Verify Eq VIH5VDD + 3VTN + 5VTPVOLVDD - VTN + VrPand

Answered: VIH 5VDD + 3VTN + 5VTP VOL VDD - VTN +…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

Verify Eq

VIH
5VDD + 3VTN + 5VTP
VOL
VDD - VTN + VrP
and

Expert Solution

Step 1

Consider a simple CMOS inverter:

The pull-up network is equivalent to pMOS and the pull-down network is equivalent to nMOS of the inverter.

Considering nMOS=off and pMOS=on, then the inverter transfer characteristic is:

Here, input voltage

$0 < V_{I} < V_{T N}$

The n channel is OFF and the p channel is ON and the output voltage, V_O = V_DD

When nMOS is saturated and pMOS is Linear:

Both transistors are ON, the input voltage is

$V_{I} > V_{T N}$

and the output voltage is such that the currents through both the transistors are equal.

Step 2

The output voltage is given by:

$V_{O} = (V_{I} + V_{T P}) + \sqrt{(V_{D D} - V_{T N} - V_{T P}) (V_{D D} - 2 V_{I} + V_{T N} - V_{T P})}$

for

$V_{I} \leq \frac{V_{D D} + V_{T N} - V_{T P}}{2}$

Now, when the input voltage is

$V_{I} = \frac{V_{D D} + \sqrt{β} V_{T N} - V_{T P}}{1 + \sqrt{β}}$

then both the transistor are saturated.

The range of output voltage is given by:

$V_{I} - V_{T N} \leq V_{O} \leq V_{I} + V_{T P}$

If the input voltage is further increased then nMOS becomes linear and pMOS is saturated. Then the range of input voltage is:

$\frac{V_{D D} + \sqrt{β} V_{T N} - V_{T P}}{1 + \sqrt{β}} < V_{I} < V_{D D} - V_{T P}$

The current in this condition is:

$\begin{array}{rcl} I_{D} & = & \frac{K_{P}}{2} {(V_{D D} - V_{I} - V_{T P})}^{2} = K_{n} [(V_{I} - V_{T N}) V_{O} - \frac{1}{2} {V_{O}}^{2}] \end{array}$

The quadratic equation we get from this is:

$\begin{array}{rcl} \frac{1}{2} \end{array} \begin{array}{rcl} V_{O} \end{array} \begin{array}{rcl} ^{2} \end{array} - \begin{array}{rcl} (V_{I} - V_{T N}) \end{array} \begin{array}{rcl} V \end{array} \begin{array}{rcl} _{O} \end{array} + \frac{{(V_{D D} - V_{I} - V_{T P})}^{2}}{2 β} = 0$

The solution to this equation is:

$V_{O} = \begin{array}{rcl} (V_{I} - V_{T N}) \end{array} - \sqrt{(\begin{array}{rcl} V \end{array} \begin{array}{rcl} \end{array})}$ _I− V_TN2-VDD − VI − VTP2β

When $β = 1$

$V_{O} = (V_{I} - V_{T N}) - \sqrt{(V_{D D} - V_{T N} - V_{T P}) (2 V_{I} - V_{D D} - V_{T N} + V_{T P})}$

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