a simple question about Mosfet

Status
Not open for further replies.
W

wang.yuanzhuo

Junior Member level 2
Joined
May 11, 2007
Messages
22
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Visit site
Activity points
1,408
i'm puzzled with a simple question.
For a simple mosfet,when Vg and Vd is fixed, Id=0, and Vs is floating. How much is the Vgs, Vth or 0, or we can't get it?
 

Thanks, i saw Razavi's book "design analog cmos IC', in figure 3.51,he said when m1 is cut off, the voltage of node "x" was 'vb-vth2'. But i think it is floating, so i am puzzled
 

I took a look at the figure 3.51 and the equations shown assumes both M1 & M2 to be in saturation, not cut off. So, your Id does not equal 0 and node x is not floating....
 

No. In the analyze of large-signal, he said "For Vin<Vth1,M1 and M2 are off, Vout=Vdd, and Vx=Vb-Vth2" , maybe here is some problem
 

Ah, I see the section you are talking about... it's referring to figure 3.50, not 3.51.....

Anyway, assuming no sub threshold conduction as the book states, and vin < Vth1, you see M1 is off. Let's say node X is 0, and Vb is > Vth2. You see that M2 will turn on due to channel forming under the gate. This will "pull up" on node X until that node voltage becomes around Vth2 lower then Vb. Once that happens, the channel goes away, M2 turns off and you have Vx ~= vb-vth2. It's not exactly Vb-vth2 since MOS doesn't really turn on/off that sharply.
 

Re: a simple question about Mosfet

as per your cnd source is floting so bet both terminal there is same potential so no current is flow and cmos is cutoff.

well Vth has expression using which it can be found... source is floating and hence Vgs cannot be determined....

thanx.
 

No... Vgs can be determined. Vgs is approximately Vth of the MOS. If Vgs was higher, there would be channel under the gate between the source and drain and source node voltage will be pulled up until the channel shuts down.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…