Understanding Current Transformers Vs Voltage Transformers

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wwhitney

Senior Member
Location
Berkeley, CA
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Hello,

I'm trying to understand current transformers a bit better. The main thing I'm unclear on is the exact difference in transformer construction or installation that makes the transformer output behave like a current source versus a voltage source. It seems like in both cases the transformer consists of a core with two coils wound around it, although in the CT case the primary coil is typically a single loop. A few related questions:

1) Typical AC electrical systems are fed by a constant voltage source (the grid). So the only difference in installation I see between a CT and a VT is that a CT primary is connected to the voltage source in series with a significant impedance (the load through which you wish to measure the current), while a VT is connected through just the impedance of the wires. Is that the key difference that makes the output behave as a current source instead of a voltage source?

2) Say I have a 120V/12V AC doorbell-type transformer, i.e. a 10:1 turns ratio (no compensation). Suppose the coils on both sides have sufficient voltage rating, if I hook up the nominal 12V side to 120VAC in series with a load, does the transformer behave like a 10:1 current transformer?

Thanks,
Wayne
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
What makes a CT measure current is the fact that the secondary circuit is effectively a short circuit.

If you open the secondary circuit of a CT, it will behave as a VT and damage its own internal insulation.

An open-secondary 400:5 CT (80:1 ratio) will act as a 1:80 VT and step voltages up into the kilo-volt range.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
What makes a CT measure current is the fact that the secondary circuit is effectively a short circuit.
I'm more familiar with small mA output CTs used in residential power monitoring than 5A output CTs used by the POCO for power metering. With a mA output CT (e.g. 300A:100mA), there is typically a burden resistor (I think on the order of 1 ohm) used to generate a mV signal which is measured via the solid state electronics.

So does your statement still stand in that context? Is 1 ohm effectively a short circuit?

Thanks,
Wayne
 

GeorgeB

ElectroHydraulics engineer (retired)
Location
Greenville SC
Occupation
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Wayne, perhaps you are confusing a CT (current transformer) as a measurement device (as IME always used) with the fact that all transformers have current as a consideration. The instrumentation analog to a CT is a PT (potential transformer). Both are, in the electrician world, low power, none I've ever seen as much as 100 Watt, usually less than 10 Watt. A control transformer might be 240/480:120 or 240:24. A doorbell transformer is roughly 120:16. In the ideal world, power is the same on "both sides" of the transformer, so a 480:120 rated 500 watts would have input current rating of about 1 amp and a secondary rating of about 4 amp.

But I think your real question is the term CT or current transformer as a "useful" device ... all they are (again, IME) used for is instrumentation of some sort. No INTENTIONAL power delivery.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
I'm more familiar with small mA output CTs used in residential power monitoring than 5A output CTs used by the POCO for power metering. With a mA output CT (e.g. 300A:100mA), there is typically a burden resistor (I think on the order of 1 ohm) used to generate a mV signal which is measured via the solid state electronics.

So does your statement still stand in that context? Is 1 ohm effectively a short circuit?

Thanks,
Wayne

Basically, yes. The secondary current in a CT system is a representation of the load current, with the metering calibrated to "show" the actual current while only having to carry the representational current.

With the previous 400:5 example, for every amp of load, the meter carries 1/80th of an amp.

Despite the relatively low current, CT secondary metering is wired with #10 to assure there is very little impedance, much less than one ohm, which keeps the voltage to almost nothing.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Wayne, perhaps you are confusing a CT (current transformer) as a measurement device (as IME always used) with the fact that all transformers have current as a consideration. The instrumentation analog to a CT is a PT (potential transformer).
OK, so I agree that CTs and PTs are used as measurement devices rather than as power transformers. But the physics is the same, right?

I have this feeling that there is a continuum of transformer behavior, and one endpoint is close to an ideal voltage source, and the other endpoint is close to an ideal current source. If so, what feature of the transformer installation or construction parameterizes this continuum?

Thanks,
Wayne
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Part of the puzzle is the impedance of the primary circuit. The transformer has its self inductance, depending on the number of turns and the size and material of the core. If this self inductance dominates the primary circuit then you have a potential transformer or a normal distribution transformer.

In a CT you want to make the primary inductance very very low so you don't interfere with the circuit being measured. The rest of the primary circuit is what limits the current through the system. You also want to make the reflected voltage drop on the primary very low, so you only have a very low impedance on the secondary, a near short circuit or a low resistance depending on how you are measuring the secondary.

I think you have your doorbell transformer example exactly right. If you put the 12V coil in series with the load, then the current through the primary coil will be 1/10 the load current...as long as the current is low enough that the self impedance of the 12V coil is not an issue. But say you have a 25VA doorbell transformer (2A secondary current) and the 'load' is a 500W heater (4A nominal current)...you will no longer see ideal current transformer like behavior.

-Jon
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
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Retired PV System Designer
The real difference is that any transformer when operated with a high load impedance compared to the source and winding impedance will, in the limiting case of increasing load, deliver an output voltage proportional to the primary voltage. To first order this is independent of load impedance.
Any transformer operated with a load impedance that is small compared to the (reflected) primary side impedance (the primary current is set by the primary load, not the distribution line impedance) will act as a constant current source independent of secondary load (burden).
In real life a transformer will only be operated at one or the other extreme and not in the middle. The middle is the relevant region during overload conditions!
In terms of design, the primary of a CT must be able to carry the design load current indefinitely and also avoid core saturation at the highest overload current that needs to be accurately measured. Except for insulation strength to ground the circuit voltage does not play a role.
As noted, if the secondary is open circuited, the derived voltage can be extreme and core saturation is almost guaranteed.
In a PT the magnetizing current during operation is comparable to the load current. In a CT the magnetizing current during operation is very small compared to the load current (in the ideal world the core flux is near zero at all times.)

PS: Cross posted with Winnie who said the same thing from a different point of view.


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wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
So the parameter I referred to in my previous post is, e.g., (secondary load impedance/primary source impedance)? That is very helpful.

Thanks,
Wayne
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
So the parameter I referred to in my previous post is, e.g., (secondary load impedance/primary source impedance)? That is very helpful.

Thanks,
Wayne
Make that reflected load impedance to allow for widely varying turns ratios and you have your parameter.

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LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
Wayne, as an illustration, if you used your transformer-in-series example, if you were to short the 120v "secondary," the 12v "primary" will appear as a short to the primary circuit.

If you put an ammeter in place of that secondary short, it will display current at the inverse of the transformer ratio, showing 1a for every 10a of load.
 

paulengr

Senior Member
There is a third or fourth “type”. There are “CTs” which have a normal”real” CT and some electronics to convert the current to a regulated 4-20 mA output or sometimes other values. Also there are lots of specialized transformers that measure nearly anything.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
There is a third or fourth “type”. There are “CTs” which have a normal”real” CT and some electronics to convert the current to a regulated 4-20 mA output or sometimes other values. Also there are lots of specialized transformers that measure nearly anything.
I would be inclined to call those packages transformer-based current sensors, not CTs.

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