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Upon a second test, I get 7.5V on the tip, about .45V on the ring, and 0 on the sleeve (makes sense).  Now my question is why the 7.5V on the tip?  That said, when you look inside at the jack connections, the tip pin should be the only thing getting a measurement.  The red wire and the copper are tied to GND.  How is there a voltage in the ring and why is the voltage so high at the tip.  Is it because there was no burden resistor tied to it?

W H A T ! ! !  You tried to use a current transformer to measure mains voltage? Was there a big blue flash and did you trip out your supply? You should have done. (Actually, I don't think you meant what you wrote.)

A current transformer obeys all the rules of any transformer. The ratio of primary voltage to secondary voltage is approximately the same as the turns ratio. So if you have 1 V appearing across the secondary, the voltage across the primary is (for a 100 A : 50 mA c.t.) 0.5 mV from the transformer action, plus whatever the IR drop in the primary winding is. Bear in mind the primary winding is a piece of wire about 21 mm long as is passes from one side of the c.t. to the other.

I don't know how many times I've written this on this site: NEVER open-circuit a current transformer. You and the c.t. only survived because the c.t. is small and has internal protection. The voltage could well be dangerous if the c.t. were any bigger or the internal protection diodes were to fail; and is certainly meaningless. The output of a current transformer is a current, it operates as a current source. Does the name not give you a clue? It will generate whatever voltage it needs to in order to drive the current it is generating. When it can no longer do that, it overloads, saturates and distorts. Read the report on the c.t. to see why you measured what you did.

Thanks again Robert.  I guess I didn't understand the open-circuit comment earlier?  Wouldn't the resistors in the multimeter make this a closed system?  Maybe I'm confused again?  I wasn't measuring the mains, just the CT around a single breaker in the house @ 1.5A or so.  I would imagine the mains come in much higher than 120V @1A, since I'm using so much more power than 180W?  I don't have access to the mains--it's behind another panel that I don't want to open.  I don't mean to do things wrong, but I am still trying to understand the entire system from the ground up.  Maybe I don't understand the layout of the CT sensor correctly.  I have read your CT sensor introduction and the SCT-013-000 reports many times.  What I don't see is a diagram of how this works entirely.  I see diagrams, but nothing shows the wire running through the core.  Inside the CT is there not only 1 wire spool with 2000 turns and the primary runs through the iron core?  Are there 2 spools?  I know about the diode on the secondary side.  Just so I'm clear, please tell me where the primary coil is and verify that if I have 120V mains @ 1.5A, then the primary coil in the CT is 120V @ 1.5A.

Also, I understood that this is a current transformer so yes current does flow away from the CT.  That said, you can't have current without a voltage, so I was trying to see how much I was measuring.  If I was measuring 7.5V on the tip, i.e the voltage off the secondary, and there was a 2000:1 turn ratio, then shouldn't the voltage in the primary be 2000*7.5=15kV?  15kV would be high considering the multimeter is giving me a 120V (plus or minus) reading at the circuit breaker.

If NOTHING is connected to the ring, why is there a voltage on the ring.  The tip and sleeve should be the only things possible to be carrying a current.  This still doesn't make sense to me.

Sorry, one more question related to this.  Say 15kV is the voltage, I'm not sure why we think this is dangerous.  Isn't the current what we should be cautious of?  I've been hit by 250kV in physics class and no problem because it was a low current.  Can this be explained some more?  Sorry for the basic questions, but I don't think I'm the only one who has them and I can't find a good solution anywhere else.  I never find a straight answer...

Wouldn't the resistors in the multimeter make this a closed system?

If you've got your multimeter set to measure current then yes.  If set to measure voltage then the multimeter impedance is very high and you can still consider it pretty much open circuit.

Just so I'm clear, please tell me where the primary coil is

The primary "coil" is the single mains wire that you clamp with the CT.   A common practice in these parts when bench testing this stuff is to actually loop that main wire so that it passes through the CT 2 or 3 times.  This allows you to measure 2x or 3x the current without actually having to switch that much load on.  In that case your primary "coil" really is a coil.

