Chapter 7, Problem 7.70P

In the common−base circuit shown in Figure P7.70, the transistor parameters are: $\beta =100$ , $V_{BE}\text{(on)=0}\text{.7V}$ , $V_A=\infty$ , $C_{\pi }=1\text{0pF}$ , and $C_{\mu }=1\text{pF}$ . (a) Determine the upper 3dB frequencies corresponding to the input and output portions of the equivalent circuit. (b) Calculate the small−signal midband voltage gain. (c) If a load capacitor $C_L=15\text{pF}$ is connected between the output and ground, determine if the upper 3dB frequency will be dominated by the $C_L$ load capacitor or by the transistor characteristics.

Figure P7.70

To determine

The upper $3\text{dB}$ frequencies corresponding to the input and output portions of the equivalent circuit.

Answer to Problem 7.70P

  $f_{H\pi }=654.7\text{MHz}$

  $f_{H\mu }=161\text{MHz}$

Explanation of Solution

Given:

The given circuit is shown below.

  

The transistor parameters

  $\beta =100$

  $V_{BE(on)}=0.7\text{V}$

  $V_A=\infty$

  $C_{\pi }=10\text{pF}$

and $C_{\mu }=1\text{pF}$

  $R_{\text{S}}=50\text{Ω}$

  $R_E=0.5\text{kΩ}$

  $R_B=100\text{kΩ}$

  $R_L=1\text{kΩ}$

  $I_E=0.5\text{mA}$

Calculation:

  $\begin{array}{l}{I}_{CQ}=\left(\frac{\beta }{\beta +1}\right)I_E\hfill \\ I_{CQ}=\left(\frac{100}{100+1}\right)(0.5)\hfill \\ \therefore I_{CQ}=0.495\text{mA}\hfill \end{array}$

Transconductance

  $g_m=\frac{I_{cQ}}{V_T}$

  $g_m=\frac{0.495}{0.026}$

  $\therefore g_m=19\text{mA}/\text{V}$

  $r_{\pi }=\frac{\beta V_T}{I_{CQ}}$

  $r_{\pi }=\frac{(100)(0.026)}{0.495}$

  $\therefore r_{\pi }=5.25\text{kΩ}$

For Input Portion:

Time constant,

  ${\tau }_{p\pi }=\left[\left(\frac{r_{\pi }}{1+\beta }\right)\parallel R_E\parallel R_s\right]C_{\pi }$

  $=\left[\left(\frac{5.25}{101}\right)\parallel 0.5\parallel 0.05\right]\times {10}^3\times 10\times {10}^{-12}$

  $=0.243\times {10}^{-9}\text{s}$

  $\begin{array}{l}\text{Upper}3\text{dB}\text{frequency,}{f}_{H\pi }=\frac{1}{2\pi {\tau }_{p\pi }}\\ =\frac{1}{2\pi \left(0.243\times {10}^{-9}\right)}\\ \therefore f_{H\pi }=654.7\text{MHz}\end{array}$

For Output portion:

Time constant,

  ${\tau }_{p\mu }=\left(R_B\parallel R_L\right)C_{\mu }$

  $=(100\parallel 1)\left(1\times {10}^{-12}\right)\times {10}^3$

  $=0.99\times {10}^{-9}\text{s}$

  $\begin{array}{l}\text{Upper}3\text{dB}\text{frequency,}{f}_{H\mu }=\frac{1}{2\pi {\tau }_{p\mu }}\\ =\frac{1}{2\pi \left(0.99\times {10}^{-9}\right)}\\ \therefore f_{H\mu }=161\text{MHz}\end{array}$

To determine

The small signal mid- band voltage gain

Answer to Problem 7.70P

  $A_v=9.14$

Explanation of Solution

Given:

The given circuit is shown below.

