Thank you, Chen _T for posting that image.
This circuit is a disaster waiting to happen.
1. The maximum permissible input voltage for the 74LS04 is 5.5 V. Driving it with 14 V will almost certainly destroy it.
2. IOH (high level output current) for 74LS04 is -0.4 mA. This current is so low it is not even on the typical graphs for the 2N3055. You might get about 40 mA in Q2. Leakage currents in the 2N3055 can be about 1 mA.
3. A real op amp at U2 can probably drive 10-15 mA before the output current limits. That will get emitter current in Q1 to 1 to 1.5 A. However diodes D4 and D4 are rated at 1 A, so the circuit should not be operated at levels which would damage those devices.
4. At 1 A C1 will charge from zero to 8.8 V in about 880 us. At 40 mA discharging C1 will take 22 ms.
5. If the function generator is not operating R2 sees 8.8 V and draws 176 mA.
6. Since the current in Q2 and C1 is <= 40 mA and the current through R2 is 176 mA, no significant charge will ever get transferred from C1 to C2.
7. Real op amps at U1 and U2 would require power supplies of 16-18 V to operate at the desired input voltages.
8. Transistors turn on faster than they turn off. For the 2N3055 at 1 A (Q1 turning on) the total time is about 0.5 us. To turn Q2 off at 40 mA probably takes 6 us. (This current is off the graph so I had to extrapolate.) So, for more than 5 us every time Q1 turns on and Q2 turns off both transistors willbe conducting simultaneously. The low current in Q2 will probably limit the amount of smoke released.
9. If Q1 is conducting, the voltage at the anode of D4 (which is also the feedback voltage) will be +0.6 V. If the circuit actually worked the way you hoped it would, the voltage at that point with Q1 off and Q2 on would be negative. Comparing it to +14 V in U1 does not seem to make sense.
Now, tell us how you think this circuit is supposed to work.
Lynn
