Chapter 11, Problem 11.82P

The BiCMOS circuit shown in Figure P11.82 is equivalent to a pnp bipolar transistor with an infinite input impedance. The bias current is $I_Q=0.5\text{mA}$ The MOS transistor parameters are $V_{TP}=-0.5\text{V},K_p=0.7\text{mA}/{\text{V}}^2,$ and $\lambda =0,$ and the BIT parameters are $\beta =180,V_{BE}(\text{ on })=0.7\text{V},$ and $V_A=\infty$ (a) Sketch the small-signal equivalent circuit. (b) Calculate the small-signal parameters for each transistor. (c) Determine the small-signal voltage gain $A_v=v_o/v_i$ for (i) $R_L=10\text{k}\Omega$ and (ii) $R_L=100\text{k}\Omega$ .

To determine

To sketch: The small-signal equivalent circuit of the given BiCMOS circuit.

Answer to Problem 11.82P

The required sketch is shown in Figure 3.

Explanation of Solution

Given:

The given diagram is shown below.

  

Calculation:

Mark the value and redraw the circuit.

The required diagram is shown in Figure 1

  

Figure 1

The small signal equivalent circuit is shown in Figure 2

  

Figure 2

Conclusion:

Therefore, the required sketch is shown in Figure 2

To determine

The small signal parameters for each of the transistor.

Answer to Problem 11.82P

The value of small signal parameters are $g_{m2}=15.78\text{mA}/\text{V}$ , $g_{m1}=0.5014\text{mA}/\text{V}$ and $r_{\pi 2}=11.41\text{k}\Omega$ .

Explanation of Solution

iven:

The given diagram is shown below.

  

Calculation:

The value of the current through the resistance $R_1$ is calculated as,

  $\begin{array}{c}{I}_{R1}=\frac{0.7\text{V}}{8\text{k}\Omega }\\ =0.0875\text{mA}\end{array}$

The expression for the resistance $I_Q$ is calculated as,

  $I_Q=I_{R1}+I_{E2}$

Substitute $0.5\text{mA}$ for $I_Q$ and $0.0875\text{mA}$ for $I_{R1}$ in the above equation.

  $\begin{array}{l}0.5\text{mA}=0.0875\text{mA}+I_{E2}\\ I_{E2}=0.4125\text{mA}\end{array}$

The value of the drain current is calculated as,

  $\begin{array}{c}{I}_{D1}=I_{R1}+I_{B2}\\ =0.002279\text{mA}\end{array}$

The expression for the collector current is given by,

  $\begin{array}{c}{I}_Q=I_D+I_C\\ =I_{R1}+I_B+I_C\\ =I_{R1}+\frac{I_B}{\beta }+I_C\end{array}$

Substitute $180$ for $\beta$ , $0.5\text{mA}$ for $I_Q$ and $0.0875\text{mA}$ for $I_{R1}$ in the above equation.

  $\begin{array}{l}0.5\text{mA}=0.0875\text{mA}+\frac{I_C}{180}+I_C\\ I_C=0.41022\text{mA}\end{array}$

The value of the base current is calculated as,

  $\begin{array}{c}{I}_{D1}=I_{R1}+I_{B2}\\ =0.0875\text{mA}+\frac{I_{C2}}{\beta }\\ =0.0875\text{mA}+\frac{0.41022\text{mA}}{180}\\ =0.08978\text{mA}\end{array}$

The transconductance of the first transistor is given by,

  $g_{m1}=2\sqrt{k_n{I}_{D1}}$

Substitute $0.08978\text{mA}$ for $I_{D1}$ and $0.7\text{mA}/{\text{V}}^2$ for ${k^{\prime }}_n$ in the above equation.

  $\begin{array}{c}{g}_{m1}=2\sqrt{\left(0.7\text{mA}/\text{V}\right)\left(0.08978\text{mA}\right)}\\ =0.5014\text{mA}/\text{V}\end{array}$

The transconductance of the first transistor is given by,

  $g_{m2}=\frac{I_{C2}}{0.026\text{V}}$

Substitute $0.41022\text{mA}$ for $I_{C2}$ in the above equation.

  $\begin{array}{c}{g}_{m2}=\frac{0.41022\text{mA}}{0.026\text{V}}\\ =15.78\text{mA}/\text{V}\end{array}$

The value of the resistance $r_{\pi 2}$ is calculated as,

  $\begin{array}{c}{r}_{\pi 2}=\frac{\beta }{g_{m2}}\\ =\frac{180}{15.78\text{mA}/\text{V}}\\ =11.41\text{k}\Omega \end{array}$

Conclusion:

Therefore, the value of small signal parameters are $g_{m2}=15.78\text{mA}/\text{V}$ , $g_{m1}=0.5014\text{mA}/\text{V}$ and $r_{\pi 2}=11.41\text{k}\Omega$ .

