Hybrid Parameter or h parameter

Transistors is a 2 part network.Input and Output of a transistor are related through 2 equation
$$V_1=h_{11}I_1+h_{12}V_2$$

$$I_2=h_{21}I_1+h_{22}V_2$$


$\left . \frac{V_1}{I_1}  \right |_{V_2=0}$ derived point impedance when output is short circuited $$h_{i},h_{ie},h_{ib}h_{ic}$$
$\left . \frac{V_1}{V_2}  \right |_{I_1=0}$ Reverse transfer voltage ratio when input is open circuit
  $$h_{r},h_{re},h_{rb}h_{rc}$$
$\left . \frac{I_2}{I_1}  \right |_{V_2=0}$ forward current ratio when output is short circuited $$h_{f},h_{fe},h_{fb}h_{fc}$$
$\left . \frac{I_2}{V_2}  \right |_{I_1=0}$ driving point output admittance when input is open circuit $$h_{o},h_{oe},h_{ob}h_{oc}$$

Equivalent circuit

For CE

$$V_b=h_{ie}I_b+h_{re}V_c$$

$$I_c=h_{fe}I_b+h_{oe}V_c$$

For CC

$$V_b=h_{ic}I_b+h_{rc}V_e$$
$$I_e=h_{fc}I_b+h_{oc}V_e$$

For CB

$$V_e=h_{ib}I_e+h_{rb}V_c$$
$$I_c=h_{fb}I_e+h_{ob}V_c$$


Analysis (using without considering source)

Current Gain
$A_i=\frac {I_L}{I_I}=\frac {-I_2}{I_1}$

$I_2=h_fI_1+h_oV_2$
$= h_fI_1+h_oI_LR_L$
$I_2 = h_fI_1-h_oI_LR_L$
$I_2(1+h+oR_L) = h_fI_1$
${\frac{I_2}{I_1} }= {\frac{h_f}{1+h_oR_L}}$
$A_I={\frac{-I_2}{I_1} }= {\frac{-h_f}{1+h_oR_L}}$
cc
$A_I= {\frac{-h_{fc}}{1+h_{oc}R_L}}$
ce
$A_I= {\frac{-h_{fe}}{1+h_{oe}R_L}}$
cb
$A_I= {\frac{-h_{fb}}{1+h_{ob}R_L}}$
$2)Input Resistance(R_i)$
$R_i= \frac{v_1}{I_1}$
$V_1=h_iI_1+h_rV_2$
$V_1=h_iI_1-h_rI_2R_L$
$\frac{V_1}{I_1}=h_i-\frac{h_rI_2R_L}{I_1}$
$$\bbox[5px,border:2px solid red] {  R_i=h_i+h_rA_IR_L }$$

ce
$R_i=h_{ie}+h_{re}A_IR_L$
cc
$R_i=h_{ic}+h_{rc}A_IR_L$
cb
$R_i=h_{ib}+h_{rb}A_IR_L$

3) $Voltage Gain (A_v)$
$A_v=\frac{V_2}{V_1}$
$V_2=-I_2R_L$
$A_I=\frac{-I_2}{I_1}$
$-I_2={A_II_1}$
=>$V_2=A_II_1R_L$
$A_v=\frac{V_2}{V_1}=\frac{A_II_1R_L}{V_1}=\frac{A_IR_L}{R_i}$

$$\bbox[5px,border:2px solid red] {  A_v=\frac{A_II_1R_L}{V_1}=\frac{A_IR_L}{R_i} }$$

4 )  $Output Resistance(R_o)$
$R_o=\frac{V_2}{I_2}$
$I_2= h_fI_1+h_oV_2$
$\frac{I_2}{V_2} = \frac{h_fI_1}{V_2}+h_o$
$Y_o= \frac{h_fI_1}{V_2}+h_o$

when $V_s=0$

$(R_s+h_i)I_1+h_rV_2=0$
$\frac{I_1}{V_2}=\frac{-h_r}{h_i+R_s}$
$Y_o= \frac{-h_fh_r}{h_i+R_s}+h_o$
$Y_o= h_o-\frac{h_fh_r}{h_i+R_s}$   
$R_o=\frac{1}{Y_o}$ 
Power Gain
$A_p=A_vA_I$   

Analysis (using with considering source)

Voltage gain

$A_{vs}=\frac{V_2}{V_s}=\frac{V_2}{V_1}\frac{V_1}{V_s}$
$A_{vs}=A_v\frac{V_1}{V_s}$

$V_1=\frac{V_sR_i}{R_i+R_s}$
=>$\frac{V_1}{V_s}=\frac{R_i}{R_i+R_s}$

$$\bbox[5px,border:2px solid red]{  A_{vs}=\frac{A_vR_i}{R_i+R_s}}$$

Current Gain


$A_{is}=\frac {I_L}{I_s}=\frac {-I_2}{I_s}=\frac {-I_2}{I_1}\frac {I_1}{I_s}$
$A_{is}=A_I\frac {I_1}{I_s}$

$I_1=I_s\frac {R_s}{R_s+R_L}$
$\frac {I_1}{I_s} = \frac {R_s}{R_s+R_L}$
$$\bbox[5px,border:2px solid red]{ A_{is} = A_I \frac {R_s}{R_s+R_L} }$$