op-amp sawtooth generator
OK, so look at the comparator: R5, R6 will produce Vcc/2 at pin 6.
To have the comparator switch, you need a little more than 2.5V at pin 5. Since the voltage at this pin is the output voltage divided down by R3, R4, you get:
Vpin5=R3/(R3+R4)*Vout≈0.61*Vout.
So: 0.5*Vcc=Vout*0.61=(Vcc-1.5)*0.61
0.11*Vcc=0.915
From this, the minimum Vcc for reliable operation is:
Vccmin=0.915/0.11=8.32V.
That means you are lucky it works at 6V.
Solve the equation now to see the actual drop of your IC:
0.11*6=x x=0.66
That means the IC you used actually has a drop of:
0.66/0.61=1.082V
To test the validity of this, reduce the voltage until the oscillation stops and measure the output DC voltage. It should be about Vcc-1.082V
So, using the LMV324 I suggested, which has a drop of only 0.1V, you can ensure reliable operation down to:
0.5*Vcc=(Vcc-0.1)*0.61
Vccmin=0.061/0.11=0.55V (theoretical limit)
In reality, you can say the circuit will operate down to 2.7V, which is the minimum Vcc specified for the LMV324.