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sunrays (Electrical) (OP)
13 Aug 08 12:42
I need to send the speed reference from Drive A to Drive B over a distance of 800 Meters. Application is not very critical from the speed accuracy point of view. A +/- 10 RPM is good enough.

Option A: Use a field bus option but length is an issue. I am given to understand that I need to use Fiber Optic Cables to over come the distance, which add to the cost.

Option B: 4-20mA signal. I need a converter at both ends to convert the 0-10V Signal of Drive A to 4-20mA and 4-20mA back to 0- 10 V at Drive B since the drives can deliver/accept only Volts. The cable proposed to be used will be shielded one.

What is the maximum distance permissible with a 4-20mA option? Would I face any loss of signal over this distance?

Thanks!
electricpete (Electrical)
13 Aug 08 13:00
Most 4-20 sources have a limit on what resistance they can push through.

If memory serves me right (it's been awhile since I worked on those), somewhere around 500 ohms was typical. That would include the connecting cable both ways and the receiving unit(s) input impedance (could be more than 1 if you string several devices in series on the same loop).

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ScottyUK (Electrical)
13 Aug 08 13:05
Depends on two things: how much voltage you have to drive the current, and how thick the conductors are.

Most drives are fairly limited on the voltage they can deliver to drive current around the loop, but since you are using a pair of V/I converters you may be able to get one with a high compliance voltage. You need to apply Ohms Law and see what size conductor you end up with. Don't forget that there is 1600m of conductor in the loop. A fibre probably isn't going to be much different in terms of installed cost than a standard cable - installation labour is often the dominant cost with smaller cables and low core-count fibres.

Can you get a wireless Fieldbus solution? Consider what happens if (when) the link is lost - is there a safety issue?

Also consider a voltage/frequency converter. The signal should survive ok over that distance using little more than a telephone cable.
  

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If we learn from our mistakes I'm getting a great education!
 

sunrays (Electrical) (OP)
13 Aug 08 13:21
I did explore the wireless solution. The two systems are not in line of sight. As per the wireless solution providers non line of sight would mean use of lower band width which and multiple repearters. This adds to the cost and will end up in an unreliable solution with signal loss to compound with.

Voltage/Frequency seems to be a workable option.

Let me discuss this with my client.

Thanks!
waross (Electrical)
13 Aug 08 13:46
There is no set length for a 4-20ma circuit.
A 4-20ma circuit has a voltage source transducers and wire resistance. Each transducer will drop some voltage. From the voltage of the power supply, subtract the voltage demands of each transducer at 20ma. The voltage left  will be the voltage available to overcome wire resistance. Use Ohm's law to find the maximum length. Lower resistance wire means longer circuits. Don't try to use 100% of the allowable resistance. The voltage ratings of the transducers may be subject to tolerances. I once had a high temp shutdown fail to operate. The loop was loaded to almost 100% of nominal capacity. The input resistor of one of the devices was 15% or 20% off spec. Given the real resistance rather than the nominal resistance, the power supply was unable to push 20 ma through the loop.
The high temp shutdown failed to operate.
Moral, test loop resistance and don't use 100% of the allowable loop resistance. The loop may fail if a device is replaced with an off spec device.

Bill
--------------------
"Why not the best?"
Jimmy Carter

dpc (Electrical)
13 Aug 08 14:28
800 meters should be very achievable.  As suggested, you need to know the maximum impedance that the transmitter can work into and then add up the impedances in your loop including your meters, inputs, etc.  Then start increasing the conductor size until you get to something that works.  If the VFD analog output has a limited capability, you can add a powered loop isolator at the VFD and increase the capacity.  

I have run 4-20 mA loops over several miles.  I think we used #10 AWG.  

 
itsmoked (Electrical)
13 Aug 08 15:25
Yep no limit.  Just wire size cost.

Keith Cress
kcress - http://www.flaminsystems.com

ozmosis (Electrical)
13 Aug 08 19:53
sunrays
One other thing. Take away what the guidelines say about maximum lengths of 4-20mA and look at the installation quality of your VFD's. A VFD, as well as controlling the speed of your motor, creates high levels of RFI. If the installation is poor-and by that I mean the ability of the high frequency interference to find a low impedance path to ground, control cables are not sat right next to motor cables, cables cross at 90Deg etc.
If these and other recommendations provided by the VFD supplier are not adhered to then 8 or 80 metres is a challenge.
  
danw2 (Industrial)
13 Aug 08 23:01
Did you discount wireless after a survey check?  Did you call a 2nd wireless provider?

