Chapter 7, Problem 7.21P

For the circuit in Figure P7.21, the transistor parameters are $\beta =120$ , $V_{BE}\text{(on)=0}\text{.7V}$ , and $V_A=50\text{V}$ . (a) Design a bias−stable circuit such that $I_{EQ}\text{=1}\text{.5mA}$ . (b) Using the results of part (a), find the small−signal midband voltage gain. (c) Determine the output resistance $R_o$ . (d) What is the lower 3dB corner frequency?

Figure P7.21

To determine

To design: The bias stable circuit for the given condition.

Explanation of Solution

Given:

The circuit is given as:

  

I_EQ =1.5mA.

The parameters of the circuit is given as:

  $\begin{array}{l}\beta =120\\ V_{BE\left(on\right)}=0.7\text{V}\\ V_A=50\text{V}\end{array}$

The design needed for the bias stability is given as:

  $R_{TH}𑨀\left(1+\beta \right)R_E$

Recalling the general rule that the circuit is considered to be stable only, if:

  $R_{TH}=\left(0.1\right)\left(1+\beta \right)R_E$

Evaluating the Thevenin resistance $R_{TH}$ :

  $R_{TH}=\left(0.1\right)\left(1+\beta \right)R_E$

Substituting the known values in the equation:

  $\begin{array}{c}{R}_{TH}=\left(0.1\right)\left(1+120\right)\left(4k\Omega \right)\\ =48.4\text{kΩ}\end{array}$

Evaluating the value of base current $I_{BQ}$ :

  $I_{BQ}=\frac{I_{EQ}}{1+\beta }$

Substituting the known values in the above equation:

  $\begin{array}{c}{I}_{BQ}=\frac{\text{1}\text{.5mA}}{\text{1+120}}\\ \text{=0}\text{.01239mA}\end{array}$

The expression for Thevenin s voltage $V_{TH}$ :

  $V_{TH}=I_{BQ}{R}_{TH}+V_{BE\left(on\right)}+I_{EQ}{R}_E$

Substituting the known values in the above equation:

  $\begin{array}{c}{V}_{TH}\text{=}\left(\text{0}\text{.01239mA}\right)\left(\text{48}\text{.4kΩ}\right)\text{+0}\text{.7V+}\left(\text{1}\text{.5mA}\right)\left(\text{4kΩ}\right)\\ \text{=0}\text{.6+0}\text{.7+6}\\ \text{=7}\text{.3V}\end{array}$

The other expression for Thevenin s voltage $V_{TH}$ :

  $V_{TH}=\frac{1}{R_1}\left(R_{TH}\right)\left(V_{CC}\right)$

Substituting the known values in the above equation:

  $\begin{array}{c}\text{7}\text{.3=}\frac{\text{1}}{{\text{R}}_{\text{1}}}\left(\text{48}\text{.4kΩ}\right)\left(\text{12}\right)\\ R_1=\frac{\left(\text{48}\text{.4kΩ}\right)\left(\text{12}\right)}{\text{7}\text{.3}}\\ \text{=79}\text{.6kΩ}\end{array}$

The value of resistance $R_2$ is evaluated as:

  $\begin{array}{c}{R}_{TH}=R_1𑨀R_2\\ =\frac{\left(R_1\right)\left(R_2\right)}{R_1+R_2}\end{array}$

Substituting the known values in the above equation:

  $\begin{array}{c}\text{48}\text{.4kΩ=}\frac{\left(\text{79}\text{.6kΩ}\right)\left(R_2\right)}{\text{79}\text{.6kΩ+}{R}_2}\\ \text{3852}{\text{.64×10}}^{\text{6}}\text{+48}{\text{.4×10}}^{\text{3}}{R}_2=\left(\text{79}{\text{.6×10}}^{\text{3}}\right)\left({\text{R}}_{\text{2}}\right)\\ \text{31}{\text{.2×10}}^{\text{2}}{R}_2=\text{3852}{\text{.64×10}}^{\text{6}}\\ R_2=\frac{\text{3852}{\text{.64×10}}^{\text{6}}}{\text{31}{\text{.2×10}}^{\text{3}}}\\ \text{=123}\text{.5kΩ}\end{array}$

Therefore, the designed values are:

  $\begin{array}{l}{R}_1=\text{79}\text{.6kΩ}\\ R_2\text{=123}\text{.5kΩ}\\ I_{BQ}\text{=0}\text{.01239mA}\end{array}$

To determine

The small signal mid band voltage gain.

Answer to Problem 7.21P

The value of voltage gain $A_v$ is 0.991.

Explanation of Solution

Given:

The circuit is given as:

  

I_EQ =1.5mA.

The parameters of the circuit is given as:

  $\begin{array}{l}\beta =120\\ V_{BE\left(on\right)}=0.7\text{V}\\ V_A=50\text{V}\end{array}$

Evaluating the value of collector current $I_{CQ}$ :

  $I_{CQ}=\beta I_{BQ}$

Substituting the known values:

  $\begin{array}{c}{I}_{CQ}=\left(\text{120}\right)\left(\text{0}\text{.01239mA}\right)\\ \text{=1}\text{.49mA}\end{array}$

Evaluating the value of $r_{\pi }$ :

  $r_{\pi }=\frac{\left(\beta \right)\left(V_T\right)}{I_{CQ}}$

Substituting the known values into the above equation:

  $\begin{array}{c}{r}_{\pi }=\frac{\left(120\right)\left(26\times {10}^{-3}\right)}{1.49\times {10}^{-3}}\\ =2.094\text{kΩ}\end{array}$

Evaluating the value of output resistance $r_0$ :

  $r_0=\frac{V_A}{I_{CQ}}$

Substituting the known values into the above equation:

  $\begin{array}{c}{r}_0=\frac{20}{1.49\times {10}^{-3}}\\ =33.56\text{kΩ}\end{array}$

Evaluating the value of voltage gain $A_v$ .

