Extending wires between Current Transformer and Panel Meeter help

Thread Starter

diebog

Joined Mar 23, 2013
223
Sounds like you have a 3 phase motor and they monitor a single phase current which is not unusual at all. For some reason I want to believe your original CT is a 50/5? Meaning with 50 Amps through the primary (the single wire through the hole) you would get 5 Amps out of the secondary when properly terminated into a burden resistance. Your analog meter serves as a burden resistance. You never want to run a CT into an open circuit. The CT needs to see a load on the secondary which we call the burden resistance. Think of it this way. A CT with a 50/5 is a transformer with a 10 to 1 ratio correct? The primary line through the doughnut hole is current but also a voltage. So if the voltage is let's say 200 volts then the CT secondary, open circuit, would have 2,000 volts without a burden resistance. My guess (and only a guess) is with the system shut down if you measure the input resistance to the current meter it will be about 0.200 Ohms. That is with nothing connected to the terminals, only your ohmmeter. So with 50 amps primary and 5 amps secondary through 0.2 Ohms the voltage will be 1.0 volt. The meter is likely a 0 to 1 volt meter scaled to read 0 to 50 Amps. The meter has an internal "shunt" which serves as a burden resistor.

The problem with extending the leads of a CT are multiple. Most CTs like what you have are designed to work into a very low burden resistance. It doesn't take much lead length to upset the balance of the burden resistance and cause large errors. Additionally there is a matter of lead length contributing to causing the transformer to go into saturation. Something I have not mentioned is if you were to terminate the CT secondary at the source (in the panel or at the CT location) you may be able to run a light gauge twisted pair to a meter but the meter now would need a high input impedance.

Here are your options with the current transducer. You are looking at a 0 to 40 Amp AC transducer. The output will be 4 to 20 mA on that model. So 0 to 40 Amps = 4 to 20 mA. You place a resistor (a good 1% resistor) in the current loop. Using a 500 Ohm resistor the voltage drop across the resistor will be 4 to 20 mA = 2 to 10 Volts and you get a scalable panel meter to read 2 to 10 volts = 0 to 40 Amps. That is how it is done.
Ron
Sorry I haven't responded in a while. I had a job that had to get done and I was working late every day to get it done so I didn't have time to do anything with this.

Yes you are right about the 50/5 CT. But the wire running through the primary is wrapped around so it basically goes through 3 times. From the research I did, that means it now becomes 25/5 correct? Which is what it was setup from the factory. So would adding more turns to the primary help out my situation? I also read adding turns to the secondary can also change the ratio. Is there any way that this might work? Or is there just no way to get these 2 wires at 50 foot which means a 100 ft loop to work with my CT and panel meter? Would adding a second CT in my control panel (which is the one 50 ft away) be worth trying?

Running new wires would be possible its just a matter of buying more wire. If this would 100% work, it might be the best way to go. Vs buying a Current Transducer and another meter to use the wires I have now. If I were to buy more wire, what AWG would be the best to go with. I know a few were mentioned above, but I don't want to buy a bunch of wire to find out I should of went a tad bigger. If you were in my situation, what would you do?


In N.A. there is Red Lion Controls or the $10.00 version on ebay.
Max.
I looked on eBay for the 10$ version but there are hundreds of them. Is there a specific one you have used? I don't even know if it will work so I dont want to get into this to much. Plus its not something that needs to be dead nuts accurate.

Your CT cable is a twisted pair, right?
Yes, inside the main cable that carries the 36 wires to and from the control panel, there are 6 groups of 6 wires and each group of 6 is all twisted together. So I am using 2 of the 6 on one of the 6 twisted wires. If that makes sense. Why do you ask? Does it make a difference?
 

