Chapter 7, Problem 7.50P

A common-emitter equivalent circuit is shown in Figure P7.50. (a) What isthe expression for the Miller capacitance? (b) Derive the expression for thevoltage gain $A_{\upsilon }\left(s\right)=V_o\left(s\right)/V_i\left(s\right)$ in terms of the Miller capacitance andother circuit parameters. (c) What is the expression for the upper 3 dBfrequency?

Figure P7.50

To determine

The expression for miller capacitance.

Answer to Problem 7.50P

The expression for miller capacitance is,

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

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

Draw the small signal equivalent circuit of figure including the equivalent miller capacitance

  

From figure, the thevenin voltage across the resistance $R_B$ is

  $V_{TH}=V_i\left(\frac{R_B}{R_B+R_S}\right)$

The equivalent thevenin resistance is,

  $R_{TH}=R_B\parallel R_{\text{S}}$

From above figure, the expression for output voltage is,

  $V_o=-g_m{V}_z{R}_L$

  $\frac{V_o}{V_{\pi }}=-g_m{R}_L$

  $A_v=-g_m{R}_L$

Determine the expression for the Miller Capacitance.

  $A_v=-g_m{R}_L$

Determine the expression for the Miller Capacitance.

  $C_M=C_{\mu }\left(1+\left|A_v\right|\right)$

Substitute $-g_m{R}_L$ for $A_v$ in the equation.

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

  $=C_{\mu }\left(1+g_m{R}_L\right)$

Thus, the expression for miller capacitance is,

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

To determine

The expression for voltage gain.

Answer to Problem 7.50P

The expression for voltage gain

  $A_v(s)=\left(\frac{-\beta R_L}{r_{\pi }+R_{eq}}\right)\left(\frac{R_B}{R_B+R_S}\right)\left[\frac{1}{1+s\left(r_{\pi }\parallel R_{eq}\right)C_i}\right]$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

Draw the small signal equivalent circuit of figure including the equivalent miller capacitance

  

In figure, the capacitors $C_x$ and $C_M$ are in parallel.

  $C_i=C_{\pi }+C_M$

The value of total capacitive reactance is, $\frac{1}{s{C}_i}$

Apply voltage division rule to write the expression for voltage $V_{\pi }$

  $V_{\pi }(s)=\frac{\left(r_{\pi }\parallel \frac{1}{s{C}_i}\right)}{\left[\left(r_{\pi }\parallel \frac{1}{s{C}_i}\right)+\left(R_B\parallel R_s\right)+r_b\right]}\left(\frac{R_B}{R_B+R_s}\right)V_i(s)$

Derive the expression for voltage gain.

  $A_v(s)=\frac{V_o(s)}{V_i(s)}$

  $A_v(s)=\frac{-g_m{R}_L{V}_{\pi }(s)}{V_i(s)}\left(\text{since}{V}_0=-g_m{V}_{\pi }{R}_L\right)$

  $\text{Substitute}\frac{\left(r_{\pi }\parallel \frac{1}{s{C}_i}\right)}{\left[\left(r_{\pi }\parallel \frac{1}{s{C}_i}\right)+\left(R_B\parallel R_s\right)+r_b\right]}\left(\frac{R_B}{R_B+R_s}\right)V_i(s)\text{for}{V}_{\pi }(s)\text{in}\text{ the}\text{ equation}$

  $A_v(s)=\left(\frac{-g_m{R}_L}{V_i(s)}\right)\frac{\left(r_{\pi }\parallel \frac{1}{s{C}_i}\right)}{\left[\left(r_{\pi }\parallel \frac{1}{s{C}_i}\right)+\left(R_B\parallel R_s\right)+r_b\right]}\left(\frac{R_B}{R_B+R_s}\right)V_i(s)$

  $=-g_m{R}_L\frac{\left(r_{\pi }\parallel \frac{1}{s{C}_i}\right)}{\left[\left(r_{\pi }\parallel \frac{1}{s{C}_i}\right)+\left(R_B\parallel R_s\right)+r_b\right]}\left(\frac{R_B}{R_B+R_s}\right)$

