Chapter 4, Problem 4.12TYU

The transistor parameters of the circuit in Figure 4.49 are $V_{TN1}=V_{TN2}=0.6\text{V}$ , $K_{n1}=1.5{\text{mA/V}}^{\text{2}}$ , $K_{n2}=2{\text{mA/V}}^{\text{2}}$ and ${\lambda }_1={\lambda }_2=0$ . (a) Find $I_{DQ1},I_{DQ2},V_{DSQ1}$ , and $V_{DSQ2}$ . (b) Determine the small−signal voltage gain. (c) Find the output resistance $R_o$ . (Ans. (a) $I_{DQ1}=0.3845\text{mA}$ , $I_{DQ2}=0.349\text{mA}$ , $V_{DSQ1}=2.31\text{V}$ , $V_{DSQ2}=7.21\text{V}$ ; (b) $A_{\upsilon }=-20.3$ ; (c) $R_o=402\Omega$ )

To determine

The drain current and the individual drain to source voltages of the transistors in NMOS cascade circuit with given transistor parameters.

Answer to Problem 4.12TYU

The quiescent drain current of the transistors are $I_{DQ1}\text{=0}\text{.3845 mA}$ , $I_{DQ2}\text{=0}\text{.349 mA}$

The drain to source voltage at Q-point for the transistors are $V_{DSQ1}\text{=2}\text{.31 V}$ $V_{DSQ2}\text{=7}\text{.21 V}$

Explanation of Solution

Given Information:

An NMOS cascade device with transistor parameters $V_{TN1}=V_{TN2}=0.6\text{ V}$ , $K_{n1}=1.5{\text{ mA/V}}^2,K_{n2}=2{\text{ mA/V}}^2$ . The channel conduction parameter is given to be zero for both transistors.

Calculation:

Consider the common source amplifier in cascade with a source follower circuit in Figure 1. Here, transistor M₁ is operated in common-source configuration and M₂ is operated in common-gate configuration.

  

The drain current and gate to source voltage of both the transistors are the same.

The drain current is given by,

  $I_D=K_n{\left(V_{GS}-V_{TN}\right)}^2$

The gate to source voltage is given by,

  $V_{GS}=\left(V_G-V_S\right)$

Considering the transistor M₁ , the gate voltage for the transistor M₁ is,

  $V_{G1}=\frac{R_2}{R_1+R_2}\left(V^{+}-V^{-}\right)+V^{-}$

Substituting the resistance values from the circuit and the bias voltage, the gate voltage is obtained as,

  $\begin{array}{c}{V}_{G1}=\frac{135}{135+383}\times \left(5-\left(-5\right)\right)+\left(-5\right)\\ =-2.39\text{V}\end{array}$

Now, the source voltage is ,

  $V_{S1}=I_{DQ1}{R}_{S1}+(-5)$

Substituting the resistance value,

  $V_{S1}=3.9\times {10}^3{I}_{DQ1}-5$

Thus, gate source voltage in terms of first transistor is ,

  $\begin{array}{c}{V}_{GSQ1}=\left(V_{G1}-V_{S1}\right)\\ =-2.39-3.9\times {10}^3{I}_{DQ1}+5\end{array}$

  $V_{GSQ1}=2.61-3.9\times {10}^3{I}_{DQ1}$ …… (2)

Substituting the expression for gate source voltage in (1), the drain current for transistor M₁ in quiescent condition is,

  $\begin{array}{c}{I}_{DQ1}=K_{n1}{\left(V_{GSQ1}-V_{TN1}\right)}^2\\ =K_{n1}{\left(2.61-3.9\times {10}^3{I}_{DQ1}-V_{TN1}\right)}^2\\ =1.5\times {10}^{-3}{\left(2.61-3.9\times {10}^3{I}_{DQ1}-0.6\right)}^2\\ =1.5\times {10}^{-3}\left(4.024-15.6468\times {10}^3{I}_{DQ1}+15.21\times {10}^6{I}_{DQ1}^2\right)\end{array}$