Reloadron

Joined Jan 15, 2015
7,501
Yes you are right about the 50/5 CT. But the wire running through the primary is wrapped around so it basically goes through 3 times. From the research I did, that means it now becomes 25/5 correct? Which is what it was setup from the factory. So would adding more turns to the primary help out my situation? I also read adding turns to the secondary can also change the ratio. Is there any way that this might work? Or is there just no way to get these 2 wires at 50 foot which means a 100 ft loop to work with my CT and panel meter? Would adding a second CT in my control panel (which is the one 50 ft away) be worth trying?
Yes, it was configured as a 25:5 likely. Here is an example of multiple turns through the primary:

60 to 5 CT 3 Turn Primary.png

Note there are 3 passes through the primary so a 60:5 becomes a 20:5. That said all we are doing is changing the ratio of the CT. This will do nothing as to allowing a longer lead length. There is no easy way around this as covered in previous links that explain the involved formulas and don't forget saturation of the CT itself as lead length contributes to that. This is why when distance becomes a factor current transducers are used. You can experiment all you like but I don't see things changing. :(

Ron
 

Thread Starter

diebog

Joined Mar 23, 2013
223
Similar to this one, ebay 251362475955
Max.
That one is for AC correct? I thought we were talking about a DC meter to read a Current Transducer if I were to change out both the CT and meter. Maybe thats just what I was thinking.

It was suggested to try a digital meter vs my analog one, are digital meters able to read correctly over a long distance? Or is it just because it can be adjusted for the current drop? Would I end up using my CT or the one the supply with the meter as shown in the 251362475955 meeter? If this would work it would be the easiest way to fix this issue.

So Ron, if I went to 2 bigger wires this would reduce the resistance and should read correctly right? You said you don't see anything changing, were you just talking about adding turns to the primary? And the picture you posted is exactly how mine is. The wire wraps around 3 times.

As for as the info I read on adding a second smaller Current Transformer in the panel (which is at the end of the 50 feet of wire, where meter is), taking those wires and with a shunt/resistor pass them through this smaller CT and configured it in such a way that its secondary output would be close to what It should be when the meter was originally connected with the 10 feet of wire. What is your thoughts on this? Is it possible, or would I just be wasting allot of time trying to dial it in?
 

Reloadron

Joined Jan 15, 2015
7,501
OK, the meter you have was designed to work with the CT you have. That meter is in effect the burden resistance.

