Controlling proportional valve with potentiometer

I just recently got my hands on two TR24-SR US proportional actuators that I plan on eventually using for automated flow control (still a ways out). For the time being I would like to control these with either a 5k or 10k pot from automation direct

What I am trying to do (and probably over thinking) is how to wire the circuit to control the valve? I currently have a 12VDC power source in my panel and would like to use that for the control leg of the actuator, I also have but havent yet installed a 24VAC transformer and a 24VDC inverter.

What Im trying to pin down exactly is,

1 - what is the simplest method for wiring this up (want 1-10VDC for control, so need padding resistors)
2 - Which power source should I use to supply the actuator, 24V AC or DC? and how would this interface with the common leg on the control signal??
3 - Does it matter which pot (5k or 10k) I use? I dont think it does but thought I'd ask

I would like to know how people respond as well. I was going to say that I don't think a simple pot is going to be enough to do what you want. I'm not 100% sure though. I will investigate a bit more and see what I can find in the mean time.

Facinerous said:

I would like to know how people respond as well. I was going to say that I don't think a simple pot is going to be enough to do what you want. I'm not 100% sure though. I will investigate a bit more and see what I can find in the mean time.

Click to expand...

Is there any reason in particular you dont think it would?

AFAIK direct proportional actuators are a different beast than floating point ones, and their control is solely based on the voltage supplied, which is why i bought the particular ones I did - hopefully that wasnt a bad assumption

There have been a couple posts about these valves in the past; check out some of the posts by kladue.

So after looking up voltage dividers you should be able to do it. Here is a link to the voltage divider calculator I used. Should help out.

http://www.raltron.com/cust/tools/voltage_divider.asp

If you need more help just let me know.

Just insert the basic circuit shown on this link into the signal side of your circuit. Good luck dude.

I would go all DC to keep any noise from the AC affecting your process signal. If your going the voltage divider, get good tight tolerance components. You may just find a good 10v power supply on eBay, some have sense inputs so you know you have 10v at the device. Keep the runs as short as possible, as voltage signals are more prone to noise than current 4-20ma signals

trimixdiver1 said:

I would go all DC to keep any noise from the AC affecting your process signal. If your going the voltage divider, get good tight tolerance components.

Click to expand...

Is there any reason I couldnt just use a single 24VDC supply for both powering the actuator and providing the signal? To provide the signal I would using padding resistors around the pot to ensure that I have a range from 1-10V on the pot, and I would tie the common together?

The reason I decided to go with a voltage divider rather than current was it was easier to source. If you know of a 4-20mA current source that I could use in the application (with manual adjustment knob) I'd love to hear about it

trimixdiver1 said:

Keep the runs as short as possible, as voltage signals are more prone to noise than current 4-20ma signals

Click to expand...

How short is short enough? Im guessing that at most the line will be 10ft, but probably less

I don't have reason to believe there would be enough noise on any single conductor to cause problem. The system you are using wouldn't induce enough voltage on the line to cause problem. If it does you can use a shielded cable to isolate the signal lines. If you only use a 24 vdc supply you will need a zener diode in parallel with the signal placed after your voltage divider to regulate the voltage to a max 10 volts. This will limit the full range of the pot though. So much of the range of the pot would in effect do nothing.

ryane said:

Is there any reason I couldnt just use a single 24VDC supply for both powering the actuator and providing the signal?

Click to expand...

You can use something like that
http://www.ebay.com/itm/New-AC-DC-t...le-Module-LM317-Out-1-25-30V-1A-/321368837721

Or if you OK with resoldering a pot (original is screwdriver type) you can have an embedded voltage indicator

http://www.ebay.com/itm/0-56-LED-di...Power-Module-voltage-Adjustable-/221392799366

If you willing to pay some extra bucks- you can get digitally (by buttons) controlled converter -
http://www.ebay.com/itm/3A-DC-DC-Di...Adjustable-buck-converter-B3603-/171417756516

ryane said:

Is there any reason I couldnt just use a single 24VDC supply for both powering the actuator and providing the signal? To provide the signal I would using padding resistors around the pot to ensure that I have a range from 1-10V on the pot, and I would tie the common together?

The reason I decided to go with a voltage divider rather than current was it was easier to source. If you know of a 4-20mA current source that I could use in the application (with manual adjustment knob) I'd love to hear about it


How short is short enough? Im guessing that at most the line will be 10ft, but probably less

Click to expand...

After a quick look at the datasheet, I think it needs a 24V/AC/ source, not DC so powering both is likely not going to work without a rectifier.