With regards voltages and currents, yes, generally it's the current that kills you, but the high voltages still hurt.  I sometimes use an old mechanic's trick to balance the two carby's in my classic car.  Pull the ignition lead on cylinder 3 and note the RPM drop, reconnect that and do the same with cylinder 4.  Each time I do it, I know it's not going to kill me, but I still don't enjoy it much.

Yes, I would agree that it's the current that will actually do the damage, this being dependent on the source voltage and the circuit's resistance.

The recognised limit for a potentially fatal AC current is around 50 mA.  To get no more 50mA from a 250 kV supply, the overall resistance would need to be at least 5 Mohms.  So if your 250 kV supply has an internal resistance of at least 5 Megs, then you could expect to survive the experience.  Anything less, and you'd be relying on poor skin contact to keep the current down.

A local gentleman of 93 recently mentioned to my wife that he has had to stop using his trusty hand-held vacuum cleaner because it was giving him electric shocks.  Hardly surprising, given the state of it!  Maybe we should start a rogue's gallery for items such as this ...

Thank you so much!  So the main wire IS the primary coil!  It makes alot more sense now.  I'll be careful not to open circuit the CT anymore.  I could hear the diode screaming at me, after I disconnected my multimeter, so I knew it probably wasn't smart, but now I know why.

Does anyone know why the CT would have a voltage in the tip. ring, and sleeve?  It was wired only to the sleeve and tip.  I'm using the SCT-013-000, just like the diagrams on the page.  According to the diagrams and text, only the tip and sleeve should be useful and I'm getting positive voltages everywhere but ground (obviously)...Thanks.

Last, what should I be expecting out of my CT sensor.  I know it's pushing current and voltage to match it.  Does it have internal resistance at all?  With a N ratio of 2000, primary voltage of 120V @ 1.5A, then shouldn't we get:

Voltage Primary (V) = 120
Voltage Secondary (V) = 240000 (240 kV)
Turns Primary = 1
Turns Secondary = 2000
Current Primary (A) = 1.5
Current Secondary (A) = 0.00075 (75 mA)

If so, wouldn't that make the CT dangerous closed AND open?

Thanks everyone!

OK, taking the last post first. It is indeed current that kills. You can obviously spend a hour or two with Mr Google reading up on how electricity kills people, and I strongly recommend you don't try any practical experiments - at least not until you get a much better grasp of electrical theory and practice. A high voltage is only safe if there is some means of limiting the current to a safe value - the human body is nothing more than a bag of salty water as far as electricity is concerned, and it's quite happy to offer very little resistance and carry it from one end to the other. If vital organs like the heart are in the way, it's bad news.

You're right, the primary winding of a c.t. is usually (but it doesn't have to be) a single "turn" - the wire passes through the core once only. The diagrams do show the primary - this is the official symbol for a c.t:

The primary is the heavy horizontal line, and the test rig circuit in the report shows the entire primary circuit. The report also has a picture of the disassembled c.t. Only half of the core is there, and obviously no primary and no body, but everything else is visible.

As I said, the c.t. is a current source. It requires a totally different way of thinking to understand current sources because you are so used to seeing and thinking in terms of voltage sources. A battery is a voltage source and it is quite happy when it is open-circuit. A current source is quite happy short-circuited, and that is more or less how a c.t. normally works. Your multimeter on the voltage range is - as far as the c.t. is concerned, an open circuit (read the manual!  - what is its input impedance - 10 MΩ or so? Compared to the normal c.t. burden of a few Ohms, that's open-circuit.) Again as I said, the voltage you measured is meaningless because the c.t. will generate whatever voltage is required to drive the current it wants to.

The voltage of the c.t. primary circuit does not affect how the c.t. works, or what its output is, at all. So you can forget all about that. (But if you want to be picky, you can calculate the primary voltage knowing the secondary voltage and turns ratio. If the secondary voltage is typically 1 V, and the turns ratio 1:2000, then the primary voltage is 0.5 mV. I really mean that - it's the voltage over the 10 mm or so of the main cable from one side of the c.t. to the other. Remember which way round the turns ratio is  - that's where you went wrong when you got 15 kV.)