1) Measuring CT:
  • The principal requirements of a measuring CT are that, for primary currents up to 120% or 125% of the rated current, its secondary current is proportional to its primary current to a degree of accuracy as defined by its “Class” and, in the case of the more accurate types, that a specified maximum phase angle displacement is not exceeded.
  • A desirable characteristic of a measuring CT is that it should “saturate” when the primary current exceeds the percentage of rated current specified as the upper limit to which the accuracy provisions apply. This means that at these higher levels of primary current the secondary current is less than proportionate. The effect of this is to reduce the extent to which any measuring device connected to the CT secondary is subjected to current Overload.
  • On the other hand the reverse is required of the protective type CT, the principal purpose of which is to provide a secondary current proportional to the primary current when it is several, or many, times the rated primary current. The measure of this characteristic is known as the “Accuracy Limit Factor” (A.L.F.).
  • A protection type CT with an A.L.F. of 10 will produce a proportional current in the secondary winding (subject to the allowable current error) with primary currents up to a maximum of 10 times the rated current.
  • It should be remembered when using a CT that where there are two or more devices to be operated by the secondary winding, they must be connected in series across the winding. This is exactly the opposite of the method used to connect two or more loads to be supplied by a voltage or power transformer where the devices are paralleled across the secondary winding.
  • With a CT, an increase in the burden will result in an increase in the CT secondary output voltage. This is automatic and necessary to maintain the current to the correct magnitude. Conversely, a reduction in the burden will result in a reduction in the CT secondary output voltage.
  • This rise in secondary voltage output with an increase in burden means that, theoretically, with infinite burden as is the case with the secondary load open circuit, an infinitely high voltage appears across the secondary terminals. For practical reasons this voltage is not infinitely high, but can be high enough to cause a breakdown in the insulation between primary and secondary windings or between either or both windings and the core. For this reason, primary current should never be allowed to flow with no load or with a high resistance load connected across the secondary winding.
  • When considering the application of a CT it should be remembered that the total burden imposed on the secondary winding is not only the sum of the burden(s) of the individual device(s) connected to the winding but that it also includes the burden imposed by the connecting cable and the resistance of the connections.
  • If, for example, the resistance of the connecting cable and the connections is 0.1 ohm and the secondary rating of the CT is 5A, the burden of the cable and connections (RI2) is 0.1 x 5 x 5 = 2.5VA. This must be added to the burden(s) of the connected device(s) when determining whether the CT has an adequately large burden rating to supply the required device(s) and the burden imposed by the connections.
  • Should the burden imposed on the CT secondary winding by the connected device(s) and the connections exceed the rated burden of the CT the CT may partly or fully saturate and therefore not have a secondary current adequately linear with the primary current.
  • The burden imposed by a given resistance in ohms [such as the resistance of a connecting cable] is proportional to the square of the rated secondary current. Therefore, where long runs of cable between CT and the connected device(s) are involved, the use of a 1A secondary CT and a 1A device rather than 5A will result in a 25-fold reduction in the burden of the connecting cables and connections. All burden ratings and calculations are at rated secondary current.
  • Because of the foregoing, when a relatively long [more than a very few meters] cable run is required to connect a CT to its burden [such as a remote ammeter] a calculation should be made to determine the cable burden. This is proportional to the “round trip” resistance, i.e. twice the resistance of the length of twin cable used. Cable tables provide information on the resistance values of different sizes of conductors at 20o C per unit length.
Here is what you can try and I guarantee no great results. I believe you mentioned a 50 foot run. You can try using some AWG 10 wire. From past experience you know what the typical current was to the motor. See what you get. Keep in mind a 50 foot run is actually 100 feet of wire as you have 50 feet there and 50 feet back (round trip). There are formulas out there for working all this out but unfortunately I don't know the burden resistance of the existing meter. All I know is 5 Amps AC will drive the meter to full scale. If your CT has three loops passing through the CT primary then the 50:5 becomes a 16.66:5 CT which is real strange. Anyway, if you can get your hands on some AWG10 give it a try but I would not be too optimistic. Heck, I hated bending and working with AWG 4. So try AWG 10 and see what you get. Heck try AWG 12 if you want. Again, situations like this where distance becomes a factor they generally use a current transducer like we discussed earlier with a panel meter display.

Ron
 

Thread Starter

diebog

Joined Mar 23, 2013
223
Well either way I have to purchase something. If your still not sure even upping the size of the wire would fix the issue, then I should probably go with the current transducer. Id hate to buy 100 feet of AWG 10 and find out it didn't change much.

I don't want to dump a ton of money into this but I also want it to work correctly. So I think based on your suggestions that I will go with the current transducer. I am just not sure what to but. I would like to save some money buy buying off eBay where there is NOS and used Transducers. Like the ones I sent in a link a few posts back. Do I need a specific CTransducer for my application or just one that is within range? When using a fluke ammeter around the wire running through the CT, I got a startup spike of around 56 or so amps. After start-up it dropped down to 7-8 amps with no load. While under load I would see around 15-20 amps with 20 amps being the max amps the manufacture suggests to stop at (which is just a matter of turning down the cut speed)

So I suppose I need a 50 amp Current Transducer. Or will the 40 amp I had linked from ebay work? If it wont, will something like http://www.ebay.com/itm/NEW-SSAC-DCSA50-DCSA-Series-Loop-Powered-Current-Transducer-0-50A-QUANTITY-/160946097890?pt=LH_DefaultDomain_0&hash=item2579228ee2 this work?

http://www.ebay.com/itm/CR-Magnetic...833?pt=LH_DefaultDomain_0&hash=item3aa3f713c9
http://www.ebay.com/itm/Simpson-Mod...380?pt=LH_DefaultDomain_0&hash=item3f1f00428c
http://www.ebay.com/itm/Hal-50-S-Current-Transducer-/261508180711?pt=LH_DefaultDomain_0&hash=item3ce31a2ae7
http://www.ebay.com/itm/NK-TECHNOLO...818?pt=LH_DefaultDomain_0&hash=item35e681eb92

That last one says it can be switched to 0-50 as shown here on this data sheet http://www.flex-core.com/pdf-files/AT.pdf

Those are just a few I found on ebay. If the 40 amp onehttp://www.ebay.com/itm/CR-Magnetic...884?pt=LH_DefaultDomain_0&hash=item54056a9cbc would work, that would be great as its only 45$ But if it wont, hopefully some of those other ones will.

