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External potentiometer to Omron VFD - newbie is stumped.

PeteDenmark

Plastic
Joined
Dec 6, 2019
Sorry to bother you with this.
I am WAY out of my element, and i'm afraid that i'll fry something, if i don't consult the experts in here

I have ordered a second hand OMRON Varispeed V7 1.5KW 400v AC 3 Phase VFD Inverter Drive (CIMR - V7AZ41P5), to run my drill press

The VFD has a digital potentiometer on the front, but i would like to have the frequency control on an external analog potentiometer.

But going through the manual, i don't even know which potentiometer(how many K-Ohms)to order.

I have gather that i have to connect the three legs of the potentiometer to the FS, FR and FC (Frequency setting power supply (FS), Frequency Reference input (FR)and Frequency reference common (FC), where the variable end of the pot (middle leg) has to go to FR.

The "frequency setting power supply" (Called FS) is 20 mA at 12v. But the Frequency reference ned 0 to +10v?
Also In the manual under the parameters for the FR it says 0 to 10VDC 20K-Ohm, but then in the normal wiring diagram, and the diagram describing the frequency reference, it has an illustration showing 2K-ohm.

I'm sure this is very basic for at lot of you, but for me, it's like staring at the green vertical lines on the screen in the matrix :).

I have attached pictures of what i have gathered are the relevant places in the manual, but if not here is the link to the complete manual:
http://www.tecdriver.com.br/arquivos/VARISPEED V7 (CIMRV7) OMRON.pdf

Could anyone please tell me what potentiometer (and maybe additional resistors??)i should buy, and how to hook it up, so i dont fry anything.

I have a pretty good idea which parameters to change in the menu, so it's just how i should physically hook the pot up, and which to buy.

Thanks :).

VFD potentiometer 1.jpg VFD potentiometer 2.jpg VFD potentiometer 3.jpg
 
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Maybe 20 k-ohms stated at the 12v 20mA frequency reference power supply, is the impedance. I have no idea what i'm talking about, but one often sees impedance stated when a powersource sends a signal that has to be picked up by something else.

Then it follows, that the pot they want used is a 2k-ohm pot. That's fair.

But what's with the 12v reference power supply, when the input terminal for frequency reference setting needs 0-10v? Will this terminal be able to handle 12v? This would then mean, that when the pot is turned 5/6 the way up, nothing will happen when i turn the pot the last 1/6 of a turn.

Ok - if that is so, and i want to rectify it, should i place a resistor at the power source?

If i get the 2k pot the manufacturer specifies (as i read it), then it follows, that if i want 10v then:

I=E/R, so 10v/2000ohm= 5mA (0,005 amps. If i then want to "lose" the 2 volts to get me from 12v to 10v, i should place a resistor at the 12v source. So R=E/I is 2v/0,005amps= 400ohms.

So if i solder a 400ohm resistor to the 12v input leg of the pot, and connect it to the VFD as specified, i should get "full range of motion" on the pot, and keep the volts to 10v max right?

I have a feeling this shouldn't be this complicated, but on all the online articles i have read, the source power is always 10v, and the pot size is clearly stated.
 
Buy a good quality 2K ohm pot (allen bradley is nice) and hook it up per the manual's diagram.

Thank you Jim. Didn't see your answer before my previous post. So just ignore the ekstra 2v, and be happy?
We don't have Allen Bradley pots here in the EU as far as i can tell, and a 120 euro 2K siemens pot is too rich for my blood. Will see what i can come up with :).
 
Can nobody explain to me why it is no problem to take power from the 12v reference power supply terminal , and feed it through a 2k pot, into a reference input that uses 0 to 10 volts as reference to set the frequency.
Omron support don't answer my emails, and i got no answers to email.
 
Can nobody explain to me why it is no problem to take power from the 12v reference power supply terminal , and feed it through a 2k pot, into a reference input that uses 0 to 10 volts as reference to set the frequency.
Omron support don't answer my emails, and i got no answers to email.

Typically a variance of 10-20% is acceptable for circuits like shown in Step 7, also the voltage driven frequency circuit is going to have some input protection from over voltage.

What concerns me is you saying it has a digital pot on the controller. Is the digital pot an electronic digital pot or a mechanical digital readout connected to a mechanical potentiometer?

Does the mechanical potentiometer require disconnection before adding the 2k pot to the Step 7 wiring?

You also said you want to have the pot mounted at the machine, keep the wire run as short as you can, use shielded wire in a range of AWG 24 to 18 (speaker wire), do not run the wire anywhere near power wires even if in metal conduit. On the VFD end of that wire ground the shield at the VFD ground point - right on it - the other end at the machine will not be connected, the machine must of course be electrically grounded. You have to be concerned about electrical noise getting into your VFD.
 
If using the 12 volt source for the 10 volt destination then place a "buffer" resistor on top of the pot.

3 legs of pot.
Bottom to low side supply, return, ground or whatever common EU term.

Center terminal or wiper out to variable input.

Top of pot to one side of buffer resistor.

Other side of resistor to 12 vdc source.

10k pot and 2k resistor equals 1k per volt.

