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CURRENT TRANSFORMER CALCULATION

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Newton Brawn

Mon Apr 23 2012, 04:16AM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Hi
Help !

I need the procedure to calculate a current transformer.
Here is the specs:

Frequency: 60Hz
Primary: 2A maximum
Current ratiio: 1:50
Secondary load resistor (burden): 50 ohms
Precision over the range of 0.2 to 2A: +- 5%
Available core material: Fe Si stell, dimensions 38 x 46 x 16 mm, cross section 16x16 mm

Any information will be very welcome
Regards
Newtom

.

Dr. Slack

Mon Apr 23 2012, 07:18AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

How is "calculate" different from "read the line that says "current ratio 1:50"" ???, subject to the maximum and tolerance lines.

You can expect that if it specifies a secondary burden resistor and a primary maximum current, the burden will limit the transformer volts so that it is operating with low enough flux for the core. You can use any lower burden resistor, right down to zero if you want, but cannot use larger than 50ohms without restricting the primary current or losing accuracy.

Steve Conner

Mon Apr 23 2012, 09:38AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

I think he wants to design one. Here's what I'd do.

2 amp, 50:1, 50 ohm burden, implies 2V output across the burden.

Use the universal transformer equation (the 4.44 one) to calculate your core's relationship between flux density and volts per turn.

Solve to find the volts per turn corresponding to some lowish flux density, say 0.8 to 1.0T. (we want to operate the core well below saturation for good accuracy)

Divide this into the 2V output voltage to find the number of secondary turns required. To get the number of primary turns, divide by 50, round up to nearest whole number. Multiply by 50 again to get the actual secondary turns required.

Newton Brawn

Mon Apr 23 2012, 01:44PM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Dr SlacK, Steve

Thank for the help.

1- Yes, I want calculate one CT. And understand how to do it !

2 - Following Steve advice, the fist linee "2 amp, 50:1, 50 ohm burden, implies 2V output across the burden:"
- If max primary current is 2A and current ratio is 1:50, the secondary current will be 2/50 = 0.04A.
- 0.04A applied on 50 ohms will produde 0.04x50= 2V.

3 - From universal transformer equation,
N=E/(4.44*Bmax*F*A)

Bmax = induction [T] = 0.8tesla
E = secondary [V] = 2volts
F = frequency [Hz] = 60hertz
A = core cross section [m2] = 0.016 * 0.016 = 0.000256square meter
Solving: N=2/(4.44*0.8*60*0.000256)
N = 37 turns at the secondary. This will be the minimum turns that produces a core induction of 0.8T.

THEM primary turns will be 37/50 = 0.74 turn... Litle difficult to do it! So, it will be practical to put one turn at primary and 50 at secondary.(Bmax=0.587T)
or
2 turns at primary and 100 turns at secondary. (Bmax=0.293T)
or
3 turns primary and 150 turns secondary (Bmax = 0.195T)
or
4 turns primary and 200 turns secodary (Bmax = 0.15T) **

any comment about above calculations will be welcome

4- the next step will be verify the CT errors for each secondary set of turns...this task lets do next day,

Thanks
Newton

** = EDITED

Forty

Mon Apr 23 2012, 05:31PM

Registered Member #3888 Joined: Sun May 15 2011, 09:50PM
Location: Erie, PA
Posts: 649

A larger core will lower the number of secondary turns you need (if you've got one)

Dr. Slack

Tue Apr 24 2012, 07:17AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Apologies for the tone of my earlier post, I'm getting dense in my old age.

In choosing whether to go 1:50 or 3:150 etc, more turns is (almost) always better. For several reasons.

1) It gives you a lower B field in the core material, which means better linearity. At power line frequencies, you have to put a LOT of turns on before you run into any capacitance issues. For dealing with 2A primary current, there's no significant difference in the resistance of 1 or 5 turns of a bit of 1mm wire stripped out of a cable.

2) It gives you the opportunity to distribute the primary turns around the core, which further reduces the error effects of non-infinite core permeability and non-unity transformer coupling.

It's for the second reason that it's best to use a toroidal core if you can, and then distribute the primary turns evenly around the circumference.

As a tip, I don't make current transformers. I go into my junk box and pull out a small mains toroidal. This is now a toridal core with a selection of pre-wound secondaries, 100s of turns for the LV windings, and 1000s for the mains winding. I put a few turns of single core insulated cable through the hole, and then measure the turns ratio I've got. Then choose a burden resistor to trim the volts/amp. It's necessary to either use the mains output, or insulate it. If you only use a low voltage secondary, the mains output can get to an unsafe voltage.

