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i can't determine this MOS operating region.

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evilguy

evilguy

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pmos operation region

Hi, if a transistor has Vgs=Vds=Va and Va<Vt. What region this Mos transistor operate? In my circuit i have one transistor. from simulation it shows the transistor threshold voltage, Vt is about 0.4V and the Vgs and Vds are 0.36V. The Id is about 20µA. thanks

-evilguy-
 

If Vgs<Vt, then you're operating in weak inversion, which is an exponential, BJT like behavior.

Your Vds is fairly large for even fully inverted MOS transistors. As long as Vds is > Vdsat, you operate in saturation and not linear region. Since Vdsat is fairly small (theoretically <0 for fully inverted equation), you don't need to worry about it. In fact, if Vdsat is more then a few Kt/q (~26mV at room), then it stops having an effect.

Here is a nice document describing sub-threshold/weak inversion operation:
**broken link removed**

Greg
 

    evilguy

    evilguy

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amriths04 said:
your trans is off.
Click to expand...
and why is that, care to exlain in detail? if off why there is drain current?
 

hi evilguy,
you had not given enough pressure to create a field between the drain and source terminals. so ideally no current would flow.

could you please upload your schematic with the results on it, so that we could know whether your transistor is a part of any other circuit or a stand-alone,
 

amriths04 said:
hi evilguy,
you had not given enough pressure to create a field between the drain and source terminals. so ideally no current would flow.

could you please upload your schematic with the results on it, so that we could know whether your transistor is a part of any other circuit or a stand-alone,
Click to expand...

there is a current. as i mention in my 1st post. the current is 20µA. what do you mean by "enough pressure"? you know, even VDS is small, there will be some current flows in the transistor. The term cutoff region i think not valid anymore when we deal with MOS model level 49 and above. i begin to agree with gszczesz that mine is in subtreshold region. But some books interpreted subtreshold region will cause low current in hundreds of nanoampere. but mine is a bit high for subtreshold region. As for the circuit, the portion of my circuit that in this kind of situation is a simple current mirror. thanks
 

I agree.. its in weak inversion... subthreshold - yes... But its not OFF...

A really good question.. got lots of new scratches on my head... :D
 

Hello,

I was wondering why the current is so large.
Can you show your graph with currents on a logarithmic y-axis v.s
Vgs on a x-axis ?
Thanks.
 

I also agree that your transistor just working in the weak inversion region,maybe the dimension was too large.then 20uA current Ids is. By the way, you say that the ckt is just a simple current mirror,whether 20uA just mirrored the Iref current?
 

kenlino said:
I also agree that your transistor just working in the weak inversion region,maybe the dimension was too large.then 20uA current Ids is. By the way, you say that the ckt is just a simple current mirror,whether 20uA just mirrored the Iref current?
Click to expand...

yes the transistor size is large. you are also right about the 20µA current is just the mirror of Iref current. The simple current mirror in my circuit receive 20µA in the Iin branch. then the transistor M1 mirror the current to M2. The transistor M1 is the one that behave like i describe above which i don't know it's operating region.
 

So your transistor M1 works in weak inversion region,don't care! But I would recommend you to scale down the transistor M1 and M2. Made the transistor size large,did you mean that you care about the noise?
 

kenlino said:
So your transistor M1 works in weak inversion region,don't care! But I would recommend you to scale down the transistor M1 and M2. Made the transistor size large,did you mean that you care about the noise?
Click to expand...

thanks about the advice, but can you tell me why you suggested this? will there be a problem if the current mirror work in that region. Of course is it desired for current mirror to work in active region but i don't thing that mine can achieve active region cause there is not much voltage headroom.
 

I see,you have not enough voltage swing.then in the instance you just let it work in the weak inversion region,as I know there would be no problem in the application.
 

kenlino said:
I see,you have not enough voltage swing.then in the instance you just let it work in the weak inversion region,as I know there would be no problem in the application.
Click to expand...
what about variation of the mirror current wrt process variations?
 

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