Input to VFD should be high impedance and not cause much loading.

Sent from my SAMSUNG-SM-G930A using Tapatalk
 
Hello Pathogen, and thank you for your answer. That puts my mind at ease :).

I assume the potentiometer on the VFD itself is a digital pot (that's what i get from the diagram anyway). The pot on the machine does not need to be disconnected, you just have to change parameters in the menu. When that is done, the onboard pot no longer responds, only the external analog pot will work.

I guess the 12v, instead of the 10v, frequency reference power supply could be explained, if the intention is to insure that the reference input gets the 10v it needs, even with longer wire runs.

But thank you for the tip, regarding short runs, and shielded wire.
In my case however, everything will be mounted on the drill press itself, so short runs, and not near the power supply.
I just wanted an external pot and run/fwd/rev, so i dont put my oily hands directly on the VFD, when drilling metal.
 
If using the 12 volt source for the 10 volt destination then place a "buffer" resistor on top of the pot.

3 legs of pot.
Bottom to low side supply, return, ground or whatever common EU term.

Center terminal or wiper out to variable input.

Top of pot to one side of buffer resistor.

Other side of resistor to 12 vdc source.

10k pot and 2k resistor equals 1k per volt.

Input to VFD should be high impedance and not cause much loading.

Sent from my SAMSUNG-SM-G930A using Tapatalk

Thank you Tony - you are very helpful.
Since i'm using a 2K pot as per the manuals recommendation, the "buffer" resistor should be 390ohm, if understand this online calculator right. Right?

Electronics 2000 | Potential Divider Calculator

I will get my hands on some resistors, and solder it to the pot, for good measure.
 
You may not need any extra resistor, since it does not seem to be called out.

the 2k ohm pot will draw only 6 mA, well below the 20 mA limit.

You want a low value potentiometer, so that the scale is generally linear. The "pot" does not need to be super quality, since you are not depending on it for a calibration. Any reasonably decent one will work The control inputs on all VFDs are isolated from the power, and do not need any special construction for your protection. You WILL want to protect it from chips, so put it in an enclosure, and hook up with (preferably) shielded wire.

If you want to add the divider resistor, I would suggest going lower in resistance, since you want to be sure the full 10V range is covered. The input will have "some" resistance, and would add some extra voltage drop to the dividing ratio. We don't know what its value is, so be conservative.

You can use a decent meter to read the range as-hooked-up, and adjust if needed. Most digital meters have 10 megohm input resistance on DC ranges, so that should not add any significant load to this circuit.

Or you can just see if you get the range on the display of the VFD that you expect.
 
I believe the issue is on that particular the power supply is 12V, and the voltage range for the speed input is 0-10V, so a voltage divider at the top would give a full 0-100 range on the pot, w/o the voltage divider you would have a 0-80 range. The schematic in the manual shows a variable resistor on the 12V supply. I would try it without, but to get the full range of the pot then a resistor could be used.

speed.jpg
 
I believe the issue is on that particular the power supply is 12V, and the voltage range for the speed input is 0-10V, so a voltage divider at the top would give a full 0-100 range on the pot, w/o the voltage divider you would have a 0-80 range. The schematic in the manual shows a variable resistor on the 12V supply. I would try it without, but to get the full range of the pot then a resistor could be used.

View attachment 272442

If the input wants 0-10V and you give it 0-12V, then it will get to full speed when you're at 10/12 of full on the pot.

This means that the pot will (probably) have a dead zone for the top 2/12 of it's rotation. I doubt the VFD will go to 120% of full speed, but you never know.

The chances are that they supply 12V so that if there is any loss in the speed-control circuit(the pot, or other device to give variable DC voltage) you will still be guaranteed to be able to get to 100% speed.

And I really doubt feeding 12V to the control input will hurt it at all.
 
The drawing shows a pot between the speed pot and the 12 supply.

There are "*" next to each with a see note statement.

What do the notes state?

My guess is the top pot sets the range of the second one, given it shows both as 2 k the top pot can cause the top of the bottom pot to be 6 to 12 volts.

The 12 volt source is 20 ma source so there may be internal magic related to the current metering that limits the effect as we would think using hard 12 volts for feeder to a 0 to 10 volt control lime makes no sense as it would top out before end of stroke.

In short...RTFM...to understand how the circuit is to work.


Sent from my SAMSUNG-SM-G930A using Tapatalk
 
So the top pot is called a trim pot and it indicates what happens without it.

What it seems to be from my reading of the text is a simple adjustment to compensate for whatever value resistor is used for the speed pot to allow for the "trimming" of the resistance to set the voltage of the top of the speed resistor to 10.

Sounds like a cheap way to do things.

So suggestion,
Suggests using up to 3 k pots so use that for speed pot.

Not knowing the impedance causes us to guess but we assume high so minimal actual loading.

Maybe 500 ohm on top with a 10 volt zener diode to clamp it at 10 volts so no fiddling with adjustments.

Too lazy to do math, resistor value may be different.

Sent from my SAMSUNG-SM-G930A using Tapatalk
 








 
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