Electronics Assignment #2 - Key PDF

 

Electronics Assignment #2 Issue Date: 12/04/2020 Due Date: 19/04/2020

Chapter 2: Diode Applications Note: you can solve on a paper using a pen and then scan and insert Weight: 5 Marks out of 100 Key Answers  Answers   Problem 1: For the circuit shown below, find the current flowing through the battery in the following cases:

(a) Assume the diodes are ideal. (b) Assume the diodes are made from silicon. (c) Repeat points (a) and (b) assuming the battery is reversed. Solution: (a) Ideal diodes: Assume D1 ON and D2 ON, which means that every diode will have VD1=0 and VD2=0, in addition, ID1>0 and ID2>0. Now solve for both currents to verify the correctness of our initial assumption. ID1 = 15 A > 0 (D1 ON) ID2 = 5 A > 0 (D2 ON)  Both diodes are ON  I30V = 15 A  A 

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(b) Silicon diodes: Assume D1 ON and D2 ON, which means that every diode will have VD1=0.7 V and VD2=0.7 V, in addition, ID1>0 and ID2>0. Now solve for both currents to verify the correctness of our initial assumption. ID1 = 14.184 A > 0 (D1 ON) ID2 = 4.417 A > 0 (D2 ON)  

Both diodes are ON



 I30V = 14.534 A  A 

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(c) Battery reversed in both ideal and silicon diode cases: -  Ideal diodes (battery revered): Assume D1 ON and D2 ON, which means that every diode will have VD1=0 and VD2=0, in addition, ID1>0 and ID2>0. Now solve for both currents to verify the correctness of our initial assumption. ID1 = -15 A < 0 (D1 OFF) opposes our initial assumption. Discard all calculations.

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 Now,   Now,

assume D1 OFF and D2 ON:  ON:  ID1 = 0 A (D1 OFF) OFF) ID2 = 3.33 A > 0 (D2 ON)  D1 OFF and D2 ON  I30V = 5 A

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- Silicon diodes (battery revered): Assume D1 ON and D2 ON, which means that every diode will have VD1=0.7 V and VD2=0.7 V, in addition, ID1>0 and ID2>0. Now solve for both currents to verify the correctness of our initial assumption.

ID1 = -15.766 A < 0 (D1 OFF) opposes our initial assumption. Discard all calculations.

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 Now,   Now,

assume D1 OFF and D2 ON:  ON:  ID1 = 0 A (D1 OFF) OFF) ID2 = 3.2 A > 0 (D2 ON)  D1 OFF and D2 ON  I30V = 4.917 A

Nodal analysis at Vx:

−    − −                     +

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 ≈    .

 

 

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Problem 2: Determine the voltage and current through 2 kΩ resistor. 

Assume all diodes are ON. Calculate for currents flowing through diodes. All the diodes are ON as the currents are positives (A  K). V2k  =  = 1.5 V I2k  =  = 0.75 mA

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Problem 3: Determine the voltage V o and currents  I 1 and  I 2.

Assume both diodes are ON. Calculate for currents flowing through both diodes. GaAs diode has a negative current (IGaAs = -0.57 mA), which means it shall be OFF. Discard all calculations.

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Assume Si Diode ON and GaAs diode OFF. Calculate for current flow through Si diode. Isi = 0.65 mA > 0, which means that Si diode is ON  Si diode ON and GaAs diode OFF.

Results:

I1 = 0.65 mA I2 = 0 Vo = 5.65 V

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Problem 4: For the diode rectifier circuit shown below:  a)  b)  c)  d) 

What is the type of this rectifier circuit? Full Wave Bridge Rectifier  Rectifier  Find the output dc voltage V o assuming the threshold voltage for any diode is 1 V? What are the states of the diodes during the operation of the circuit? Find the dc current value and its direction in the load resistor?

b) Vdc of output voltage V o = - 0.636 (Vm - 2VK) = - 0.636 (60 – 2(1)) = -36.888 V c) During the positive half cycle, D2 and D3 are conducting while D1 and D4 are blocking. During the negative half cycle, D2 and D3 are blocking while D1 and D4 are conducting. d) Idc = Vdc / R = -36.888 / 4 k = - 9.222 mA (The positive current direction is shown below)

v ii  (t) (t) = 60sin( ωt) 4 k   Ω  

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Problem 5: For the Zener voltage regulator circuit shown below: a)  Determine the value of R required to obtain 75 mA Zener current assuming V i = 15 V and  R L = 500 .. b)  Determine the maximum load current above which the circuit will not be able act as a voltage regulator assuming V i = 15 V and R = 250 .. c)  Determine the maximum input voltage above which the Zener diode will not be able to  and R = 250 .. withstand the current flow through it assuming R L = 125 .  and R

V ii   

V   = 4 V  Z  P Z mW  ZM M  = 500 mW

a)  IL = 4 V / 0.5 k = 8 mA IR = 75 mA + 8 mA = 83 mA R = (15 – 4) / 83 m = 132.5 Ω  b)  IZ = 0, so the whole current from the source will flow through R and R L, taking into consideration that V L = VZ = 4 V ILmax = (15 – 4) / 250 = 44 mA  mA  c)  IZ = IZmax = PZmax / VZ = 500 mW / 4 V = 125 mA IRmax = (Vimax – VZ ) / R = IZmax + IL  IL = VL / RL = VZ / RL = 4 V / 125 Ω = 32 mA   (Vimax – 4 ) / 250 = 125 mA + 32 mA = 157 mA  Vimax = 43.25 V

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