## Solution for CMOS Digital Integrated Circuits Analysis and Design 3RD Edition Chapter 5, Problem 1

by Sung-Mo, Kang and Yusuf Leblebici
77 Solutions 13 Chapters 28313 Studied ISBN: 9780072460537 5 (1)

# Chapter 5, Problem Exercise_Problems 1 : 5.1Design a resistive-load inverter with R = 2...

5.1Design a resistive-load inverter with R = 2 kΩ, such that VOL = 0.05 V. The enhancement-type nMOS driver transistor has the following parameters:

VDD = 1.1 V

VT0 = 0.52 V

γ = 0 V1/2

λ = 0

µnCox = 216 µA/V2

(a) Determine the required aspect ratio, W/L.

(b) Determine VIL and VIH.

(c) Determine noise margins NML and NMH.

## Step-By-Step Solution

5.1Design a resistive-load inverter with R = 2 kΩ, such that VOL = 0.05 V. The enhancement-type nMOS driver transistor has the following parameters:

VDD = 1.1 V

VT0 = 0.52 V

γ = 0 V1/2

λ = 0

µnCox = 216 µA/V2

(a) Determine the required aspect ratio, W/L.

(b) Determine VIL and VIH.

(c) Determine noise margins NML and NMH.

SOLUTION :

(a)When Vout = VOL, Vin = VDD = 1.2V, the driver transistor operates in linear region. Using Eq.(5.12)

Since VOL is small, VOL/EcLn can be ignored. Using (5.17)

Solve for W/L,

(b)When Vin = VIL, driver transistor operates in saturation region.

Since Vin is slightly higher than VT0, Vin-VTO can be ignored. Also vsat=EC/2. Using Eq.(5.21)

Taking derivative with respect to Vin and set

Thus,

When Vin = VIH, driver transistor operates in linear region. Using Eq.(5.27) repeated here :

Differentiating both sides with respect to Vin and set ,

Plug in back to Eq. (5.27) repeated above to solve for Vout

(c)NML = VIL – VOL = 0.546 – 0.05 = 0.496V

NMH = VOH – VIH = 1.1 – 0.772 = 0.328V