Assignment 1(Fall 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 50

Due Date: 19/10/2006

Q.1. For the circuit shown in Figure 1, find the currents I1, I2, I3, and I4. Sol. First we will label the diagram First we will assume that entering current s at Nodes will be negative while leaving currents will be positive At Node D:

I4 + 4A – 2A = 0 ⇒ I4 = -2A ----- (A) At Node C: I3 – I4 -7A = 0 From (A) we have

I3 – (-2A) -7A = 0 ⇒ I3 + 2A -7A = 0

⇒ I3 = 5A ----- (B) At Node B: I2 + 7A + 3A = 0

⇒ I2 = -10A ----- (C) At Node A: -I1 – I2 + 2A = 0 From (C) we have

-I1 – (-10A) + 2A = 0 ⇒ I1 + 10A + 2A = 0

⇒ I1 = 12A

Q.2. You are given the network shown in Figure 2. Find VX as indicted in the diagram.

Sol. First we will label the diagram

Using KVL , starting at A , going Clockwise, using ∑drops = 0 -12 + (1)(15) -24 + VX + 9 + 12 = 0 VX = 12-15+24-21 = 36-36 VX = 0

Q.3. You are given the circuit shown in Figure 3. Find the voltages V1, V2 and V3.

Sol. First we will label the diagram

For V1

Using KVL , starting at A , going Clockwise - 24 + V1 + 10 +12 = 0 V1 = 24-22 = 2V For V2

Using KVL , starting at A , going Clockwise - 24 + V1 + V2 = 0 V2 = 24- V1

= 24-2 = 22V For V2

Using KVL , starting at A , going Clockwise - V2+ V3 + 12 = 0 V3 = 22- 12 = 10V

Q.4. You are given the network shown in Figure 4. Find REQ and Io. Draw the circuit diagram of each step otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. First we will label the diagram 6Ω and 12 Ω are in parallel so, 6Ω|| 12Ω =4Ω And 80Ω and 80Ω are in parallel so, 80Ω|| 80Ω =40Ω Now we will redraw the circuit diagram

IO=40/12.09 = 3.309A

------ Good Luck -----

Assignment 2(Fall 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 06/11/2006

Q.1. For the circuit shown in Figure 1, Determine VAB. Sol. VCA = 12x11 / 11+6.8 = 7.146V VCB = 12x120 / 120+0.22 = 11.978 = 12V VCA + VAB + VBC = 0 -------- (A) Where VBC = - VCB put in (A) VCA + VAB - VCB = 0 VAB + 7.416 – 12V = 0 VAB = 4.58V Q.2. First Identify and label each node in the network. Use nodal analysis to find voltage at each node in the network given below. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. First we will identify and nodes and labeled them,

KCL equation at Node V1 will be, (V1 - 20)/30 + V1/20 + (V1 -V2)/10 = 0 11V1 - 6V2 = 40 ------------(A)

KCL equation at Node Vb will be, (V2 - V1)/10 +V2 /30 + V2 /30 = 0

-3V1 + 5V2 = 0 ------------(B) Solving (A) and (B) leads to V1 = 5.41 V, V2 = 3.24V

Q.3. First Identify and label each node in the network. Use nodal analysis to find Voltage at each node and Current I in the network given below. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. First we will identify and nodes and labeled them,

At Node V1 V1/200 + (V1 - V2)/5 + (V1-V3/100 = 0 43V1 – 40V2 -2V3 = 0 -------- (A) Constraint equation V3-V2 = 50V 0V1+V2-V3 = -50V ------- (B) Now we will write KCL equation at Super Node,

(V2 - V1)/5 + 0.2 + V3/50 + (V3- V1)/100 = 0 -21V1 + 20V2 + 3V3= 0 ----------- (C) Solving (A) (B) and (C) simultaneously we have , V1 =-45.23V, V2= -48.69V, V3= 1.31V At Node V2 using KCL we have, (V2- V1)/5 +0.2 + I = 0 I = (-45.23 + 48.69)/5 - 0.2 I = 0.492A

------ Good Luck -----

Assignment 3(Fall 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 15/11/2006

Q.1. First Identify and label each node in the network. Use nodal analysis to find voltage at each node in the network given below and also find out Current IO through 10KΩ. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value. Sol. First we will identify nodes and labeled them, KCL equation at Node V1 will be, V1 – V2 V1