Once we get a Current Transducer picked out, what type of a panel meter will work? There is a square hole in the panel where the old one sat which looks like this


I can always cut the hole a little bigger, but cant go smaller obviously. Would that meter posted above work?
 

ebeowulf17

Joined Aug 12, 2014
3,307
Does the trouble caused by extra wire length come from something more than simply its higher resistance? I thought the losses here were simply a function of voltage drop due to higher resistance.

If that's the case, going with larger wire would seem like the simplest solution, and the online wire resistance calculators should make sizing calculations pretty straightforward. I came up with 15 gauge just like whoever originally mentioned the idea earlier in this thread. Since 15 is unusual, I figured 14 would be easier to get, and if that actually made the resistance too low, you could then add resistance as needed to fine tune your readings.

Again, all of that is assuming the problem here is simply a function of resistance (which I don't know for sure, and would love to learn to understand better.) If the effect of wire length and gauge on a current transformer is more complicated than that, I could be way off base.
 

Reloadron

Joined Jan 15, 2015
7,501
The existing meter panel cutout should be based on a standard known as DIN IEC 61554:2002-08. Panel cutouts should be standardized. For example:

1/2 DIN = 7.28" x 3.64" [184 mm x 92 mm]
1/4 DIN = 3.64" x 3.64" [92 mm x 92 mm]
1/8 DIN = 3.64" x 1.78" [92 mm x 45 mm]
1/16 DIN = 1.78" x 1.78" [45 mm x 45 mm]
1/32 DIN = 0.874" x 1.77" [22.5 mm x 45 mm]

My guess is your panel has a 1/4 DIN cutout so you want a meter that will fit a 1/4 DIN panel cutout. I would run with this transducer which you linked to. I would run the same number of primary loops, these transducers accept primary loops just like a CT does. This is a rough of what your circuit would look like:

Current Transducer.png
The resistor can be 250 or 500 ohms which will yield 1 to 5 volts or 2 to 10 volts respectively. That or you can get a 4 to 20 mA process meter. The meter you want is called a process meter commonly. They are programmable for voltage and current inputs. Most I have worked with were 1/8 DIN but there must be some 1/4 DIN sizes out there from China for a cheap price.

Ron
 

Thread Starter

diebog

Joined Mar 23, 2013
223

Reloadron

Joined Jan 15, 2015
7,501
That will work. Overkill but it will work. :) Also it is configurable for 4 to 20 mA so no need for a resistor as seen in my drawing.

Ron
 

Thread Starter

diebog

Joined Mar 23, 2013
223
Ya its way more than I need, but I cant find any cheaper models. On ebay anyway. So no other panel meter will work huh? I have to get a "process meter" in order to work with that transducer I am buying. I really don't understand what the difference is besides its programmable. As far as search terms, I was just using "process panel meter". Would there be any other terms or names people call these meters? That 10$ one that was linked above wont work right?
 

Reloadron

Joined Jan 15, 2015
7,501
The reason they are called a "process meter" is because they can be used to control a process. You can try looking for a "process monitor meter". There is no shortage of meters like this to be had including simple small loop powered meters. The problem is your 1/4 DIN hole. The Simpson F35 and F45 meters come to mind but they are 1/8 DIN like most meters of this type. Red Lion is one of the few players who make the larger 1/4 DIN sizes. The $125 Red Lion you linked to will work fine.

Ron
 

Thread Starter

diebog

Joined Mar 23, 2013
223
So I have been looking and looking for just a simple green 34 or 5 digit process meter and I am not having any luck. Most all of them are for temperature control as well. I guess just get one like that and ignore the temp part? I just seems like I'm buying this meter that can do all this stuff and has a bunch of buttons but all I need it to do is display the amps. And that's another thing I like it to have an "A" for amps just so people know what it is. So a meter like this one http://www.alliedelec.com/1/products/356679-1-16-din-4-digit-lcd-w-backlight-current-meter-mfr-part-ma501-110v-cu.html will not work? It looks like its adjustable, and would fit my panel.