Line of sight is not absolutely essential with full 1 watt ISM band 900Mhz freq hopping (US situation) wireless.  If there's some structural steel with openings (as opposed to solid steel walled buildings in the direct path) to bounce the signal, then 800m could very well be doable with  wired in/wired out wireless.

Dan
SteveWag (Civil/Environmental)
14 Aug 08 9:44
I was maintaining a 4 to 20 ma. loop that was over 5 miles long, one way. The main portion of the distance was two independent conductors, overhead on glass insulators. I measured loop resistance once a month to detect bad splices. 800 meters should be no problem, #18 wire has a resistance of less than 7 ohms per 1000 feet, # 16 has less than 5 per 1000.  # 16 would have a loop resistance of 27 ohms and # 18, 37 ohms. This resistance is trivial. Be very aware that if the ma. signal is anywhere near the VFD load conductors there is a high chance of induced noise. Best to use steel conduits for both.
Steve
 
electricpete (Electrical)
14 Aug 08 10:54
On the surface, it sounds like you are suggesting that  the resistance associated with 5 miles each way of cable as small as 18AWG is "trivial" in a 4-20 milliamp circuit.  I would respectfully disagree with that suggestion.

5 miles each way = 10 miles round trip = 50,000 ft.
For 18AWG use your number  37 ohms/1000 ft.
Resistance = 37 ohms/1000ft * 50,000 ft = 1850 ohms.

Now, we expect the circuit to carry 20 milliamps.  
V = IR = 0.020A * 1850Ohms = 37 volts used by the wire alone.

A 0-1V input device has input impedance of 1V/20milliamps = 50 ohms.   A 0-5V input device has input impedance of 5V/20milliamps = 250 ohms.   (1850 ohms  doesn't sound trivial compared to 250, does it?)

We need 1-5 voltage for just a single receiving device.   Let's say 5 volts.

And we need approx 12vdc left accross the transducer for it to operate correctly.

Power supply requirement is 12vdc + 5VDC + 37VDC = 55VDC.

Typical power supply might be 24vdc.  What happens when you use a 24vdc power supply  to drive the above system?  You couldn't even push the current through the wire alone, let alone leave enough voltage for the transducer to operate correctly.

But , let's assume there is a special power supply, even though you haven't said a special power supply is needed for these long distances.  Give it the full 55vdc power supply needed to power the full 20 milliamps through all that resistance.

Now what happens when we reduce the signal to 4 milliamps?   The (37 + 5)  = 42 drops to 42*(4/20) = 8.5.   The voltage accross the transducer is now 55 - 8.5 = 46.5VDC.

What happens when you put 46.5VDC on a transducer?

http://celesco.com/faq/420.htm

Quote:

Warning! the absolute maximum our sensors should ever receive is 40 volts--anything greater and you'll have toast.
I think you'll need a special transducer to go along with your special power supply.
 

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SteveWag (Civil/Environmental)
14 Aug 08 11:19
No, No, the 5 mile loop was No. 10, mostly overhead, connecting a power plant with a well level monitor. I was talking about the originator of this topic when I mentioned # 16 and # 18 conductors in the 800 meter loop. 1600 meters is 5250 feet of wire and will have about 40 ohms resistance, depending on stranding and temperature, and a voltage drop of less than 1 volt.
Steve
 
electricpete (Electrical)
14 Aug 08 11:23
Sorry.   Others in this thread have expressed an opinion that there is "no limit".  I thought you were expressing the same view.
My apologies.  

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waross (Electrical)
14 Aug 08 11:41
You have given a good explanation and description of the limits of a 4-20ma loop, electricpete.
I am happy to overlook the misunderstanding that prompted your reply.

Bill
--------------------
"Why not the best?"
Jimmy Carter

ScottyUK (Electrical)
14 Aug 08 12:24
Well, theoretically there isn't a limit provided the conductor resistance is kept low enough. From any practical perspective there are reasonable maximum sizes of cable which could be used before economic considerations drive the choice away from 4-20mA. I haven't yet seen one of our instrument engineers asks for 500mm2 LV power cable because he needs to run an especially long loop: it would work ok but the cost would be outrageous. Just because something which works in theory doesn't make it a good solution.
  

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If we learn from our mistakes I'm getting a great education!
 

