EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: ELS122 on May 02, 2023, 06:58:20 am
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Is the output voltage, open circuit, the turns ratio x the resistivity voltage drop in the primary wire?
If so, how does one sense the secondary when the primary voltage drop is small, i.e if I want to measure 100ma flowing trough rebar? the voltage drop would be very small and so would the secondary voltage?
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Usually the turn ratio of a current transformer is very big, so big that if you forget the current-transformer (CT) can self damage from overvoltage. For example, the powerlines (those for electric energy distribution) have CTs at their ends, to measure the current through the wires and/or to trip various protections against overload. Those CT can pop/catch-fire/explode if the secondary goes open circuit (most common example is during test, if the secondary turn is forgot disconnected)
To answer the question, yes, the Vsecondary is still the Vprimary * Turns_ratio (and at the same time there is a similar relation for current), no matter it is called a CT or a VT (voltage transformer). The transformation ratio stays the same. CT or VT is more about the function we assign for a transformer, the transformer doesn't "know" what we call it, and the physics based on which a transformer works remain the same (thus the corresponding formulas stay the same, too).
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when the primary voltage drop is small, i.e if I want to measure 100ma flowing trough rebar? the voltage drop would be very small and so would the secondary voltage?
Maybe rig up a dummy CT that only has a primary and measure the AC voltage across that and multiply it by the turns ratio that it would have if there was a secondary. That should give you some idea of the open cct secondary voltage.
The fact that there is only 100mA flowing in the conductor doesn't really matter if the CT primary has enough inductance and it's reactance doesn't reduce the measured current. In fact, in that situation there would probably be no limit on the primary voltage either until the core saturated.
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Is the output voltage, open circuit, the turns ratio x the resistivity voltage drop in the primary wire?
If so, how does one sense the secondary when the primary voltage drop is small, i.e if I want to measure 100ma flowing trough rebar? the voltage drop would be very small and so would the secondary voltage?
A current transformer is also a voltage transformer so the secondary voltage is the total voltage applied to the primary times the turns ratio. People often forget that the short length of wire that goes through CT core is just part of a larger circuit back through the supply mains etc so there is always one complete turn in the primary side. It is the voltage developed across that complete turn that needs to taken into account.
For example, if you are using a 100:1 current transformer on a 230V circuit the the open circuit secondary voltage could theoretically be 23kV although in practice this will be limited to a much lower figure by saturation of the transformer core.
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Is the output voltage, open circuit, the turns ratio x the resistivity voltage drop in the primary wire?
If so, how does one sense the secondary when the primary voltage drop is small, i.e if I want to measure 100ma flowing trough rebar? the voltage drop would be very small and so would the secondary voltage?
A current transformer is also a voltage transformer so the secondary voltage is the total voltage applied to the primary times the turns ratio. People often forget that the short length of wire that goes through CT core is just part of a larger circuit back through the supply mains etc so there is always one complete turn in the primary side. It is the voltage developed across that complete turn that needs to taken into account.
For example, if you are using a 100:1 current transformer on a 230V circuit the the open circuit secondary voltage could theoretically be 23kV although in practice this will be limited to a much lower figure by saturation of the transformer core.
But the voltage across the primary winding will be 0 if there's no current path.
So if you're measuring a 230V AC line, unless you're measuring a "short circuit" across the 230V line, the primary voltage wont be equivalent to 230V, instead (since a comment confirmed this) the voltage will be the voltage drop in the wire, that will be the equivalent primary voltage.
It's probably annoying to calculate the voltage too since... over what distance is the primary winding? ;D
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Illustration of how it looks like after a CT explodes (see the pic on the second page) https://download.atlantis-press.com/article/25850187.pdf (https://download.atlantis-press.com/article/25850187.pdf) , or this video (random pick of first search result):
Why Current Transformer Blasted ? Reason of Blast ? Nitesh kumar (In English)
https://www.youtube.com/watch?v=26wS2GvuRZY (https://www.youtube.com/watch?v=26wS2GvuRZY)
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Is the output voltage, open circuit, the turns ratio x the resistivity voltage drop in the primary wire?
I think the resistance of the transformer primary is not the main factor. For an ideal transformer, electrical power in equals electrical power out. Any wire resistance makes the transformer non-ideal. Electrical power is converted to heat in the resistive windings. Any voltage drop across the resistance of the primary winding is not what should be used in simple Vin * Iin = Vout * Iout or turns ratio type calculation.
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You cannot compare the operation of a "normal" voltage transformer to a CT.
The primary/secondary voltage is not even in play in a CT, and a "primary voltage" does not exist.
