Question

Show that the voltage gain, A_V, of the amplifier is given by `A_v = (beta_(ac) R_1)/r_i`where β_ac is the current gain, R_L is the load resistance and r_i is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain?

Answer 1

The expression for voltage gain of the transistor in CE configuration is:

`Av =V_0/V_2 = (-beta_(ac) R_L)/r`

β_ac → ac current gain

R_L → Load resistance

r = R_B + r_i

r_i → Input resistance

R_B → Base resistance

Current gain of the transistor will decrease if the base is made thicker because current gain, `beta = I_c/I_b`

If the base of an n-p-n transistor is made thicker, then more and more electrons will recombine with the p-type material of the base. This results in a decrease in collector current I_c. Furthermore, I_b also increases.

Hence, ac current gain `(beta = I_c/I_b)`decreases.

Finding expression for voltage gain of the amplifier:

Applying Kirchhoff’ law to the output loop,

V_CC = V_CE + I_CR_C

V_BB = V_BE + I_BR_B

v_i≠ 0

Then, V_BB + v_i = V_BE + I_BR_B + ΔI_B (R_B + r_i)

`because r_i = ((DeltaV_(BE))/(DeltaI_B))_(V_(CE))`

`therefore v_i = Delta I_B (R_B + r_i)`

`= rDelta I_B`

`beta_(ac) = (Delta I_c)/(DeltaI_B)  = i_c/i_b`

It is current gain denoted by A_i.

Change I_C due to change in I_B causes a change in V_CE and the voltage drop across resistor R_C, because V_CC is fixed.

` therefore DeltaV_("CC") = DeltaV_(CE)+R_cDeltaI_c=0`

`=> DeltaV_(CE)= -R_CDeltaI_C`

Change in V_CE is the o/p voltage V₀.

`therefore V_0 =DeltaV_(CE) = -beta_(ac)R_cDeltaI_B`

Voltage gain of amplifier

`A_v =V_0/V_i = (DeltaV_(CE))/(rDeltaI_B)`

`=-beta_(ac)=R_c/r`

Negative sign represents that the o/p voltage is in opposite phase to i/p voltage.