+ = 2 mA 10 k 10 k V1 – V2 + V1

= 2 mA 10 k V1 – V2 + V1 = (10 k)(2 mA) 2V1 – V2 = 20 … (A) KCL equation at Node V2 will be,

V2 V2 – V1

+ 2Ix =

10 k 10 k Here

V1

Ix = 10 k Substituting the value of Ix

V2 2V1 V2 – V1

+ = 10 k 10 k 10 k V2 2V1 V2 – V1

+ - = 0 10 k 10 k 10 k V2 + 2V1 – [V2 – V1] = 0

10 k V2 + 2V1 – [V2 – V1] = 0 V2 + 2V1 – V2 + V1 = 0 V1 = 0 V Substituting the value of V1 in eq. (A) 2[0 V] – V2 = 20 V2 = -20 V

V2

I0 = 10 k

-20 V I0 = 10 k I0 = -2 mA Q.2. Use Mesh analysis to find VO in the given network. Identify and label each mesh otherwise you will lose your marks. Draw and labeled complete circuit diagram. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. First we will identify mesh and labeled them,

From Mesh I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 3I1 + 6[I1 – I2] = 12 3I1 + 6I1 – 6I2 = 12 9I1 – 6I2 = 12 -----------------(A) I2 = 2 mA Substituting the value of I2 in Equation (A) 9I1 – 6[2 mA] = 12 9I1 – 12 = 12 9I1 = 24 I1 = 2.667 mA V0 = 2I2 + 6[I2 – I1] V0 = 2I2 + 6I2 – 6I1

V0 = 8I2 – 6I1

Substituting the values of I1 & I2

V0 = 8[2 ] – 6[2.667 ] V0 = 16 – 16.002 V0 = -0.002 V

Q.3. Use Mesh analysis to find VO in the given network. Identify and label each mesh otherwise you will lose your marks. Draw and labeled complete circuit diagram. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. First we will identify mesh and labeled them,

From Mesh I2: I2 = -2 mA From Mesh I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 1[I1 – I2] + 1I1 = 12 1I1 – 1I2 + 1I1 = 12 2I1 – 1I2 = 12 Substituting the value of I2

2I1 – 1[-2] = 12 2I1 + 2 = 12 2I1 = 10 I1 = 5 mA From Mesh I3: According to KVL Sum of all the voltage rise = sum of all the voltage drop 1I3 + 1[I3 – I2] = -12 1I3 + 1I3 – 1I2 = -12 2I3 – 1I2 = -12 Substituting the value of I2

2I3 – 1 [-2 mA] = -12 2I3 + 2 = -12 2I3 = -12-2 2I3 = -14 I3 = -7 mA

V0 = 1 [I3 – I2] + 1 [I1 – I2] V0 = 1 [-7 – (-2)] + 1 [5 – (-2)] V0 = 1 [-5] + 1 [7] V0 = 2 V

------ Good Luck -----

Assignment 4(Fall 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 50

Due Date: 19/11/2006

Q.1. First Identify and label each node in the network. Use Nodal analysis to find out voltage at each node in the network given below and also find out Current IO and Voltage Vo in the given network. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value. Sol. First we will identify nodes and labeled them, Write KCL equations.

Apply KCL on node V1:

040

)120(20

410

100 2111 =+−−

+−

+− VVvVV o ------------ (A)

20 Vv = ------- (B)

Sub (B) into (A): 040

)120(20

410

100 21211 =+−−

+−

+− VVVVV

----------- (C)

Apply KCL at node V2:

80

2 20

Vii o =+ ---------- (D)

40

120210

+−=

VVi -------------- (E)

Substitute (E) into (D): 8040

1203 221 VVV

=⎟⎠

⎞⎜⎝

⎛ +− ---------------- (F)

Solving (C) and (F) vo = V2 = -1344V, V1 = -1688V so, io = (V1 –V2 + 120)/40 = -5.6A

Q.2. First Define all nodes and Identify and label each Mesh in the network. Use Mesh analysis to find Current IO and Voltage Vo in the given network. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value. Sol. First define all the meshes and assign mesh current variables in each mesh

Observe any mesh which the mesh current can be known by observation.

No mesh can be solved by observation in this case.

