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a)

To begin with, IB is assumed to be negligible. Hence, we have

 

RB(Q1) = R1 + R2

  R1 + R2

R1 = 72Ω

R2 = 63Ω

 

Therefore, RB(Q1)  = 72 x 63

 72 + 63

 

         = 34.067Ω

 

(i) VB =        R2     x  VCC

  R1 + R2 

           VB =       63        x 18

     72 + 63

                       VB =    8.4V

 

We then apply Kirchhoff’s Voltage Law

 

VB – IB(Q1)RB(Q1)  – VBE(Q1)  – IB(Q1)RE(Q1)  =  0

VBE(Q1) = VB – 2V

VBE(Q1) = 8.4V – 2V

VBE(Q1) = 0.7V

 

= 2.7 – IB(Q1)RB(Q1) – 0.7 – IB(Q1) x (1 + β) x 1A

 

IB(Q1) =          2.7 – 0.7

  RB(Q1) + 1000(1 + β)

IB(Q1) =                  2

   34.067+151 x 1000  

IB(Q1) =  0.1324µA

 

Now,

        VB(Q1) = 2.7 – IB(Q1) x RB(Q1)

         VB(Q1)  = 2.7 – (0.1324 x 10-6) x 34.067

         VB(Q1)  = 2.7V

 

 

(ii) VE(Q1)  = IE(Q1)  x RE(Q1)

 

But IC(Q1) = IB(Q1)  x β = 0.1324 x 10-6 x 150

       IC(Q1) = 0.1986 x 10-6 A

       IC(Q1) = 0.1986µA

 

IE(Q1) = IB(Q1) (1 + β)

         = 0.1324 x 10-6 x (1 +β)

         = 0.1324 x 10-6 x (1 + 150)

         = 0.1999 x 10-6

         = 0.2µA

 

Therefore, VE(Q1) = 0.2 x 10-6 x 100

                     VE(Q1) = 0.2mV

 

 

(iii)  

                 IE(Q1) = IB(Q1) (1 + β) 

         = 0.1324 x 10-6 x (1 +β)

         = 0.1324 x 10-6 x (1 + 150)

         = 0.1999 x 10-6

         = 0.2µA

 

(iv) VCC - IC(Q1)RC = VC(Q1)

18 – (0.1986 x 10-6 x 48) = VC(Q1)

VC(Q1) = 17.9999V = 18V

VC(Q1) = VB(Q2) = 18V

 

 

(v) From the solution in (iv) we can see that VC(Q1) = VB(Q2). Therefore, the value for VB(Q2) is 18V.

(vi) VB(Q2) - VBE(Q1) = VE(Q2)

 VE(Q2) = 18 – 0.7

VE(Q2) = 17.3V

 

 

(vii) VE(Q2) = R3 + IE(Q2)

IE(Q2)  =VE(Q2) - R3

IE(Q2)  = 17.3 – 1000

IE(Q2)  = 0.0173A

 

 

Also, we find out that VC(Q2) = VCC =18V.

 

(viii) IC(Q2)  =  β(Q2)    x  IE(Q2)

1 +β(Q2)

                                 = 100     x 0.0173

                                   100 + 1

          = 17.128 x 10-3A

                                 = 17.128mA

 

 

 

(b)

 

The emitter follower is omitted and the output from the collector of Q1 is capacitively coupled to the 250Ω load, RL.

 

Vout  = -β x Ib (Rc //Rl)

 

Ib(Q1) = Vs

rTH

rTH = VT

       Ib(Q1)

rTH = 26x10-6

             0.1324 x 10-6

rTH = 196.375kΩ

 

 

 

 

Ib(Q1) = Vs

rTH

Ib(Q1) = 15 x 10-3

            196375

 

Ib(Q1) = 7.64 x 10-8A

 

 

 

Vout = -150 x 7.64 x10-8 x (48 x 250)

                                                (48 + 250)

 

Vout = -4.615 x 10-4

Vout   = -461.5µV

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i. Problem Description/Formulation ii. Solution Approach and Calculations
iii. Results and Discussion

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1. Use the figure below to solve following questions Eniter Folower CE Anier -33 150 V- 16 m, 4-12 2500 (a) Solve the following de quantities i. VB(Q1) ii. VEQ1) iii. IEQ1) iv. VeQ) VB(Q2) vi. VE(Q2) V. vii. IEQ2) viii. VeQ2) (b) Suppose that the emitter follower is omitted and the output from the collector of Q1 is capacitively coupled to the 2502 load, RL. What is the output voltage across the 2502 load?