  

The transistor parameters

  $\beta =100$

  $V_{BE(on)}=0.7\text{V}$

  $V_A=\infty$

  $C_{\pi }=10\text{pF}$

and $C_{\mu }=1\text{pF}$

  $R_{\text{S}}=50\text{Ω}$

  $R_E=0.5\text{kΩ}$

  $R_B=100\text{kΩ}$

  $R_L=1\text{kΩ}$

  $I_E=0.5\text{mA}$

Calculation:

  $\begin{array}{l}{I}_{CQ}=\left(\frac{\beta }{\beta +1}\right)I_E\hfill \\ I_{CQ}=\left(\frac{100}{100+1}\right)(0.5)\hfill \\ \therefore I_{CQ}=0.495\text{mA}\hfill \end{array}$

Transconductance

  $g_m=\frac{I_{cQ}}{V_T}$

  $g_m=\frac{0.495}{0.026}$

  $\therefore g_m=19\text{mA}/\text{V}$

  $r_{\pi }=\frac{\beta V_T}{I_{CQ}}$

  $r_{\pi }=\frac{(100)(0.026)}{0.495}$

  $\therefore r_{\pi }=5.25\text{kΩ}$

  $\begin{array}{l}{V}_o=-g_m{V}_{\pi }\left(R_B\parallel R_L\right)\hfill \\ g_m{V}_{\pi }+\frac{V_n}{r_n}+\frac{V_x}{R_E}+\frac{v_i-\left(-V_n\right)}{R_s}=0\hfill \\ V_{\pi }\left[g_m+\frac{1}{r_n}+\frac{1}{R_E}+\frac{1}{R_s}\right]=\frac{-v_i}{R_s}\hfill \end{array}$

  $\begin{array}{l}{V}_{\pi }\left[19+\frac{1}{5.25}+\frac{1}{0.5}+\frac{1}{0.05}\right]=\frac{-v_i}{0.05}\hfill \\ V_{\pi }(41.19)=-v_i(20)\hfill \\ \therefore v_o=-g_m{V}_z\left(R_B\parallel R_L\right)\hfill \end{array}$

  $\begin{array}{l}=-(19)(-0.4856)(100\parallel 1)v_i\hfill \\ \frac{v_o}{v_i}=9.14\hfill \\ \text{voltage gain,}{A}_v=\frac{v_o}{v_i}=9.14\hfill \\ \therefore A_v=9.14\hfill \end{array}$

To determine

To identify: Whether the upper $3\text{dB}$ frequency will be dominated by the C_L load capacitor or by the transistor characteristics.

Answer to Problem 7.70P

Yes, it is dominated by C_L

Explanation of Solution

Given:

The given circuit is shown below.

  

The transistor parameters

  $\beta =100$

  $V_{BE(on)}=0.7\text{V}$

  $V_A=\infty$

  $C_{\pi }=10\text{pF}$

and $C_{\mu }=1\text{pF}$

  $R_{\text{S}}=50\text{Ω}$

  $R_E=0.5\text{kΩ}$

  $R_B=100\text{kΩ}$

  $R_L=1\text{kΩ}$

  $I_E=0.5\text{mA}$

  $\begin{array}{l}{I}_{CQ}=\left(\frac{\beta }{\beta +1}\right)I_E\hfill \\ I_{CQ}=\left(\frac{100}{100+1}\right)(0.5)\hfill \\ \therefore I_{CQ}=0.495\text{mA}\hfill \end{array}$

Transconductance

  $g_m=\frac{I_{cQ}}{V_T}$

  $g_m=\frac{0.495}{0.026}$

  $\therefore g_m=19\text{mA}/\text{V}$

  $r_{\pi }=\frac{\beta V_T}{I_{CQ}}$

  $r_{\pi }=\frac{(100)(0.026)}{0.495}$

  $\therefore r_{\pi }=5.25\text{kΩ}$

Calculation:

For $C_L=15\text{pF}$

Time constant,

  $\tau =C_L\left(R_L\parallel R_B\right)$

  $=\left(15\times {10}^{-12}\right)(1\parallel 100){10}^3$

  $=14.85\times {10}^{-9}\text{s}$

Upper $3\text{dB}$ frequency, $f=\frac{1}{2\pi \tau }$

  $=\frac{1}{2\pi \left(14.85\times {10}^{-9}\right)}$

  $=10.7\text{MHz}$

  $\therefore f=10.7\text{MHz}$

Since $f<f_{H\mu },3dB$ frequency dominated by $C_L$