design and implement metal detector

The circuit is an oscillator and the way it keeps oscillating is due to positive feedback. Can u help me to determine how the circuit work and the phase shift of the circuit? thanks

Remember CIVIL. Vertical vector = Ic, so Vc lags by 90 degs (inductive off resonant circuit), Vc across Cc-e, causes Ie to lead by 90 Degs, so is in phase with Ic and away it goes!!!!!
can't see how this will detect metal?
Frank

thanks for expaination, can u explain more how the transistor work, the function of Cc, Cc-e and Ce?

Cc,brings the inductor to resonance (almost). A series tuned circuit will have a low impedance off resonance, as the the frequency increases the impedance remains inductive but rises and becomes more resistive as the frequency approaches resonance. At resonance, it appears resistive, then it becomes less resistive and more capacitive as the frequency continues to increase. Cc might not be required if the gain of the transistor is high enough/ the Q of the inductor is great enough. Remember its the change in collector voltage that drives the feed back current Ie, so low voltage , low Ie, low/absent oscillations. Cc-e, feeds back Vc as Ie / tunes the inductor. Ce actually bypasses some Ie to earth, so lowers output. Cicuit should work without it. It does remove the effect of the Re, so increases the Q of the tuned circuit as it appears in parallel with Cc-e to the oscillations.
Frank

hi Frank,
did u make a mistake here? this tuned circuit is in parallel, not series

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hi Frank,
did u make a mistake here? this tuned circuit is in parallel, not series

Did you not read what I wrote. The "parallel" tuned circuit MUST act as an inductor, or the phase shifts are not correct. So de facto its not at resonance, so it is not a parallel resonance circuit. As I said the effect of the parallel Cc and L are to act like a high value resistor with a high value inductor across it. You can get the circuit to act the same by removing Cc and using a much higher value of inductor. It just a lot more work and expense.
Frank

thank you, Frank. Do you think this circuit have ability to detect metal? My idea about the working of this circuit is explained like this:
The transistor is only turned on when the voltage across the capacitor is negative on the bottom plate. This is when the voltage on the capacitor is added to the voltage of the supply and passed through the 1n capacitor across the transistor to reduce the voltage on the emitter. The transistor turns ON and delivers a short burst of energy to the tuned circuit.
The 4n7 on the emitter charges slightly during this action and the voltage on the emitter rises to turn the transistor OFF.
The charge on the capacitor is passed to the coil and it produces expanding magnetic flux. The capacitor runs out of charge and the coil collapses. The collapsing magnetic flux produces a voltage in the opposite direction and this is passed to the capacitor.
During this part of the cycle the voltage on the capacitor is not of the correct polarity to turn the transistor on and it remains OFF.
This is the part of the cycle when very little load is placed on the tuned circuit and the voltage produced by the coil can be higher than the applied voltage. The capacitor then delivers its charge to the coil and the cycle repeats.