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You need to know that the SCT-013-060 is a 100 A : 50 mA current transformer - but with an internal "burden" resistor. A current transformer is like the more familiar voltage transformer, but as the name implies, it deals with currents, not voltages. so if you feed 100 A into the primary winding - and that is actually the cable you clip it onto - you get 50 mA out. If you want a voltage out, and we do, you need to feed that current into a resistor, which for CTs has a special name, the "burden". (Because it loads the CT.)

So to take your -060 CT, if you feed 60 A into it, it produces a current of 50 mA × 60 / 100 = 30 mA. Now, Ohm's Law says that to produce a voltage of 1 V, you need a resistor with a resistance of 1 V / 30 mA = 33.33 Ω. That then is the value of the burden resistor inside your CT.

But you want less voltage than the 1 V your CT gives you. What voltage do you want? Your input is 1.8 V peak-peak, so the rms value is that divided by 2√2, or 0.6364 V. Therefore doing the maths for the resistor value again, you want a burden resistor of 12.73 Ω. The hard way to get this is to take your CT apart and change the resistor inside. The easy way is to add a second burden resistor in parallel outside of the CT, and the convenient place is either inside the plug or on your breadboard. What value should that second burden resistor be? Clearly not 12 Ω. The formula for parallel resistors is 1/Result = 1/R1 + 1/R2. The result is 12.73, R1 is 33.33, therefore R2, your new parallel resistor is 20.6 Ω.

But you can't buy one of those, you might be able to buy a 20 Ω resistor, and that's fine, otherwise, you'll need to go to the next smaller value, 18 Ω.

That's solved the problem with too much signal. But where does your 0.9 V "bias" come from? That's fairly easy: you find a pin with VCC_ADC on it, or VCC_1V8 (pin 32 on the expansion header I think) and pin 34 is the GND that goes with it, and you connect your voltage divider, the two 100 kΩ resistors that you've drawn (& R1 & R2 on my 'Building Blocks' diagram) to those pins.

Therefore, you should have exactly the same circuit diagram as the 'Building Blocks' one, but your second parallel burden resistor ("Burden 33 Ω") is 18 or 20 Ω (not 33 Ω) and the built-in burden resistor is inside the CT. Arduino GND => Pin 34, Arduino 5 V d.c. => Pin 32, Arduino Input => Pin 39 (or another input - your choice). C1 should be about 10 μF

Note: You can only do this - adding a parallel resistor to reduce the voltage - with a current transformer. You cannot do it with a voltage transformer, like the ac adapter.

Mike, the circuit in your esboco2 diagram appears to be a voltage sensor.  The purpose of R1 and R2 is to reduce the amplitude of the AC signal that is seen by the Atmega processor. To complete the circuit, the top end of R1 needs to go to one of the analog ports of the processor. 

The current sensor is similar and normally has fewer components.  By changing the value of the burden resistor, the signal from the CT can be adjusted in amplitude so that it is suitable for sampling by the processor. 

I had thought of the paralel resistor, but then I wonder, where do I feed the 0.9V for the offset ? would it pass through the second resistor and influence the output in an undesired way ? where do I feed it ?

the burden is built inside the sensor. Only doing parallel maths, as the other answer sugested

I think Robin has confused the issue by mentioning voltage sensors. The easy way is as I wrote above.

You should use exactly the same circuit diagram as the 'Building Blocks' one, but your second parallel burden resistor ("Burden 33 Ω") is 18 or 20 Ω (not 33 Ω) and the built-in burden resistor is inside the CT - obviously not shown in that diagram as we always use the -000 version.

The offset voltage/current, whichever way you wish to view it, flows in the parallel combination of CT secondary and burden resistor/resistors anyway, and won't affect operation in the slightest.

Is there a way to calculate the second burden resistor to obtain the ratio from 1.4(1V * sqrt(2)) peak to 0.9. As I don't know the number of turns inside the sensor nor the built in resistor value.

And I know you've said it's OK, but would the second burden and the built-in one act in series with the voltage divider (that provides the 0.9 offset) and cause strange behaviour ?

(I'm asking cause I don't have the material to test it yet and THANKS A LOT for your explanations)

"Is there a way to calculate..." Not for you. But I could do that calculation because we are almost certain that the only difference between the various SCT-013 versions is whether they have the internal burden resistor and what value it is. I think you'd better read through my post (Thu, 16/07/2015 - 20:10) again and carefully follow the calculations and reasoning.

"would the second burden..." I've already said no to this. Look, I'm a Chartered Electrical Engineer, I've been doing electrical engineering formally since I started my student apprenticeship in 1967, I think there's a reasonable chance that I've not missed anything. But if you get a copy of LTSpice or another similar simulation program, you can verify it for yourself.

I'm trying to make sense of all you said. And I assume in the first answer when you said the final burden value I want is 12.73 (" Therefore doing the maths for the resistor value again, you want a burden resistor of 12.73 Ω") that, that value is to transform this sensor into a 100 A : 50mA with 0.9V at peak, is that correct ?

If I still wanted it to be 60 A : 30 mA with 0.9V at peak (to get a better resolution) I would have to use a combined burden of 21.21 Ω ( = 0,6364V / 0,03), right ? (with the "second burden" of 58.35 Ω in paralel with the built-in 33.33 Ω to achieve the 21.21 Ω)

I think Robin has confused the issue by mentioning voltage sensors.

I hope not.  My concern was that the initial post was asking for advice about the current sensor, yet the attached drawing appears to be for a voltage sensor.  I just wanted to make sure that the correspondent appreciated the difference between these two types.

Robin:

I thought Mike wanted to put a standard voltage divider across his CT with the internal burden, i.e. using exactly the same circuit arrangement as the ac adapter input, but on a voltage output CT. That would have worked, but why use two resistors when one will do the job.

Mike:

If you had a Shop Standard SCT-013-000, which is a 100 A : 50 mA CT, then to get 1.8 V p-p (= 0.9 V peak, = 0.6364 V rms) across the burden resistor, the value of that resistor would be R = V / I = 0.6364 / 0.050 = 12.73 Ω as we said.

"If I still wanted it to be 60 A : 30 mA with 0.9V at peak..."
Yes, you followed the logic and the maths and that's all correct.

You can't buy that resistor, but 56 Ω will be close enough. You can do the sums backwards if you like to determine your maximum current with that - 60.9 A.

OK, Robert, I can see that interpretation now. 

Without the (internal) burden being shown, that squiggly coil didn't look to me like a CT.