What formulas and calculations are needed to ensure a 500/5A current transformer does not saturate?

Question

I'm working with a 500/5A current transformer (CT) and need to make sure it doesn't go into saturation during operation. The core of the CT is made of mu-metal with the following specifications:

• Number of turns \$N\$: 100
• Wire wound on core: AWG 18
• External diameter \$De\$: 70 mm
• Internal diameter \$Di\$: 45 mm
• Thickness \$h\$: 35 mm
• Relative permeability \$\mu_r\$: 45000
• Saturation flux density \$B_{\text{sat}}\$: 0.8 T
• Primary current frequency \$f\$: 50 Hz

The calculations performed to verify this CT date back to 1999 and I would like to understand if the procedure applied is correct and unfortunately I struggle to find the formulas online.

Going back over the calculations performed, I have calculated the following:

Inductance per square loop\$A_L\$: $$ A_L = \frac{\mu_0 \mu_r A_{\text{e}}}{l_e} $$

Calculate the effective section of the core: $$ A_{\text{e}} = \frac{De - Di}{2} * h = 437.5 \text{ mm}^2 = 4.375 \cdot 10^{-4} \text{ m}^2 $$

Calculate the effective length of the core: $$ l_{\text{e}} = \frac{De + Di}{2} \cdot \pi = 180.64 \text{ mm} \approx 0.180 \text{ m} $$

I replace the values ​​and find the value of \$A_L\$: $$ A_L = \frac{\mu_0 \mu_r A_{\text{e}}}{l_e} = \frac{4\pi \cdot10^{-7} \cdot 45000 \cdot 4.375 \cdot10^{-4} } { 0.180} = 1.374 \cdot 10^{-4} \frac{H}{N^2} \approx 137 \frac{\text{ $\mu$}H}{N^2} $$

Now that I know the value of \$A_L\$ I can calculate the value of the inductance: $$ L = A_L \cdot N^2 $$

Substituting I get: $$ L = 1.374 \cdot 10^{-4} \cdot 100^2 \approx 1.37 H $$

The CT has a bar on the primary through which up to 500A of current can pass at 50Hz. A resistor of approximately 200 ohms is connected to the secondary.

Below is a picture very similar to the CT I would like to test:

Calculations carried out in 1999 verified that the core does not go into saturation in the case of both DC and AC at 50Hz.

In the case of AC at 50Hz, two cases were assumed:

• The case in which the secondary was short-circuited.
• The case in which the secondary had a known resistance of 200 ohms.

For both cases, the maximum current that would bring the core to saturation was calculated by knowing its \$B_{\text{sat}}\$ of 0.8T

I would need the formulas required to complete the verification and would like to ask for an opinion as to whether the required verifications are correct and sufficient to proceed with the physical realisation of the CT.

Thank you all for your help and advice!

P.S. I do not have the original data sheet of the muMetal core, but I can provide a fairly similar one (see last page):

Magnetisation curves

Answer 1

I am working with a current transformer (CT) with a rating of 500/5A, and I need to ensure that it does not go into saturation during its operation.

No formulas needed. Make sure the primary current does not exceed 500 A. Make sure the secondary load is either a short-circuit, or has a lower resistance than the upper limit specified in the CT data sheet.

There are two routes to express the core flux in other measurables.

1. One is the net amount of current needed to create the flux, also called the magnetising current. Note that this is the difference between the primary and secondary ampere turns, not simply the primary current alone
2. The other is the voltage generated by the flux.

The H field needed for the saturation B field Bsat is \$H_{sat}=\frac{B_{sat}}{\mu_0 \mu_r}=\frac{I_{sat}}{L}\$.

which plugging in the numbers gives Isat = 2.5 A

Note that this is the difference between the primary ampere.turns and the secondary ampere.turns. This is why the CT is operated with the secondary shorted, so that the secondary current can flow and cancel almost all of the primary current.

Unfortunately, the secondary resistance is not given, so even with a short circuit secondary load, we don't know how close the currents will come to cancelling, so cannot calculate the actual core B field for any given primary current.

The voltage generated by the flux is given by the core area and rate of change. The core is 12.5 mm wide, 35 mm deep, so has an area of 437 μ m2. At a peak flux of 0.8 T, that's a flux of 350 μW. The peak rate of change is 2πf that, so at 50 Hz, the peak voltage generated by a peak flux of 0.8 T is about 110 mV per turn, or 11 V peak in the 100 turn secondary, or 7.8 V rms.

Normally you would operate a CT a long way below saturation flux for accuracy. Operating into a short circuit load keeps the voltage, and hence the flux, down. Note that this voltage includes the IR drop in the secondary, so there is a finite flux even working into a short circuit.

It's rather easier to measure the secondary voltage and add on the secondary IR drop than it is to estimate the difference between primary and secondary ampere.turns, so the voltage method is preferred to estimate the flux your CT is running at.

The CT data sheet will, or should, give you an upper limit for the resistance of the measurement load, which will limit the voltage developed. This will have been adjusted for the secondary resistance, and for the maximum core flux for accuracy which will be rather less than the saturation flux.

Answer 2

Calculations carried out in 1999 verified that the core does not go into saturation in the case of both DC and AC at 50 Hz.

Considering that the primary winding has only one turn with Lp = 137 uH, the core will start to saturate already at a modest DC current:

Isat = (Bsat x Ae x N) / Lp = 2.55 A

You can confirm Isat by measurement. A power supply in CC mode runs a variable DC current through the copper bar primary and a LCR meter with low excitation voltage monitors when the inductance of the secondary starts collapsing.

