Chapter 7, Problem 7.49P

Consider the circuit in Figure P7.49. Calculate the impedance seen by thesignalsource $V_i$ at(a) $f=1\text{kHz}$ ,(b) $f=10\text{kHz}$ ,(c) $f=100\text{kHz}$ ,and(d) $f=1\text{MHz}$ .

Figure P7.49

To determine

The impedance seen by the signal source V1 at the given frequency.

Answer to Problem 7.49P

The impedance $Z_i$ for $f=1\text{kHz}$ is 2700- j3 .55.

Explanation of Solution

Given:

The circuit diagram is given as:

  

The given data:

The value of the resistor $r_{b}is200\Omega$ .

The value of the resistor $r_{\pi }is2.5\text{kΩ}$ .

The value of the resistor $R_{L}is2.\text{5kΩ}$ .

The value of the capacitor $C_{\pi }is10\text{pF}$ .

The value of the capacitor $C_{\mu }is0.8\text{pF}$ .

The value of the trans-conductance $g_{m}is0.04$ .

The frequency, $f=1\text{kHz}$ .

Evaluating the miller capacitance:

  $C_M=C_{\mu }\left(1+g_m{R}_L\right)$

Substituting the known values:

  $\begin{array}{c}{C}_M=\left(0.8\times {10}^{-12}\right)\left(1+0.04\left(2500\right)\right)\\ =80.8\times {10}^{-12}\\ =80.\text{8pF}\end{array}$

Evaluating the value of the capacitance $C_i$ :

  $C_i=C_{\pi }+C_M$

Substituting the known values to the above equation:

  $\begin{array}{c}{C}_i=10pF+80.8pF\\ =90.8\text{pF}\end{array}$

Redrawing the given diagram after some changes:

  

Evaluaating the impedance $Z_i$

  $\begin{array}{c}{Z}_i=200+\left(2500\right)𑨈\left(\frac{1}{j2\pi f\left(90.8\times {10}^{-12}\right)}\right)\\ =200+\frac{\left(2500\right)\left(\frac{1}{j2\pi f\left(90.8\times {10}^{-12}\right)}\right)}{\left(2500\right)\left(\frac{1}{j2\pi f\left(90.8\times {10}^{-12}\right)}\right)}\\ =200+\frac{\left(2500\right)}{1+\left(2500\right)j2\pi f\left(90.8\times {10}^{-12}\right)}\\ =200+\frac{\left(2500\right)}{1+jf\left(1.42\times {10}^{-6}\right)}\\ =200+\frac{\left(2500\right)\left[1+jf\left(1.42\times {10}^{-6}\right)\right]}{\left[1+jf\left(1.42\times {10}^{-6}\right)\right]\left[1-jf\left(1.42\times {10}^{-6}\right)\right]}\\ Z_i=200+\frac{\left(2500\right)\left[1-jf\left(1.42\times {10}^{-6}\right)\right]}{\left[1+f^2\left(2\times {10}^{-12}\right)\right]}\mathrm{...........}\left(1\right)\end{array}$

Since, $f=1\text{kHz}$ :

Evaluating the impedance $Z_i$ :

Substitute $1\text{kHz}$ for f in $Z_i$ :

  $\begin{array}{l}{Z}_i=200+\frac{\left(2500\right)\left[1-j\left({10}^3\right)\left(1.42\times {10}^{-6}\right)\right]}{\left[1+{\left({10}^3\right)}^2\left(2\times {10}^{-12}\right)\right]}\\ =200+\frac{\left(2500\right)\left[1-j\left(1.42\times {10}^{-3}\right)\right]}{\left[1+\left(2\times {10}^{-6}\right)\right]}\\ =200+2500-j3.55\\ =2700-j3.55\end{array}$

Hence, the impedance $Z_i$ for $f=1\text{kHz}$ is 2700- j3 .55.

To determine

The impedance seen by the signal source V1 at the given frequency.

Answer to Problem 7.49P

The impedance $Z_i$ for $f=1\text{0kHz}$ is 2700-j35.5.

Explanation of Solution

Given:

The circuit diagram is given as:

  

The given data:

The value of the resistor $r_{b}is200\Omega$ .

The value of the resistor $r_{\pi }is2.5\text{kΩ}$ .

The value of the resistor $R_{L}is2.\text{5kΩ}$ .

The value of the capacitor $C_{\pi }is10\text{pF}$ .

The value of the capacitor $C_{\mu }is0.8\text{pF}$ .

The value of the trans-conductance $g_{m}is0.04$ .

The frequency $f=1\text{0kHz}$ .

