电子科大罗萍模拟集成电路期末试题3

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电子科技大学二零零 四 至二零零 五 学年第 二 学期期 中
《微电子电路设计》课程考试题（ 120 分钟） 考试形式： 闭卷 考试日期 2005年 5 月 日
一 二 三 四 五 六 七 八 九 十 总分 评卷教师
1. Identify the source, drain, gate and bulk terminals, and find the current I in the
transistors in the following Figure. Assume 2'
/25V A K n
μ=，V TN =0.75V . (16pts) (problem4.3)
Solution:
(a) V V V V V V V V V S D DS S G GS 2.0)2.0(0,
5=--=-==-=
A V A V V V V L W K I I DS DS TN G n
DS μμ2082.022.075.02.51102522'
=⎪⎭⎫
⎝
⎛--=⎪⎭⎫ ⎝⎛--== (b) V V V V V V V V V S D DS S G GS 2.0)2.0(0,
2.5)2.0(5=--=-==--=-=
A V A V V V V L W K I I DS DS TN G n
DS μμ2182.022.075.02.51102522'
-=⎪⎭⎫
⎝
⎛--=⎪⎭⎫ ⎝⎛---=-=
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2. Design the bias circuit in the following figure to give the Q-point of mA I C 10= and V V EC 3= if the transistor current gain is 75=F β and V V BE 7.0=.What is the Q-point if the current gain of the transistor is actually 40? (15 pts) ( problem 5.62 )
Solution:
Ω
→Ω=-==Ω→Ω=-===+=
=-=+-=183.171333.0)7.03( ,68069110013)310( 13.101075
76
1 ,10)(10k mA
V
I -V V R mA V R mA mA I I
I R I R I I V B EB EC B C C F F F
C
E C E C B C EC ββα
0)5()(6801800007.05=--+---B C B I I I
mA I I A V
I B F C B 108.8,
7.202)680(41180007.010===Ω
+-=
βμ
V mA V V EC 35.4680)311.8(10=Ω-= )35.4,11.8(:int V mA po Q -⇒
3. Find V OH , V OL and the power dissipation (for v o =V OL ) for the logic inverter with the saturated load in
the following figure. Assume 0=γ, and 2
'/25V A K n μ=，V V TN 1=. (16 pts) (Problem7.12m)
Solution:
V V V V For TN DD OH 6.216.3,0=-=-==γ
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DSS DSL OL I I V For =,,
OL OL n OL n V V K V K ⎪⎭⎫ ⎝
⎛--⎪⎭⎫ ⎝⎛=--⇒216.215)16.3(212'2', V V OL 1927.0=⇒ A I DD
μ22.36)11927.06.3(2
12102526=--⨯=-
W A V P μμ39.130)22.36)(6.3(==
Checking: A I DD μ22.361927.0)2/1927.016.2(152
10256=--⎪⎭
⎫
⎝⎛⨯=
-
4. Design a CMOS logic gate that implements the logic function ADE ABC Y += based on the CMOS
inverter. Select the transistor sizes with the same delay as that of the reference inverter whose
15=⎪⎭⎫
⎝⎛P L W ,12=⎪⎭⎫ ⎝⎛L L W . ( 20 pts).
Solution:
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5. Find the poles and zeros of the following transfer function, and draw its asymptote Bode plot of the magnitude. (17pts)
)
2000210)(10000)((10102()
50(10)(2
632
8++++⨯++=s s s s s s s s A V Solution ：
Zeros: s z1=0, s z2=-50
Poles: s=-10, s=-200, s=-1000, s=-1000, s=-10000
6. Draw the output voltage waveform for the circuit in the following figure for the triangular input waveform shown. (16pts)
Solution:。