Mumbai University > EXTC > Sem 3 > Analog Electronics I

Marks: 12 M

Year: May 2014

• Common base amplifier have relatively low input impedance and a high output impedance
• Current gain of amplifier is less than one but voltage gain is quite large

$$\text{Fig1 Common Base amplifier}$$

• For AC equivalent circuit, capacitors will be short circuited and DC sources will be grounded.

$$\text{Fig2 AC equivalent circuit of common base amplifier}$$

• Now for small signal analysis, we will draw $r_e$ model. In $r_e$ model transistor is replaced by resistance on input inside and current source on output side

$$\text{Fig3 re model of common base amplifier}$$

1. Input impedance:

   From the diagram we can conclude that voltage across resistor $R_E$ and $r_e$ is same i.e. $V_i$ hence they are in parallel

   $Z_i=R_E||r_e \\ Since \ R_E \gt \gt r_e \\ \boxed{Z_i≈r_e}$

2. Voltage gain:

   $A_v=\dfrac{V_o}{V_i} \\ V_o=-I_oR_L \\ I_o=\dfrac{I_CR_C}{R_C+R_L} \\ V_o=-\dfrac{I_CR_CR_L}{R_C+R_L} \\ V_o=-I_C(R_C||R_L) \\ V_o=-\alpha I_E()R_C||R_L) \\ V_i=-I_Er_E \\ A_V=\dfrac{-\alpha I_E(R_C||R_L)}{-I_Er_E} \\ A_V=\dfrac{\alpha(R_C||R_L)}{r_E}$

• Sine AV is positive quantity; $V_o$ and $V_i$ are in phase.

$$\boxed{A_V=\dfrac{\alpha(R_C||R_L)}{r_E}}$$

1. Current Gain

   $A_i=\dfrac{I_o}{I_i} \\ A_i=\dfrac{-\dfrac{I_CR_C}{R_C+RL}}{I_i} \\ A_i=\dfrac{-\alpha I_E\dfrac{R_C}{R_C+R_L}}{I_i} \\ I_E=\dfrac{-I_iR_E}{R_E+r_e} \\ A_i=\dfrac{-\alpha\dfrac{-I_iRE \ R_C}{R_E+r_eR_C+R_L}}{I_i} \\ \boxed{A_i=\alpha\dfrac{R_E}{R_E+r_e}\dfrac{R_C}{R_C+R_L}}$

• Current gain of common base circuit is less than one. Since $A_i$ is positive quantity, $I_O$ and $I_i$ are in phase.

1. Output impedance

   Make input zero i.e. $V_i=0$ hence $I_E=0$ Hence $I_C=0$

$$\text{Fig4. Determining ZO for common base amplifier}$$

$$\boxed{Z_O=R_C||R_L}$$