Assume the circuit above, answer the following questions 1. How can the amount o

Question

Assume the circuit above, answer the following questions

1. How can the amount of feedback (gain) be varied useing a potentiometer of a value equation RE without affecting Q-point?

isn't varying RE will change the Q-point anyways? (not sure)

2. how can high frequency be reduced usin external component?

(no clue)

3. Why is the calculation value of high frequency measured analyltically is diffrent than the circuit built by hand?

I said because of the inner (internal) capacitance values of the BJT , they are not all the same. (not sure)

Please give answers. thanks

Explanation / Answer

Hi here are the answers!!! 1) The Q point will be effected if the value of RE is varied to obtain the gain because the Q point will be calculated in the DC analysis !!! and in the DC analysis the capacitor is open circuit because the frequency is 0; and the reactance is 0; so either vary the load because the gain in the bypassed case is -gmRL; 2) Using a capacitor in the output stage ,ie between the Vo and ground will vary the high frequency value!!! 3) The frequency measured analytically will be different from the practical ones because of the fact that there will be internal parasitic capacitances in the BJTs like the Cpi and C(mu).

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