# Current transformer not sensing currents

I have made a current sense transformer to sense primary current maximum 2A(of my electric iron) and secondary output is 1A but there is no current in the secondary. I have used a iron "Nut" as a torroidal core(I suspect this may be the problem) and used normal wire with plastic insulation for winding. I took 2 turns for secondary, and burden resistor is 1.4 Ohm. As a newbie I am getting problems, experts please guide me.

update: Thank you experts for helping now I can develop a CT for me.

Current transformers normally have hundreds to thousands of turns in their secondaries, since the idea is usually to turn a sensed current into a measurable voltage.

Keep in mind that a current transformer is also a voltage transformer. With a 1:2 ratio, the output voltage will only be at most twice the input voltage. In this case, the input voltage is simply the voltage drop caused by the primary current interacting with the impedance of your 1-turn primary winding, which is very tiny.

Just to put some numbers to this, let's say that your nut has an inductance on the order of 10 µH. At 50 Hz, this will have an impedance of 2πfL = 3 milliohms. Your 2A primary current will produce a voltage drop of about 6 mV, and your secondary will produce a maximum open-circuit voltage of 12 mV.

• Thank you for providing this knowledge of CT, I was unknown of it. actually I found secondary winding by using formula for secondary Is=Ip(Np/Ns). So by rearranging I found secondary turns. So for Ip=2A, NP=1, and Is=1A I found the Ns, is this correct? or else then which method to follow to calculate secondary turns? and is iron nut good for CT application? Commented Jan 31, 2015 at 15:52
• You would get 1A in your secondary only if the total resistance of the secondary circuit (including the secondary wire itself) is much less than 12 milliohms, by at least an order of magnitude, preferably more. Just as a point of reference, one inch of AWG24 (0.5 mm) copper wire has a resistance of about 2 milliohms. If you were trying to measure the current with a typical multimeter, its internal resistance is probably on the order of 100 milliohms, which is far too much. Commented Jan 31, 2015 at 17:55

The primary winding is your live feed to your iron and this has a small inductance possibly in the range of 10uH to 1mH but with a small steel nut it will act as a partial shorted turn as well as exhibiting the desired ferrous magnetic properties so I could envisage your primary magnetizing inductance being in the order of 1uH.

This will have a reactive impedance at 50Hz of 0.314 milli ohms. This in itself isn't a hindrance to making a CT because the whole idea about a CT is that the burden resistor divided by the turns ratio squared appears in parallel with the magentization inductance and, for a CT to work effectively, the equivalent primary impedance of the burden resistor must be significantly lower than the reactance.

Here is the development of the idea in circuit form: -

Unfortunately this isn't going to happen with your design - you have a turns ratio of 2:1 and a burden value of 1.4 ohms. This burden, when referred to the primary looks like a resistance of 1.4/4 ohms i.e. 0.35 ohms - this is many times bigger than the magnetization reactance so your signal will be pitifully small.

If you wound 100 turns on a ferrite ring core this would be better - your primary inductance will be higher (due to the ferrite not acting like a shorted turn) so inductance might be 10uH for a small core and possibly 100uH for a bigger core. At 10uH the reactance is 3.14 milli-ohms - remember the reflected impedance of the burden must be significantly smaller than this.

With 1.4 ohms and 100 turns, the reflected burden on the primary is now 0.14 milliohms and this is looking better because it's starting to be significantly smaller than the 3.14 milliohms primary reactance.

Hope you get the general idea by now.

Most of the primary current is now passing thru the burden reflected thru to the primary (as opposed to the mag reactance) and for 1 amp drawn, I would expect to see about 14mV across the burden.

• Now I got general idea. And I am using multi stranded wire to wind secondary, is this correct? or should use single strand (enameled copper)? and will you please explain how you calculated 100 turns? Commented Jan 31, 2015 at 16:02
• @MrunalAhirrao - I've added a picture. The 100 turn secondary is incidental - it has to be high enough to make the burden look much lower than the magnetization inductance when referring to the primary winding. Stranded or solid wire makes no difference. Commented Jan 31, 2015 at 16:55
• Now I got it.This information was not given in the tutorial I have seen. I am using 1 inch iron nut, so is the inductance 10uH is correct for this size? and is it correct to use iron nut? Commented Feb 1, 2015 at 4:30
• And by taking secondary winding 100 turns how we know that output current will be 1A(any formula?) because by using formula Is=Ip(Np/Ns), Is=2A(1/100)=0.02A. Commented Feb 1, 2015 at 4:45
• For N = 100 and Ip = 2A, output current is 20mA. If you believe you can get currents like 1A from a CT for an input current of 2A, forget about it if you want accuracy. Commented Feb 1, 2015 at 8:53