Newton Brawn

Tue Apr 24 2012, 09:15PM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Hi Dr Slack:

Thanks to return and thanks too for your comments.
I agree 100% with you to choose 3:150 winding , with more sec turns and lower B, resulting in better linearity.(better presision)
Also a ring core will give better wire distribution around the core.
My core is common E & I core, grain non oriented Fe Si stell, that one used in a old 10VA 110/12V, 60Hz supply.
See the B/H curve in the begining of thread.
The dimensions of the core will be shown soon. This will allow you follow the calculations, and more inportant, provide your comments
Regards
Newton

Newton Brawn

Tue Apr 24 2012, 09:26PM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Hi
Lets try to check the precision of the 2:100 winding CT:
First I calculate the length of the secondary wire, that is 100 turns #27 AWG, as 9.2 meter.
Based in a wire table, the #27AWG resistance of 1 meter is 0.196 ohms, so the secondary resistance is 1.88 ohms. Lets say 1.90 ohms.

The second step is calculate the core reluctance, and for the secondary side, the magnetizing inductance and the magnetizing reactance.

Core reluctance R = lm/(4*pi*0.0000001*ur*A)
R= 0.092/(4*pi*0.0000001*5500*0.000256)
R= 51996 At/Wb
Were:
lm = core magnetic path, as 0.092 meter
ur = core relative permeability, as 5500, from the B/H graphic
A = core cross section, as 16mmx16mm or 0.000256 m2.

Them the magnetizing inductance over 100 turns, Lm;
Lm = N2/R = 100*100/51996
Lm = 0.192 henries

Were N = number of secondary turns, 100

The magnetizing reactance over the secondary, Xmag :
Xmag = 2*pi*F*Lm = 2*pi*60*0.192
Xmag = 73.38 ohms

Were F= frequency, as 60Hz.

Then we have a R + R + L circuit as shown in the drawing bellow.

The total load Zsec on the secondary winding may be calculated as:

Rb+Rw = 50+1.9 >>> 1/(Rb+Rw)>>> 0.0193
Xmag = 73.38j >>> 1/Xmag >>> - 0.0136j

Ysec = ............................... 0.0193 -0.0136J
Zsec = 1/Ysec = 1/(0.0193 - 0.0136j) = 34.62+24.39j
|Zmag| = 42.35 ohms

As the secondary current is 0.04A, and secondary impedance of 42.35 ohms, the secondary voltage will be 1.694V.

The 1.694V applied on the secondary wire resistance and burden will result in a current of 1.694/51.9 = 0.0326A. This is the current delivered to the 50 ohm burden.

The 2:100 winding current error will be (0.0326 - 0.04)/0.04 = -18.5%

Conclusion:
The error is too much.

Lets try the 3:150 winding (next time)

I will return soon

Regards

.

1335302769 3343 FT137441 Ct Secondary L R Circuit

Dr. Slack

Wed Apr 25 2012, 07:31PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

As far as I can see, your calculations agree with mine to within engineering roundoff, though I tend to take a more "from the ground up" approach. It's not clear to me from a quick, or even not so quick read-through, whether all of your terms have been combined correctly, so allow me to set it out slightly differently, and I hope in a way that's easier to follow.

Derive the primary inductance of the core with one turn on it (the AL factor)
1 amp in 1 turn gives an H field of 1/lm A/m
which gives a B field of u0.ur/lm
which gives a flux of A.u0.ur/lm
which gives a flux linkage per amp (definition of self inductance) of A.u0.ur/lm

which when you plug in the values gives 1/51996 Wb/A.

For a transformer with a turns ratio of N, the burden resistor will look like R/N^2 at the primary. With your figures of 50:1 and 50ohms, that's 20mohm, or a conductance of 50 amps/volt.

This reflected load will be in parallel with the primary inductance, which is another way of arriving at the core magnetising current. The primary has two turns, giving a primary inductance of 4/51996 = 77uH. At 60Hz, this has an impedance of 2.60.pi.77u = 29mjohm, or a conductance of 34.5j Amps/volt.

The primary voltage will drive a current through both of these conductances. The load current will be in phase with the applied voltage. The magnetising current will be in quadrature, so they will add at right angles. This is entirely consistent with adding parallel conductances where one is real and the other imaginary.