SteveWag (Civil/Environmental)
14 Aug 08 13:02
Well, lets see what it takes to build a very long 4 – 20ma. loop. But where to? The moon must be ruled out as the cost of the Earth-end rotary coupler would be pretty high, not to mention the tree huggers' objections. (No coupler needed at the Moon-end but a cord reel would be good there, less weight than on Earth) So let's keep the loop on Earth. Let's go from one side of the Earth to the other. Now the diameter of the earth is a little less than 8,000 miles so I'll call the circumference 25,000 miles and round up to 30,000 miles to allow for right-of-way acquisition issues, hills, savages, and other unknowns. I propose a 24 volt supply, I have some and they could act as spares. My transmitter measures light intensity and requires a minimum of 9 volts when producing 20 ma.  My chart recorder has a 250 ohm input and drops 5 volts at 20 ma. That means 10 volts, max, loss in the loop conductors. R=E/I = 10/0.02 = 500 ohms loop resistance. 30,000 miles is about 160,000,000 feet, so, 500 ohms/160,000,000feet = 0.000004 ohms/foot or 0.004/1000. 0000 copper has 0.05 ohms/1000 feet, so, 0.05/0.004 = 13 paralleled conductors, say 15 and allow space in the conduit for 40. Now 0000 wire weighs 641 lbs/1000 feet, for 15 conductors say 10,000 pounds or 20,000 pounds per 1000 feet for both supply and return. Total copper used is about 20,000# x 160,000 = 3200 million pounds of copper. I would install 15 x 2 conductors with additional space for 5 x 2 conductors if needed for route changes and splice loss.
Steve
 
electricpete (Electrical)
14 Aug 08 13:48
Good one.  At any rate, I withdraw any objection to anything anyone said at any time.

I recreated now where my approx 500 ohm number from my very first post came from.   24vdc power supply.  12vdc transducer minimum leaves 12vdc for the rest of the circuit.  12vdc/0.02A = 600 ohms.  Round down to the nearest round number(500) as correction factor to account for engineer's fading memory.

Let's go back to the original post:

Quote:

Option B: 4-20mA signal. I need a converter at both ends to convert the 0-10V Signal of Drive A to 4-20mA and 4-20mA back to 0- 10 V at Drive B since the drives can deliver/accept only Volts. The cable proposed to be used will be shielded one.
I assume 24 volt power supply in drive A

Drive B will take 10 volt at 20 milliamp, leaving 14 volts for the cable and the transducer A.

I assume transducer A requires around 12 volts, and that leaves 2 volts for cable.

The calculations provided above 1600m total ft of 18AWG at conservatively elevated temperature gives approx  40  ohms or 0.8 volts

So, of 2 volts available, the cable voltage drop takes up 0.8 volts and we still have some margin.  If we tripled the length of the cable, we would be in a situation where we would need to start looking at bigger cable or higher voltage supply.

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SteveWag (Civil/Environmental)
14 Aug 08 13:53
I just had some spare time over lunch.
Steve
electricpete (Electrical)
14 Aug 08 13:56
And by the way, thanks for your comments Bill. I needed that after sticking my foot in my mouth.

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peebee (Electrical)
14 Aug 08 17:43
That's pretty good for a civil/environmental guy.
roydm (Industrial)
15 Aug 08 0:08
Some of the posters seem to think this is a two wire loop. Drive A puts out 0-10  Volts, Drive B needs 0-10 Volts
In order to use a current loop
   At Drive A you will need a 4 wire 0-10/0-20 transducer. Either way you will only loose a few volts e.g. 5 (not 12 as you would with a 2 wire transmitter) so assuming you just have a 500 Ohm resistor at B you have 9 V for line loss.
   If you use a loop powered 0-10/4-20 transducer at A you would also need a transducer at B but this would be 250 Ohms so now you have 14 Volts for line loss. But now you can also use a higher power supply voltage e.g. 48 V giving you 38 for line losses.
Not all transducers are 250 Ohms some are 50 Ohm or less.

Another option would be fiber optics thus eliminating any noise problem and need for conduit.
http://weedinstrument.com/product_groups/fiber/analog_system.html

Roy
electricpete (Electrical)
15 Aug 08 1:15
Thanks rodym.  When I worked with transmitters, they were all remote 2-wire pressure / flow transmitters that get all their power from remote power supply through those two signal wires.  I guess if the transmitter is in a control panel like a vfd, then it is usually 4 wire with direct input from both a power supply and a transducer?

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sreid (Electrical)
15 Aug 08 9:35
Did anyone notice that 18 ga wire has a resistance of about 6.4 Ohms/1000 ft, not 37 Ohms?
SteveWag (Civil/Environmental)
15 Aug 08 10:34
From my post of 14 Aug 08 9:44,
 
"800 meters should be no problem, #18 wire has a resistance of less than 7 ohms per 1000 feet, # 16 has less than 5 per 1000"

Steve
 
electricpete (Electrical)
15 Aug 08 13:18
This distance we are talking about was 1600m.  Round numbers 5000 ft.  5*6.4 =32. Conservative allowance made for temperature higher than reference (20C) and other factors.   

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