On rearranging the above equation, the final quadratic equation is obtained as

  $\begin{array}{l}1.5\times {10}^{-3}\left(4.024-15.6468\times {10}^3{I}_{DQ1}+15.21\times {10}^6{I}_{DQ1}^2\right)-I_{DQ1}=0\\ 22.815\times {10}^3{I}_{DQ1}^2-24.47I_{DQ1}+0.006036=0\\ I_{DQ1}=0.3845\text{ mA, 0}\text{.688 mA}\end{array}$

Since the transistor is in saturation, the lower value among the two is considered. Hence, the drain current for the first transistor is,

  $I_{DQ1}=0.3845\text{ mA}$

Now, the drain to source voltage for the transistor M₁ can be expressed as,

  $V_{DSQ1}=V^{+}-V^{-}-I_{DQ1}\left(R_{D1}+R_{S1}\right)$

Substituting the values of parameters,

  $\begin{array}{c}{V}_{DSQ1}=5-\left(-5\right)-0.3845\times {10}^{-3}\left(16.1\times {10}^3+3.9\times {10}^3\right)\\ =2.31\text{ V}\end{array}$

Considering the transistor M₂, the gate voltage is same as the drain voltage of transistor M₁, given by

  $\begin{array}{c}{V}_{G2}=V_{D1}\\ =V^{+}-I_{DQ1}{R}_{D1}\\ =5-0.3845\times {10}^{-3}\times 16.1\times {10}^3\\ =-1.19045\text{ V}\end{array}$

The source voltage is given by,

  $\begin{array}{c}{V}_{S2}=-I_{DQ2}{R}_{S2}+V^{-}\\ =-5-8\times {10}^3{I}_{DQ2}\end{array}$

The gate source voltage is therefore,

  $\begin{array}{c}{V}_{GSQ2}=\left(V_{G2}-V_{S2}\right)\\ =-1.19045+5+8\times {10}^3{I}_{DQ2}\\ =3.80955+8\times {10}^3{I}_{DQ2}\end{array}$

The drain current for transistor M₂ is given by,

  $\begin{array}{c}{I}_{DQ2}=K_{n2}{\left(V_{GSQ2}-V_{TN2}\right)}^2\\ =K_{n2}{\left(3.80955+8\times {10}^3{I}_{DQ2}-V_{TN2}\right)}^2\\ =2\times {10}^{-3}{\left(3.80955+8\times {10}^3{I}_{DQ2}-0.6\right)}^2\\ =2\times {10}^{-3}\left(10.3012-51.3528\times {10}^3{I}_{DQ2}+64\times {10}^6{I}_{DQ2}^2\right)\end{array}$

Solving the above expression, the final quadratic equation is obtained as,

  $\begin{array}{c}{I}_{DQ2}=2\times {10}^{-3}\left(10.3012-51.3528\times {10}^3{I}_{DQ2}+64\times {10}^6{I}_{DQ2}^2\right)\\ =128\times {10}^3{I}_{DQ2}^2-102.7056I_{DQ2}+20.6024\times {10}^{-3}\end{array}$

Thus, the equation is given by,

  $128\times {10}^3{I}_{DQ2}^2-103.7056I_{DQ2}+20.6024\times {10}^{-3}=0$

  $I_{DQ2}=0.3489\text{ mA, 0}\text{.4612 mA}$

Since the transistor is in saturation, the lower value among the two is considered. Hence, the drain current for the second transistor is,

  $\begin{array}{c}{I}_{DQ2}=0.3489\text{ mA}\\ =0.349\text{ mA}\end{array}$

Now, the drain to source voltage for the transistor M₂ at Q-point can be expressed as,

  $V_{DSQ2}=V^{+}-V^{-}-I_{DQ2}{R}_{S2}$

Substituting the values of parameters,

  $\begin{array}{c}{V}_{DSQ2}=5-\left(-5\right)-0.349\times {10}^{-3}\times 8\times {10}^3\\ =7.21\text{ V}\end{array}$

To determine

The voltage gain of an NMOS cascade circuit with given transistor parameters.