If that's a no go, I found these two on eBay and I thought about asking how much for one. http://www.ebay.com/itm/131102878671 Here are the specs on it. http://www.omega.com/pptst/CNI_Series.html

This one is a bit more but may work to. http://www.ebay.com/itm/251636004151 Reason I was looking at the last two is they can change between green red and amber. Green is what Id prefer as the other panels displays are green as well.

I also found they have 1/16 to 1/32 din adapters. So I could get something like that. But that's a few more hours searching 1/32 din meters. I also may be able to cut the opening wider to accommodate an 1/8 din meter but its next to another display and I don't think there is much room. Ill double check tomorrow. Then I would have allot of options. Like this one on eBay http://www.ebay.com/itm/141585872383
 

Reloadron

Joined Jan 15, 2015
7,501
Be careful when choosing a process meter. Some that are designed for temperature will only do temperature. You want to make sure the meter you choose is designed for inputs like 0 to 20 mA. 4 to 20 mA. Another problem is most are 1/8 DIN. The Red Lion you linked to several post back will work fine with the CR Magnetics current transducer you were looking at. I have on occasion placed a new 1/8 DIN meter in a 1/4 DIN hole cutout making up (fabricating) new mounting. Just a matter of how creative and decorative you get with the hole.

Ron
 

Thread Starter

diebog

Joined Mar 23, 2013
223
I measured my panel today and It looks like I have room to cut it longer and use the 1/8 din meters. So that should help out allot. Now just to find a decent priced, green simple process meter.


 

Thread Starter

diebog

Joined Mar 23, 2013
223
So I have been looking for some time now for a meter that i like and that didn't have a bunch of options I don't need. I ran into this one which looks pretty cool and can be configured via USB and a PC to set everything. I just want to make sure that it will work before I ordered it. Can you see any reason why it wouldn't? Also they offer positive an negative LCD, if you were choosing, which would you go with? Here are a few links: The last two show the positive and neg versions
http://www.rammeter.com/media/specsheets/Trumeter_APM_Process_Specs.pdf
http://www.rammeter.com/media/specsheets/Trumeter_APM_Process_Specs.pdf
http://www.rammeter.com/trumeter-apm-advanced-panel-meter-w-dual-display-process-meter.php
http://panelmeters.weschler.com/ite...-trumeter-advanced-panel-meters/apm-proc-apo?
http://www.nesslegrove.com/Trumeter-APM-PROC-APO-Process-Meter-Positive-LCD-p/apm-proc-apo.htm
http://www.nesslegrove.com/Trumeter-APM-PROC-ANO-Process-Meter-Negative-LCD-p/apm-proc-ano.htm

I realize I will have to trim the opening a tad to make this work as its not 1/16 din. But that is fine.
 

Thread Starter

diebog

Joined Mar 23, 2013
223
Will I have to add any resistors with this meter? As far as hooking it all up, you said to stay with 3 wraps through the primary on the transducer I bought, and the transducer is loop powered so I dont need to bring 24v there, but the meter needs 24v which I can tap into in the panel. The only thing I don't really know is wiring the meter to the transducer as well as configuring the meter to take the voltage from the transducer and have the meter read out the amps correctly. I guess I will need to read the manual that it comes with real good. Is there any tips or suggestions you have? Any pointers would be great. I don't have the meter yet, I ordered it 2 days ago, but will post back when I get it in my hands.
 

Reloadron

Joined Jan 15, 2015
7,501
Was out of town for several days, sorry slow in my reply.

The first link in your post #37 shows the meter has a 4 to 20 mA input which working with your sensor we looked at will work fine. The meter includes software to set it up. I believe the sensor was a 0 to 50 Amp sensor so 0 to 50 Amps is 4 to 20 mA. That goes directly to the meter from the sensor in a current loop. Using the meter's software you scale the meter. The original CT had three passes through the primary (doughnut hole). You can make that the same. Then change the scaling accordingly.

Ron
 
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