What exists is a primary (normally just a conductor through the centre of a toroid coil) that will carry a certain AC current.
This will induce current into the secondary winding through the magnetic field at a ratio of Np/Ns. If this current can't flow anywhere, the voltage can rise to destructive levels (imagine ignition coil or flyback transformer).
That's why you always need a "burden" resistor on the secondary, both to let the current flow, and to measure the voltage produced by the primary current.
A very crude way of describing it is that voltage transformers are "voluntarily" magnetized, whereas CTs are "force" magnetized.
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A perfect CT would be a perfect current source. That means the attempt to make even a small current flow in its unloaded secondary would produce an infinite voltage across the gap. The voltage you get with a real CT is a measure of how imperfect that CT is. While they may be far from perfect, that voltage can be pretty big with real transformers. If they flash over the results can be quite destructive to the CT, and a flash might start a fire. A lot of people worry about having a suitable burden resistor, or cascaded CT, in place at all times, and having one directly attached to the CT, so its never missing, is a good safety move. However, it is a current source. Shorting the leads is also safe and harmless. Just never use switching in the secondary that could leave it open circuit for even an instant.
Some people see the spec of a CT quote a specific burden resistance, and think there is something very special about that value. Actually its just the maximum resistance found to give good linearity, and phase response. I've never met a CT that behaves worse when you use a lower resistance, but most people want the maximum voltage they can get across the burden resistor, to simplify what is downstream. So, the specified resistance, or something close, it usually the right choice.
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A perfect CT would be a perfect current source.
I disagree. That would be a Current-Controlled Current-Source (CCCS), which is a different device than a CT.
The CT is a transformer, based on two coupled coils, and often with only one turn in primary (can have more than one turn in primary, as well).
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Here my explanation (see photo). Like all VoltageTransformer CurrentTransformer will act as a ideal voltage transformer if kept far from saturating the core ("low" voltage input).
A VT has thousands of turns permitting high input voltages. A CT has just a few primary turns. This means that the working voltage (before saturation) is low (millivolts!)
*a CT is accurate when the magnetization current is negligible (I1 - I1' = 0, I2=I1*ratio).
a VT is accurate when the voltage drop over the winding impedance is negligible (V1 - V1' = 0, V1=V2*ratio).
*A CT is built in a way to have correct current ratio and V2=V1/ratio will never be correct due to high winding impedance (the fact that the primary is left floating in air will increase Zwinding by a lot).
*The typical values used in the example are for industrial CTs (5A secondary current). For other CTs (1mA secondary current) parameters will be different...
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A current transformer still acts like a transformer. So there is a primary voltage related to the secondary voltage. For the primary voltage this is however only the inductive part. The resistive drop is on top if this. So a higher resistance at the primary does not directly effect the secondary voltage. This is the same with a normal transformer: the resistance of the windings adds extra voltage drop that is not included in the ideal transformer equation. So with a current flow part of the voltage is lost and both sides.
In the normal use of a CT the secondary current and the burden resistor limit the secondry voltage to a relatively low voltage (e.g. 1 V or 100 mV). So the primary voltage is also rather small (e.g. 1 V / 1000 = 1 mV). So one usually does not care much about the primary side current drop.
With a burden resistor the secondary voltage is proportional to the current and the impedance on the primary will also looks much like a resistor of size (burnden resistor + winding resistance) times the winding ratio squared plus the normal resistance of the primary.
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A current transformer still acts like a transformer. So there is a primary voltage related to the secondary voltage. For the primary voltage this is however only the inductive part. The resistive drop is on top if this. So a higher resistance at the primary does not directly effect the secondary voltage. This is the same with a normal transformer: the resistance of the windings adds extra voltage drop that is not included in the ideal transformer equation. So with a current flow part of the voltage is lost and both sides.
In the normal use of a CT the secondary current and the burden resistor limit the secondry voltage to a relatively low voltage (e.g. 1 V or 100 mV). So the primary voltage is also rather small (e.g. 1 V / 1000 = 1 mV). So one usually does not care much about the primary side current drop.
With a burden resistor the secondary voltage is proportional to the current and the impedance on the primary will also looks much like a resistor of size (burnden resistor + winding resistance) times the winding ratio squared plus the normal resistance of the primary.
Ooooh... now I get it!
They effective primary voltage is the voltage drop from the impedance at the given frequency.
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A perfect CT would be a perfect current source.
I disagree. That would be a Current-Controlled Current-Source (CCCS), which is a different device than a CT.
The CT is a transformer, based on two coupled coils, and often with only one turn in primary (can have more than one turn in primary, as well).
I should have said the output of a perfect CT would be a perfect current source.
