Chapter 10, Problem D10.90P

Design a bipolar cascode amplifier with a cascode active load similar tothat in Figure P10.89 except the amplifying transistors are to be pnp andthe load transistors are to be npn. Bias the circuit at $V^{+}=10V$ and incorporate a reference current of $I_{REF}=200\mu A$ . If all transistors arematched with $\beta =100$ and $V_A=60V$ , determine the small-signal voltage gain.

To determine

The design of a circuit for a given specifications.

To find: The small-signal voltage gain $A_v=\frac{v_o}{v_i}$ that all transistors are matched.

Answer to Problem D10.90P

The design is shown in Figure 2.

  $A_v=-228475$

Explanation of Solution

Given:

  $\begin{array}{l}{V}^{+}=10\text{V}\\ V_A=60\text{V}\\ I_{REF}=200\text{μA}\\ \beta =100\end{array}$

Calculation:

The given circuit is,

  

Figure 1

The equivalent designed circuit for active load amplifier is as shown,

  

Figure 2

The expression for input voltage $V_i$ is,

  $g_{m1}{V}_i=\frac{V_{\pi 2}}{r_{\pi 2}}+\frac{V_{\pi 2}}{r_{o1}}+g_{m2}{V}_{\pi 2}+\frac{V_o-(-V_{\pi 2})}{r_{o2}}\left(1\right)$

  $g_{m1}$ is transcondutance of transistor $Q_1$

  $g_{m2}$ is transcondutance of transistor $Q_2$

  $r_{o1}$ is output resistance of transistor $Q_1$

  $r_{o2}$ is output resistance of transistor $Q_2$

  $r_{\pi 2}$ is hybrid resistor parameter of transistor $Q_2$

  $V_o$ is output voltage

  $V_{\pi 2}$ is hybrid voltage parameter of transistor $Q_2$

  $\begin{array}{l}{g}_{m1}{V}_i=V_{\pi 2}\left(\frac{1}{r_{\pi 2}}+\frac{1}{r_{o1}}+g_{m2}+\frac{V_o}{r_{o2}}+\frac{1}{r_{o2}}\right)\\ g_{m1}{V}_i=V_{\pi 2}\left(\frac{1}{r_{\pi 2}}+\frac{1}{r_{o1}}+g_{m2}+\frac{1}{r_{o2}}\right)+\frac{V_o}{r_{o2}}\left(2\right)\end{array}$

Now expression for output resistance is,

  $\frac{1}{r_o}=\frac{1}{r_{o1}}+\frac{1}{r_{o2}}\left(3\right)$

Put equation (3) in equation (2).

  $g_{m1}{V}_i=V_{\pi 2}\left(\frac{1}{r_{\pi 2}}+g_{m2}+\frac{1}{r_o}\right)+\frac{V_o}{r_{o2}}\left(4\right)$

Consider $g_m\gg \frac{1}{r_o}$

  $g_{m1}{V}_i=V_{\pi 2}\left(\frac{1}{r_{\pi 2}}+g_{m2}\right)+\frac{V_o}{r_{o2}}\left(5\right)$

Now expression for hybrid resistor parameter.

  $r_{\pi 2}=\frac{\beta }{g_{m2}}$ substitute in equation (5).

  $\begin{array}{l}{g}_{m1}{V}_i=V_{\pi 2}\left(\frac{g_{m2}}{\beta }+g_{m2}\right)+\frac{V_o}{r_{o2}}\\ g_{m1}{V}_i=V_{\pi 2}{g}_{m2}\left(\frac{1+\beta }{\beta }\right)+\frac{V_o}{r_{o2}}\\ g_{m1}{V}_i=V_{\pi 2}\left(\frac{1+\beta }{\beta /g_{m2}}\right)+\frac{V_o}{r_{o2}}\\ g_{m1}{V}_i=V_{\pi 2}\left(\frac{1+\beta }{r_{\pi 2}}\right)+\frac{V_o}{r_{o2}}\left(6\right)\end{array}$

Now output voltage will be,

  $\begin{array}{l}\frac{V_o}{R_{o3}}+\frac{V_o-(-V_{\pi 2})}{r_{o2}}+g_{m2}{V}_{\pi 2}=0\\ V_o\left(\frac{1}{R_{o3}}+\frac{1}{r_{o2}}+\frac{V_{\pi 2}}{r_{o2}}+g_{m2}{V}_{\pi 2}\right)=0\\ V_o\left(\frac{1}{R_{o3}}+\frac{1}{r_{o2}}\right)+V_{\pi 2}\left(\frac{1}{r_{o2}}+g_{m2}\right)=0\left(7\right)\end{array}$

Consider $g_m\gg \frac{1}{r_o}$

  $V_{\pi 2}=-\frac{V_o}{g_{m2}}\left(\frac{1}{R_{o3}}+\frac{1}{r_{o2}}\right)\left(8\right)$

Put value from equation (8) to equation (6).

  $\begin{array}{l}{g}_{m1}{V}_i=-\frac{V_o}{g_{m2}}\left(\frac{1}{R_{o3}}+\frac{1}{r_{o2}}\right)\left(\frac{1+\beta }{r_{\pi 2}}\right)+\frac{V_o}{r_{o2}}\\ g_{m1}{V}_i=\frac{-V_o}{R_{o3}}\left(\frac{1+\beta }{\beta }\right)\\ \frac{V_o}{V_i}=-g_{m1}{R}_{o3}\left(\frac{1+\beta }{\beta }\right)\left(9\right)\end{array}$

Put $R_{o3}=\beta r_{o3}$ in equation (9)

  $\begin{array}{l}{R}_{o3}=\beta r_{o3}\\ =100\times 300k\Omega \\ =30\text{M}\Omega \end{array}$

  $\begin{array}{l}\frac{V_o}{V_i}=\frac{-g_{m1}{\beta }^2{r}_{o3}}{1+\beta }\\ A_v=\frac{V_o}{V_i}=\frac{-g_{m1}{\beta }^2{r}_{o3}}{1+\beta }\end{array}$

Now

  $g_{m1}=\frac{I_o}{V_T}=\frac{I_{REF}}{V_T}$

Substitute the value for $\frac{I_{REF}}{V_T}$ ,

  $\begin{array}{l}{g}_{m1}=\frac{{\text{200×10}}^{\text{-6}}\text{A}}{\text{0}\text{.026V}}\\ g_{m1}=7.69\text{2mA/V}\end{array}$

Now

  $\begin{array}{l}{r}_{o3}=\frac{V_A}{I_o}\\ r_{o3}=\frac{60}{200\times {10}^{-6}A}\\ =300k\Omega \end{array}$

  $A_v=\frac{V_o}{V_i}=\frac{-g_{m1}{\beta }^2{r}_{o3}}{1+\beta }\left(10\right)$

Substitute the derived values in equation (10).

  $\begin{array}{l}{A}_v=\frac{\text{-7}{\text{.692mA/V×(100)}}^{\text{2}}\text{×300kΩ}}{\text{1+100}}\\ A_v=\frac{\text{-7}{\text{.692mA/V×(100)}}^{\text{2}}{\text{×300×10}}^{\text{3}}\text{Ω}}{\text{101}}\end{array}$

  $A_v=-228475$