Solutions to Homework # 1

(1/19/04-1/23/04)

1.10 V_FS = 10 V
      Step Size = 10 V / 2¹² bits  = 10 V / 4096 bits  = 2.44 mV
      Voltage Corresponding to LSB  = 2.44 mV
      Voltage Corresponding to MSB = 2.44mV * 2¹¹ = 5 V
      For given Binary Code
      (100100100100)₂ = 2¹¹ + 2⁸ + 2⁵  + 2²  = (2340)₁₀
      V_o = 10V * 2340/ 4096 = 5.7128 V


1.12 Step Size = 5 V /2 ⁸ bits  = 5 V / 256 bits = 19.53 mV / bit
        Voltage corresponding to LSB = 19.53mV
        2.77 V/ 19.53 mV/bits   = 142 bits
       (142)₁₀ = 128 + 8 + 4 + 2 = (10001110)₂


1.16 I_B = dc component  = 0.002A
i_b = ac component  = 0.002(cos 1000t) A

1.26 (a)


R₂ and R₃ are connected in parallel

R_P = R₂*R₃/ (R₂ + R₃) = 2.2k * 18k / (2.2k + 18k)  = 1.96k

Now this R_P and R₁ are in series combination and hence applying the “voltage divider rule” as follows:

V₁ = V_dc * R₁/(R₁ + R_P) = 10 * 4.7k ( 4.7k + 1.96k) = 7.06 V

V₂ = V_dc – V₁ = 10- 7.06 = 2.94 V

I = V_dc / (R₁ + R_P) = 10 / (4.7k + 1.96k) = 1.5 mA

Now applying the “current divider rule” to the parallel circuit

I₂ = I * R₃ / (R₂ + R₃) = 1.5 * 18k ( 2.2k + 18k ) = 1.34 mA

I₃ = I * R₂ / (R₂ + R₃) = 1.5 * 2.2k ( 2.2k + 18k ) = 0.164 mA


1.22





i = V_S/ R₁ = - V_S / 100k

V_TH = V_O = -β * i  * R2  = β * V_S * R2 / R1 = 150 * 39k * V_S/ 100k

= 58.5 * V_S

To find R_TH ,



 V_S = 0 V and hence i = 0
Therefore B* i = 0
and hence the current dependent current source can be replaced by an open circuit as shown above
and hence R_TH = R₂ = 39k

Therefore, Thevenin's equivalent circuit is shown below