hey everyone, thank you for the help so far. one more things to ask, how thus the mosfet working in this circuit, like specifically how the mosfet function and interface with op-amp. how can we know that specific current will flow through the mosfet when 1V is applied at the gate to source. in simply form, what is the main function of mosfet in this circuit.

The MOSFET simply passes current based on its gate voltage. The Op Amp tells the MOSFET how hard to turn on based on the difference between the feedback voltage (the MOSFET current times the 1 Ohm resistor) and the reference voltage 1V. The output of the op amp will keep rising, turning the MOSFET on harder and harder, until the voltage drop across the 1 Ohm resistor is exactly the same as the voltage on the + input.

the second question, if the resistor R1 which is 1ohms is change to 2ohms thus the value of current change at R2?. next can u guys explain the current flow through dc source to mosfet and to the node at the source of the mosfet and inverting input of op-amp. and from the node to the R1 and ground.

As long as the reference voltage is fixed at 1V, the current flow through the sense resistor has to produce the same voltage. If the sense resistor is set to 2 Ohms, it takes 500 mA to create the 1V feedback.

The simple way to think about this is to separate the MOSFET gate issue from the drain-source issue. The drain-source current wants to be 1A. At some level of gate voltage the MOSFET will deliver that amount of current. We don't know what the gate voltage has to be and we don't care. That's the problem of the op amp. If will put out as much voltage as necessary to get the MOSFET to turn on hard enough to get 1V across the sense resistor.

Remember, the open loop gain of an op amp is HUGE! If there is a voltage difference at the input terminals, the output will change enough to reduce the difference to 0V. That's why these circuits oscillate. The output voltage goes all the way to the op amp rail, the MOSFET turns on harder, the sense voltage increases, the op amp swings all the way to the other rail and the MOSFET turns off. This is the very definition of oscillation. That's why that little extra capacitor slows things down. The math of what it is doing is a little hairy but basically, it is anticipating a change in the sense voltage before it happens based on how much the op amp output changes. It does this to provide a little extra sense voltage (or less voltage if the op amp is going low) in anticipation of the larger change across the sense resistor.

the third question, how to determine if we have to put value of Rs at the input of voltage reference? if not stated in the datasheet. secondly if we have to put Rs how to determine the value of Rs. noted that im using ltc1798.

That resistor on the + input is just there to limit any fault current that might get involved with the input. It's value is pretty much irrelevant. No current flows into either of the op amp inputs (ideal op amp laws). As a result, there is no voltage drop across that resistor under any reasonable condition. Since there is no voltage drop and no current flowing, the value isn't important. There really is SOME current flow into the + input so keep the value reasonable - 1k seems ok, 100 Ohms seems ok, 1 gigaohm doesn't seem ok.