Also, generating current from a 12V source is really simple. Just hook up a 600-ohm resistor and a 2400-ohm potentiometer/rheostat in series and you'll have a 4-20mA current generator.

Edit: After reexamining the datasheet, it looks like it puts a 500-ohm resistance across the current source so what I suggested above won't work. Just use a voltage divider.

Double edit: I also just saw in the datasheet that you can use 24VDC. So ignore my entire post.

There is a note on the datasheet that says actuators may be powered by 24vdc.

Duely noted with the current generator. Good info to remember for me.

This would be much easier to do if you used a simple opamp. then you could use the offset pins to set the 2volt offset, and the gain to set the top end

with only a pot and some resistors, this would work. It would depend on the input impedance of the control circuit as that will effect overall resistance and change the voltage dividers a little. So you might have to play with the values a little.



you could do it with two LM317 as well. hard wire one of them for 10VDC and use the output from that LM317 to input of the other one. use the pot to control the output of LM317 #2. the bottom end of the 1.7VDC of the regulator might be close enough to 2volts to work.

just be aware that LM317 get stupid hot! make sure you heat sink them well. if they get too hot they tend to go up in voltage.

It wants 24 V ac or dc to operate it so whichever is most convenient.

It wants 2 - 10 VDC as a control input which can be obtained by applying 2 - 10 VDC or 4 - 20 ma through a 500 Ω resistor. Input impedance is 100 K so the most it will draw from the control circuit is 10/100K = 0.1 ma and thus your control source doesn't need to be too stiff. Lets say you have a regulated 20V supply and want 20 ma through the voltage divider (half or a quarter of this should be plenty). It's total resistance would have to be 20/20 x 10^-3 = 1000 Ω. A fixed resistor connected to the power supply of 500 Ω will drop 500*20x10^-3 = 10 volts. The drop across the pot would be 8 V so its resistance would be 8/2E-2 =400Ω and a fixed resistor between the bottom of the pot of 100 Ω would produce the remaining 2 volts drop. Bottom of the 100 Ω resistor goes to the common pin and the wiper of the pot to the control input. You can scale things around to any values you need depending on available power supply (must be 10 V or greater) and the available components.

You could use a transistor or op amp in emitter follower configuration but given the high input impedance of the device you don't need to.

ryane said:

Is there any reason I couldnt just use a single 24VDC supply for both powering the actuator and providing the signal? To provide the signal I would using padding resistors around the pot to ensure that I have a range from 1-10V on the pot, and I would tie the common together?

The reason I decided to go with a voltage divider rather than current was it was easier to source. If you know of a 4-20mA current source that I could use in the application (with manual adjustment knob) I'd love to hear about it



How short is short enough? Im guessing that at most the line will be 10ft, but probably less

Click to expand...

I would say less than 50ft, more importantly, don't run signal or analog wires with power wiring and if they must cross in close proximity, cross at 90deg. Little details help when your chasing gremlins later.

If concerned about noise use the 4 - 20 configuration (that's what it's for). Duality makes it just as easy to wicker a 4 - 20 ma loop as a 2 - 10 volt one. In the 4 - 20 configuration the device impedance is 500 Ω (but note that you have to supply the 500 Ω resistor). At minimum you need to supply 4 ma to the loop. From a 24 volt (regulated) supply you would need R = 24/4E-3 = 6K. As the regulated power supply will have very low impedance and the valve controller has 500 Ω you will need 5500Ω in the loop for minimum opening. At max opening the loop current must be 20 ma so the required loop resistance is R = 24/2E-2 or 1200 Ω. With the controller supplying 500 of those that means 700 in the external circuit. Thus you need a 700 Ω resistor connected to the power supply + in series with a 5500-700 = 4800 Ω pot. With one end of the pot connected to the 700 Ω resistor the wiper would be connected to one end of the 500 Ω resistor and the control terminal is connected to that same point. The other end of the 500 Ω resistor is connected to the controller common terminal and the power supply negative.

In this configuration you can make the control loop as long as you want. To be precise use a resistor of value 700 minus the wire loop resistance (should only be a couple of Ω unless you are controlling this from a neighbor's back yard).

