Re: [問題] MOS電路問題
看板Electronics作者deathcustom (litron-intl)時間12年前 (2012/03/25 01:28)推噓2(2推 0噓 6→)留言8則, 2人參與討論串2/2 (看更多)
※ 引述《ziizi (ziizi)》之銘言:
: -6
: λ=0.1 ,W/L=20/0.18,μnCox=200*10 ,Vth=0.4 求小訊號模型.
: ┌────┬─VDD=1.8V
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: 這學期修電子學碰到的問題,希望知道的大大能幫幫我~"~
: 這題我列的方程式如下:
: 2
: ID=1/2 * μnCox *W/L*(Vgs-Vth)*(1+λVds) -(1)
: ID= (VDD-Vds)/2k -(2)
: Vds=Vgs -(3)
: 列了三個方程式 要把ID給解出來 才能算小訊號電阻與ro
: 有問過學校老師,他是說要用疊代來解,可是我不管怎麼疊到最後都會發散.
: 希望板上的大大幫幫我~ 如果遇到這題,該如何解?
: 能否使用疊代的方法來解呢?
: 解答給的答案是ID=588μA,Vgs=Vds=0.623V,gm=4.961mS,ro=16.996KΩ
function:
f(V) = 1.8 - V - IR = 0
= 1.8 - V - 2k*0.5*200u*(20/0.18)(V-0.4)^2(1+0.1V)
= 1.8 - V - (4/0.18)*(V-0.4)^2 *(1+0.1V)
一開始先忽略channel length求出(解二次式)
Vgs ~ 0.63V
I = 0.5*200u*(20/0.18)*0.23^2 = 587uA, Vs = IR = 1.174
考慮ro
I = 587*(1+0.063) = 624uA => IR = 1.248V
f(V) = 1.8 - 0.63 - IR = -0.078
V1 = 0.63
法一:Newton Approximation
f(V1)/(V1-V2) = m = f'(V1)
V2 = V1 - f(V1)/f'(V1)
f'(V) = -1 - (4/0.18)[2(V-0.4)(1+0.1V) + 0.1(V-0.4)^2]
f'(V1) = -1 - (4/0.18)[2*0.23*1.063 + 0.1*0.23^2]
~ -12
V2 = 0.63 - (-0.078)/(-12) = 0.6235
I(V2) = (2m/0.18)*0.2235^2*(1+0.06235) = 589.63uA
I(V2)R = 1.179V
f(V2) = 1.8 - 0.6235 - 1.179 = -0.0025
V3 = V2 - f(V2)/f'(V2) = 0.6235 - (-0.0025)/(-11.67) = 0.6233
I(V3) = (2m/0.18)*0.2233^2*1.06233 = 588.565uA
I(V3)R = 1.17713
f(V3) = 1.8 - 0.6233 - 1.17713 = -0.00043
以上是疊代的流程,基本上因為f(V3) < 1mV,所以可以忽略了
我們可以取V = V3 = 0.6233或是 0.623
法二:insertion
V1 = 0.63 => IR = 1.248 => Vgs = 0.552 => err = Vgs - V1 = -0.078
=> V太大, I太大
V2 = 0.62 => I = (2m/0.18)*0.22^2*1.062 = 571.12uA
IR = 1.142 => Vgs = 0.658 => err = 0.038
內插得V3 = 0.6233
I = 588.565uA
IR = 1.17713 => Vgs = 0.62287 => err = -0.00043
......
如果你一開始選錯方程式就......
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※ 編輯: deathcustom 來自: 218.166.196.61 (03/25 03:12)
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