How do I connect PTs and CTs for 400A 3ph service

Unfortunately every CT manufacture does it different. So connect everything through a shorting block. With 2 wires from each CT to the shorting block. Place the shorting block in an hot accessible area and install everything. Remember to fuse the wiring to the meter to local codes.

If you screwed up and got the wiring backwards, no harm just install the shorting screws and reverse the wiring to the meter. Your done.

If you do it the cheap way, with a common for all the CT's then you are faced with a shut down and reinstalling the wiring. The shorting block is

20-30 bucks and is a life saver. Your aware that you can not install CT's on a powered circuit with out them being shorted,,,,,,, you knew that right?

If this meter is for utility accuracy make sure that the CT's and PT's are utility grade IE less than 3-5% accuracy. 5% is the limit for utility grade out here in the west, your meter will add 1-3% as well. CT's used for tripping circuit breakers can be as high as 10-15%. You do not mention the voltage, over a 1000 volt you need a different grade of CT's and PT's. The clearances can be interesting above a 1000 volts. Most installations I have done below 1000 volts do not use PT's. Be safe and be careful.

| If you do it the cheap way, with a common for all the CT's then you are | faced with a shut down and reinstalling the wiring. The shorting block is | 20-30 bucks and is a life saver. Your aware that you can not install CT's on | a powered circuit with out them being shorted,,,,,,, you knew that right?

And once you have the CTs installed and shorted, how do you go about using them? Do you now un-short them? While the power is still up?

Are you seriously asking these questions? If so, never go near a CT circuit.

You never leave the secondary of a CT open when the primary is energized. You use a shorting block to short the secondary when you need to remove a meter (for example). Once the meter is installed, you simply remove the shorting bar from the block. This can be done safely with the circuit energized.

Charles Perry P.E.

If you have to ask, then you better get someone knowledgeable to do this for you. CTs produce lethal voltage across an open secondary when current flows through the primary side. The shorting bar prevents this until the meter is connected.

Ben Miller

Charles, You can help me out on this:

CT=============================Burden=meter

That is the way it is installed now - about

100 feet between the CT's (one on each phase, 30 KW 208 rooftop mounted heater) and the burden resistors. My knee-jerk reaction is that that is wrong, and it should be:

CT=Burden=============================meter

I don't remember the values, but the burdens were sized right to keep the maximum V low and were oversized for dissipation - 50 watts when the max was something like 5 watts - as I recall. The question is about the placement. Is it legitimate to have that distance? If a burden opens ... well, I don't want to think about what could happen, I'd rather limit the "physical exposure" to the much shorter path. Am I all wet?

Thanks!

| Are you seriously asking these questions? If so, never go near a CT | circuit. | | You never leave the secondary of a CT open when the primary is energized. | You use a shorting block to short the secondary when you need to remove a | meter (for example). Once the meter is installed, you simply remove the | shorting bar from the block. This can be done safely with the circuit | energized.

I know it needs shorted. The question was whether the step of unshorting it so the meter functions can be done safely with power on. If there is a danger there that shorting mitigates, are you saying a meter will do it, too?

Just how much voltage is coming off that thing (for some given current and CT you specify)?

Well, that's two people that misunderstood my question. So I guess it's my fault for wording it badly. So I will try it from another angle:

Why is it safe to unshort the secondary once the meter is attached?

And: what is the voltage?

Hire someone who knows what they are doing!!!!!!!!!!!!!!!!!

The meter acts as the"short" when it is in the circuit, but you should hire someone who knows what they are doing.

The voltage across an open CT secondary can be thousands of volts, but you should hire someone who knows what they are doing.

Charles Perry P.E.

I think the burden resistor is best where it is presently located. You will have much less interference than if you had the resistor further away. There would be a slight increase in safety by burdening close to the CT but your readings will likely be more accurate with the burden where it is.

Personally, I don't use many meters that require a burden to measure current. Most of what I use are meters that take 5A input current.

Charles Perry P.E.

Because the meter provides the low resistance path for the CT secondary current.

Normally? A few volts at most. For example, if the meter burden is 0.5 Ohm, then the voltage at full load (5A secondary current) will be 2.5 volts.

