Sir,
1) suppose I fix the output to 20V and that will be the non inverting input of opamp and on inverting input we have R2 and R5. In what way current sense resistance R3 is related to R2 and R5???
I will carry out simulation.
You can set the voltage to anything you want. When the output current becomes great enough, there is a voltage across R3. The bottom of R3 is connected to op amp pin 3, and the top of R3 is connected through the voltage divider to pin 2. When the drop across R3 is great enough, the voltage at pin 2 becomes slightly greater than the voltage at pin 3, and so the output of the op amp ramps down more and more until the output falls. When the output falls, it will fall to a level that allows just enough current through R3 to maintain the condition of the voltages at pin 2 and pin 3. That means the circuit is not in current regulation and so the voltage may fall a little, down to 19 volts or even less if the output current is higher.
If you were to set the pot to put out just 4 amps, then the output will put out 4 amps, and the voltage will be less than the voltage set point of 20v.
Resistor R7 also sets the gain of the op amp part of the circuit. It looks like the designer did not want to the gain too high, just high enough to get it to work well.
If you really want to understand this really well, look into a linear voltage regulator using an op amp as the control element. Check out the voltages that come on the inputs and outputs. In fact, you may gain insight by looking at just a voltage follower using an op amp. That works in a similar manner. As the output ramps up, the two voltages at the two inputs become nearly equal, and that stops the output from ramping up, and that's when it can be said to be regulating the output voltage. A voltage follower is the simplest example of this kind of op amp action.
For example, when the circuit starts up the output could be assumed to be at 0 volts, and then when say 2v is applied to the non inverting input, the output begins to ramp up. Eventually the inverting input becomes nearly equal to the non inverting input, and that means the output stops ramping up. It will ramp up to nearly 2 volts when this happens, then stop, holding the output at 2 volts.
The simplest way to look at this is as the following expression shows:
Vout=A*(vp-vn)
where
A is the internal gain (typically 100000),
vp is the non inverting input which is the input voltage (say 2v),
vn is the inverting input, which is connected directly to the output.
This means the above comes out to be:
Vout=A*(Vin-Vout)
Solving this explicitly for Vout we end up with:
Vout=(Vin*A)/(A+1)
Now in pure theory, we take the limit as A goes toward infinity and get:
Vout=Vin
and so this shows how we get the same output as the input. However, in real life we don't really have an infinite gain, so we stick with:
Vout=Vin*A/(A+1)
Now going back to that farther above:
Vout=A*(Vin-Vout)
if we start with Vout=1.999 (not quite 2.000 volts yet) we then have this:
Vout=A*(2-1.999)
or just:
Vout=A*(0.001)
and now with a gain of 1999 (just as an example) we have:
Vout=1999*0.001
or
Vout=1.999
and so that means the circuit had reached the equilibrium point because:
1.999=1999*0.001
It's good to think of this process as though the output could not ramp up quickly though as it takes some time. As time progresses, eventually it would get to the point where the output was 1.999 volts and that would mean we had reached the regulation point and it will stay at that level.
This might seem a little strange to you at first, but it makes sense when you go over it and try to understand what happens when the circuit is first turned on. One of the main properties that comes into play here is the slew rate which limits how fast the output can ramp up to get to the right voltage so the feedback and internal gain end up maintaining that output at the same level thereafter.
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