What would be the ball park or typical impedance of a current transformer, say somewhere between 200:5 and 4000:5? I would think it would be quite low because there shouldn't be that many turns in the winding and it needs to handle 5 amps.
Do you really want someone to field this for you ?
with what's given, from me you will probably just get a formula using (20:1::800:1) ~ .oo5., unless you tell us what wire gauge you are to use in your Tr.s so we can throw in wire resistance and consider Z.
I don't go through this for the heck of it }:) what's in it for me ?
ops, make that.oo5 back into a (5) i thought i saw 5ma.
later
It might help if we knew what exactly you are trying to do.
Charles Perry P.E.
(ac or dc)? you should have this
AC Only Transformer Impedance Ratio:
Zp/Zs + (Np=B2/Ns) OR Zp =3D ZsT=B2
where T is your primary to secondary turns ratio
A current transformer obeys the same rule as any other transformer.
The impedance of the secondary is divided by the SQUARE of the turns ratio and effectively inserted into the loop you are trying to monitor.
If you have a 50 mV shunt on the secondary side rated, say, for 1 amp, the secondary impedance would be 1/20 ohm plus the resistance of the winding (which may be MUCH higher). But any reasonable turns ratio when SQUARED will reduce the transformed impedance to the measured loop down to "it doesn't matter" range.
| It might help if we knew what exactly you are trying to do.
I wouldn't want to distract you :-)
I'm thinking about a way to get low voltage by running them in reverse. Apply voltage to the terminals, and induce current in the wire passing through the window. Obviously I'd want to limit the current on the terminals to no more than 5 amps.
| A current transformer obeys the same rule as any other transformer. | | The impedance of the secondary is divided by the SQUARE of the turns ratio | and effectively inserted into the loop you are trying to monitor. | | If you have a 50 mV shunt on the secondary side rated, say, for 1 amp, the | secondary impedance would be 1/20 ohm plus the resistance of the winding | (which may be MUCH higher). But any reasonable turns ratio when SQUARED | will reduce the transformed impedance to the measured loop down to "it | doesn't matter" range.
If I apply a voltage to the terminals of the CT, what voltage should that be to draw a current of no more than 5 amps? So this will require knowing what the impedance is of the circuit through the window. I'm going to try to make the window circuit be as low impedance as possible.
Lots of people build things like Tesla coils to produce as high a voltage as they can. I am studying the idea of doing the opposite to get as high a current as I can, obviously with a very low voltage, and a very low impedance system.
We do this frequently. We simply use a 240V variac as the input to the secondary side of the CT. You need VERY big conductor(s) on the primary side. The impedence of your primary wire is reflected onto the secondary based on the turns ratio. It probably would not hurt to have an ammeter in the secondary to make sure you do not exceed the rating. Although, it is very very hard to drive as much as 5 amps in most cases due to the impedence of the primary. Also, keep in mind that if you open circuit the CT, even for a short time, you have probably ruined it.
Charles Perry P.E.
OK.
Most electric "spot welders" used that technique. (Also, "soldering guns.") The primary of the transformer is MANY turns of relatively fine wire. The primary is excited by mains voltage (120/240) or by some kind of capacitor storage circuit. The secondary is a SINGLE TURN of heavy copper with a flexible section and a small gap. The gap is filled with the material that is supposed to be welded.
If you have some spare change, get yourself a spot welder from a discount place like Harbor Freight or buy a cheap soldering gun from a place that sells junky imported "tools." You can see your idea implemented in fact and then you can determine whether you want to scale up.
Yep--depending on your duty cycle.
|> | A current transformer obeys the same rule as any other transformer. |> | |> | The impedance of the secondary is divided by the SQUARE of the turns |> ratio |> | and effectively inserted into the loop you are trying to monitor. |> | |> | If you have a 50 mV shunt on the secondary side rated, say, for 1 amp, |> the |> | secondary impedance would be 1/20 ohm plus the resistance of the winding |> | (which may be MUCH higher). But any reasonable turns ratio when |> SQUARED |> | will reduce the transformed impedance to the measured loop down to "it |> | doesn't matter" range. |>
|> If I apply a voltage to the terminals of the CT, what voltage should that |> be to draw a current of no more than 5 amps? So this will require knowing |> what the impedance is of the circuit through the window. I'm going to try |> to make the window circuit be as low impedance as possible. |>
|> Lots of people build things like Tesla coils to produce as high a voltage |> as they can. I am studying the idea of doing the opposite to get as high |> a current as I can, obviously with a very low voltage, and a very low |> impedance system. |>
| We do this frequently. We simply use a 240V variac as the input to the | secondary side of the CT. You need VERY big conductor(s) on the primary | side. The impedence of your primary wire is reflected onto the secondary | based on the turns ratio. It probably would not hurt to have an ammeter in | the secondary to make sure you do not exceed the rating. Although, it is | very very hard to drive as much as 5 amps in most cases due to the impedence | of the primary. Also, keep in mind that if you open circuit the CT, even | for a short time, you have probably ruined it.
