超大规模集成电路第二次作业2016秋

Assignment 2

1. Consider a 180 nm technology. Compute the value of V DSAT for the NMOS and PMOS

device assuming V GS = 1.8 V , V Tn = 0.5 V , V Tp = -0.5 V , L = 200 nm.

Solution ：

（1）、针对于NMOS ，由V V V V Tn GS GT 3.1=-=可知道，GT V 值很大，且对于短沟道器件（勾道长度=0.2um ），速度饱和因子V κ明显小于1，相比于长沟道器件往

往工作于饱和区，所以由公式可知：GT GT DAST V V V )(κ=，其中)/(11

)(L V V cn GT GT ξκ+=，

当处于速度饱和时V L cn 2.1102.010666=⨯⨯⨯=--ξ，所以V V DASTN 62.0)

2.1/

3.1(13.1=+=。 （2）、同理针对于PMOS ，V V V V Tp GS GT 3.1=-=，同（1），

V L cp 8.4102.0102466=⨯⨯⨯=--ξ，得到V V DASTP 02.1)

8.4/3.1(13.1=+=。

2. Compute the NMOS and PMOS saturation currents per micron of width for a 130 nm

technology and the ratio of I DSA TN /I DSATP . Assume a channel length of 100 nm, t ox =22 Å, V Tn = 0.4 V , V Tp = -0.4V , V GS = 1.2 V . Use νsat = 8*106 cm/s.

Solution ：

根据公式)(DAST GT ox sat DSAT V V W C v I -=，)

/(1)(L V V V V V c GT GT GT GT DAST ξκ+==，合并得到：L

V V V V W C v I c T GS T GS ox sat DSAT ξ+--=)()(2，同时cm F t C ox ox ox ox /105.3,13-⨯==εε， ,4.21.024,6.01.06V L V L cp cn =⨯==⨯=ξξ所以：

um uA L V V V V C v W I cn Tn GS Tn GS ox sat DSATN /5856

.04.02.1)4.02.1(102.2105.3108)()(2

317102=+--⨯⨯⨯⨯⨯=+--=--ξ um uA L V V V V C v W I cp Tp GS Tp GS ox sat DSATP /2564

.24.02.1)4.02.1(102.2105.3108)()(2

317102=+--⨯⨯⨯⨯⨯=+--=--ξ 所以最后：

285.2256

585==DSATP DSATN I I 。

3. Compute the overlap capacitance, C ol , for a 130 nm technology with t POLY /t OX = 100 and a lateral diffusion of 10 nm.

Solution ：

将100/=OX POLY t t ，带入到边缘扩散电容公式：

um fF t t C ox POLY ox

f /1.0)1001ln(14.31054.32)1(ln 213

≈+⨯⨯=+=-πε，如果：nm L D 10=，则um fF L C C D ox ov /15.0≈⨯=横向扩散电容，所以整个覆盖电容为：

um fF um fF um fF C ol /25.0/15.0/1.0=+=.

4. Compute the channel capacitance in th e

cutoff, resistive, and saturation regions required for a PMOS device with t ox =22 Å and W = 400 nm, L = 100 nm.

Solution ：

um fF um F L t L C C ox ox

ox g /6.1/106.11.010

2.2105.315317=⨯=⨯⨯⨯===---ε，所以总电容是：fF um um fF W C C g G 64.04.0/6.1=⨯==，所以在各个区中计算如下：

（1）截止区中：fF fF C C C GB GD GS 32.02/64.0,0,0====；

（2）线性区中：0,2/64.0,2/64.0===GB GD GS C fF C fF C ；

（3）饱和区中：0,0,43.03

2===⨯=GB GD G GS C C fF C C ；

5. Compute the junction capacitance for a 130 nm technology.

(a) Find Φ0 and C jo for an n +p junction with N D = 1020 cm -3 and N A = 3*1017 cm -3. (b) W = 400 nm, L = 100 nm, x j = 50 nm, and the diffusion extension L S is 300 nm. Find C diff in unit of fF for V D = 0 V and V D = -1.2 V . Assume C j = C’jsw .

Solution ：

（a ）由题中条件可知：V n N N q kT i D A 1))

105.1(10103ln(026.0ln 21020

1720=⨯⨯⨯==φ， 由公式可知道：

217

191400/6.10

.12103106.11085.87.1122um fF qN N N N N q C A si D A D A B si j ≈⨯⨯⨯⨯⨯⨯⨯=≈+=--φεφε （b ）总共的结电容PERIMETER C AREA C C jsw j diff ⨯+⨯=，由于'

jsw C C j =，所以