$π$-model parameters
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Mumbai University > EXTC > Sem 3 > Analog Electronics 1

Marks: 5 M

Year: Nov 2016

1. Calculation of $I_B, I_C $and $V_{CE}$

• Calculation for $I_B$

  $R_{th} =R_1||R_2$

  $R_{th} =100k||33k$

  $R_{th}=24.8kΩ$

  $E_{TH}=\frac{Vcc×R2}{R1+R2}$

  $E_{TH}=4.46V$

Fig1 Simplified circuit

• Applying KVL to base emitter loop,

  $-E_{TH}-I_B R_{th} -V_{BE}-I_ER_E=0$

  Put $I_E= (β+1)I_B$

  $E_{TH}-I_B R_{th} -V_{BE}-(β+1)I_B R_E=0$

  $I_B=\frac{E_{TH}-V_{BE}}{R_E (β+1)+R_{th}}$

  $I_B=\frac{4.46-0.7}{1k(200+1)+24.8k}$

  $I_B=16.65µA$

• Calculation for $I_E$ and $I_C$

  $I_E=(β+1)I_B$

  $I_E=201×16.65µA=3.346mA$

  $I_C=βI_B=200×16.65µA=3.33mA$

• Calculation for $V_{CE}$

  • Applying KVL to base emitter loop,

  $V_{CE}=V_{CC}I_CR_C-I_ER_E$

  $V_{CE}=18-(3.33m×2k)-(3.346m×1k)$

  $V_{CE}=7.994V≈8V$

2. Calculation of hybrid parameters

• Transconductance

  $g_m=\frac{I _C}{V_T} =\frac{3.33mA}{26mV} =0.12S/120mS$

• $rπ=\frac{V_T}{I _B} =\frac{26mV}{16.65µA}=1.56kΩ$

• $ro=\frac{V_{CE}+V_A}{I_C} =\frac{8+100}{3.33m A}=32.43kΩ$

$$\boxed{gm=120mS , rπ=1.56kΩ , ro=32.43kΩ}$$