q*******n
发帖数: 52 | |
le
发帖数: 190 | 2 I guess the gain is 0.5*gm*(rop||...),
the feedback nullifies the gm generator of the right pmos follower.
how about 18?
【在 q*******n 的大作中提到】
: http://stevenic.blogspot.com/2010/04/some-on-site-interview-questions-fuding.html
: problem #15. Thanks.
|
H***F
发帖数: 2501 | |
l****o
发帖数: 184 | 4 For #9,
Gain=Gm*Ro=gm/Id*Va (early voltage)
gm=2*Id/Vds,sat
Gain=2*Va/Vds,sat
Current increase, Vds,sat increases, Gain decrease.
As for BJT, Gm/Id is constant (1/Vt)
Thus, gain does not change. |
q*******n
发帖数: 52 | 5 can you explain more?
for 18, is the answer 5v * (c1/(c1+c2)) ?
【在 le 的大作中提到】
: I guess the gain is 0.5*gm*(rop||...),
: the feedback nullifies the gm generator of the right pmos follower.
: how about 18?
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l****r
发帖数: 97 | 6 Final settled value is c1/(c1+c2).
The question asks for waveform at B.
Assume voltage is 0 at B at the begining. Key is to realize the switch
resistance is not 0 and voltage at across a cap is continuous.
【在 q*******n 的大作中提到】
: can you explain more?
: for 18, is the answer 5v * (c1/(c1+c2)) ?
|
le
发帖数: 190 | 7 I am not quite sure. Remove the source follower, and replace the right PMOS
with a rop. Then the gain is from vin to the output of the current mirror.
The transconductance is 0.5*gm, and then find the output resistance...
【在 q*******n 的大作中提到】
: can you explain more?
: for 18, is the answer 5v * (c1/(c1+c2)) ?
|
ET
发帖数: 10701 | 8 #15确定没画错？
input transistor是pmos with pmos current mirror load..
【在 q*******n 的大作中提到】
: http://stevenic.blogspot.com/2010/04/some-on-site-interview-questions-fuding.html
: problem #15. Thanks.
|
ET
发帖数: 10701 | 9 #18哪里来的5V?
那不是1v吗？
【在 q*******n 的大作中提到】
: can you explain more?
: for 18, is the answer 5v * (c1/(c1+c2)) ?
|
ET
发帖数: 10701 | 10 #18.. linear ramp from 0 to c1/(c1+c2)
charge balance between c1 & c2.
i think so..
【在 q*******n 的大作中提到】
: can you explain more?
: for 18, is the answer 5v * (c1/(c1+c2)) ?
|
|
|
ET
发帖数: 10701 | 11 small signal gain,
avo = gm(ro1//ro2)
gm = sqrt(2*k*w/l*Id) ， id = iss/2, iss就是下面那current source generated..
ro1 = 1/(lamma*Id)= r02
gm & ro cancel sqrt(id), left sqrt(id)在分母
所以id 增加，gain减少。
【在 H***F 的大作中提到】
: #9为啥gain降低？电流增大
:
: http://stevenic.blogspot.com/2010/04/some-on-site-interview-questions-fuding.html
: problem #15. Thanks.
|
le
发帖数: 190 | 12 this is very basic
..
【在 ET 的大作中提到】
: small signal gain,
: avo = gm(ro1//ro2)
: gm = sqrt(2*k*w/l*Id) ， id = iss/2, iss就是下面那current source generated..
: ro1 = 1/(lamma*Id)= r02
: gm & ro cancel sqrt(id), left sqrt(id)在分母
: 所以id 增加，gain减少。
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l****o
发帖数: 184 | 13 今天我问了一下我老板， 他也觉得画的不对，
【在 ET 的大作中提到】
: #15确定没画错？
: input transistor是pmos with pmos current mirror load..
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ET
发帖数: 10701 | 14 pmos 做为input transistor对这个topology不太make sense.
因为pmos 的source都希望接在most positive supply.
另外biasing pmos也困难。因为对pmos,current draw from vdd, 对nmos, sink curre
nt to ground.
【在 l****o 的大作中提到】
: 今天我问了一下我老板， 他也觉得画的不对，
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q*******n
发帖数: 52 | 15 对不起，没看清，是1v.
【在 ET 的大作中提到】
: #18哪里来的5V?
: 那不是1v吗？
|
q*******n
发帖数: 52 | 16 I think if the switch is ideal, Ron=0, then instantly finish charge balance.
If Ron is not zero, then classical exponential rising with time constant of
Ron *(c1//c2)?
【在 ET 的大作中提到】
: #18.. linear ramp from 0 to c1/(c1+c2)
: charge balance between c1 & c2.
: i think so..
|
q*******n
发帖数: 52 | 17 谢谢各位回复，我感觉也是画错了.
curre
【在 ET 的大作中提到】
: pmos 做为input transistor对这个topology不太make sense.
: 因为pmos 的source都希望接在most positive supply.
: 另外biasing pmos也困难。因为对pmos,current draw from vdd, 对nmos, sink curre
: nt to ground.
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s*****o
发帖数: 22187 | 18 我觉得：
VB=C1/(C1+C2)*(1-exp(-t/tau)), where tau=Ron*C1*C2/(C1+C2);
如果你的c1//c2是指c1,c2串联，那应该是对的:D
balance.
of
【在 q*******n 的大作中提到】
: I think if the switch is ideal, Ron=0, then instantly finish charge balance.
: If Ron is not zero, then classical exponential rising with time constant of
: Ron *(c1//c2)?
|
q*******n
发帖数: 52 | 19 是, 这样表示有些歧意. 谢谢.
【在 s*****o 的大作中提到】
: 我觉得：
: VB=C1/(C1+C2)*(1-exp(-t/tau)), where tau=Ron*C1*C2/(C1+C2);
: 如果你的c1//c2是指c1,c2串联，那应该是对的:D
:
: balance.
: of
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d********g
发帖数: 11948 | 20 gain = gm * ro
ro = 1/(lamda*id)
gm = 2id/(vdsat)
gain = 2/(lamda*vdsat)
id increases, vdast increases, gain decreases
【在 H***F 的大作中提到】
: #9为啥gain降低？电流增大
:
: http://stevenic.blogspot.com/2010/04/some-on-site-interview-questions-fuding.html
: problem #15. Thanks.
|
d********g
发帖数: 11948 | 21 那如果是nmos input 画对的话呢？ feedback 在哪里
curre
【在 ET 的大作中提到】
: pmos 做为input transistor对这个topology不太make sense.
: 因为pmos 的source都希望接在most positive supply.
: 另外biasing pmos也困难。因为对pmos,current draw from vdd, 对nmos, sink curre
: nt to ground.
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