" resistor inside the Keyence PLC (resistors are present only on the R050x outputs).

With this setup, there is a risk of blowing the inputs to the Stepper Controller if the -CW or -CCW lead is accidentally shorted to ground.
Two ways to avoid this.
1) Add two 1.6K,.5W Metal Film resistors. Prefer 1% or 2%. One in series with the +CW, the other in series with +CCW. Use 050x outputs.
2) Use a 5 volt supply, and the 050x outputs.

Technical:
You will have 3 devices in series.
1.6K Ohm resistor (inside PLC) - 220 Ohm resistor (inside Stepper Controller) - LED (inside Stepper Controller opto isolator)

Ohms law calculations. LED will be at a fixed voltage, about 2.2 volts. With that in mind, we calculate current across resistors using 1820 total Ohms, and 21.8 Volts total voltage across resistors.
Current = 12 mA
LED = 2.16 Volts
1.6K resistor = 19.2 Volts
220 resistor = 2.64 Volts

Andreik · Jun 28, 2012

keithkyll said:
Yes. There will be 5 volts at the Stepper Controller. The other 19 volts will be dropped across the "Current Limiting" resistor inside the Keyence PLC (resistors are present only on the R050x outputs).

With this setup, there is a risk of blowing the inputs to the Stepper Controller if the -CW or -CCW lead is accidentally shorted to ground.
Two ways to avoid this.
1) Add two 1.6K,.5W Metal Film resistors. Prefer 1% or 2%. One in series with the +CW, the other in series with +CCW. Use 050x outputs.
2) Use a 5 volt supply, and the 050x outputs.

Technical:
You will have 3 devices in series.
1.6K Ohm resistor (inside PLC) - 220 Ohm resistor (inside Stepper Controller) - LED (inside Stepper Controller opto isolator)

Ohms law calculations. LED will be at a fixed voltage, about 2.2 volts. With that in mind, we calculate current across resistors using 1820 total Ohms, and 21.8 Volts total voltage across resistors.
Current = 12 mA
LED = 2.16 Volts
1.6K resistor = 19.2 Volts
220 resistor = 2.64 Volts

Hmmh. So, the bottom line is I wouldn't need the 5VDC Power Supply anymore right? We're using Voltage Divider I see. Thanks

keithkyll · Jun 28, 2012

Yes and yes.

Andreik · Jun 28, 2012

keithkyll said:
Yes and yes.

So this now attached will be the actual wiring? Thanks

keithkyll · Jun 28, 2012

No. You have double 1.6K resistors.
1) On PLC, change R502 to 502.
2) On CCW-, remove 1.6K resistor.

Stepper Motor 5V Pulse

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