Back to the basics again. The CT you had if I recall correctly was a 200:5 CT. Therefore the actual ratio is 200 / 5 = 40 so in reality it has a ratio of 40:1. The fact that it has a 200 amp rating means it will handle 200 amps of primary current. Therefore 40 amps through the primary yields 1 amp through the secondary. 200 amps through the primary is 5 amps through the secondary. Therefore 0 to 200 amps results in 0 to 5 amps.
So we add lets say a 1.0 Ohm shunt or burden resistor across the secondary. There will be a voltage drop across that resistor. With 0 amps through the primary of the CT the output will be 0 amps through 1 ohm for 0 volts. With 200 amps through the CT primary the secondary current will be 5 amps through 1 ohm for 5 volts. V = I * R so 1 * 5 = 5 volts. Thus the span is 0 to 200 amps = 0 to 5 volts. This is basic ohms law.
The capacitors you see in the original Atmel drawing are used for coupling. They are 47 uF electrolytic capacitors. They block any DC component level. The value is chosen based on the 50 Hz. line frequency. All of that is well covered on page 8 of the AVR pdf. The formula is shown. Everything is explained in detail.
The CT used in the drawing has a max current of 275 amps. It should be noted that in the circuit they never exceed 10 amps. That is it! The circuit is designed around a max current of 10 amps. The fact that the CT is rated for a max of 275 amps means nothing! With a 2500:1 ratio and 10 amps the CT secondary would be 10 / 2500 = .004 amps or 4 mA. They use a 68 Ohm shunt so .004 * 68 = .272 volts. The amplifier at this point has a gain of 1.25 so we get 1.25 * .272 = .34 volts RMS. That becomes .34 * 1.414 = .48076 volts peak and .48076 * 2 = .962 volts peak to peak.
The micro controller is a 10 bit ADC using a 1.1 volt reference. The ADC input needs to be kept below 1.1 volts which is what is done.
OK, so the max current is 10 amps. You want 100 amps? Then change Rshunt to .1 * 68 = 6.8 ohms. You want 200 amps? Then change Rshunt to .5 * 6.8 = 3.4 ohms.
You need to start understanding the basic fundamentals and apply them.
You have a current that is converted to a small voltage proportional to that current. Matters not if that small voltage comes as a result of a CT or a plain current shunt. You work with that small voltage and signal condition it.
The AVR in this case uses a 1.1 volt reference. The A to D is a 10 bit A/D converter. Ten bits means 2 ^ 10 or that an input of 1.1 volts = 1024 bits. When 1.1 volts is applied to the A/D input we get an analog input of 1024 bits. The input to the analog channel can not exceed 1.1 volts or the input will over range.
If the instructor suggest using a shunt I suggest you read up and understand high side and low side current sensing with a basic shunt.
I am not about to do this assignment. I did all my homework over 40 years ago. More to your question:
I've question about old way (using ct) as i've change the given CT using 5:200 instead of 1:2500, I've change the shunt or burden resistor to 1 ohm, but the question is, do i have to change the whole analog current front i.e capacitor value and resister value? if so, then how do i know the value for these components? capacitor and Resistors? pleas do have a look of the analog current front of the avr465
First, the ratios are 200:5 which is 40:1 and 2500:1 they are not 5:200 or 1:2500. Please get it right! May seem silly but it is very important. The value of Rshunt has nothing to do with the cap value or the resistors that determine the gain of the amplifier.
Ron