Thanks to everyone that contributed one way or the other. I now understand based on @WBahn explanation

 

Pls explain with this diagram

The circuit fragment shown, along with the amplifier power connections, is an incomplete example of how to bias an amplifier to provide an undistorted output when operated in a single supply mode. The text should have explained that. It is very important to read and understand the text,not just look at the pictures.

 

It's bias as a comparator not as an amplifier. Can you explain yours with a diagram?

The circuit fragment shown, along with the amplifier power connections, is an incomplete example of how to bias an amplifier to provide an undistorted output when operated in a single supply mode. The text should have explained that. It is very important to read and understand the text,not just look at the pictures.

O

 

The circuit fragment shown, along with the amplifier power connections, is an incomplete example of how to bias an amplifier to provide an undistorted output when operated in a single supply mode.

There is no negative feedback to reduce the circuit gain below the device's open-loop gain. The device is operating as a comparator, not a linear amplifier.

ak

 

There is no negative feedback to reduce the circuit gain below the device's open-loop gain. The deevice is operating as a comparator, not a linear amplifier.

ak

It is an incomplete circuit fragment in an inadequate textbook.

 

The opamp used in the instructional is a TL071 that has Jfet inputs with an extremely low input bias current of typically 65pA.

 

While the circuit will operate as a comparator, it is an example of how an opamp is biased to serve as an amplifier when it is operated with no negative supply voltage. And typically an opamp does not make a good comparator.

 

If you read the words under the circuit image in post #4, you will see that it says "Op-amp Comparator Circuit". Without the accompanying text, the instructional value of the circuit cannot be determined.

ak

 

The opamp in figure 4 has no negative feedback and has a voltage gain close to one million times then its output switches as high or as low as it can go without ever being linear. The oscillating opamp in the project has positive feedback to make it switch even faster. But for simply alternating groups of LEDs switching speed does not matter.

 

Please note that the basic question that the TS was interested in has nothing to do with a particular opamp or even opamps in general. The question was simply why can't you choose just any resistor value and how do you choose specific resistor values when a design, conceptually, only depends on the ratio of resistors.

 

Ok, that circuit segment was presented as a comparator. Yes I should have clicked on it and read the brief description.
Now about the question of why some resistance values are chosen, the explanation will be lost on those who have not learned the details.
Op amps have two forms, one is theoretical and idealized, while acttual physical opamps are much more complex. Real parts are never perfect, even when they are very good. That is a major reality. Real opamps have input bias currents and leakage currents and offset voltages and internal impedance. Because of those currents, and because some resistance values would lead to undesired voltages being developed or excessive currents flowing. So while the ration of Rf to Ri sets the gain of an opamp, the desired current output determines the value of Rf. The textbook neglects to mention that detail.
For the very simple voltage divider setting the reference voltage for the comparator, it is theimput current that the opamp requires that decides what value the resistors to have.
I hope that explanation provides a bit of insight.

 

I understand the circuit operation but I find it difficult as how they arrive at those resistors values. Can't we use 10ohms instead of 10k ohms for example in biasing the non inverting pin using voltage divider?

The order of magnitude (10 ohms vs 10k) determines current. Current matters in circuits.

In your comparator example, the non-inverting input is being biased to Vcc/2. If we choose Vcc=10V for convenience, using 10k resistors would give a divider current of 0.5mA and each resistor would dissipate 2.5mW. If you used 10 ohm resistors, divider current would be 500mA and each resistor would dissipate 2.5W.