Write KVL equations: For Mesh I:

04)20)(()10(100 211 =+−++− oVIII ---------------- (1)

)80(30 IV = ----------------- (2)

Sub. (2) into (1): 0)80)((4)20)(()10(100 3211 =+−++− IIII --------- (3) There is a CCCS between mesh 2 and 3, so Mesh 2 and Mesh 3 form a SuperMesh: ]

KVL For SuperMesh :

0)80)((120)40()20)((4 32120 =+−+−+− IIIIV ---------- (4)

Sub. (2) into (4):

0)80)((120)40()20)(()80)((4 32123 =+−+−+− IIIII ------- (5)

Constraint equation: 2302 III −= Since,

, so 20 II = 2322 III −= ---------- (6)

Solve (3), (5) and (6), i1 = 178.8A, i2 = -5.6A, i3 = -16.8A, vo = 80(i3) = -1344V

Q.3. Use Mesh analysis to find VO in the given network. Identify and label each mesh otherwise you will lose your marks. Draw and labeled complete circuit diagram. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. First define all the meshes and assign mesh current variables in each mesh From Mesh I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 2000I1 + 4000[I1 – I2] = 2000Ix

2000I1 + 4000I1 – 4000I2 = 2000Ix

6000I1 – 4000I2 = 2000Ix

Here I2 = Ix

6000I1 – 4000I2 = 2000[I2] 6000I1 – 4000I2 - 2000I2 = 0 6000I1 – 6000I2 = 0 6000[I1 – I2] = 0

I1 – I2 = 0 ……….… (A) From Mesh I2: According to KVL Sum of all the voltage drop = sum of all the voltage rise 2000I2 + 4000[I2 – I1] = 6 2000I2 + 4000I2 – 4000I1 = 6 6000I2 – 4000I1 = 6 2[3000I2 – 2000I1] = 6 3000I2 – 2000I1 = 3 From equation (A) I1 = I2

3000I2 – 2000[I2] = 3 3000I2 – 2000I2 = 3 1000I2 = 3 I2 = 3 mA

According to ohm’s Law V0 = (I2)(2 k) V0 = (3 mA)(2 k) V0 = 6 V

------ Good Luck -----

Assignment 5(Fall 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 28/12/2006

Q.1. Apply Superposition to the circuit given below to find IO. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value. Sol. Now we will turn off current source and replace it with an open circuit and only Voltage source is acting in the above circuit Series combination = 8 kΩ + 2 kΩ + 6 kΩ = 16 kΩ

Parallel combination

12 kΩ × 16 kΩ

= 12 kΩ + 16 kΩ Fig (A)

= 6.857 kΩ

According to voltage divider rule

6.857 k V4k = × 40 V

4 k + 6.587 k V4k = 25.263 V V4k = VAB

From fig. (A)

VAB

I0' = 12 k 25.263 V I0' = 12 k I0' =2.1 mA Now we will turn off a Voltage source and replace it with a short circuit and only Current source is acting

Parallel combination 12 kΩ × 4 kΩ

= 12 kΩ + 4 kΩ

72 k × k

= 3 kΩ Series combination = 3 kΩ + 8 kΩ = 11 kΩ

According to current divider rule

8 k I1 = × 10 mA

11 k + 8 k I1 = 4.21 mA According to ohm’s Law V3k = (I1)(3k) V3k = (4.21mA)(3k) V3k = 12.63 V V3k = VAB

From fig. (B)

VAB

I0" = 12 k 12.63 V I0" = 12 k I0" = 1.05mA Hence I0 = I0' - I0" I0 = 2.1 mA - 1.05 mA I0 = 1.05 mA Q.2. Apply Superposition to the circuit given below to find VO. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value. Sol. Now we will turn off current source and replace it with an open circuit and only Voltage source is acting in the above circuit

Parallel combination Series combination

6 kΩ × 6 kΩ

= = 3 kΩ +6 kΩ 6 kΩ + 6 kΩ

= 3 kΩ = 9kΩ

Parallel combination

6 kΩ × 9 kΩ

= 6 kΩ + 9 kΩ

54 k × k

= = 3.6kΩ 15 k

According to KVL Sum of voltage drop = sum of voltage rise 6000I + 3600I = 2 9600I = 2 I = 0.208 mA According to ohm’s Law V0′ = (I)(6 k) V0′ = (0.208 mA)(6 k) V0′ = 1.248 V Now we will turn off a Voltage source and replace it with a short circuit and only Current source is acting Technique I

