What are the noise margins for the circuit ? +3.3 VM13м, +MsFigure P6.57

As we know VTN=0.6 VVDD=2.5 VKn'=100 μA/V2WLs=41WLL=21 The maximum input voltage accept as low input…

Answered: +3.3 V M13 м, + Ms Figure P6.57

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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What are the noise margins for the circuit ?

Expert Solution

Step 1

As we know

$\begin{array}{rcl} V_{T N} & = & 0.6 V \\ V_{D D} & = & 2.5 V \\ K_{n}^{'} & = & 100 {μA/V}^{2} \\ {(\frac{W}{L})}_{s} & = & \frac{4}{1} \\ {(\frac{W}{L})}_{L} & = & \frac{2}{1} \end{array}$

The maximum input voltage accept as low input logic level is given by

$\begin{array}{rcl} V_{I L} & = & V_{T N S} \\ = & 0.6 V \end{array}$

The output voltage across an input voltage $V_{I L}$ is

$\begin{array}{rcl} V_{O H} & = & V_{D D} - V_{T N L} \\ = & 2.5 - 0.6 \\ = & 1.9 V \end{array}$

Step 2

The output voltage across an input voltage $V_{I H}$ is

$V_{O L} = \frac{V_{D D} - V_{T N L}}{\sqrt{1 + 3 K_{R}}}$

$\begin{array}{rcl} K_{R} & = & \frac{{(\frac{W}{L})}_{s}}{{(\frac{W}{L})}_{L}} \\ = & \frac{\frac{4}{1}}{\frac{2}{1}} \\ = & 8 \end{array}$

Hence,

$\begin{array}{rcl} V_{O L} & = & \frac{2.5 - 0.6}{\sqrt{1 + 3 \times 8}} \\ = & \frac{1.9}{5} \\ = & 0.38 V \end{array}$

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