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That could be said only when the secondary is short-circuited. Otherwise, a current source has infinite impedance, while a transformer does not, a controlled current-source is not bidirectional and does not conserve energy/power, to name the biggest differences.
A CT is just like any other transformer, only that it is used with its secondary short-circuited.
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A very crude way of describing it is that voltage transformers are "voluntarily" magnetized, whereas CTs are "force" magnetized.
Yes, that's a confusing difference which many people forget (or doesn't know).
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A very crude way of describing it is that voltage transformers are "voluntarily" magnetized, whereas CTs are "force" magnetized.
Yes, that's a confusing difference which many people forget (or doesn't know).
I don't think this is a helpful description - more confusing.
A current transformer is designed to have a small magnetizing current, just like a normal transformer. Like with a normal transformer the magnization follows the voltage (from A*dB/dt = U) and is little directly effected by the load current. There is a link to the current with the CT because the secondary load is normally a fixed resistor, but that is indirect via the voltage. One can also have a nonlinar burden to a CT, e.g. with an added rectifier and in this case the secondary voltage is the more relevent part for the magnetization, not the secondry current. The assuption that the "magenization" follows the current is used with a rogowski coil or hall effect senstors.
A difference is that a normal transfromer is usually used with a fixed primary voltage. This way the decription as forced magnetization is more fitting to normal transformer as the fixed voltage forces a certail level of magnitization and thus the link from voltage to magnetization level and thus a maximum primary voltage set by the saturation.
With a CT the use is essentially always with a fixed secondary resistance and the voltage variing with the current to measure.
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A very crude way of describing it is that voltage transformers are "voluntarily" magnetized, whereas CTs are "force" magnetized.
Yes, that's a confusing difference which many people forget (or doesn't know).
I'm not familiar with the magnetized-transformer term, nor with the meaning of "voluntarily"/"force"-magnetized. What is that about, what does it means?
Later edit:
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My question was posted at the same time (3 seconds later) with the Kleinstein's explanation of what is the voluntarily/forced-magnetised about. It's clear now, thank you.
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I'm not familiar with the magnetized-transformer term, nor with the meaning of "voluntarily"/"force"-magnetized. What is that about, what does it means?
If you are asking me then sorry, I want partisipate in a long discussion of terms now. I guess it is fine when people use some known terms to describe unknown terms. But I'll try to explain used term. I guess 'magnetised transformer' here means that there is a magnetic field somewhere.
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With a normal transformer, as at the input of a linear AC to DC PSU, a known constant(ish) voltage is applied to the primary. The primary current is then mainly dependent on the load on the secondary.
In a current transformer the primary current is what is to be measured. That current should be affected as little as possible by the measuring device. Whatever is sourcing the current almost completely sets the primary current.
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Just out of curiosity I tested a CT and worked as VT better than I expected!
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All transformer have the same principle of function. They just acting like an impedance transformer. It means that it receive power with one impedance on input coil and out power with another impedance on secondary coil. So, ideal transformer has the same power on input and output.
Power P = U*I, so if impedance is changed it leads to a different voltage and current values, this is how voltage or current transformation happens in transformer.
Also note that impedance here is not DC resistance, but AC resistance due to inductance of the coil. Such AC resistance usually named impedance.
Of course a real transformer has some DC resistance of the coil, but this is parasitic resistance which is not involved in transformation process, it just lead to a some power loss, so the output power will be a little lower than input power. This power difference is spent on heating the windings and the core of transformer.
The transformer impedance ratio is equals to its coils inductance ratio.
The transformer voltage ratio is equals to its coils turns ratio.
The transformer impedance ratio is a square of its voltage ratio.
So, if transformer has turns ratio is 1:3, it means that it's voltage ratio also 1:3.
And also it means that it's impedance ratio is 1:9 and it's inductance ratio is also 1:9.
Note, if you want to measure transformer impedance ratio, you're needs to measure it's coils inductance, not resistance. Then, when you know transformer impedance ratio you can calculate it's voltage ratio just by get square root from impedance ratio. The result voltage ratio also represents transfromer turns ratio.
Use working AC frequency for inductance measurements, because real transformer parameters varying depends on the frequency.
Open secondary coil means that there is no load and ideal transformer should produce about twice higher voltage than with properly matched load impedance (which leads to the best power transfer efficiency through transformer). But since real transformer is not ideal, such voltage may have lower value.
Also note that a voltage meter also has some input impedance and it affects measurements result.
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Just out of curiosity I tested a CT and worked as VT better than I expected!
Yes, until the primary voltage is minuscule enough, CT will work as VT too.