A question about the book writed by Razavi

[wjxcom]

Hi all, from the book "Design of Analog CMOS Integrated Circuits“writed by Razavi, in page 122, under the figure 4.30, there have a sentence:"To calulate the gain from.......writing ID1=gm1*(Vin,CM-Vp), ID2=gm2*(Vin,CM-Vp)" I think this two equation is not right. We know ID=1/2*K*(Vgs-Vt)^2, and we calulate that the gm should be K*(Vgs-Vt), i.e. gm=K*(Vgs-Vt). so I think ID should be 1/2*K*(Vgs-Vt), i.e. ID=1/2*gm*(Vgs-Vt), i.e. ID1=1/2*gm1*(Vin,CM-Vp) rather than ID1=gm1*(Vin,CM-Vp) that writed by Razavi

各位,我有一个问题: 在Razavi的书中，第122页的图4.30下面一段，Razavi写道：To calulate the gain from.......writing ID1=gm1*(Vin,CM-Vp), ID2=gm2*(Vin,CM-Vp)。我对这个公式有些怀疑：根据饱和区电流公式：ID=1/2*K*(Vgs-Vt)^2，我们可以得到跨导gm=K*(Vgs-Vt)。所以漏电流应该是ID=1/2*gm*(Vgs-Vt), 而不是ID1=gm1*(Vin,CM-Vp)。

I do not know if my idea is correct. I think I can get help from here.

 

[wanily1983]

"ID=1/2*gm*(Vgs-Vt)" is not correct to used in small signal analysis, ID=gm*Vgs is the small signal current in MOS. Please read the chapter 2 of Razavi 's book.

 

[timer]

Hi,
Razavi uses ID for both large and small signal currents. this is a little confusing sometimes. the gm in the equation clearly implies that the curret he is referring to is the small signal current.
hope this helps.

 

[nuiscet]

Gm is an parameter that shows an ability converting the voltage between Gate and source to the current (in the small signal condition), that is just what Gm means:Gm=dIds/dvgs

 

[zenisle]

good question