  $Z_i=200+\frac{\left(2500\right)\left[1-jf\left(1.42\times {10}^{-6}\right)\right]}{\left[1+f^2\left(2\times {10}^{-12}\right)\right]}\mathrm{...........}\left(1\right)$

Consider the following data

  $f=1\text{0kHz}$

Calculate the impedance $Z_i$ .

Substitute $f=1\text{0kHz}$ for f in $Z_i$ .

  $\begin{array}{l}{Z}_i=200+\frac{\left(2500\right)\left[1-j\left({10}^4\right)\left(1.42\times {10}^{-6}\right)\right]}{\left[1+{\left({10}^4\right)}^2\left(2\times {10}^{-12}\right)\right]}\\ =200+\frac{\left(2500\right)\left[1-j\left(1.42\times {10}^{-2}\right)\right]}{\left[1+\left(2\times {10}^{-4}\right)\right]}\\ =200+2500-j35.5\\ =2700-j35.5\end{array}$

Therefore , the impedance $Z_i$ for $f=1\text{0kHz}$ is 2700-j35.5.

To determine

The impedance seen by the signal source V1 at the given frequency.

Answer to Problem 7.49P

The impedance $Z_i$ for $f=100kHz$ is $2651-j348$ .

Explanation of Solution

Given:

The circuit diagram is given as:

  

The given data:

The value of the resistor $r_{b}is200\Omega$ .

The value of the resistor $r_{\pi }is2.5\text{kΩ}$ .

The value of the resistor $R_{L}is2.\text{5kΩ}$ .

The value of the capacitor $C_{\pi }is10\text{pF}$ .

The value of the capacitor $C_{\mu }is0.8\text{pF}$ .

The value of the trans-conductance $g_{m}is0.04$ .

The frequency $f=10\text{0kHz}$ .

  $Z_i=200+\frac{\left(2500\right)\left[1-jf\left(1.42\times {10}^{-6}\right)\right]}{\left[1+f^2\left(2\times {10}^{-12}\right)\right]}\mathrm{...........}\left(1\right)$

Consider the following data:

  $f=10\text{0kHz}$

Calculate the impedance $Z_i$ .

Substitute $f=10\text{0kHz}$ for f in $Z_i$ .

  $\begin{array}{l}{Z}_i=200+\frac{\left(2500\right)\left[1-j\left({10}^5\right)\left(1.42\times {10}^{-6}\right)\right]}{\left[1+{\left({10}^5\right)}^2\left(2\times {10}^{-12}\right)\right]}\\ =200+\frac{\left(2500\right)\left[1-j\left(0.142\right)\right]}{\left[1+\left(2\times {10}^{-2}\right)\right]}\\ =200+2451-j348\\ =2651-j348\end{array}$

Therefore, the impedance $Z_i$ for $f=100kHz$ is $2651-j348$ .

To determine

The impedance seen by the signal source V1 at the given frequency.

Answer to Problem 7.49P

the impedance $Z_i$ for $f=1MHz$ is $\mathrm{1.33.34}-j1183.4$ .

Explanation of Solution

Given:

The circuit diagram is given as:

  

The given data:

The value of the resistor $r_{b}is200\Omega$ .

The value of the resistor $r_{\pi }is2.5\text{kΩ}$ .

The value of the resistor $R_{L}is2.\text{5kΩ}$ .

The value of the capacitor $C_{\pi }is10\text{pF}$ .

The value of the capacitor $C_{\mu }is0.8\text{pF}$ .

The value of the trans-conductance $g_{m}is0.04$ .

The frequency $f=\text{1MHz}$ .

  $Z_i=200+\frac{\left(2500\right)\left[1-jf\left(1.42\times {10}^{-6}\right)\right]}{\left[1+f^2\left(2\times {10}^{-12}\right)\right]}\mathrm{...........}\left(1\right)$

Consider the following data:

  $f=\text{1MHz}$

Calculate the impedance $Z_i$

Substitute $f=\text{1MHz}$ for f in $Z_i$ .

  $\begin{array}{l}{Z}_i=200+\frac{\left(2500\right)\left[1-j\left({10}^6\right)\left(1.42\times {10}^{-6}\right)\right]}{\left[1+{\left({10}^6\right)}^2\left(2\times {10}^{-12}\right)\right]}\\ =200+\frac{\left(2500\right)\left[1-j\left(1.42\right)\right]}{\left[1+2\right]}\\ =200+833.34-j1183.4\\ =\mathrm{1.33.34}-j1183.4\end{array}$

Therefore, the impedance $Z_i$ for $f=1MHz$ is $\mathrm{1.33.34}-j1183.4$ .