Answer to Question #191016 in Electrical Engineering for Gowtham

Question

Consider the single stage amplifier, values of h-parameters are hie=15kΩ, hfe=(150-X1), hre=15˟10-4 and hoe=50µA/V. Rs=15Ω, Calculate Av, Ai, Ri, Zo and also derive the relations which you used.

Answer 1

$A_V=\frac{v_o}{v_s}$

$v_0= \frac{-h_{fe}}{h_{oe}}I_b \implies I_b=\frac{-h_{oe}}{h_{fe}}V_0$

$I_b=\frac{V_s-h_{re}V_0}{R_s+h_{ie}}$

$\frac{-h_{oe}}{h_{fe}}V_0(R_s+h_{ie})=v_s-h_{re}v_0$

$v_0(\frac{h_{re}h_{fe}-h_{oe}(R_s+h_{ie}}{h_{fe}})=v_s$

$\frac{v_0}{v_s}=(\frac{h_{fe}}{h_{re}h_{fe}-h_{oe}(R_s+h_{ie}})$

$A_v=(\frac{h_{fe}}{h_{re}h_{fe}-h_{oe}(R_s+h_{ie}})$

$A_v=\frac{149}{1.3*10^{-4}+149-50*10^{-6}(0.013+0+1.3)}=7718.46 V/V$

$A_i=\frac{I_0}{I_b} \implies I_0=-hf_eI_b \implies \frac{I_0}{I_b}=-hf_eI_b=-149 \frac{A}{A}$

$R_i{n}=\frac{V_{in}}{i_b}$

$V_{in}=h_{ie}ib+h_{re}v_0$

$v_0= \frac{-h_{fe}}{hoe}I_n$

$v_{in}=[h_{ie}-\frac{h_{fe}h_{re}}{h_{oe}}]I_n$

$\frac{v_{in}}{I_n}=R_{in}=h_{ie}-\frac{h_{fe}h_{re}}{h_{oe}}$

$R_{in}=1.3-\frac{14.9*1.3*0.0001}{50*0.000001}=0.912 k\Omega$