By open circuit I assume you mean cutting or removing the wire going through the window (since this is a case of feeding the CT terminals). What I am wanting to do is find that open circuit impedance so I can at least initially drive it at a safe level so it won't exceed 5 amps. But I guess I could use an external current limiter like a resistor.
| You are using it as a voltage transformer. While it can carry more, it is | probably safest to limit the input to the VA burden rating. An input | voltage that is, say , 2 times the VA/I value may result in saturation and | high exciting current. This depends on the quality of the transformer but | except for short term use, I would avoid that. | IF you have a 5000:5 CT , reversed it is a 1000:1 voltage (step down) | transformer. Suppose that its VA rating is 25VA. This would limit you to | about 5V at 5A. The secondary output voltage, ignoring internal impedance | would be about 0.005V at a current of 5000A (25VA - and actually it would be | a bit less) The maximum low voltage impedance would be 1 micro ohm. This | would appear as 1 ohm on the 5V, 5A side. | You probably wouldn't be able to get this- depends on the internal | impedance which is normally quite small. One thing, unlike normal use of a | CT, the open circuit voltage would be safe, and I see no reason for any | damage to the CT as long as you stay within the current and VA limits. | However, there may be other transformers which are more suited to this.
At 25VA, it might not all that useful, then, considering the cost. The step-down transformers I have seen just don't have the big ratios, though, to get into the high kA range. I was thinking several CT's with the meter terminals in series, but parallel bus bars through the windows. Picture bus bars formed as a narrow ladder with CT's on each rung (and more runs and CT's for more current). The challenge is keeping the impedance very low. Plus there's a hazard in backfeeding one open CT from all the others. I guess paralleling transformers in general has scaling issues like that.
I've wondered what kind of equipment is used to verify that circuit breakers will indeed reliably open their interruption current ratings into the 100 kA range.
There aren't very many labs left in the world that can do the type of testing you are talking about. Our lab in Lenox MA
It just hasn't been economical for utilities/companies to maintain those kinds of testing capabilities, with the exception of breaker manufacturers. It is certainly not something you are going to want to try to build on your own.
Charles Perry P.E.
that
manufacturers.
I don't know about a high current test facility at BC Hydro (my juice supplier) but there is one atached to the EHV lab that HydroQuebec has. As I recall, this will handle 100KA and is set up for synthetic tests- say a
100KA fault and a recovery voltage of nominal 700-1500KV after arc extinction. Both factors are needed for realistic testing.
Of course! As soon as I posted that message, I realized I had thought HydroQuebec but typed BCHydro! Oh well, it has been a long week. There are other labs with the capability. I have a list of them at work.
Charles Perry P.E.
| There aren't very many labs left in the world that can do the type of | testing you are talking about. Our lab in Lenox MA |
I'm not interested in trying to build such a test lab. But I am curious what goes into the making of one. What would breaker manufacturers like Cutler-Hammer, GE, Siemens, and Square-D use to test their 600 volt class breakers? And what would UL use to indepedently verify the interruption ratings?
And then there are test procedures. How many units would need to be tested for valid sampling? How many tests would have to be performed on a single unit? Would the test involve closing a bolted fault on the load side of an energized and closed breaker? Would that test be limited to a cold breaker, or would it be applied to a breaker running full rated current and voltage for a few hours? Would the test involve closing the breaker onto an already bolted fault? Would the test involve energizing a breaker already closed onto a bolted fault? Would power factors less than unity be tested at fault current levels? Would every combination of phase faulting be tested?
What kinds of testing would be appropriate for a newly constructed home to verify that the electrical work meets specifications (and not just passes inspection)? If I have 200 amp service and a 200 amp main breaker, I should expect that if I run test loads above 200 amps I should have a main trip within certain times at certain current levels based on the manufacturer trip curves. At what point could such testing risk damaging the system?
As of a few years ago, Allen-Bradley had a high-current low-voltage facility across the street from its Milwaukee factory. I don't recall the detailed specs, but they could do their own testing on NEMA ac and dc devices up to at least 65,000 AIC. I don't know if the facility is still there, but I witnessed tests there in 1997.
|> I don't know about a high current test facility at BC Hydro (my juice |> supplier) but there is one atached to the EHV lab that HydroQuebec has. As |> I |> recall, this will handle 100KA and is set up for synthetic tests- say a |> 100KA fault and a recovery voltage of nominal 700-1500KV after arc |> extinction. |> Both factors are needed for realistic testing. |> -- |> Don Kelly |> snipped-for-privacy@peeshaw.ca |> remove the urine to answer | | Of course! As soon as I posted that message, I realized I had thought | HydroQuebec but typed BCHydro! Oh well, it has been a long week. There are | other labs with the capability. I have a list of them at work.
I found this one in India:
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