can someone explain what does the highlighted part does to me? why is there a need to half the Vcc into non-inverting input of op amp? What does R1 does? and how do i know the duty cycle after comparator U1B ? i replaced POT1 with LDR and Resistor in series, i want to know what is the duty cycle of the pulse. what i learn in last time the circuit suppose to look like
there is R1 and R2 and therefore i can calculate frequency , duty cyle, period. but what happens when there is only 1 R1 left? can someone explain that part to me and the equation needed to calculate the frequency, duty cyle and the derivation of it?

 

why is there a need to half the Vcc into non-inverting input of op amp?

To set the operating point at 1/2 the supply voltage.
If you had dual (plus and minus) supplies you wouldn't need R2 and R3.

What does R1 does?

I it provides positive feedback (hysteresis) to the voltage at the (+) input.
Thus C1 will charge through R4 until the upper voltage is reached at the (+) input, at which point the voltage at the n (-) input exceeds that at the (+) input and the output goes low.
C1 now discharges through R4 until the lower voltage at the (+) input is reached, causing the output to go back high.
This cycle repeats, giving the exponential-sawtooth signal out across C1.

what is the duty cycle of the pulse

The voltage at U1B's (+) input is compared with the voltage from C1.
If it's higher, the output of U1B is high, and if it's lower the output is lower.
For the values of R1, R2, and R3 shown for the oscillator, the sawtooth goes from about 2/3 the amps peak output voltage to 1/3 that voltage.
From that you can determine the duty cycle for a given voltage U1B's (+) input.

Below is an LTspice simulation of the circuit with LM324 op amps, showing its operation for a varying ramp at the comparator's input:

 

To set the operating point at 1/2 the supply voltage.
If you had dual (plus and minus) supplies you wouldn't need R2 and R3.
I it provides positive feedback (hysteresis) to the voltage at the (+) input.
Thus C1 will charge through R4 until the upper voltage is reached at the (+) input, at which point the voltage at the n (-) input exceeds that at the (+) input and the output goes low.
C1 now discharges through R4 until the lower voltage at the (+) input is reached, causing the output to go back high.
This cycle repeats, giving the exponential-sawtooth signal out across C1.
The voltage at U1B's (+) input is compared with the voltage from C1.
If it's higher, the output of U1B is high, and if it's lower the output is lower.
For the values of R1, R2, and R3 shown for the oscillator, the sawtooth goes from about 2/3 the amps peak output voltage to 1/3 that voltage.
From that you can determine the duty cycle for a given voltage U1B's (+) input.

Below is an LTspice simulation of the circuit with LM324 op amps, showing its operation for a varying ramp at the comparator's input:

View attachment 129199

THANKS!! U HELPED ALOT!`