To determine

The value of small signal voltage gain for the given values of load resistor.

Answer to Problem 11.82P

The value of the small signal voltage gain for $R_L=10\text{k}\Omega$ is $0.99736$ and for $R_L=100\text{k}\Omega$ is $0.99936$ .

Explanation of Solution

Given:

The given diagram is shown below.

  

The given values of load resistor is $R_L=10\text{k}\Omega$ and $R_L=100\text{k}\Omega$

Calculation:

The expression for the voltage $V_O$ is calculated as,

  $V_{\pi }=g_{m1}{V}_{sg}\left(\frac{R_1{r}_{\pi 2}}{R_1+r_{\pi 2}}\right)$

The expression for the gate to source voltage is given by,

  $V_{sg}=V_O-V_i$

The expression to determine the value of the output voltage is given by,

  $\begin{array}{l}{V}_O=-\left(g_{m1}{V}_{sg}+g_{m2}{V}_{\pi }\right)R_L\\ V_O=-\left(g_{m1}\left(V_O-V_i\right)+g_{m2}{g}_{m1}\left(V_O-V_i\right)\left(\frac{R_1{r}_{\pi 2}}{R_1+r_{\pi 2}}\right)\right)R_L\\ -V_O=\left(V_O-V_i\right)\left(g_{m1}+g_{m2}{g}_{m1}\left(\frac{R_1{r}_{\pi 2}}{R_1+r_{\pi 2}}\right)\right)R_L\\ \frac{V_O}{V_i}=\frac{g_{m1}\left[1+g_{m2}\left(\frac{R_1{r}_{\pi 2}}{R_1+r_{\pi 2}}\right)\right]R_L}{1+g_{m1}\left[1+g_{m2}\left(\frac{R_1{r}_{\pi 2}}{R_1+r_{\pi 2}}\right)\right]R_L}\end{array}$ …. (1)

Substitute $8\text{k}\Omega$ for $R_1$ , $0.5014\text{mA}/\text{V}$ for $g_{m1}$ , $15.78\text{mA}/\text{V}$ for $g_{m2}$ , $11.41\text{k}\Omega$ for $r_{\pi 2}$ and $10\text{k}\Omega$ for $R_L$ in the above equation.

  $\begin{array}{c}\frac{V_O}{V_i}=\frac{\left(0.5014\text{mA}/\text{V}\right)\left[1+\left(15.78\text{mA}/\text{V}\right)\left(\frac{\left(8\text{k}\Omega \right)\left(11.41\text{k}\Omega \right)}{8\text{k}\Omega +11.41\text{k}\Omega }\right)\right]\left(10\text{k}\Omega \right)}{1+0.5014\text{mA}/\text{V}\left[1+\left(15.78\text{mA}/\text{V}\right)\left(\frac{\left(8\text{k}\Omega \right)\left(11.41\text{k}\Omega \right)}{8\text{k}\Omega +11.41\text{k}\Omega }\right)\right]\left(10\text{k}\Omega \right)}\\ =0.99736\end{array}$

Substitute $8\text{k}\Omega$ for $R_1$ , $0.5014\text{mA}/\text{V}$ for $g_{m1}$ , $15.78\text{mA}/\text{V}$ for $g_{m2}$ , $11.41\text{k}\Omega$ for $r_{\pi 2}$ and $100\text{k}\Omega$ for $R_L$ in equation (1).

  $\begin{array}{c}\frac{V_O}{V_i}=\frac{\left(0.5014\text{mA}/\text{V}\right)\left[1+\left(15.78\text{mA}/\text{V}\right)\left(\frac{\left(8\text{k}\Omega \right)\left(11.41\text{k}\Omega \right)}{8\text{k}\Omega +11.41\text{k}\Omega }\right)\right]\left(100\text{k}\Omega \right)}{1+0.5014\text{mA}/\text{V}\left[1+\left(15.78\text{mA}/\text{V}\right)\left(\frac{\left(8\text{k}\Omega \right)\left(11.41\text{k}\Omega \right)}{8\text{k}\Omega +11.41\text{k}\Omega }\right)\right]\left(100\text{k}\Omega \right)}\\ =0.99936\end{array}$

Conclusion:

Therefore, the value of the small signal voltage gain for $R_1=10\text{k}\Omega$ is $0.99736$ and for $R_1=100\text{k}\Omega$ is $0.99936$ .