《CMOS集成電路設(shè)計答案.doc》由會員分享，可在線閱讀，更多相關(guān)《CMOS集成電路設(shè)計答案.doc（21頁珍藏版）》請在裝配圖網(wǎng)上搜索。

2.5）a) λ=0.1, =0.45, =0.9, =0.7 , , So, The above equation is valid for , ie So , And for 2.5)b) for , S and D exchange their roles. (absalutate value) The above equations are valid for Then the direction of current is reversed. for , device operates in the triode region . for , device goes into saturation again . So , 2.5)c) S and D exchange their roles . Device is in saturation region , so Device turns off when and never turns on again . So , ; x<0.3 ；otherwise Then , ; x<0.3 2.5)d) D and S exchange their roles for ; Device remains in the saturation region until , then device goes into the triode region . for : for : S and D exchange their roles again,when for ,device operates into the triode region So , : : 2.5)e) for , so device is in saturation region. so these equations are valid upon the edge of triode region,i.e. Above ,device is in the triode region. ; This problem has been considered only for in which schich-Hodges E9. is valid for . 2.7)a D and S exchange their roles for device is off for device is in the saturation region for device is in the triode region 2.7)b D and S exchange their roles for device is in triode for device is in the saturation region doesn’t depend on and it is consfant for 2.7)c for device is in triode for device is in the satution region is constant for (it doesn’t depend on ) 2.7)d for device is off then device turns on and goes up until ,then device enters triode region for and This is good for for Now by subsfifufly , and in①and②we have: { { or: ③ ⑨ From equation ③ and ⑨ and can be solued in farms of and . Now if ,we have: ⑤{ { { , ⑤ 5.9)a 5.9)b 5.9)c 5.9)d 5.9)e 6.9)a (i) at low fraguency , is like virtual ground (ii) At high fraguency, the quiralent circuit if 6.9)c (i) At low frequancy , the equivalent circuit The impedance @ (ii) at high frequency ▁ the inpedance looking into The inpedance @ 6.9)d (i) at low frequancy ,the equivalent circuit is if KCL @ , : if given , (a) (b) , (c) Assuming all transistors are in saturation, we have When we have assumed and Thus , When the circuit turns on , inifically both M5 and H6 are off and and rise together, i.e.,. When reaches , is also near .Thus,M6 and H5 turn on almost simultanecusly . The surge in the drain current ef M5 turns the rest of the circuit on . As increases further, begins to drop if M6 is turned on sidffreiently became the voltage gain of H6 and exceels unity. For high value of , can be lower than . Since , we solve the quadratic equatiom: This value is substifuted in the other condition: To give the condition for furning off H5.