***Please show all steps & formulas, as well as, explain the reasoning behind th

Question

***Please show all steps & formulas, as well as, explain the reasoning behind the solution.

***Please no blurry or dark pictures, & no sloppy writing so I can read the solution.

***Thank you for help in solving this problem.

Best,

Problem 6.3 The transistor amplifier shown (Figure 6-30) is a high-bandwidth, common-emitter configuration used to amplify a video signal. A small-signal video input drives from a 75 source impedance and is terminated with a 75 termination resistor. The transistor is biased with a DC power supply Vpc - 5.7 volts. For the MPSH20 transistor you may assume r, = 20(2, AC beta h,-DC beta hFE-25, C, = 0.9 pF andfr 630 MHz. Make reasonable approximations, state them, and justify them. Assume room temperatur kTIq 26 millivolts e is 25°C and hence

Explanation / Answer

Common –emitter amplifier:

      In this type of amplifier the base terminal serves as the input and the collector terminal serves as output and the emiltter is commen to both base and collector. Common emitter amplifier give an inverteg output and a very high gain that vary from one transistor to another.

Given data:

Vcc=10V        RL=470 ohms       Vdc=5.7V      Re=1K ohms     Rs=75 ohms        Rr=75 ohms

Beta=25

Solution:

Beta= Ic/Ib

Therfore Ic(max) = Vcc/Rl

                                 = 10/470 = 0.021A

Generally the Q-point of the amplifier is with zero input signal, applied to the base. So the collector sits about half-way along the load line. Therefore, the collector current is given as:

Ic(Q) = (Vcc/2)/RL

          = (10/2)/470 = 0.106A

We know that

Beta = Ic/Ib

Therefore       Ib = Ic/beta

                         Ib   = 0.106/25 = 0.00424 A

Collector base voltage is given as

Vcb = Vb-Vc

Therefore Vb = Vbe and Vc=Vce=Vcc

Therefore Vc=10 V

And    Vb = Vcc/Rr

                = 10/75 = 0.133V

Therefore

Vcb= Vb-Vc

      = 0.133 – 10 = -9.867 V

The voltage gain depends almost exclusively on the ratio of resistors, Therefore voltage gain is given as:

Av=- RL/Re

   = -470/1K =- 0.47

The bandwidth of the common emitter amplifier is low due to the high capacitance.

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