I found this circuit some where in the internet. I am trying to find how they got the value Vref as 10.5v. Can anybody please give me how they compute the vref as 10.5?

 

Presumably by adjusting the 20k potentiometer until the required value was obtained .

 

The 20kΩ is variable resistor wired as a voltage divider also known as a potentiometer or pot for short.
You manually adjust the pot until you get the desired voltage,

 

If it is not the potentiometer and if we assume it as 20k resistor. Then what will be the Vref. Can we compute it?
If yes how?

The 20kΩ is variable resistor wired as a voltage divider also known as a potentiometer or pot for short.
You manually adjust the pot until you get the desired voltage,

 

You would then simply be tying the op-amp positive input to Vcc and the circuit wouldn't function as intended.
Perhaps you need to better understand the overall circuit operation.

 

There is nothing to compute. You adjust the pot until you get 10.5V.

Google voltage divider.

 

OK understood. I thought op amp is acting as comparator without any reference to the output. And if v+ is greater than v- then Vout=Vcc. If v- is greater then Vout=Vee. Then the current ic is adjusted by resistor (10ohm).
So by the way you all explained can I assume Vref as 10v instead 10.5v. And adjust potentiometer to get 10v.

 

There is nothing to compute. You adjust the pot until you get 10.5V.

Google voltage divider.

If I use voltage divider.
Vref = (Vcc × 20k)/(20k+10)
= 11.99v if vcc is 12. Is it right?

 

No. The opamp is not acting as a comparator. The output is fed back to the inverting input which makes it a unity-gain voltage follower.

 

If I use voltage divider.
Vref = (Vcc × 20k)/(20k+10)
= 11.99v if vcc is 12. Is it right?

No. RE is not part of the voltage divider.

The total resistance of the pot is 20k.
Part of the pot is R (let us assume the lower part). The upper portion is 20k - R.

Vref = Vcc x R /20k

 

Thank you so much for clearing my doubts. I really love this forum and it's people.

 

OK understood. I thought op amp is acting as comparator without any reference to the output.

It's hard to see how this thought is reasonable given that the inverting input of the opamp is tied to a node whose voltage depends on the output.

And if v+ is greater than v- then Vout=Vcc. If v- is greater then Vout=Vee. Then the current ic is adjusted by resistor (10ohm).

If that's the case, then what would the current be?

So by the way you all explained can I assume Vref as 10v instead 10.5v. And adjust potentiometer to get 10v.

Why do you think you can assume it is 10V instead of the 10.5V that it is stated as being adjusted to?

If you want to adjust it so that it is 10V, then do so. That will result in 2V being dropped across the 10Ω resistor and, thus, 200mA of current in the LED.

 

If I use voltage divider.
Vref = (Vcc × 20k)/(20k+10)
= 11.99v if vcc is 12. Is it right?

You can't just pick the first to resistances you see and throw them into an equation and expect the result to have any meaning whatsoever.

You need to understand what the voltage divider equation means and when it does and doesn't apply. It applies when you have two resistances that are in series. The 20kΩ pot and 10Ω resistor are not in series.

The potentiometer is effectively two resistors in series, one between the top of the device and the wiper and the other between the wiper and the bottom of the device. They are effectively in series because only a negligible amount of current flows in/out the wiper terminal. The sum of the two resistors is equal to 20kΩ. The position of the wiper merely determines how much of that appears above the wiper and how much appears below.

 

I was mistaken. I thought they are in seeies . iam in very learning stage.

 

I have design for buck driver for led. Can I use that for laser diode. What makes difference in design between led and laser diode if Vf and current requirement of led and laser diode is same?

 

I was mistaken. I thought they are in seeies . iam in very learning stage.

That's fine -- making mistakes is how we learn best and having our mistakes pointed out to us is how we learn best the fastest.