0
883views
A 50mV signal with a 600$\Omega$ source resistance is applied to circuit in Fig .Calculate $v_{o}$ for: (a)CE circuit operation $V_{s}$ at transistor base $R_{g}$ bypassed (b)CB circuit $V_{s}$

enter image description here

1 Answer
0
0views

(a) CE circuit

Voltage gain,

$A_{v}=\frac{h_{fe}(R_{C}||R_{L})}{h_{ie}}$

$\frac{100(5.6k\Omega||33k\Omega)}{1.5k\Omega}$=319.......(i)

Device input impedance

$Z_{0}=g_{ic}=1k\Omega$...........(ii)

Circuit input impedance

$Z_{i}=Z_{b}||R_{1}||R_{2}$

$1k\Omega||100k\Omega||47k\Omega$

=970$\Omega$ .........(iii)

Input voltage

$u_{i}\frac{v_{s}\times Z_{i}}{r_{s}+Z_{i}}$

=$\frac{50mV\times 970\Omega}{600\Omega+970\Omega}$=30.9mV ..........(iv)

Output voltage

$v_{0}=A_{v}\times v_{i}=319\times30.9mV$

=99.v ........(v)

(b)CB circuit

Voltage gain

$A_{v}=\frac{h_{fe}(R_{C}||R_{L})}{h_{ie}}=319$ ......(vi)

From Table 4.6 Device input impedance

$Z_{e}=\frac{h_{fe}}{1+h_{fe}}=\frac{1.5k\Omega}{1+100}$=14.85\Omega.....(vii)

Circuit input impedance,

Z_{i}=Z_{e}||R_{E}

=14.85\Omega||5.6k\Omega

14.81$\Omega$......(viii)

Input voltage

$v_{i}\frac{V_{s} \times Z_{i}}{r_{s}+Z_{i}}$

$\frac{50mV \times 14.81\Omega}{600\Omega+14.81\Omega}$=1.2mV....(ix)

Output voltage

$v_{o}=A_{v} \times v_{i}$

319 \times 1.2 mV=383 mV.....(x)

Please log in to add an answer.