Open? The CT at full load is trying to produce a 5A secondary current. With an infinite resistance, the voltage will theoretically go to infinity trying to push the current through the circuit. The actual voltage is limited by the magnetic characteristics of the CT core, but it could be from one to several thousand volts.

I don't think any of us misunderstood the question. We are just sensitive to the hazard that exists for an untrained individual, and are showing concern for your safety. People are electrocuted by CTs.

Ben Miller

Thanks!

The question you ask makes it clear you do not understand how to use CT's. That is not said as an insult, but rather to underscore what other's have told you: you need to get someone who knows CT usage to help you out. I'll try to shed some light on it - but *please* don't mess with these unless you know before hand exactly what will happen.

Think of the whole circuit, in "electronic" detail:

CT=======Meter is, electronically, the secondary of a transformer with a resistance connected across it, so it can be redrawn as CT=====R

The CT is designed to achieve a specific current (NOT a specific voltage as in a "regular" transformer) in the secondary when a specific current flows in the primary. This concept sometimes causes those learning about CT's some difficulty, but it is critical to understand it: a current will exist in the secondary with or without a load connected to the secondary.

Now, that current transformer will produce some specific current when a specific current flows in the primary. Lets use the 2.5 amps as the current produced. What value of R is required to keep the voltage safe? Assume whatever voltages you want, within reason, and work ohms law to get a feel for how the voltage changes depending on the resistance. Say, for example, the resistance is 1 ohm. The voltage therefore has to be 2.5 volts. Now assume that the resistance is 11 megohm, as it would be if you wrongly set the meter (DMM) to the voltage scale. Remember, E = IR, so E would equal 2.5 amps times 11,000,000 ohms!

If you use the meter properly, it would be set on the amps scale, where a low resistance shunt is connected in the circuit. That shunt is why it is safe to remove the shorting bar with the meter properly connected.

There is another circuit to consider:

CT====BurdenResistor====meter.

In the above case, the burden resistor keeps the voltage low, and the meter can be, and is, used on the voltage scale.

Bottom line - you never want a current transformer on an energized circuit without either a proper load or a shorting bar connected across the secondary. An open, or improperly loaded secondary circuit on a CT is a definite "NO-NO".

The output of a CT is a specific value of current for a specific input current - the secondary CURRENT is proportional to the primary CURRENT. The primary voltage is fixed, and the primary current varies. The secondary current varies when the primary current varies. The secondary voltage depends on the secondary current AND the resistance connected across that secondary.

See, that's the thing. When using CT's *you* need to select the voltage. You either make it low enough where you don't care about it at all (with a shorting bar), or you select it and design it into the system by proper selection of the burden resistor. You need to understand this and be able to do it before you can mess with the things safely.

I've been looking through my collection of "scary aftermath" photographs for a few shots of the destructive results of failed/improperly serviced/applied CT installations that I thought I had saved. Unfortunately I can't locate them.

Perhaps if anyone has access to photographic documentation of the destructiveness of such a failure, and could post them to alt.binaries.schematics.electronic it could emphasize, and illustrate the seriousness of uninformed service/experimentation involving CTs that many of you have explained so well.

Louis--

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Ok while we are on the subject heres my next question and thanks to everybody who answered my last one:

If I had a 600/347Y 400A service and Lets say you have CTs at 400:5 and PTs at 360:120 connected like the picture on the left in this link.

The above values of transformers is what I have been made aware is commonly used for this type of service.

The CT ratio is 80:1 and the PT ratio is 3:1 This would mean that the CTs would deliver 0-5A to the meter current coils and the PTs would supply 115.6V to the voltage coils of the meter. I'm guessing the meter would be specifically calibrated to use

115.6V instead of 120V such as you would expect on a 480V service. Let me know if I am out to lunch on that one.

Next, since the CT secondaries are Y connected, and so are the PT secondaries, the meter multiplier would be the product of the two ratios or in this case 80x3=240. So everything on the meter's register would have to be multiplied by 240 right?

To send me e-mail remove the sevens from my address.

Chrisd

serviced/applied

OK. And what was the mechanism of the failure in your experience.

