## Solution for CMOS Digital Integrated Circuits Analysis and Design 3RD Edition Chapter 7, Problem 2

by Sung-Mo, Kang and Yusuf Leblebici
77 Solutions 13 Chapters 29233 Studied ISBN: 9780072460537 5 (1)

# Chapter 7, Problem Exercise_Problems 2 : 7.2 Calculate the equivalent W/L of...

7.2

Calculate the equivalent W/L of the two nMOSTs with W1/L and W2/L connected in series. For simplicity, neglect the body effect, i.e., the threshold voltages of individual transistors are constant and do not depend on the source voltages. Although this is not true in reality, such an assumption is necessary for simple analysis with a reasonably good approximation.

7.3Analytical expressions for Vth(logic) have been derived in Chapter 7 for the CMOS

NOR2 gate. Now consider the CMOS NAND2 gate for the following cases and use

kp = kn = 100 uA/V2:

•  two inputs switching simultaneously

•  top nMOS switching while the bottom nMOS’s gate is tied to VDD

•  top nMOS gate is tied to VDD and the gate input of the bottom nMOS is changing

(a)Derive an analytical expression for Vth corresponding to the first case. Also find the Vth value for the first case for VDD = 1.2 V when the magnitudes of the threshold voltages are VTn = 0.53 V, VTp = –0.51 V with = 0.

.

## Step-By-Step Solution

7.2

Calculate the equivalent W/L of the two nMOSTs with W1/L and W2/L connected in series. For simplicity, neglect the body effect, i.e., the threshold voltages of individual transistors are constant and do not depend on the source voltages. Although this is not true in reality, such an assumption is necessary for simple analysis with a reasonably good approximation.

SOLUTION:

We know that IDS=IDS1=IDS2, assuming all transistors are operating in linear region, neglecting body effect,

Rewriting the first equation, we obtain,

rearranging the above equation,

rearranging the above equation,

where, α, β < 1.

If we compare this result with long channel transistor case, assuming W1=W2=2μm, α=β=0.9.

Long-channel:

Short-channel:

Therefore, stacked transistor logic with short channel device does not need to be increased as much as in the long-channel case.