The circuit shown below is supplied with an input sinusoidal voltage vs of ampli
ID: 1300280 • Letter: T
Question
The circuit shown below is supplied with an input sinusoidal voltage vs of amplitude 0.1 volts and of variable frequency Using the parameter values given next to the circuit diagram. the phase difference between the input voltage vs and the output voltage vo as the frequenc approaches OHz is determined as nearest to which of the following five values Voc = 12 Volts DC vs 0.1 .cos(est) Rs 470 ?ms Rb 9.5k RbS Rc C2 Rs Cl RLvo C10.1F C2 is very large (C2>>>C1) om100(/volt) O degrees 45 degrees O 90 degrees 230degrees O 180 degreesExplanation / Answer
Here the option is (D).As it is a common emitter configuration.Here the output is taken against the collector rather than the emitter side with any resistance rather it is shorted and grounded .While calculating the voltage gain of this instrument and amplifier we will find that here the current flows in a direction opposite to the natural direction of flow current .When an AC signal is applied to the transistor amplifier it causes the base voltage VB to fluctuate in value at the AC signal. The positive half of the applied signal will cause an increase in the value of VB this turn will increase the base current IB and cause a corresponding increase in emitter current IE and collector current IC. As a result, the collector emitter voltage will be reduced because of the increase voltage drop across RL. The negative alternation of an AC signal will cause a decrease in IB this action then causes a corresponding decrease in IE through RL. The output signal of a common- emitter amplifier is therefore 180 degrees out of phase with the input signal. And the parameters given in the question are not required to find the phse shift as it is fully 180 phase shift because of negative direction of flow of current to the output.
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