Is my current transformer actually outputting half a million volts?

[SparkyTD]

I'm building a fancy smart electricity meter, and I'm using one of those cheap Chinese SCT-013 current transformer clamps to monitor a live 230VAC mains wire. If I understand correctly, the primary "winding" of the CT is the wire that I clamp it on, and the secondary is a built in coil. So the winding ratio is my case is 1:1800. The peak voltage of RMS 230V is about ~325 volts. If I multiply that by 1800, I get 585,000 volts peak on the output of the current transformer. Is this really possible, or am I misunderstanding something here? The burden resistor on the CT's output is just a small SMD resistor. Is there actually half a million volts flowing through that tiny little resistor?

[bdunham7]

This is why you never allow CT outputs to go open circuit!  If you do, the voltage will climb towards that amount until something gives, presumably at a much lower voltage.  Once you put a load on them, the voltage drops to whatever is required to put 1/1800 of the primary current through whatever you've loaded the secondary with.

[bob91343]

The point is that there is no voltage input to the clamp-on input transformer.  It senses the current.  You could compare it to an old style ammeter which is sensitive to current but you won't apply mains voltage to it.

The voltage drop of a current transformer is proportional not only to the input current, but to the load resistance.  If said load resistance should open, then we have a recipe for disaster.  Under normal operation, the voltage across the input (technically called the burden) is supposed to be very small.

[srb1954]

I'm building a fancy smart electricity meter, and I'm using one of those cheap Chinese SCT-013 current transformer clamps to monitor a live 230VAC mains wire. If I understand correctly, the primary "winding" of the CT is the wire that I clamp it on, and the secondary is a built in coil. So the winding ratio is my case is 1:1800. The peak voltage of RMS 230V is about ~325 volts. If I multiply that by 1800, I get 585,000 volts peak on the output of the current transformer. Is this really possible, or am I misunderstanding something here? The burden resistor on the CT's output is just a small SMD resistor. Is there actually half a million volts flowing through that tiny little resistor?

Although the open-circuit secondary voltage can theoretically reach very high voltages in practice it will be limited by the saturation of the transformer core material. Knowing the cross sectional area of the core and and the number of turns you could work out the maximum secondary voltage without saturation. Once the core saturates there could be a lot of heat dissipated in the core and windings, potentiallly damaging the transformer.

It is always best to have some form of protection e.g. back-to-back Zener diodes across the secondary to ensure that an open-circuit secondary condition never occurs.

[IanB]

The burden resistor on the CT's output is just a small SMD resistor. Is there actually half a million volts flowing through that tiny little resistor?

No, because voltage does not flow through resistors (or anything), current flows through resistors.

In the case of a current transformer, a small current flows through the resistor, which causes a small voltage to appear across the resistor, which can be measured.

Since (voltage) = (current) x (resistance), and since the resistance is known, the voltage is proportional to the current. Measure the voltage and you can work out the current; multiply the current by 1800x and you can deduce the current flowing in the mains wire.

[ejeffrey]

The primary voltage is not the 230 VAC of the power line but only the drop across the "winding" inductance.  This explains one reason we use current transformers.  If you used a shunt resistor instead you would probably need to have some millivolts of voltage drop to get a good measurement even with precision amplifiers and no interference considerations.  That ends up dissipating a lot of power.  With a current transformer the voltage drop is ideally the output voltage divided by the turns ratio which can be made much smaller.

[ledtester]

Perhaps these responses to the following SE question will help:

https://electronics.stackexchange.com/questions/31072/difference-between-current-transformers-and-power-transformers

The primary voltage is not the 230 VAC of the power line but only the drop across the "winding" inductance.
...

This is basically what is discussed in the comments of the second response of the above SE question:

[Kleinstein]

I'm building a fancy smart electricity meter, and I'm using one of those cheap Chinese SCT-013 current transformer clamps to monitor a live 230VAC mains wire. If I understand correctly, the primary "winding" of the CT is the wire that I clamp it on, and the secondary is a built in coil. So the winding ratio is my case is 1:1800. The peak voltage of RMS 230V is about ~325 volts. If I multiply that by 1800, I get 585,000 volts peak on the output of the current transformer. Is this really possible, or am I misunderstanding something here? The burden resistor on the CT's output is just a small SMD resistor. Is there actually half a million volts flowing through that tiny little resistor?

Although the open-circuit secondary voltage can theoretically reach very high voltages in practice it will be limited by the saturation of the transformer core material. Knowing the cross sectional area of the core and and the number of turns you could work out the maximum secondary voltage without saturation. Once the core saturates there could be a lot of heat dissipated in the core and windings, potentiallly damaging the transformer.

It is always best to have some form of protection e.g. back-to-back Zener diodes across the secondary to ensure that an open-circuit secondary condition never occurs.

Nothing really bad happens when the current transformer goes into saturation. There is just less induced voltage and less inductance seen by the main current. So no extra heat dissipation. Saturation can be a problem if there is a superimposed DC current and the CT will than no longer be accurate, especially with a higher voltage at the sense resistor.

The extra diodes may be just 2 back to back diodes or a bridge rectifier (shorted DC) used are 2 pairs of back to back diodes, as the voltage at the sense resistor should be low, to avoid saturation of the core and keep the accuracy high. The diodes are more to protect the circuit to read the voltage at the sense resistor in case of a large current spike, e.g. with short or mains spike (e.g. from lightning).