According to current divider rule

6 k I = × 12 mA 6 k + 6 k I = 6 mA According to ohm’s Law V3k = (I)(3 k) V3k = (6 mA)(3 k) V3k = 18 V Technique II When only current source is acting Constraint Equation:

( )12 1mA− − − − − − − − − − − −3 2I I− =

Super Mesh

( ) ( )

( )3 3 1 2 1

3 2 1

6 3 6 0

3 2 3 0 2

I I I I I

I I I

+ − + − =

⇒ + − = − − − − − − − − −For Mesh I

( ) ( )

( )1 1 2 1 3

1 2 3

6 6 3 0

5 2 3 0 3

I I I I I

I I I

+ − + − =

⇒ − − = − − − − − − − − − − −

From equation (1) putting the value of 3I in equation ( 2)and(3)

( )2 1 2

1 2

3 36 3 2 03 5 36 4

I I II I+ − + =

⇒ − = − − − − − − − − − − −

( )

( )1 2 2

1 2

35 2 12 0

5 3 12 5

in equationI I I

I I− − − =

⇒ − = − − − − − − − − − − − − − −

Multiplying equation (4) by 3 and (5) by 5 and subtracting

1 2

1 2

1

25 15 609 15 36

3

I II I

I mA

− =− =

⇒ = −

From (5) and (1)

( )

2

2

3

3

15 3 129

9 123

II mA

II mA

− − =⇒ = −

− − =

⇒ =

Voltage drop due to current source is given by VO = (3)(6) = 18 volts. V3k = VAB

V3k = V0″ V0 = V0′ + V0″ V0 = 1.248 V + 18 V V0 = 19.248 V

Q.3. Apply Source Transformation on the circuit given below to find VO. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Sol.

6mA current source and 6kW resistance is in parallel so convert it into voltage source

According to ohm’s Law V = (6 mA)(6 k) V = 36 V Using mesh analysis I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 6000I1 + 4000[I1 – I2] = 39 6000I1 + 4000I1 – 4000I2 = 39 10000I1 – 4000I2 = 39 … (A) I2: According to KVL Sum of all the voltage drop = sum of all the voltage rise 12000I2 + 4000[I2 – I1] = 12 12000I2 + 4000I2 – 4000I1 = 12 16000I2 – 4000I1 = 12 Rearranging -4000I1 + 16000I2 = 12 … (B) Multiplying equation (A) by (4) 40000I1 – 16000I2 = 156 … (C) Adding equations (B) & (C) -4000I1 + 16000I2 = 12 40000I1 – 16000I2 = 156 36000I1 = 168 I1 = 4.667 mA Substituting the value of I1 in equation (B) -4000[4.667 mA] + 16000I2 = 12 -18.668 + 16I2 = 12 16I2 = 30.667 I2 = 1.91 mA I4k = I1 – I2

I4k = 4.667 mA – 1.91 mA I4k = 2.757 mA According to ohm’s Law V0 = (I4k)(4 k) V0 = (2.757 mA)(4 k)

------ Good Luck -----

Assignment 6(Fall 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 45

Due Date: 08/01/2007

Q.1. Find VO in the network given below using Thévenin’s theorem. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value. Sol.

Using Nodal analysis At Node I: V1/16 + (V1-V2)/4 -3=0 5V1 -4V2= 48 --------- (A) At Node II: V2 – V1/4 + (V2 -12)/5 -3=0 5V1 - 9V2= -48 --------- (B) Solving A and B we have VTh=V2= 19.2

THEVENIN’S EQUIVALENT: Vo = 10/(10+5) x 19.2V = 12.8V

Q.2. Find VTH and RTH in the network given below using Thévenin’s theorem. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Sol.

Load resistance is already removed ,

Calculate VTH

Using Nodal analysis At Node I: V1/40 + (V1-40)/10 + (V1-V2)/20 -3=0 7V1 -2V2= 40 --------- (A) At Node II: V2 – V1/20 + 3=0 V1 = V2 -60 --------- (B) Solving A and B we have V1= 32 V , V2 = 92V and VTH=V2= 92V

Q.3. Find VO in the network given below using Thévenin’s theorem. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Sol. Removing load resistance RL= 1KΩ

In the above circuit Ix = 0 because current is not completing its path due to open circuit so, value of dependent current source is also zero.

