Question

Derive the expression for the voltage gain of a transistor amplifier in CE configuration in terms of the  load resistance R_L, current gain a β_a and input resistance. 
Explain why input and output voltages are in opposite phase.

Answer 1

To operate the transistor as an amplifier it is necessary to fix its operating point somewhere in the middle of its active region. If we fix the value of V_BB corresponding to a point in the middle of the linear part of the transfer curve then the dc base current I_B would be constant and corresponding collector current I_C  will also be constant. The dc voltage V_CE = V_CC - I_CR_C would also remain constant. The operating values of V_CE and  I_B determine the operating point, of the amplifier. 

If a small sinusoidal voltage with amplitude vs is superposed on the dc base bias by connecting the source of that signal in series with the V_BB supply,  then the base current will have sinusoidal variations  superimposed on the value of I_B. As a consequence the collector current also will have sinusoidal  variations superimposed on the value of I_C, producing in turn corresponding change in the value of V_O.  We can measure the ac variations across the input and output terminals by blocking the dc voltages by large capacitors. 

In the discription of the amplifier given above we have not considered any ac signal. In general, amplifiers are used to amplify alternating signals. Now let us superimpose an ac input signal v_i (to be amplified) on  the bias V_BB (dc) as shown in figure. The output is taken between the collector and the ground. 

The working of an amplifier can be easily understood, if we first assume that v_i = 0. Then applying  Kirchhoff’s law to the output loop, we get 

V_cc = V_CE + I_c R_L ——— (i) 

Likewise, the input loop gives

V_BB = V_BE + I_B R_B ——— (ii) 

When v_i is not zero, we get 

V_BB + v_i = (V_BE + ΔV_BE) + (I_B + ΔI_B)R_B 

_{Or ,} V_BB + v_i = V_BE + I_B R_B + ΔV_BE + ΔI_BR_B  

The change in V_BE can be related to the input resistance r_i and the change in I_B as Δ V_BE = ΔI_B r_i  

So we get, V_BB + v_i = V_BE + I_B R_B + ΔI_B(R_B + r_i) —— (iii) 

Subtracting (ii) from (iii) we get

v_i = ΔI_B (R_B + r_i) = r ΔI_B ——— (iv) 

The change in I_B causes a change in I_c. We define a parameter β_ac, which is similar to the β_dc, as 

`beta_ac = (Delta I_c)/(Delta I_B) = (i_c)/(i_b) `  ——— (v)

which is also known as the ac current gain A_i. Usually β_ac, is close to β_dc in the linear region of the output characteristics.

The change in I_c due to a change in I_B causes a change in V_CE and the voltage drop across the resistor R_L because V_CC is fixed. 

These changes can be given by Eq. (i) as

ΔV_CC = ΔV_CE + R_L ΔI_C = 0  

or ΔV_CE = –R_L ΔI_C 

The change in VCE is the output voltage v0. From Eq. (v), we get 

v₀ = ΔV_CE = - β_ac R_L Δl_B  ——— (vi)

The voltage gain of the amplifier is 

`A_v = (v_0)/(v_i) = (Delta V_(CE))/(rDelta I_B) = (-beta _(ac) R_L)/r`       {using eq. (iv) and (vi) }

The negative sign represents that output voltage is opposite in phase with the input voltage.