A BJT transistor amplifier shown in Fig. 11.27 has RE = RC = 1 kΩ, RS = 600, RL = 2 k and transistor parameters as β = 100 and rπ = 1 kΩ. Determine the values of CC1, CC2 and CE needed to obtain fL = 50 Hz.

Question

A BJT transistor amplifier shown in Fig. 11.27 has [latex]R_{E} = R_{C}[/latex] = 1 kΩ , [latex]R_{S} = 600 , R_{L}[/latex] = 2 kΩ and transistor parameters as β= 100 and [latex]r_{\pi}[/latex] = 1 kΩ . Determine the values of [latex]C_{C1}, C_{C2} and C_{E}[/latex] needed to obtain [latex]f_{L}[/latex] = 50 Hz.

Accepted Answer

[latex]\omega _{1p} = 2 \pi f_{L} = 100 \pi rad/s[/latex]

We know that [latex]\omega _{1p}= \frac{1}{C ^{′}_{E}(R_{s}+r_{\pi})^{’}}[/latex]

Therefore, [latex]C ^{′}_{E}= \frac{1}{\omega _{1p}(R_{s}+r_{\pi})}[/latex]

[latex]C ^{′}_{E}= \frac{1}{100\pi imes 1.6}= 1.99 \mu F[/latex]

We know that [latex]C_{E} = (1 + \beta )C ^{′}_{E}[/latex]

[latex]C_{E} = 101 × 1.99 × 10^{–6} = 201 \mu F[/latex]

We know that [latex]\omega _{1l}= \frac{\omega _{1p}}{10}= \frac{100\pi }{10}= 10\pi = \frac{1}{(R_{s}+R_{\pi })C_{C1}}[/latex]

[latex]C_{C1}= \frac{1}{10\pi imes 1.6}= 19.9 \mu F[/latex]

We choose [latex]\omega _{12}= \frac{\omega _{1p}}{20}= \frac{100\pi }{20}= 5\pi .[/latex]

Therefore, [latex]\omega _{12}= 5\pi = \frac{1}{(R_{C}+R_{L })C_{C2}}[/latex]

Hence, [latex]C_{C2}= \frac{1}{\omega _{12}(R_{C}+R_{L })}= \frac{1}{5\pi imes 3 imes 10^{3} }= 21.23 \mu F [/latex]

Since [latex]\omega_{12}[/latex] is inversely proportional to [latex]C_{C2}[/latex], we can choose any value [latex]C_{C2} > 21.23 \mu F[/latex]. Hence, let us choose [latex]C_{C2} = 22 \mu F[/latex].

Question 11.15: A BJT transistor amplifier shown in Fig. 11.27 has RE = RC =...

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