EMWTK

To bad. Although one blown up panel looks about the same as another blown up panel.

>

Not necessarily! Some are a lot more dramatic than others, and some are purely caused by "pilot" error and not equipment failure. You know, the fingers/tools where they don't belong, or haven't the expertise necessary to complete the task without a catastrophic result syndrome!

The two CT failures I can clearly remember (and used to have pictures of) were a result of the secondary circuit being opened while the switchgear was energized, and under load. One was due to poorly secured termination connections verified by multiple remaining loose connections that weren't destroyed, and deemed to have never been tightened properly. The other was caused by an individual that removed a secondary connection lead without proper preparation, and training. The end result in both cases was switchgear that required considerable man hours to repair before being returned to service. Amazingly there wasn't any personal injury in either case (with the possible exception of reputations).

A PT failure (the assumption due to lack of material left to evaluate) I used to have pictures of was assumed to have been caused by a technician attempting to change the PT while the switchgear was energized. The resulting damage was exacerbated by improper over current protection. The switchgear in this case had to be replaced. It was deemed costlier to repair than replace. The amazing thing was that the technician was burned, but wasn't killed! Unfortunately he had little recollection of the events just prior to the fireworks so little was gleaned other than proper safety procedure wasn't followed. The summation by the investigation team was that most likely the service wasn't properly deenergized, and bonded to ground before work was commenced. It never was clear why the technician was attempting to replace the PT. If someone knew they weren't talking.

Louis--

********************************************* Remove the two fish in address to respond

|> | Are you seriously asking these questions? If so, never go near a CT |> | circuit. |> | |> | You never leave the secondary of a CT open when the primary is | energized. |> | You use a shorting block to short the secondary when you need to remove | a |> | meter (for example). Once the meter is installed, you simply remove the |> | shorting bar from the block. This can be done safely with the circuit |> | energized. |>

|> I know it needs shorted. The question was whether the step of unshorting |> it so the meter functions can be done safely with power on. If there is |> a danger there that shorting mitigates, are you saying a meter will do it, |> too? |>

|> Just how much voltage is coming off that thing (for some given current and |> CT you specify)? |>

| Hire someone who knows what they are doing!!!!!!!!!!!!!!!!!

Lose the 'tude, dude!!!!!!!!!!!!!!!!!

Only a small percentage of people think that because there is something that someone doesn't know, and knows enough to ask, that they should NOT be told. You appear to be in that small percentage. Of course, you are not obligated to be the one to answer ... just ignore if you wish. But instead, your attitude is that someone who doesn't know shouldn't have an answer. OTOH, maybe you don't know enough about this yourself to even give a valid answer if you wanted to. I am not the one installing one, but if I were, I'll be sure to NOT hire YOU.

| The meter acts as the"short" when it is in the circuit, but you should hire | someone who knows what they are doing.

You didn't answer the question I was asking. Did you mix up the answer and answer a question someone else asked?

| The voltage across an open CT secondary can be thousands of volts, but you | should hire someone who knows what they are doing.

You didn't answer the question I was asking. Did you mix up the answer and answer a question someone else asked?

| Charles Perry P.E.

It's a good thing for you that the engineers license doesn't take into consideration one's personal attitudes.

I don't think it is safe to remove the shorting block with the power on without special safety steps. Somehow, I bet you don't do any.

1) What was the voltage level in the primary?

2) Did the insulation in the CT break down?

And what are the specs typical used to measure those characteristics? Impedance?

| I don't think any of us misunderstood the question. We are just sensitive to | the hazard that exists for an untrained individual, and are showing concern | for your safety. People are electrocuted by CTs.

"Don't let a little training get in the way of an education" ... paraphrased quote by Ben Franklin.

Are you saying that no hazard exists for those who are trained? Based on the numbers of trained people I see listed killed every year from various electrical accidents, I'd say that training might well be inadequate. Or maybe there's just a lack of education to simply _understand_ what's going on.

Whether you actually understood my question or not I cannot be sure of. I speculate that you did not because of the fact that it went unanswered while you were attempting to answer something. No matter, anyway; you didn't provide any helpful information at all. Someone eventually will.