VTH = 6 V Now for RTH we will short the open terminal of the Thévenin’s circuit and calculate the Isc and then divide VTH with Isc to calculate RTH

Applying KCL at node V Sum of all the currents leaving the junction = sum of all the currents entering the junction V V + 6 + = 2Ix

1 k 2 k Here V + 6 Ix = 2 k V V + 6 V + 6 + = 2 1 k 2 k 2 k

V V + 6 V + 6 + = 1 k 2 k 1 k V V + 6 V + 6 + - = 0 1 k 2 k 1 k 2V + V + 6 – 2[V + 6] = 0 2 k 2V + V + 6 – 2[V + 6] = 0 2V + V + 6 – 2V - 12 = 0 V = 6 V

Now 6 + 6 Ix = 2 k Ix = 6 mA = ISC

VTH

RTH = ISC

6 V RTH = 6 mA RTH = 1 kΩ

THEVENIN’S EQUIVALENT:

According to Voltage divider rule: 1 k V0 = × 6 V 1 k + 1 k V0 = 3 V

------ Good Luck -----

Assignment 7(Fall 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 50

Due Date: 17/01/2007

Q.1. Find IO in the network given below using Norton’s theorem. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value. Sol. ISC =?

Using mesh analysis: Loop I1: According to KVL Sum of all the voltage drop = sum of all the voltage rise 2I1 = 12 I1 = 6 mA Loop I2:

According to KVL Sum of all the voltage drop = sum of all the voltage rise 4000I2 = 4 I2 = 1 mA ISC = I1 – I2

ISC = 6 mA – 1 mA ISC = 5 mA Parallel combination:

2 kΩ × 4 kΩ =

2 kΩ + 4 kΩ

8 k × k = 6 k RN = 1.334 kΩ NORTON’S EQUIVALENT:

According to current divider rule: 1.334 k I2k = × 5 mA 2 k + 1.334 k I2k = 2 mA = I0

Q.2. Find IO in the network given below using Norton’s theorem. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Sol. Applying KCL at node V1

V1 – V2 V1 – V2 V1

+ + = 0 6 k 3 k 2 k V1 – V2 + 2V1 – 2V2 + 3V1

= 0 6 k V1 – V2 + 2V1 – 2V2 + 3V1 = 0 6V1 – 3V2 = 0 2V1 – V2 = 0 Here V2 = 12 V 2V1 – 12 = 0 V1 = 6 V V2 – V1

I1 = 3 k 12 V – 6 V I1 = 3 k I1 = 2 mA

V2 I2 = 4 k 12 V I2 = 4 k I2 = 3 mA According to KCL ISC = I1 + I2

ISC = 2 mA + 3 mA ISC = 5 mA RN =?

Parallel combination:

2 kΩ × 6 kΩ =

2 kΩ + 6 kΩ 12 k × k

= 8 k = 1.5 kΩ

Series combination = 1.5 kΩ + 3 kΩ = 4.5 kΩ Parallel combination:

4.5 kΩ × 4 kΩ =

4.5 kΩ + 4 kΩ

18 k × k = 8.5 k = 2.118 kΩ RN = 2.118 kΩ NORTON’S EQUIVALENT:

According to current divider rule: 2.118 k I2k = × 5 mA 2.118 k + 2 k I2k = 2.572 mA = I0

Q.3. The circuit shown in figure below utilize four identical diodes having n=1 and Is =10-14

A. Find the value to obtain an output voltage Vo=3V, if a current of 1mA is drawn away from the output terminal by a load, what is the change in output voltage.

Sol. For V

0 =3V, so the voltage drop across each diode is 3/4 V= 0.75V

Thus I must be

I = Is ev/nVT

VT= 25 mV=0.025 volts

= 10-14

e0.75/0.025

=106.86mA If a current of 1mA is drawn away from the terminals by means of a load, the current though the diodes reduces to 106.86 – 1 = 105.86 Thus the voltage across each diode changes by I=105.68mA VD = 0.025 ln(105.68 x 10-3 / 10-14)

VD = 0.749V So ∆V =0.75V- 0.749V =0.0010 The total decrease in V

0

V0 = 4 x 0.0010

= 0.0040v

Q.4. Determine the dc load voltage for the circuit shown below.