At AC (f = 50 Hz) the max. RMS (sinus) voltage Vs across the secondary 100 turn winding is key and needs to be limited by choosing the correct burden resistor:

Vs <= (2 x Pi x Ae x f x Bsat x N) / SQR(2) <= 7.77 V

A 200 Ohm burden resistor Rb for a max. AC current through the primary of Ip <= 500 A is obviously out of question. With Is <= 5 A (100:1 turn ratio) flowing through the secondary, the burden resistor would be Rb <= 1.55 Ohm, including the DC/AC resistance of the secondary winding. The DC resistance Rdc of the secondary can easily be measured and at f = 50 Hz it is safe to assume Rac = Rdc (for AWG 18 copper wire).

Answer 3

Calculations carried out in 1999 verified that the core does not go into saturation in the case of both DC and AC at 50Hz.

In the case of DC, we can use Ampere's Circuital Law, which tells us that the line integral of the magnetic field strength along some closed curve is equal to the total current passing through any surface enclosed by that closed curve.

If we choose our closed curve to be a circle within the body of the toroid, with a center on the axis of the toroid, then the field strength will be constant along the circle, and inversely proportional to the circumference of the circle. The field strength will thus be maximal in a circle that is on the inner diameter of the toroid.

(Typically, the closed curve that is chosen is a circle that passes through the center of the cross-section of the toroid. Since the values of the permeability of the core material and of the maximum flux density are not know precisely, such a choice of the closed curve gives adequately accurate results. However, I believe choosing a circle on the inner diameter of the toroid is a better choice, especially if the inner diameter and outer diameter are significantly divergent.)

Because we are considering the case where the line current is DC, there will be no transformer action, and no secondary current. Thus, the total current linked by our circle will just be the line current.

The ratio between the flux density in our circle, and the current passing through the donut hole of the toroid will be given by

$$\frac{B}{I_p} = \frac{45000 \times 4\pi \times 10^{-7} }{2\pi \times 45 \times 10^{-3}} = 0.2 \; \text{H}/\text{m}^2 = 0.2 \; \text{T}/\text{A}$$

Since your maximum flux density is said to be 0.8T, by my calculations, the maximum DC current that can pass through the CT without saturating the core is 4 A.

Of course there is a possibility that I made a mistake and you should verify for yourself.

In the case of 50 Hz AC, we do not need to consider the case where the secondary is short circuited. If the core does not saturate with 500 A line current and a 200 ohm burden, it will not saturate with a shorted secondary.

We will again use Ampere's Circuital Law (ACL), even though it is not exact for AC magnetic circuits. It will be close enough. The equation to use if you want an exact solution is Ampere's Circuital Law with Maxwell's Correction. However, that law adds complexity to the calculations, and since the values for other parameters are known only to an approximation, there is not a great deal of benefit to be had by using the more exact equation in this case.

In the AC case, there is both current flowing through the line, and current flowing in the opposite direction through each of the turns of the secondary. If the CT were "ideal", the secondary turns times the secondary current would be exactly equal to the primary (line) current. Thus the total current through the surface bounded by our circle would be 0, and there would be no flux in the toroid. So clearly, we must not assume that the transformer is ideal. Rather, we assume that the transformer consists of of two coupled inductors.

The formula for the inductance factor for a toroidal core with rectangular cross section is:

$$A_L = \frac{\mu h}{2\pi}\text{ln}\left(\frac{o.d.}{i.d.}\right)$$

Plugging the values of \$\mu = 45000 \times 4\pi \times 10^{-7}\$, \$h=0.035\$ and \$\text{ln}(o.d./i.d.) = 0.442\$ into the equation we get

\$A_L = 139.2 \;\mu\text{H}/\text{turn}^2\$

This is very close to the value of \$ A_L = 1.374 \times 10^{-4}\$ H per turns squared found in the original post using the formula for \$A_L\$ for solenoids.

Assuming ideal inductors, and a coupling factor of 1, the inductance of the primary would then be 139.2 \$\mu\$H, the inductance of the secondary would be 1.392 H, and the mutual inductance would be 13.92 mH.

For coupled inductors the phasor voltage of the secondary in terms of the phasor currents of the primary and secondary are:

$$v_s = j\omega Mi_p - j\omega L_Si_s$$

But

$$v_s = i_sR_B$$

where \$R_B\$ is the burden resistance.

this gives us

$$i_s = \frac{j\omega M}{R_B + j\omega L_s}i_p$$

The net current that magnetizes the core is

$$i_m = i_p - Ni_s$$

With a little algebra,

$$i_m = i_p\left(\frac{R_B}{R_B+j\omega L_s}\right)$$

and

$$\frac{|i_m|}{|i_p|} = \frac{R_B}{\sqrt{R_B^2 + \omega^2L_s^2}}$$

Plugging in \$R_B=200\$, and \$\omega L_s = 437.3\$ gives

$$\frac{|i_m|}{|i_p|} = 0.416$$

If I have not made a mistake, it seems that to keep the peak magnetization current less than or equal to 4 A, with a 200 \$\Omega\$ burden, the peak primary current (total) would need to be less than \$4 / 0.416 = 9.62\$ A, which is far below the 500 A specified in the original post.

This is not surprising. My guess is that either the specification for the burden resistor, or for the secondary current is incorrect. The power dissipated by the burden resistor is \$I^2R\$. For a current of 5 A, and a burden resistor of 200 \$\Omega\$, this would be \$25*200 = 5000\$ W. Although there are resistors of that size (for example brake resistors in machinery) it does suggest that one take a second look at the current and burden resistor specs.