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calculate bandwidth for two stages RC coupled CE amplifier.
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Solution:

$\beta_1 = 100, \beta_2 = 150, r\pi_1 = rb\pi_2 = 1.3 k \Omega$

g$m_1$ = g$m_2$ = 50 mA/v

c$\pi_1$ = c$\pi_2$ = 15 pf , c$\mu$ = c$\mu_2$ = 1 pf

enter image description here

1) Low Frequency Response :

Low frequency small signal equivalent circuit.

enter image description here

Step 1.

Calculate R$b_1$ R$b_2$

rb1 = 22k || 47k = 14.99 k$\Omega$ = 15k$\Omega$

rb2 = 22k || 1147k = 14.99k$\Omega$ = 15k$\Omega$

step 2.

Calculate low frequency time constants Tc1, Tc2, Tc3, Tce, Tce2

a) Tc1 = (RS + $RE_1$ || r$\pi_1$) C1

= (100 + (15k ||1.3k) 10$\mu$f

= 0.01296s

= 12.96 ms

b) Tc2 = [RC1 + (RB2 || r$\pi_2$)] $c_2$

= [8.2k + (15k || 1.3k)] 5$\mu$

= 46.98 s

c) Tc3 = (Rc + Rl) C3

= (8.2k + 5k) 10$\mu$

= 132ms

d) Tce1 = [$\frac{r\pi_1 + (RS || Rb1)}{(1+\beta}$ ||RE] CE1

= [$\frac{1.3k+(0.1 || 15k)}{1+100}$ || 5k] 50 $\beta$f

= 0.698 ms = 6.98 x $10^-4$s

e) $Tce_2$ = [[$\frac{(r\pi_2+(R4||RB2)}{(1+\beta_2)}$] || $Re_2$ ] CE

= [$\frac({1.3k+(8.2k||15k)}{(1+150)})$ || 5k ] 50$\mu$f

= 1.74 x $10^-3$ s

step 3.

$F_L$ corresponds to the smallest of all the constants calculated thus,

FL = $\frac{1}{2\pi cce1}$ = $\frac{1}{2\pi (6.98x10_-4)}$

= 228.015 Hz $\approx$ 228 Hz

2) High Frequency Response

high frequency equivalent circuit of the given amplifier is,

enter image description here

To find time constants of all capacitors

a) Time constant of c$\pi$1

Assume c$\mu$1, c$\beta$2, c$\pi$2 open circuited and Vs = 0 the effective resistance seen by c$\pi$1 is given by,

enter image description here

R$\pi$1 = r$\pi$1 ||(Rs||RB1)

= 1.3k || (0.1k || 15k)

= 0.092k$\Omega$

$\zeta\pi_1$ = R$\pi_1$ x C$\pi_1$

= 0.092k x 15$\mu$

= 1.38 x $10^-9$ s

b) Time constant of C$\pi_2$

Assume C$\mu_1$ C$\mu_2$ and C$\pi_1$ open circuit and Vs = 0.

Then effective resistance seen C$\pi_2$

$\zeta\pi_2$ = R$\pi_2$ x C$\pi_2$

R$\pi_2$ = R$c_2$ || R$B_2$ || r$\pi_2$

= 8.2k || 15k || 1.3 k

= 1.043 k $\Omega$

$\zeta\pi_2$ = 1.043 k $\Omega$ x 15pf

= 1.57 x $10_-8$ seconds

c) Time Constant Of C$\mu_1$

Assume c$\pi_1$, c$\pi_2$ and C$\mu_2$

open circuit and Vs = OV. then the effective resistance by c$\mu_1$ will be R$\mu_1$

ri = rs || RB, || r$\pi_1$ = 0.1k || 1.3k || 15k = 92.3 k $\Omega$

RL = Rc1 || RB2 || r$\pi_2$ = 8.2k || 15k || 1.3k = 1044$\Omega$

enter image description here

R$\mu_1$ = RL + Ri (1+gmRL)

= 1044 + 92.3 (1+50x $10^-3$x 1044)

= 5.95 k$\Omega$

$\zeta\mu_1$ = R$\mu_1$ x c$\mu_1$

= 5.95 x $10^3$ x 1 x $10^-12$

d) Time Constant Of C$\mu_2$

c$\pi_1$, c$\pi_2$, c$\mu_1$ be open circuit and Vs = 0

enter image description here

r$\mu_2$ = R$L^1$ + Ri (1 + gmR$L^1$

R$L^1$ = R$c_2$ || $R_L$ 8.2k || 5k = 3.1 k $\Omega$

Ri = R$\mu$ || $RB_2$ || r$\pi_2$ 8.2k || 15k || 1.3k = 1044$\Omega$

$\therefore$ R$\mu_2$ 3.1k + 1.044k (1+50 x $10^-3$ x 3.1k $\Omega$

= 165.96k$\Omega$

$\zeta\mu_2$ R$\mu_2$ x C$\mu_2$

= 165.96k x 1 x $10^-12$ F

= 1.66 x $10^-7$ seconds

fH will correspond to the largest of all time constant calculated for high frequency circuit.

fH = $\frac{1}{2\pi\zeta\mu_2}$

= $\frac{1}{2\pi (1.66x10^-7)}$

fH = 958.76 KHz

Bandwidth = fH = FL

= 958.76 KHz - 228 Hz

= 958.53 KHz.

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