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1、浙江大学电气工程学院 模拟与数模混合集成电路模拟与数模混合集成电路 吴晓波，赵梦恋 2014-2015 学年夏学期 2015 年 06 月 6-1. Assume that W/L ratios of Figure 6.1 are (W/L)1 = 2m/1m and (W/L)2 = (W/L)3 = (W/L)4 =1m/1m. Find the dc value of vIN that will give a dc current in M1 of 110A. Calculate the small signal voltage gain and output resistance us
2、ing the parameters of Table 6.2. Assume =0. vOUT VDD M1 M2M3M4 vIN100A Figure 6.1 Solution: 2 111 2 1 () () 2 12 110(110 )(0.7) 21 1.7 DNinTH in in W IKVV L V VV 34 213 111 222 66 2 22 100 10 (/)1102110 6.96/ (/)50110 111 31.6 2(/) 2 50 101 10 10 DD DDD mND v mPD out m PD IIA IIIA gKWLI AV V gKWLI R
3、K gKWLI (For Tab.4.1 Vin=1.61V, Av=-8.77V/V, Rout=36.3K) 6.2 Calculate the differential transconductance gmd and the differential voltage gain Av of an n-channel input differential amplifier shown in Figure 6.2 , with the parameters shown in table 6.2. Consider Iss=100A, and W1/L1=W2/L2=W3/L3=W4/L4=
4、1. Assuming all the channel lengths are equal to 1m, and VDD=5V. If W1/L1=W2/L2=10W3/L3=10W4/L4=10, repeat the calculation M1M2 M3M4 VDD Vb VG1VG2 Fig 5.1 M5 Figure 6.2 Solution: 1234 12 1 22 2424 ) 1 104.9 2 23.31 / () mdmmnSS mm v dsdsSS WWWW a LLLL W gggKIS L gg AV V ggI 1234 12 1 22 2424 ) 10101
5、0 331.7 2 73.71 / () mdmmnSS mm v dsdsSS WWWW b LLLL W gggKIS L gg AV V ggI 6.3 Calculate the maximum(VIC(max) and the minimum input common-mode voltages (VIC(min), and the input common mode voltage range (ICMR) of an n-channel input differential amplifier shown in Figure 6.2, with the parameters sh
6、own in table 6.2. Assume all MOSFETs are in saturation, all the (W/L)s are equal to 10m/1m, ISS=10A, and VDD=5V. Solution: 6.4 Find the value of the unloaded differential-transconductance, gmd, and the unloaded differential-voltage gain, Av, for the p-channel input differential amplifier of Figure 6
7、.4 when ISS=10A and ISS=1A. What is the slew rate of the differential amplifier if a 100 pF capacitor is attached to the output? Assuming W1/L1=W2/L2=W3/L3=W4/L4=1, and all the channel lengths are equal to 1m. Use the transistor parameters of Table 6.2. VDD VG1VG2 Vb M1M2 M3M4 M5 Figure 6.4 Solution
8、: a) Given ISS=10A, 12 11 2412 22.36 2 49.69/ () mdmm mm v dsdsSS gggS gg AV V ggI Given ISS=1A 12 11 2412 7.07 2 157.11 / () mdmm mm v dsdsSS gggS gg AV V ggI 6.5 The circuit shown in Figure 6.5 called a folded-current mirror differential amplifier and is useful for low values of power supply. Assu
9、me that all W/L values of each transistor is 100. Using the parameters shown in table 6.2, a) Find the maximum input common mode voltage, VIC(max) and the minimum input common mode voltage, VIC(min). Keep all transistors in saturation for this problem. b) What is the input common mode voltage range,
10、 ICMR? c) Find the small signal voltage gain, vout/vin, if vin = v1 - v2. Figure 6.5 Solution: 1(max)3 2 50 0.70.71.495 110 100 GSTN vVVV b) 1(max)1(min) 1.4950.93020.5648 ICMR VvvV c.) 1247 (/ / /)116.5 / vmooo AgrrrV V b) Table 6.1 NMOS 模型 LEVEL=1 VTO=0.7 GAMMA=0.45 PHI=0.9 PSUB=9e+14 LD=0.08e-6 U
11、O=350 LAMBDA=0.1 TOX=9e-9 PB=0.9 CJ=0.56e-3 CJSW=0.35e-11 MJ=0.45 MJSW=0.2 CGDO=0.4e-9 JS=1.0e-8 PMOS 模型 LEVEL=1 VTO=-0.8 GAMMA=0.4 PHI=0.8 PSUB=5e+14 LD=0.09e-6 UO=100 LAMBDA=0.2 TOX=9e-9 PB=0.9 CJ=0.94e-3 CJSW=0.32e-11 MJ=0.5 MJSW=0.3 CGDO=0.3e-9 JS=0.5e-8 上表给出的是 0.5m 工艺 level 1 MOS SPICE 模型参数的典型值
12、，其中的参数定义如下： VTO: VSB=0 时的阈值电压 （单位：V） GAMMA: 体效应系数 （单位：V1/2） PHI: 2F （单位：V） TOX: 栅氧厚度 （单位：m） NSUB: 衬底掺杂浓度 （单位：cm-3） LD: 源/漏侧扩散长度 （单位：m） UO: 沟道迁移率 （单位：cm2/(v/s)） LAMBDA: 沟道长度调制系数 （单位：V-1） CJ: 单位面积的源/漏结电容 （单位：F/m2） CJSW: 单位长度的源/漏侧壁结电容 （单位：F/m） PB: 源/漏结内建电势 （单位：V） MJ: CJ 公式中的幂指数 （无单位） MJSW: CJSW 等式中的幂指数 （无单位） CGDO: 单位宽度的栅/漏交叠电容 （单位：F/m） CGSO: 单位宽度的栅/源交叠电容 （单位：F/m） JS: 源/漏结单位面积的漏电流 （单位：A/m2） Table 6.2
