[SOLVED] AC Current Measuring -ADC- with Current Transformer H.ow convert the Ac to Dc ?

I'm not familiar with Proteus, but you could try using the X-Y feature, taking the difference between the 2 phases of the mains source ( eg. A+B, internally inverting one of them ).

There is no problem simulating this.

E-design said:

There is no problem simulating this.

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Thank you very much working the simulation right now

E-design said:

You can use the circuit below and scale it to your needs. It uses a 2.5 V offset so that the zero of your current measurement will be at 2.5 VDC. You can easily handle that in code to get your maximum and minimum readings.

The burden resistor that KlausST talked about is R1 in the diagram. It converts the current produced by the CT, to a voltage. The bandwidth of the circuit is about 300 Hz.

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Hello
How the design in the given circuit ?
I mean its calculation ?
Is this calculation is right way ?

Most of the circuits in google telling that There should be a resistor to the ground ? It is easy to calculate the gain ?

How the design in the given circuit...

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If you do not have a ground, the input is floating and if R1 and R2 are not exactly equal, you will be getting lots of noise. Finally, the output have no reference (unless your supply is bipolar).

I found Klaus's original design (without a rectifier diode) very appealing: you need to sample the ADC values 3 times at equal interval and you will get both frequency and peak values of the sine wave after removing the bias Vref/2. You can do a lot with the simple design but you need to know how to write a small program in C.

thannara123 said:

How the design in the given circuit ?
I mean its calculation ?
[...]Is this calculation is right way ?

Most of the circuits in google telling that There should be a resistor to the ground ? It is easy to calculate the gain ?

Click to expand...

This is exactly as the circuit from post #8.

First of all, you want R1/R2=Rf/R3 because that improves the CMRR. Usually it is made R1=R2 and Rf=R3 (see post #8).
That being said, the output voltage is the following:
v_out(t)=Rf/R1*V_differential+V_REF*R2/(R2+R3) *(1+Rf/R1)
Another point is the input resistance seen by the differential voltage, which is R1+R2 (=2*R1 if R1=R2)

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E-design said:

It uses a 2.5 V offset so that the zero of your current measurement will be at 2.5 VDC. You can easily handle that in code to get your maximum and minimum readings.

Click to expand...

Or he can add a DC blocking capacitor and do not need to do anything.

Most of the circuits in google telling that There should be a resistor to the ground...

Click to expand...

Google is right (unsurprisingly): consider the op-amp is ideal and V3 is absent.

Voltage applied at V2 will go through R2 but cannot go anywhere because the opamp has infinite input impedance.

Voltage applied at V1 will go through R1 and Rf but there is no ground return (unless the load is connected to ground).

Consider Vref is connected to ground. Current at the input at V2 can now run to ground. But V1 is still problematic (unless the output is loaded to ground),

If V3 is connected to Vref, the tie point must be able to source/sink some small current without affecting the potentials.

The problem will be worse for opamp run with a single supply (as the output is low impedance, the noise will be low) because the current lacks no path to return.

Very useful post
Thannara123 did you found extact values for your design?
Can you share your design for learning purpose

Yes offcourse ,i will post it . when i reach at home .

abc_de said:

Very useful post
Thannara123 did you found extact values for your design?
Can you share your design for learning purpose

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sorry for late i am giving the circuit which i have worked

Hi,

Can you share your design for learning purpose

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The circuit may satisfactory work for the OP.
But I don't think this is good design to learn from.

The circuit lacks from several issues, mostly discussed above.
I assume a lot of resistor values are found by trial and error - this is not what one should learn

--> I recommend to follow standard rules for designing a "differential amplifier circuit".

Klaus