Answer to Problem 4.12TYU

The voltage gain is given by $A_v=-20$

Explanation of Solution

Given Information:

:An NMOS cascade device with transistor parameters $V_{TN1}=V_{TN2}=0.6\text{ V}$ , $K_{n1}=1.5{\text{ mA/V}}^2,K_{n2}=2{\text{ mA/V}}^2$ . The channel conduction parameter is given to be zero for both transistors.

Calculation:

Consider the common source amplifier in cascade with a source follower circuit in Figure 1. Here, transistor M₁ is operated in common-source configuration and M₂ is operated in common-gate configuration.

The voltage gain of the circuit is expressed as,

  $\begin{array}{c}{A}_v=\frac{V_o}{V_i}\\ =-\frac{g_{m1}{g}_{m2}{R}_{D1}\left(R_{S2}\left|\right|R_L\right)}{1+g_{m2}\left(R_{S2}\left|\right|R_L\right)}\frac{R_i}{R_i+R_{Si}}\end{array}$

Here, $R_i$ is the resistance at the input side which is ,

  $\begin{array}{c}{R}_i=R_1\left|\right|R_2\\ =\frac{383\times 135}{383+135}\\ =99.81\\ \simeq 100\text{ k}\Omega \end{array}$

Now, the transconductance of the amplifier is given by,

  $g_{m1}=2\sqrt{K_{n1}{I}_{D1}}$

Considering quiescent value of drain current,

  $\begin{array}{c}{g}_{m1}=2\sqrt{1.5\times {10}^{-3}\times 0.3845\times {10}^{-3}}\\ =1.51\text{ mA/V}\end{array}$

Similarly, the transconductance of the second transistor is,

  $\begin{array}{c}{g}_{m2}=2\sqrt{K_{n2}{I}_{D2}}\\ =2\sqrt{2\times {10}^{-3}\times 0.349\times {10}^{-3}}\\ =1.67\text{ mA/V}\end{array}$

Substituting the transconductance values and the resistor values, the voltage gain is given by,

  $\begin{array}{c}{A}_v=-\frac{1.51\times 1.67\times {10}^{-6}\times 16.1\times {10}^3\left(8\times {10}^3\left|\right|4\times {10}^3\right)}{1+1.67\times {10}^{-3}\left(8\times {10}^3\left|\right|4\times {10}^3\right)}\frac{100\times {10}^3}{100\times {10}^3+4\times {10}^3}\\ =\frac{40.59937\times 2.667}{1+26.887\times 2.667\times {10}^3}\times 0.9615\\ \approx -20\end{array}$

To determine

The output resistance of an NMOS cascade circuit with given transistor parameters.

Answer to Problem 4.12TYU

The output resistance is given by $R_o\text{=557}\text{.1}\Omega$

Explanation of Solution

Given Information:

An NMOS cascade device with transistor parameters $V_{TN1}=V_{TN2}=0.6\text{ V}$ , $K_{n1}=1.5{\text{ mA/V}}^2,K_{n2}=2{\text{ mA/V}}^2$ . The channel conduction parameter is given to be zero for both transistors.

Calculation:

The output resistance of the circuit is that of the output resistance of the emitter follower circuit which is low. It can be deduced from the small signal equivalent circuit shown below.

  

As it appears in the circuit, the output resistance of the circuit excluding the load resistance is obtained by considering the Kirchoff’s current law at the output node x which is,

  $\begin{array}{c}{I}_x=\frac{V_{gs2}}{R_{S2}}+g_{m2}{V}_{gs2}\\ =V_{gs2}\left(\frac{1}{R_{S2}}+g_{m2}\right)\end{array}$

This implies, the output resistance is given by,

  $R_o=R_{S2}\left|\right|\frac{1}{g_{m2}}$

Substituting the resistance and transconductance value,

  $\begin{array}{c}{R}_o=\frac{1}{\frac{1}{8\times {10}^3}+0.00167}\\ =557.10\Omega \end{array}$