[RoGeorge]

I'm building a fancy smart electricity meter, and I'm using one of those cheap Chinese SCT-013 current transformer clamps to monitor a live 230VAC mains wire. If I understand correctly, the primary "winding" of the CT is the wire that I clamp it on, and the secondary is a built in coil. So the winding ratio is my case is 1:1800. The peak voltage of RMS 230V is about ~325 volts. If I multiply that by 1800, I get 585,000 volts peak on the output of the current transformer. Is this really possible, or am I misunderstanding something here? The burden resistor on the CT's output is just a small SMD resistor. Is there actually half a million volts flowing through that tiny little resistor?

In theory, yes, but only if it were to measure a 230V on the short piece of wire passing through the hole of the secondary coil, which is not possible for your setup.

In practice, the primary is just a piece of thick wire, so unless your CT is used in the power grid industry, in a domestic circuit there will never be enough current to sustain a 230V drop on the primary wire.

Also in practice, never let a CT's secondary winding as open circuit.  That can pop/explode your CT, or at least pierce the isolation between turns, and ruin your CT and the attached measuring instruments.

Another thing to take care is that coils (so transformers, too) act like energy reservoirs, just like capacitors.  Therefore, if not operated properly (or if something fails/breaks) they can release all the stored energy very fast, so at huge powers that can do real damage.

[Vtile]

All of above are pretty much ignoring the concept. As OP is asking is that really 500kV etc. shows the typical thinking in "voltage biased" world view. Stop it!

As name suggest it is a current transformer. If we return a bit to normal piecefull voltage biased world and think how typical "ideal" power transformer behaves - it gives rated voltage and outputs any amount of current to archieve the rated voltage... Back to current world.. Current tranny gives rated current and outputs ANY voltage to archieve that current, it can be 500kV or 50MV depending practical constructional limits of the device.

The current transformer is ideal current source, while typical power transformer is ideal voltage source.

There is a reason why outlet gives 230Vac and not ie. 10Aav.

[ejeffrey]

All of above are pretty much ignoring the concept. As OP is asking is that really 500kV etc. shows the typical thinking in "voltage biased" world view. Stop it!

As name suggest it is a current transformer. If we return a bit to normal piecefull voltage biased world and think how typical "ideal" power transformer behaves - it gives rated voltage and outputs any amount of current to archieve the rated voltage... Back to current world.. Current tranny gives rated current and outputs ANY voltage to archieve that current, it can be 500kV or 50MV depending practical constructional limits of the device.

The transformer doesn't know or care if we call it a voltage or a current transformer.  In practice both the current and the voltage are transformed by the turns ratio, you just have to measure them correctly and take into account leakage inductance, winding resistance, and other factors.

Since the OP wanted to know about the voltage, considering the voltage ratio is appropriate.  You just have to realize that the input voltage is the burden voltage, i.e., the voltage drop across the transformer not the circuit voltage.

[Kleinstein]

The current transformer does not even see the other side of the mains voltage. It does not know if the current flows back to ground (N) or another phase in a 2 or 3 phase consumer.

One can use the voltage ration the other way aroud. If the ouput from the CT is 180 mV (as a reasonable reading near the full scale) the 1:1800 truns ration means that the load would see 0.1 mV drop from the inductive part of the CT. So the burden voltage on the input side is usually very low.

[Zero999]

As mentioned above, if the burden resistor goes open circuit, then the transformer will saturate. You won't get 500kV, but high voltages, up to say 1kV are still possible. Someone at work buggered my old OWON oscilloscope up because they used a burden resistor with too lower power rating. The resistor smoked, went open circuit, causing a high voltage do go into the oscilloscope's input, resulting in more destruction. Fortunately the company bought me a new one and one for the department, since they're not expensive.

[Vtile]

All of above are pretty much ignoring the concept. As OP is asking is that really 500kV etc. shows the typical thinking in "voltage biased" world view. Stop it!

As name suggest it is a current transformer. If we return a bit to normal piecefull voltage biased world and think how typical "ideal" power transformer behaves - it gives rated voltage and outputs any amount of current to archieve the rated voltage... Back to current world.. Current tranny gives rated current and outputs ANY voltage to archieve that current, it can be 500kV or 50MV depending practical constructional limits of the device.

The transformer doesn't know or care if we call it a voltage or a current transformer.  In practice both the current and the voltage are transformed by the turns ratio, you just have to measure them correctly and take into account leakage inductance, winding resistance, and other factors.

Since the OP wanted to know about the voltage, considering the voltage ratio is appropriate.  You just have to realize that the input voltage is the burden voltage, i.e., the voltage drop across the transformer not the circuit voltage.

Yes.

You and others have made great at describing the fine details, but only scratched the surface of the actual concept. That is the ideal voltage source and the ideal current source as those are the two magic search prahes for ie. Wikipedia.

You can argue of course that real deal is not ideal, but ideal concepts aren't supposed to describe the fine details of unideal components, their meaning is in to open phenomenoms and giving tools for thinking and study.

That said as OP is after a fancy power meter, he/she needs also take account the additional phase shift of the CURRENT transformer.