why Vgs is less than Vth here in my case

[prcken]

i am just wondering why this diff-pair of the vgs of the input devices is less than vth, i think the voltage at the source must be equal to vg-vth.


my opamp works good in transient, how could this happen? it's not reasonable

 

[erikl]

... the vgs of the input devices is less than vth, i think the voltage at the source must be equal to vg-vth.
... how could this happen? it's not reasonable

Sure it is: your input diff pair (with Veff=vgs-vth = -80mV) is operating in weak inversion mode, which is quite normal for MOSFETs with low Ids and high W/L ratios (input stages). See the following figure which is valid for a 0.18µm process:

 

[xplorin]

With technology scaling and shrinking device sizes, subthreshold condition is becoming an efficient operating region where device functions. It is of importance in analog circuits while in digital circuits it is viewed as a parasitic leakage with no current flowing. It is perfectly okay for a transistor to work in weak inversion region as long as your functionality is correct

 

[prcken]

Sure it is: your input diff pair (with Veff=vgs-vth = -80mV) is operating in weak inversion mode, which is quite normal for MOSFETs with low Ids and high W/L ratios (input stages).

Hmmmm, quite interesting.
i didn't notice when i designed at early stage until i checked noise, then i found this issue in my circuit.
my W/L=160µ/0.5µ in a 130nm process. i only checked vds>vdsat at early stage, i thought this diff-pair was in saturation region, while it is not.
I found in this inversion region, gm can be saturated, because i tried to increase W to get more gm, but seems less increase when i increase the W more.
does that mean if i want to increase gm, i have to increase current.

 

[erikl]

... does that mean if i want to increase gm, i have to increase current.

Yes. gm/Id gets its maximum in weak inversion = subthreshold region, however.

 

[prcken]

Yes. gm/Id gets its maximum in weak inversion = subthreshold region, however.

Thank you.
Maybe that's part of the reason why the sampling clock of my switched-cap opamp can't run at 10MHz even the small signal unity frequency looks good(>100MHz).

 

[erikl]

Maybe that's part of the reason why the sampling clock of my switched-cap opamp can't run at 10MHz even the small signal unity frequency looks good(>100MHz).

Try and grant it some more current.

i thought this diff-pair was in saturation region, while it is not.

To get enough gm from the input stage, you have to shift it into saturation region by decreasing its Vgs.

 

[prcken]

Try and grant it some more current.

To get enough gm from the input stage, you have to shift it into saturation region by decreasing its Vgs.

i can't control overdrive or Vgs directly. what i can control directly only tail current and W/L of the device.

if i increase current I, fix W/L. by gm=√2µCoxWI/L, gm increases. but then from gm=µCoxW(Vgs-Vth)/L, my Vgs-Vth is increaes too since W/L is fixed. why?

 

[erikl]

if i increase current I ... my Vgs-Vth is increaes too since W/L is fixed. why?

Just study the output characteristic (Ids vs. Vds, Vgs as parameter): Increased Ids needs larger Vgs (for comparable Vds).

 

[prcken]

yes, but you said by decreasing vgs to get gm in saturation in the reply above, so i am confused

 

[erikl]

yes, but you said by decreasing vgs to get gm in saturation in the reply above, so i am confused

Kehan, this was meant for constant Ids. Again, use the a.m. output characteristic: draw a horizontal line (= constant Ids) to get out from the linear region (above you said

i thought this diff-pair was in saturation region, while it is not.

... so I think you wanted to get out of the linear into the saturation region, which I also recommended to increase gm.

Now if you follow the horizontal constant Ids line to the right, from the linear into the saturation region - and, by this, to increasing Vds - you'll see that you need lower Vgs, isn't it? See these example curves:

 

[prcken]

Yes, I got it, that should be Ids fixed. moving to the right to decrease vgs in that picture is another intuitive way to understand gm=2βIds/(vgs-Vth)
sometimes, choose which of those three gm equations is confusing.

 

[nikhadelokesh]

my W/L=160µ/0.5µ in a 130nm process.

How it can be 130nm process with L=0.5µ?

 

[prcken]

How it can be 130nm process with L=0.5µ?

why not? 0.5µm>130nm, correct?