# Current Transformer with no burden resistor?

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#### MikeMl

##### Well-Known Member
What happens if the secondary of a ferrite-core current transformer (single wire through the donut hole, multiple turns on the secondary) is left open (or is terminated with a high-z like a scope input)?

With the secondary open, what is the impedance reflected into the primary circuit?

#### crutschow

##### Well-Known Member
In theory the reflected impedance is the secondary load impedance times the square of the turns ratio.

If the secondary is open then the theoretical secondary voltage equals the primary supply voltage times the turns ratio.

#### ronsimpson

##### Well-Known Member
With out a resistor you don't have a CT. You have a voltage transformer.
Much like removing the o.o1 ohm resistor in your current meter.

#### MikeMl

##### Well-Known Member
In theory the reflected impedance is the secondary load impedance times the square of the turns ratio.
So if the primary is 1T, and the secondary is 32T, and the secondary is terminated w/ 1MegΩ, then the impedance reflected into the primary would be 1e6*(1/32)^2 =976Ω?

#### schmitt trigger

##### Well-Known Member
Mike, your calculation is correct....that is, until the transformer core saturates and then the impedance just drops completely.

Since CTs are designed to present a very low burden, this essentially means that after a couple hundred millivolts (reflected to the primary) it will go into saturation.

#### JonSea

##### Well-Known Member
We used to use an amp clamp meter as a current sensor. This particular device allowed removing the meter assembly entirely and connecting to the now-open CT coil. Lesson learned the hard way - with only a high impedance load of an instrument input across the coil, the output voltage could easily exceed a couple hundred volts! If I recall correctly, a 1k resistor across the coil kept the output to reasonable levels.

#### MikeMl

##### Well-Known Member
Mike, your calculation is correct....that is, until the transformer core saturates and then the impedance just drops completely.

Since CTs are designed to present a very low burden, this essentially means that after a couple hundred millivolts (reflected to the primary) it will go into saturation.
Imagine a coax feedline between a 7MHz, 100W transmitter and a 50Ω antenna. Current in the feedline would be sqrt(100/50) = 1.4A .
Run the center conductor of the coax through the hole of a ferrite core, and wind ~30 turns on it as a secondary. What would be the burden resistor on the secondary side have to be so as not to materially upset the impedance in a 50Ω system.

The goal is to create an AC voltage across the burden resistor that is proportional to the feedline current.

#### ronsimpson

##### Well-Known Member
A 30:1 transformer gives a 900:1 resistor ratio.
A 900 ohm resistor puts 1 ohm on your 50 antenna. Not good. You might have 2 watts loss.
A 90 ohm resistor adds 0.1 ohm in you coax. Is that ok?

#### MikeMl

##### Well-Known Member
...
A 900 ohm resistor puts 1 ohm on your 50 antenna... You might have 2 watts loss.
That's barely acceptable. Say we use 90Ω. The reflected impedance would be 90/(30^2) = 90/900 = 0.1Ω. The secondary current would be 1.4/30 = 47mArms, so the voltage across the scope input would be 90*0.047 = 4.2Vrms.

This topic came up in our 40m round-table this morning. I had actually never thought it through before. The point of the discussion was to deflect the vertical axis of an o'scope with a voltage signal proportional to the antenna current. The proposed solution was the C.T. I described earlier, but terminated only by the nominal 1Meg scope input impedance. I said, "don't you need a burden resistor?", but couldn't recall off the top what its value should be...

So without considering core saturation, a 1MegΩ load on the secondary, the 30 turn transformer would present a whooping 1e6/(30^2) = 1E6/900 = 1.1K Ω resistance in series with the 50Ω antenna. Not Good!

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#### ronsimpson

##### Well-Known Member
When you make the transformer, you will have inductance in parallel with your 90 ohm resistor. This makes a high pass filter. At your 7mhz if the inductance is 90 ohms the voltage will be 0.5 of what it should be. So your inductance should be high enough that it does not effect your reading much.

On the other end; there is capacitance across the winding that short out very high frequency signals. If you have too many turns, the high frequency role off comes down.

#### Tony Stewart

##### Well-Known Member
if you break 1A primary AC current with. a switch and CT with no load, it may generate a secondary arc that destroys probe 9M resistor, scope 300V limit and CT insulation.

#### ronsimpson

##### Well-Known Member
I am more concerned with what the transmitter will do with 1mH in series with the load. SWR problem.
I put a resistor on the CT, no switch, no long wires.