  $A_v=\frac{\left(1+\beta \right)\left(r_0𑨀R_E𑨀R_L\right)}{r_{\pi }\left(1+\beta \right)\left(r_0𑨀R_E𑨀R_L\right)}$

Substituting the known values into the above equation:

  $\begin{array}{c}{A}_v=\frac{\left(1+120\right)\left(33.56k\Omega 𑨀4k\Omega 𑨀4k\Omega \right)}{2.094\times {10}^3\left(1+120\right)\left(33.56k\Omega 𑨀4k\Omega 𑨀4k\Omega \right)}\\ =\frac{\left(121\right)\left(33.56k\Omega 𑨀\left(\frac{\left(4k\Omega \right)\left(4k\Omega \right)}{4k\Omega +4k\Omega }\right)\right)}{2.094\times {10}^3+\left(121\right)\left(33.56k\Omega 𑨀\left(\frac{\left(4k\Omega \right)\left(4k\Omega \right)}{4k\Omega +4k\Omega }\right)\right)}\\ =\frac{\left(121\right)\left(33.56k\Omega 𑨀2k\Omega \right)}{2.094\times {10}^3+\left(121\right)\left(33.56k\Omega 𑨀2k\Omega \right)}\\ =\frac{\left(121\left(\frac{\left(33.56k\Omega 𑨀2k\Omega \right)}{\left(33.56k\Omega +2k\Omega \right)}\right)\right)}{2.094\times {10}^3+\left(121\left(\frac{\left(33.56k\Omega 𑨀2k\Omega \right)}{\left(33.56k\Omega +2k\Omega \right)}\right)\right)}\\ =\frac{\left(121\right)\left(1.89k\Omega \right)}{2.094\times {10}^3+\left(121\right)\left(1.89k\Omega \right)}\\ =\frac{228.69k\Omega }{230.784k\Omega }\\ =0.991\end{array}$

Therefore, the value of voltage gain $A_v$ is 0.991.

To determine

The value of the output resistance.

Answer to Problem 7.21P

The value of output resistance $R_0$ is $17.23\Omega$ .

Explanation of Solution

Given:

The circuit is given as:

  

I_EQ =1.5mA.

The parameters of the circuit is given as:

  $\begin{array}{l}\beta =120\\ V_{BE\left(on\right)}=0.7\text{V}\\ V_A=50\text{V}\end{array}$

Evaluating the output resistance $R_0$ :

  $R_0=R_E𑨀r_0𑨀\frac{r_{\pi }}{1+\beta }$

Substituting the known values into the above equation:

  $\begin{array}{c}{R}_0=\left(\text{4kΩ}\right)𑨀\left(\text{33}\text{.56kΩ}\right)𑨀\frac{\text{2}\text{.094kΩ}}{\text{1+120}}\\ =\left(\text{4kΩ}\right)𑨀\left(\text{33}\text{.56kΩ}\right)𑨀\frac{\text{2}\text{.094kΩ}}{\text{121}}\\ =\left(\text{4kΩ}\right)𑨀\left(\text{33}\text{.56kΩ}\right)𑨀\left(\text{17}\text{.31}\right)\\ =\frac{\left(\text{4kΩ}\right)\left(\text{33}\text{.56kΩ}\right)}{\left(\text{4kΩ}\right)\text{+}\left(\text{33}\text{.56kΩ}\right)}𑨀\left(\text{17}\text{.31}\right)\\ \text{=}\left(\text{3}\text{.574kΩ}\right)𑨀\left(\text{17}\text{.31}\right)\\ \text{=}\frac{\left(\text{3}\text{.574kΩ}\right)\left(\text{17}\text{.31}\right)}{\left(\text{3}\text{.574kΩ}\right)\text{+}\left(\text{17}\text{.31}\right)}\\ \text{=17}\text{.23Ω}\end{array}$

Therefore, the value of output resistance $R_0$ is $17.23\Omega$ .

To determine

The lower 3dB corner frequency for the given circuit.

Answer to Problem 7.21P

The value of lower corner frequency is 19.81Hz.

Explanation of Solution

Given:

The circuit is given as:

  

I_EQ =1.5mA.

The parameters of the circuit is given as:

  $\begin{array}{l}\beta =120\\ V_{BE\left(on\right)}=0.7\text{V}\\ V_A=50\text{V}\end{array}$

Evaluating the value of lower 3dB corner frequency:

  $f_L=\frac{1}{2\pi \left(R_0+R_L\right)C_{C2}}$

Substituting the known values to the above equation:

  $\begin{array}{c}{f}_L=\frac{1}{2\pi \left(17.23+4\times {10}^3\right)\left(2\times {10}^{-6}\right)}\\ =19.81\text{Hz}\end{array}$

Hence, the value of lower corner frequency is 19.81Hz.