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Derive the equation of differential gain, common mode gain and CMRR of differential amplifier.

Subject: CMOS VLSI Design

Topic: Differential Amplifiers

Difficulty: Medium

cvd • 1.6k  views
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For dual input ac input signal is applied at both the input terminal of transistors and output is measured at both collector with respect to ground due to balanced output as shown below:

enter image description here

Figure 1: Circuit diagram of DIBO Differential Amplifier

Differential mode analysis:

enter image description here

Since two input are same in amplifier but out of phase.

Let, $Vs1 = Vs/2$

$Vs2 = -Vs/2$

$Vd = Vs1 - Vs2$

$Vd = \frac{Vs}{2} - \frac{(-Vs)}{2} = \frac{2Vs}{2} = Vs$

$Vc = \frac{Vs1 + Vs2}{2} = 0$

Due to $Vs1$, ac current $Ie1$ will pass through emitter of T1, similarly $Ie2$ results due to $Vs2$. Both $Ie1$ & $Ie2$ will be same in amplitude but out off phase. Hence net ac current passing through $R_E$ becomes 0. Due to this ac voltage drop across $R_E$ becomes zero.

enter image description here

Figure 2: A.C. equivalent circuit using re model.

$Vo1 = Io1 \times R_C$

$Vo2 = Io2 \times R_C$

$Io1 = - \beta \times ib1$

$Io2 = - \beta \times ib2$

$Vo1 = - \beta \times ib1 \times R_C$..........(1)

$V02 = - \beta \times ib2 \times R_C$...........(2)

Apply KVL at input side,

$Vs1 - R_S \times ib1 - \beta r_e \times ib1 = 0$

$ib1 = \frac{Vs1}{R_S + \beta r_e}$

$Vs2 - R_S \times ib2 - \beta r_e \times ib2 = 0$

$ib2 = \frac{Vs2}{R_S + \beta r_e}$

Substitute value of $ib1, ib2$ in equation (1) and (2),

$Vo1 = - \frac{\beta R_C Vs1}{R_S + \beta r_e}$

$Vo2 = - \frac{\beta R_C Vs2}{R_S + \beta r_e}$

$V_O = Vo1 - Vo2 = -\frac{\beta R_C}{R_S + \beta r_e}[Vs1 - Vs2]$

But, $V_d = Vs1-Vs2$

$V_O = -\frac{\beta R_C}{R_S + \beta r_e}V_d $

$A_d = \frac{V_O}{V_d} = -\frac{\beta R_C}{R_S + \beta r_e}$

If $R_S = 0 $ then,

$A_d = -\frac{ R_C}{r_e}$

Common mode analysis:

In common mode $Vs1 = Vs2 = Vs / 2$

$V_d = Vs1 - Vs2 = 0$

$V_C = \frac{Vs1 + Vs2}{2} = \frac{Vs}{2}$

Due to Vs1 ac emitter current, Ie1 passes through emitter terminal of T1 and Ie2 due to Vs2. But Ie1 and Ie2 both are same in amplitude and same in phase.

Hence net ac current passing through $R_E$ will be $2I_e$

AC equivalent circuit becomes,

enter image description here

Figure 3: AC equivalent circuit

Practically there exists difference between Vo1 and Vo2 due to mismatching of transistor $\beta$ but during ac analysis in common mode, only one output is measured which is assumed to be balanced.

Equivalent circuit using re-model.

enter image description here

Figure 4: Equivalent circuit using re-model

$V_O = I_O \times R_C$

But, $I_O = - \beta i_b$

$V_O = - \beta \times R_C \times i_b$ .............(3)

Apply KVL at input side,

$Vs1 - R_S \times i_b - \beta r_e \times i_b - (1+ \beta)i_b \times 2R_E = 0 $

$i_b = \frac{Vs1}{R_S + \beta r_e + (1+ \beta)2R_E}$

Since $(1+\beta) \approx \beta$

$i_b = \frac{Vs1}{R_S + \beta( r_e + 2R_E)}$

Substitute value of $i_b$ in equation (3),

$V_O = - \frac{\beta \times R_C \times Vs1}{R_S + \beta( r_e + 2R_E)}$

But, $Vs1 = Vs = V_C$

$A_C = \frac{V_O}{V_C} = - \frac{\beta \times R_C}{R_S + \beta( r_e + 2R_E)}$

If $R_S = 0$ then,

$A_C = - \frac{R_C}{ r_e + 2R_E} $

CMRR:

$CMRR = \frac{A_d}{A_C} = - \frac{R_C}{r_e} \times - \frac{r_e + 2R_E}{R_C}$

$CMRR = \frac{r_e + 2R_E}{r_e}$

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