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In a class C amplifier (Figure A), every time the transistor turns on in every "pulse", the collector current becomes saturated. This charges the capacitor to its maximum (Vcc). This capacitor will always get charged to its maximum, independent of the input signal frequency.

So why does the output voltage reduce when the frequency is below or above its own specific resonating frequency? (Figure B)

I mean, whether the input frequency is in resonance or not, the capacitor will always get charged to the fullest when the transistor turns on, thus swinging the tank circuit to peak Vcc. So why is the output voltage damped when it is beyond the resonance?

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  • \$\begingroup\$ Does the capacitor really get charged fully regardless of the frequency? I think not. \$\endgroup\$ – JRE May 3 at 10:48
  • \$\begingroup\$ In every pulse, collector current will be saturated(Vce=0 volts) then the capacitor will get whole the Vcc voltage, and I think that will happen regardless of input frequency \$\endgroup\$ – Iwatani Naofumi May 3 at 11:02
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In a common emitter amplifier, the gain will be given by the the collector impedance and load impedance:

\$A_v = \frac{Z_{collector}//R_L}{r_e}\$, where \$r_e = \frac{26mV}{I_E}\$

Given that, a tank circuit reaches its maximum impedance at its resonant frequency. In other words, \$Z_{collector}\$ is maximum at \$f_r\$.

For frequencies below or above \$f_r\$, its impedance is lower and decreases as long as the frequency moves further from \$f_r\$. So the parallel \$Z_{collector}//R_L\$, and thus Gain \$A_v\$, is low for any frequency different of \$f_r\$.

Therefore, you have a tuned amplifier which only amplifies signals of a certain frequency and "ignores" other frequencies.

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I mean, whether the input frequency is in resonance or not, the capacitor will always get charged to the fullest when the transistor turns on

No it doesn't - consider case 1 where the frequency is lower than resonance - the capacitor will start to acquire charge and continue to charge up to the point where the inductor removes that charge from the capacitor by progressively shorting it out over time. In this scenario the inductor wins because it is more dominant at frequencies below resonance.

Case 2 where the frequency is higher than resonance - the transistor won't properly "saturate" because the impedance of the capacitor is very low and the capacitor needs a flow of current for some non-zero period of time to acquire anything like the full Vcc range.

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