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Design a single stage BJT amplifier with a split emitter resistance with the fol

ID: 1411579 • Letter: D

Question

Design a single stage BJT amplifier with a split emitter resistance with the following constraints:

-Use a 2N2222 transistor with beta of 200               -Power Supply (Vcc): 7V

-An input impedance of 5k ohm     -Midband Voltage gain of -12 V/V

-A Load resistance of 2.1k ohm     -The Input signal has a Rsignal of 400 ohm

-Choose the coupling and bypass capacitors to obtain an F(low)-3db of 100 HZ

-The Capacitor and resistors should have commericial values

The total harmonic distortion of the output vltage should be less than 1% when the amplifier produces the maximum swing out voltage. The minimum peal-to-peak out voltage swing should be 40%of the voltage power supply.Obtain the frequency response of the amplifier to verify the midband voltage gain, F(low)-3db and F(high)-3db

P.s: please find the value of both emitter resistors.

Sample Circuit:

7 2 R2 Rc C2 RB1 R6 C1 Cc Q1 R5 5 Rsig Cb NPN V1 R3 RB2 R4 Re1 R7 C3

Explanation / Answer

value of coupling capacitors = 5 * 10-6/1000 = 5 nF

value of by pass capacitors = 10 * 10-6/1000 = 10 nF

=>   the frequency response of the amplifier =    flat frequency response range of frequencies from 20 Hz to 20 kHz.

The coupling and bypass capacitors cause the fall of the signal in the low frequency response of the amplifier because their impedance becomes large at low frequencies.

In the mid frequency range large capacitors are short circuits and the stray capacitors are open circuits, so that no capacitance appears in the mid frequency range. Hence the mid band frequency gain is maximum.

At the high frequencies, the bypass and coupling capacitors are replaced by short circuits. The stray capacitors and the transistor determine the response.

=> The minimum peal-to-peak out voltage swing   = 0.4 * 7 = 2.8 V

=>   value of both emitter resistors   =   250 ohm and   300 ohm .

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