  $=-g_m{R}_L\left(\frac{R_B}{R_B+R_s}\right)\left(\frac{\frac{r_{\pi }\left(\frac{1}{s{C}_i}\right)}{r_{\pi }+\frac{1}{s{C}_i}}}{\frac{r_{\pi }\left(\frac{1}{s{C}_1}\right)}{r_{\pi }+\frac{1}{s{C}_i}}+\left(R_s\parallel R_s\right)+r_b}\right)$

  $=-g_m{R}_L\left(\frac{R_B}{R_B+R_s}\right)\left[\frac{\frac{r_{\pi }}{s{C}_r{r}_{\pi }+1}}{\frac{r_{\pi }}{s{C}_r{r}_{\pi }+1}+\left(R_B\parallel R_s\right)+r_b}\right]$

Further simplification as follows,

  $A_v(s)=-g_m{R}_L\left(\frac{R_B}{R_B+R_s}\right)\left[\frac{r_n}{r_n+\left(1+s{r}_n{C}_i\right)\left(R_B\parallel R_s+r_b\right)}\right]$

Consider

  $R_{ey}=\left(R_B\parallel R_s\right)+r_b$

Therefore,

  $A_v(s)=-g_m{R}_L\left(\frac{R_B}{R_B+R_s}\right)\left[\frac{r_{\pi }}{r_{\pi }+\left(1+s{r}_{\pi }{C}_i\right)\left(R_{eq}\right)}\right](\text{since}{R}_{eq}=(R_B\parallel R_S))$

  $=-g_m{R}_L\left(\frac{R_B}{R_B+R_s}\right)\left[\frac{r_F}{r_{\pi }+R_{eq}+s{r}_{\pi }{R}_{eq}{C}_i}\right]$

  $=-\left(g_m{r}_{\pi }\right)R_L\left(\frac{R_B}{R_B+R_s}\right)\left[\frac{1}{\left(r_{\pi }+R_{aq}\right)\left(1+s\left(\frac{r_{\pi }{R}_{eq}}{r_{\pi }+R_{er}}\right)C_i\right)}\right]$

  $=-\beta R_L\left(\frac{R_B}{R_B+R_S}\right)\left[\frac{1}{\left(r_{\pi }+R_{\alpha q}\right)\left[1+s\left(r_{\pi }\parallel R_{eq}\right)C_i\right]}\right]\left(\text{since}\beta =g_m{r}_{\pi }\right)$

  $=\left(\frac{-\beta R_L}{r_{\pi }+R_{eq}}\right)\left(\frac{R_B}{R_B+R_S}\right)\left[\frac{1}{1+s\left(r_{\pi }\parallel R_{eq}\right)C_i}\right]$

Thus, the expression for voltage gain is,

  $A_v(s)=\left(\frac{-\beta R_L}{r_{\pi }+R_{eq}}\right)\left(\frac{R_B}{R_B+R_S}\right)\left[\frac{1}{1+s\left(r_{\pi }\parallel R_{eq}\right)C_i}\right]$

To determine

The expression for upper $3\text{dB}$ frequency

Answer to Problem 7.50P

The expression for upper $3\text{dB}$ frequency is,$f_H=\frac{1}{2\pi \left(r_{\pi }\parallel R_{eq}\right)C_i}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Calculation:

The expression for voltage gain is,

  $A_v(s)=\left(\frac{-\beta R_L}{r_{\pi }+R_{eq}}\right)\left(\frac{R_B}{R_B+R_S}\right)\left[\frac{1}{1+s\left(r_{\pi }\parallel R_{eq}\right)C_i}\right]$

From the voltage gain expression, the time constant is,

  $\tau =\left(r_{\pi }\parallel R_{eq}\right)C_i$

Determine the expression for upper $3\text{dB}$ frequency.

  $f_H=\frac{1}{2\pi (\tau )}$

Substitute $\left(r_{\pi }\parallel R_{eq}\right)C_i$ for $\tau$ in the equation.

  $f_H=\frac{1}{2\pi \left(r_{\pi }\parallel R_{cq}\right)C_i}$

Thus, the expression for upper $3\text{dB}$ frequency is,

  $f_H=\frac{1}{2\pi \left(r_{\pi }\parallel R_{eq}\right)C_i}$