Why you are measuring a voltage on the plug ring when there's no connection is because there is a connection. You can't see it, there's not a component there but the construction of the plug means there is a tiny capacitor connecting each metal part to every other. The very high input impedance of your meter means that a very small current flows through that capacitor and it's enough for your meter to register. There's only one sensible measurement you can make with your meter and a c.t.: put it on a current range and connect one probe to the plug sleeve and the other to the plug tip, and read the current. (Safety warning: immediately you have done this, take your meter off the current range in case you forget and try to measure a voltage.)

[Edit - 3 posts while I was typing - why did I bother????]

squareone: Your maths is wrong. Try again: 1.5 A / 2000 != 75 mA

Remember, a transformer - any transformer - does not create power. The power is transferred from primary to secondary intact barring a small loss. So primary voltage × primary current ≈ secondary voltage × secondary current. You can always use that as a check on your sums.

And the secondary resistance - can't you measure it (when the c.t. primary is not energised of course)? If you see anything wildly different to 100 Ω, there's something wrong.

Robin, what's wrong with that? - that the local tip wouldn't sort out? I sincerely hope you've persuaded him that a new one would be a good idea - and the sooner the better.

Robert, thanks for the post.  I'm going to kick back and try to absorb this.  I have a working version of the CT sensor and voltage measurement.  I was just trying to figure out if I could back calculate how everything was working.  Needless to say I won't be measuring a CT open circuited again.  

I understand how the CT works now, though I clearly have some work to do.  Can someone please confirm that I did the maths right here:

Vp/Vs=Is/Ip=Np/Ns

Those equations led me to my calculations earlier.  Sorry if they are wrong, but it's what I found googling.  I did measure 100 ohms across the unenergized secondary, as you said.  Didn't even think to try that.  Thanks for the idea about the capacitor on the tip of the CT as I didn't think about that.  It makes sense though now as you have 2 conductors separated by a insulator...

I would love to see a CT breakaway, I guess.  Maybe I'll take one of the CTs apart to see what's there myself.

Vp/Vs=Is/Ip=Np/Ns

Correct - if you ignore losses (I can be pedantic at times!). You should read 0.75 mA (750 µA) at 1.5 A, not 75. What you'd done earlier was use the turns ratio in the wrong direction for the voltage. I've never tried this, but you should be able to feed 1 v into the secondary and read 0.5 mV across the primary.

Maybe I'll take one of the CTs apart

The YHDC? It's quite easy and non-destructive - just open the c.t., spot the two plastic clips that hook over the side of the bobbin and gently push them both away together, at the same time push on the cable to lift the core and pcb out. Note: needs 3 hands or 2 and a vice.

Thanks again sir!

My id is ********@gmail.com.Please suggest me wt do i do to learn hardwares.I'm working as an embedded software engineer.Please suggest me about hardware.I realized to ask you after i saw your solutions which u had given for others.

I learned the theory by studying at University and at the same time doing a student apprenticeship in industry at a firm making industrial controls. I started in 1967, I've been in the electrical engineering industry since then. If you are not in the industry - I don't really know. There is obviously now a resource that just did not exist when I started nor for a long time after - the Internet. I haven't really thought about it before. 

You must be careful what you read on the Web - there is some quite simply dangerous information, and a lot that is not of very good quality. The manufactures' application notes are a very reliable source - they have to be because they are what the industry bases its designs on.

You say you work on embedded software - what is it embedded in, where does the hardware come from? Is there someone in your office who looks after it - if so they are the best people to ask. If your firm buys the hardware in, who from? Talk to your supplier and ask them about it - they should be happy to talk to you as the more you know, the likelier it is that you and your firm will buy from them!

Or there's this site. There's quite a lot about hardware here - all the designs are published as open source.

The best way in the end is do it. Buy the components and some basic tools and test gear, build your own version of a known good design and experiment. At the same time study the theory and work out for yourself exactly why it works when it works, and then when it doesn't work you can work out why it doesn't. And that's how you learn.