Sol. Primary voltage is in rms so we can calculate the peak voltage

V1(pk)

= Vrms

/0.707

= 45/0.707 = 63.65V

pk Primary voltage and turn ratio is known so we can determine the secondary voltage

V2(pk)

=(N2/N1)V1pk

=1/5 (63.65) =12.73V

pk Now load voltage can be calculated as

VL(pk)

= V2/2 – 0.7

= 12.73/2 -0.7 = 5.67V

pk Now dc value of the voltage can be calculated as

Vave

= 2VL(pk)

/ Π

= 0.636VL(pk)

= 3.609 Vdc = 0.636(5.67)

------ Good Luck -----

Assignment 8(Fall 2006) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 35

Due Date: 27/01/2007

Q.1. For the circuit shown in the figure below, both diodes are identical, conducting 10mA at 0.7V and 100mA at 0.8V. Find the value of R for which V=80mV Sol. It is told that D1 ,D2 conduct 10mA @ 0.7V and 100mA @ 0.8V We know that lni2/i1 = V2-V1/nVT nVT = V2-V1 / ln (i2/i1) = 0.8-0.7 / ln10 = 0.0434 From the above figure i1+ i2 = 10mA -------- (I) V = Ri1 Where V=80mV 0.08 = Ri1 -------- (II) i1 = Is e V1/0.0434 -------- (A)

i2 = Is e V2/0.0434 ---------- (B) where, V2= V1 + 0.08 put this value in (B)

i2 = Is e (V1 + 0.08)/0.0434 = Is eV1 /0.0434 (e0.08/0.0434) From (A) we have i2 = i1 6.3175 From eq(I) we have, i1 + 6.3175i1 = 10 i1 = 10/7.3175 = 1.366mA From eq (II) we have, R = 80mV/ 1.36mA = 58.8Ω

Q.2. For the circuits shown in figure below, using the constant voltage-drop (V D =0.7V) diode model, find the voltages and currents indicated. Sol.

(A) Diode FB by inspection,

I = 3-(-3 +0.7)/10 = 530 µA =0.53mA

Diode FB , so replace by 0.7V source ⇒ V=3 – (0.53mA) (10k)= -2.3V

(B) Diode RB by inspection,

Diode ≡ Open circuit ⇒ V=+3V I = 0 (same as ideal diode)

(C) Diode On by inspection, 0.7V drop

I = 3- 0.7) –(-3)/10 = 530 µA =0.53mA

⇒ V= -3 + (10k)(0.53mA)= 2.3V

(D) Diode Off by inspection,

Diode ≡ Open circuit ⇒ V=-3V I = 0 (same as ideal diode)

Q.3. Measurement of VBE and two terminal current taken on a number of npn transistors are tabulated below. For each, calculate the missing current value as well as α, β, and Is as indicated by the table. Transistor a b c d e VBE(mV) 690 690 580 780 820 Ic(mA) 1.000 1.000 10.10 IB(mA) 50 7 120 1050 IE(mA) 1.070 1.137 75.00 α β Is Sol.

a) IE = IC + IB = 51mA

α = IC/IE = 1.00/51 = 0.0196

β = IC/IB = 1.0/50 = 0.02

Ic = Is e VBE / V

T

IS = Ic e -VBE / V

T = 10-3 e --690

/ 25

= 1.03x10-15

b) IB = IE - IC = 0.070mA

α = IC/IE = 1.00/1.07 = 0.935

β = IC/IB = 1.0/0.7 =14.3

Ic = Is e VBE / V

T

IS = Ic e -VBE / V

T = 10-3 e --690

/ 25

= 1.03x10-15

c) IC = IE - IB = -5.863

α = IC/IE = -5.863/1.137 = -5.156

β = IC/IB = -5.863/7 = -0.8375

Ic = Is e VBE / V

T

IS = Ic e -VBE / V

T = -4.33 x10-10

d) IE = IC + IB = 130.10mA

α = IC/IE = 10.1/130.01 = 0.0776

β = IC/IB = 10.1/120 = 0.0841

Ic = Is e VBE / V

T

IS = Ic e -VBE / V

T = 10.1x 10-3 e - 780

/ 25

= 2.84x10-16

e) IC = IE - IB = -975mA

α = IC/IE = -975/75 = -13

β = IC/IB = -975/1050 = 0.9285

Ic = Is e VBE / V

T

IS = Ic e -VBE / V

T = -975 e -820

/ 25

= -5.5x10-12

------ Good Luck -----