Thanks for all the input from everybody

This is what Im going to do


This will give me a range of 1-10V on the pot, while the valve will begin to actuate at 2V

I will then run an additional 24V lead and common out to the actuator, so 4 wires going to the valve

ajdelange said:

If concerned about noise use the 4 - 20 configuration (that's what it's for). Duality makes it just as easy to wicker a 4 - 20 ma loop as a 2 - 10 volt one. In the 4 - 20 configuration the device impedance is 500 Ω (but note that you have to supply the 500 Ω resistor). At minimum you need to supply 4 ma to the loop. From a 24 volt (regulated) supply you would need R = 24/4E-3 = 6K. As the regulated power supply will have very low impedance and the valve controller has 500 Ω you will need 5500Ω in the loop for minimum opening. At max opening the loop current must be 20 ma so the required loop resistance is R = 24/2E-2 or 1200 Ω. With the controller supplying 500 of those that means 700 in the external circuit. Thus you need a 700 Ω resistor connected to the power supply + in series with a 5500-700 = 4800 Ω pot. With one end of the pot connected to the 700 Ω resistor the wiper would be connected to one end of the 500 Ω resistor and the control terminal is connected to that same point. The other end of the 500 Ω resistor is connected to the controller common terminal and the power supply negative.

Click to expand...

Do you think it has to be 4-20mA or do you think they just used it as an example and you can use whatever current-resistor combo as long as you create a 2-10v drop across that resistor at the control lead?

bu_gee said:

Do you think it has to be 4-20mA or do you think they just used it as an example and you can use whatever current-resistor combo as long as you create a 2-10v drop across that resistor at the control lead?

Click to expand...

4-20ma is a common process analog signal used in PLCs. The only benefit over voltage is its ability to be used in longer runs and less crosstalk. For our purposes a voltage signal is fine.

bu_gee said:

Do you think it has to be 4-20mA or do you think they just used it as an example and you can use whatever current-resistor combo as long as you create a 2-10v drop across that resistor at the control lead?

Click to expand...

The latter but clearly the 2 - 10V control range was chosen so that one could set up at 4-20 loop conveniently (with a 500 Ω resistor). 4-20 is an industry standard where you need analogue control at a distance. For example, in a water treatment plant you have multiple sensors and actuators (e.g. pH sensor, valve like this one for dosing lime milk etc..) dispersed over several acres all controlled from a central control room. Many industrial sensors read out with 4 - 20 loops (in addition to RS-422, Modbus, ethernet...) and many industrial controllers have 4 -20 loops to control valves like these.

Look at the diagram in #16. Move the 555 Ω across the +/- terminals and run the loop from the wiper to the + terminal. You now have a 1.8 - 2.9 current loop. It's still essentially the same circuit. Change the resistor and pot sizes and it becomes 4 - 20. It's still a voltage divider circuit but now you have 500 Ω impedance at the end of the loop instead of the 100K of the controller itself.

I use the Belimo valves with a 24VDC supply and they work as well as when fed with 24VAC. The diagram in #16 is almost right, loose the 555 ohm resistor and you should be there. Be prepared for the 90 second open - close time, not a bad thing with automation, but you will have to be patient waiting for large changes. The other thing to know is the "dead angle" (number of degrees valve has to move before flow starts) which varies by valve size, 2.5 volts will not equal flow start, usually is closer to 3 volts.
Most common control method with these valves is the 4-20MA loop and 500 ohm resistor, but finding cheap precision 500 ohm resistors is a bit of a challenge, so for simplicity of parts the voltage divider setup parts would be easier to find.

As pointed out in Nos. 17 and 19 the size of the current sense resistor is arbitrary unless a 4 - 20 loop is desired as it would be if using off the shelf control equipment with 4 - 20 output. But the controller here is a power supply with pot and current limiting resistor. Using a 555 Ω sense resistor creates a 3.6 - 18 ma loop rather than a 4 - 20 ma loop but it's still a current loop. Also, as pointed out in No. 19, it is, at the same time, a voltage divider but one with low impedance across the valve terminals which confers a degree of noise immunity not that that should be at issue in this application.

kladue said:

The other thing to know is the "dead angle" (number of degrees valve has to move before flow starts) which varies by valve size, 2.5 volts will not equal flow start, usually is closer to 3 volts.

Click to expand...

This is good information, I will adjust the padding resistors accordingly, but will first play with this on my own

As set up in my last post, I added the 555 and 7.7k resistors to ensure a min/max voltage on the pot signal of 1-10V (on/off) . If the valves dont open enough for flow to begin till 3V, I might adjust the resistors to give me a tighter V range.

By using the padding resistors all Im doing is trying to get the largest amount of control-ability using the pot (no dead range on the pot)

But, my guess is you will still need it to go down to 2v or it wont close all the way