So adding 34.5j and 50 gives us 60.7amps/volt at the primary. This means the primary current, at least in terms of magnitudes, will divide 34.5/60.7 to magnetising current and 50/60.7 to output current, which as you rightly say in your previous post, is 18% low.

I think this derivation is a little clearer, and it's also obvious what to do to improve things, and by how much. The problem is the low ratio between load conductance and magnetising current conductance. Three routes are available

a) Increase the load conductance, so reduce the burden resistor

If you can tolerate a lower burden, then that will improve the conductance ratio at the primary. Without presenting all of the maths, and neglecting the secondary resistance which is small but will increase the error slightly, the error for a 2 turn primary versus varying burden resistor is

50ohms 18%
40ohms 13%
30ohms 8%
20ohms 4%
10ohms 1%

This demonstrates why current transformers are frequently used into a short circuit, rather than into a finite resistance.

b) Decrease the magnetising conductance, so increase the absolute number of turns

As the primary inductance increases as the square of the primary turns, it doesn't take many more turns to improve the conductance ratio. The error for a 50ohm burden on 50:1 ratio is

2 turns 18%
3 turns 4.5%
4 turns 1.5%
5 turns 0.6%

c) Don't sweat it, just use a small number of turns

The error added by the non-infinite core permeability is a constant ratio error, so is really only scaling. The amount of error is fairly small at the moment, and variations in the permeability over time and temperature will be diluted significantly. You can stay within +/- 5% of the 82% gain with a permeability variation from 4700 to 6400, would the core permeability be this stable over time and temperature? Of course, with either a or b approaches, you could tolerate a much larger deviation in permeability for the same transfer error.

Newton Brawn

Thu Apr 26 2012, 12:05PM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Dr Slack;

Thanks for the lecture !
It is nice when we get the same result trying 2 differrent approches done by 2 different people from different places...

You solve the problem transfering the circuit to the transformer primary and I solve the circuit at secondary side. The result was the same, confirming the methods.

Following bellow is the solution for the 4 primary turns /200 secondary turns. I have considered the secondary winding resistance increased to 3.53 ohms and a magnetic core relative permeance reduced to 4000, This is due the more turns at secondary and reduction of ur due reduction of B at magnetic core.

PRECISION OF 4:200 WINDING CT:
The length of the secondary winding wire, that is 200 turns #27 AWG, is 18.4 meter.
Based in a wire table, the #27AWG resistance of 1 meter is 0.196 ohms, so the 18.4 meter secondary resistance is 3.53 ohm.

The magnetic core reluctance R may be calculate as:
Core reluctance R = lm/(4*pi*0.0000001*ur*A)
R= 0.092/(4*pi*0.0000001*4000*0.000256)
R= 71495 At/Wb
Were:
lm = core magnetic path, as 0.092 meter
ur = core relative permeability, as 4000, from the B/H graphic
A = core cross section, as 16mmx16mm or 0.000256 m2.

Them the magnetizing inductance over 200 turns, Lm;
Lm = N2/R = 200*200/71495
Lm = 0.559 henries
Were N = number of secondary turns, 200

The magnetizing reactance over the secondary, Xmag :
Xmag = 2*pi*F*Lm = 2*pi*60*0.559
Xmag = 211j ohms
Were F = frequency, as 60Hz.

Then we have a R + R + L circuit as shown in the drawing, and the total load Zsec on the secondary may be calculated as:
Rbw = Rb+Rw = 50+3.53= 53.53 ohm. >> Ybw = 1/53.53 = 0.0187 A/V
Xmag = 211j >> Ymag = 1/Xmag = 1/211j = -0.00474j A/V

Ysec = Ybw +Ymag ........................ 0.0187 -0.00474J A/V
Zsec = 1/Ysec = 1/(0.0187 â€“0.00474j) = 50.29+12.76j V/A
|Zsec| = 51.89 ohms

As the secondary current is 0.04A, and secondary impedance of 51.89 ohms, the secondary voltage will be 2.075V.

The 2.075V applied on the secondary wire resistance and burden will result in a current of 2.075/53.53 = 0.03877A. This is the current delivered to the 50 ohm burden .

The 4:200 winding current error will be (0.03877 - 0.04)/0.04 = 3.1%

CONCLUSION:
The CT transformer with 4 turns primary and 200 turns secondary satisfy the conditions of +- 5% error .

Note: I have done the same procedure for the 3/300 turn winding, resulting in a error of 6.8%.

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