#### Tony Stewart

##### Well-Known Member
I am more concerned with what the transmitter will do with 1mH in series with the load. SWR problem.
I put a resistor on the CT, no switch, no long wires.
of course, SWR problem may overheat driver. RF vs AC power, each has risk problems.

#### MikeMl

##### Well-Known Member
So here is a hack. Note I reduced the burden to 50Ω (BNC coax termination). This suggests that any ferrite core that would make a secondary inductance of 25uH to~400uH with 30turns on it would be usable. I would like the CT to work from ~1Mhz to ~30MHz. The loss in L1 is insignificant in a 50Ω system. Green trace= 25uH, Violet trace =400uH

The high-freq roll-off is due to the reactance of the shunt capacitance C1 relative to R1. The low freq roll-off is due to the reactance of L2 relative to R1.

I'm not sure where the 1mH in the primary came from? I figure a max of 400uH/(30^2) = 444nH, less than 1/2000 of 1mH.

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#### JimB

##### Super Moderator
For a low frequency 50/60Hz current transformer, the accepted wisdom is that if a current passes through the primary while there is lo load on the secondary, then the transformer will be damaged and the world will come to an end (worst case scenario).

I have never considered the case of an RF current transformer, as used in an ndB coupler, such as this home made 30dB coupler of mine:

So I did some tests using a signal generator (HP8640B), my 30dB coupler and my home made RF power meter which is based on an AD8307.

I set the sig gen to 10MHz 0dBm, and connected via a 1metre long coax cable to the power meter. The power meter read -0.3dBm

I then inserted the 30db coupler at the power meter end of the cable. The coupler was terminated with 50R on its sample port. The power meter again read -0.3dBm.

I then changed the load on the coupler sample port, and got the following results:

Open circuit -4.3 dBm
Short circuit -0.3 dBm
100R -0.3dBm

Clearly something odd is happening when the couple sample port is open circuit.

To investigate further, I used an AIM4170 Antenna Analyser (a simple form of a Vector Network Analyser), connected up like this:

Sweeping the frequency from 3 to 33 MHz gave the following results:

Looking directly at the input of the power meter, a nice flat VSWR / Return Loss from 3 to 33 MHz.

With the 30dB coupler at the input of the power meter, again a reasonably flat VSWR / Return Loss from 3 to 33 MHz.

With the 30dB coupler sample port un terminated (open circuit), something serious is happening in the 10MHz region.

With the 30dB coupler sample port terminated with a short circuit, we are back to the fairly flat response.

And similar results when terminated with 100R.

It is quite by chance that I started the experiments with the signal generator giving 10MHz, I thought that it was a low enough frequency to give good results when looking at the 30dB coupler sample port with a 200MHz scope.
I have not included my scope test results here, but I did see quite large voltages when the coupler was unterminated.
What I did notice was the voltage on the sample port was very dependant on what was connected to the port, even just putting a BNC "Tee piece" could make a visible difference. I guess the small capacitance of the T piece was enough to affect the high impedance of the unterminated sample port.

JimB

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#### ronsimpson

##### Well-Known Member
Note I reduced the burden to 50Ω (BNC coax termination).
I like that. You can connect directly to the scope input with a coax. No probe.
The frequency sweep shows what I expected.

#### ronsimpson

##### Well-Known Member
With the 30dB coupler sample port un terminated (open circuit), something serious is happening in the 10MHz region.
What I would expect.

#### MikeMl

##### Well-Known Member
So I set the secondary inductance to 100uH, and then show what happens with different values of Burden resistor; 50Ω (grn), 1K (yel) and 1meg (red). Top plot pane is power delivered to the load, bottom pane is the output voltage.

As shown previously, the CT works well with B=50Ω; nice flat response and the load gets close to 100W. (grn traces)

With the burden being only the 1meg input impedance of the scope shunted by a bit of cable capacitance it resonates with the secondary inductance and we get a huge voltage out at the resonant frequency. Note the power delivered to the load has a huge notch at the resonant frequency. Isn't this what Jim measured near 10Mhz? Since the values I picked for C1 and L2 were a wag on my part, I'd say the sim comes remarkably close to what he measures in a real circuit.

The 1K (yel traces) is some where between the two cases above, still showing resonance, but with a much lower Q.

Thanks guys, I learned a lot from this...