b********n
发帖数: 181 | 1 见图
答案我在图里抄了，从书后面的附录里抄的。无奈没找到分析过程
我直观的看，也就能看出来Rin是1/gm^2/rds
请问大牛，这个3.25是怎么算的 |
h*******a
发帖数: 184 | |
b********n
发帖数: 181 | 3 Allen的第三版
新题，所以第二版的solution manual上找不到
【在 h*******a 的大作中提到】
: 好奇是那本书上的
|
h*******a
发帖数: 184 | 4 我整出2.5。allen那书CMOSanalog circuit design第二版好像没有这题。 |
b********n
发帖数: 181 | 5 直观看的还是笔算的？
直观看的能讲下么
【在 h*******a 的大作中提到】
: 我整出2.5。allen那书CMOSanalog circuit design第二版好像没有这题。
|
h*******a
发帖数: 184 | 6 是用笔算的（呵呵，好久没算了）不过没那么复杂，但还不知道对否。 |
b********n
发帖数: 181 | 7 我笔算了一下，跟你一样，也是2.5。
allen 习题答案错误不少，我就不多想了
看来还是得动手画图，我在那儿试图直观的看，看半天没看出个所以然来。
画张小图就清楚了。
多谢。
【在 h*******a 的大作中提到】
: 是用笔算的（呵呵，好久没算了）不过没那么复杂，但还不知道对否。
|
b********n
发帖数: 181 | 8 赶紧买第三版吧，多了点东西。
【在 h*******a 的大作中提到】
: 我整出2.5。allen那书CMOSanalog circuit design第二版好像没有这题。
|
y*m
发帖数: 144 | 9 without feedback, Rin is about 1/gm(1+2/3)=5/(3gm)
so Rin equals 5/(3gm*gm*ron).
approxmation is made here, maybe with hand calculation, the weighting factor
is not exact 5/3 but it is not the point of this question, I guess.
【在 b********n 的大作中提到】
: 见图
: 答案我在图里抄了，从书后面的附录里抄的。无奈没找到分析过程
: 我直观的看，也就能看出来Rin是1/gm^2/rds
: 请问大牛，这个3.25是怎么算的
|
s*****l
发帖数: 19 | 10 我整出来也是2.5。
Loop gain 是 2/3*gmN*rN.
factor
【在 y*m 的大作中提到】
: without feedback, Rin is about 1/gm(1+2/3)=5/(3gm)
: so Rin equals 5/(3gm*gm*ron).
: approxmation is made here, maybe with hand calculation, the weighting factor
: is not exact 5/3 but it is not the point of this question, I guess.
|
s*****n
发帖数: 2 | 11 1) Without feedback, assume the impedance looking up at the source of M3 is
RD, then the impedance looking at the source of M3 is
Rin = (RD+Ro3)/(1+gm3*Ro3) = (RD+Ro3)/(gm3*Ro3)
2) M1 senses the Iout (Iout*RD) and feedback the current to input node gm1*
RD*Iout, loop gain T = gm1*RD
3) shunt feedback at the input --> Rin_eff = (RD+Ro3)/(gm3*Ro3*gm1*RD) (
ignore the 1+T factor)
4) Now look at the RD, 坑爹, depends on the load impedance:from the fig,
assume no loading,
RD = Rdsp || (gmn*Rdsn^2+Rdsp)/(1+gmpRdsp)=Rdsp||Rdsn = 2/3 *Rdsn
5) Rin_eff = (2/3 +1)Rdsn/(gmn^2*Rdsn*2/3*Rdsn) = 2.5/(Rdsn*gmn^2)
In the real application, the load impedance should be much lower than gmn*
Rdsn^2 in order to take the output current, then the Rin_eff will be
different from (5). RD should be in order of 1/gmp --> Rin_eff is about 1/(
2gmn)
Rout is roughly 3/2* Rdsp |
b********n
发帖数: 181 | 12 glad to see so many people trying this one
is
【在 s*****n 的大作中提到】
: 1) Without feedback, assume the impedance looking up at the source of M3 is
: RD, then the impedance looking at the source of M3 is
: Rin = (RD+Ro3)/(1+gm3*Ro3) = (RD+Ro3)/(gm3*Ro3)
: 2) M1 senses the Iout (Iout*RD) and feedback the current to input node gm1*
: RD*Iout, loop gain T = gm1*RD
: 3) shunt feedback at the input --> Rin_eff = (RD+Ro3)/(gm3*Ro3*gm1*RD) (
: ignore the 1+T factor)
: 4) Now look at the RD, 坑爹, depends on the load impedance:from the fig,
: assume no loading,
: RD = Rdsp || (gmn*Rdsn^2+Rdsp)/(1+gmpRdsp)=Rdsp||Rdsn = 2/3 *Rdsn
|
