written 6.8 years ago by | modified 22 months ago by |

**Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits**

**Marks:** 10M

**Year:** Dec 2016

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For the given circuit find Icq, VCEq, Vc & Ve.

written 6.8 years ago by | modified 22 months ago by |

**Mumbai University > Electronics Engineering > Sem 4 > Discrete Electronic Circuits**

**Marks:** 10M

**Year:** Dec 2016

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written 6.8 years ago by |

**Given:** $\beta$ = 100, Vbe = 0.7V

For DC analysis, all the coupling capacitor acts as open circuit as shown in below circuit diagram. We need to find equivalent circuit at base terminal (Rth & Vth).

**Figure 1: Circuit diagram without coupling capacitor and equivalent circuit diagram**

Vth can be found using voltage divider method:

$Vth = \frac{10 \times 12.2k}{12.2k + 56k}$

$Vth = 1.78 V$

$Rth = 12..2k || 56k = 10.01 k \Omega$

Apply KVL from base to emitter junction,

$Vth - Rth I_B - Vbe - R_EI_E = 0 $ ----(1)

where $I_E = I_C + I_B = \beta I_B + I_B = I_B(1+\beta)$

Substitute in equation (1)

$Vth - Rth I_B - Vbe - R_EI_B(1+\beta) = 0 $

$I_B = \frac{Vth - Vbe}{Rth + R_E(1+\beta)}$

$I_B = \frac{1.78 - 0.7}{10.01k + (1+100)0.4k}$

**$I_B = 21.42 \mu A $**

$Icq = \beta I_B = 100 \times 21.42 \mu A$

**$Icq = 2.142 mA$**

Apply KVL from Vcc to Ground through collector to emitter,

$Vcc - I_C R_C - Vceq - I_E R_E = 0$ ----(2)

$I_E = I_B + I_C = 2.142 mA + 21.42 \mu A$

**$I_E = 2.16 mA$**

Substitute $I_E$ value in equation (2),

$Vcc - I_C R_C - I_E R_E = Vceq$

$10 - 2k\Omega \times 2.142mA - 2.16mA \times 0.4k\Omega = Vceq$

**$Vceq = 4.852V$**

$V_C = Vcc - I_C R_C$

**$V_C = 10 - 2.142mA \times 2k\Omega = 5.716V$**

**$V_E = I_E R_E = 2.16mA \times 0.4k \Omega = 0.864$**

**Answers:**

- $Icq = 2.142mA$
- $Vceq = 4.852V$
- $V_C = 5.716V$
- $V_E = 0.864V$

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