a basic question about cascode amplifer

[nibo_mmx]

I have a simple question.

in the following ckt, when Vin is smaller than Vth1, M1, M2 will all be off. Vx will equal to Vb-Vth2 if neglect the subthreshold effect.

I cannot understand why Vx=Vb-Vth2.

Thanks for your explanation

 

[khouly]

if we nglect the body effect ,
the transistor cannot be on , so if u assumed that it is on the edge , so VGS , must be lower or equal Vth
so vb - vx = at least vth
so vx ≤ vb -vth

khouly

 

[nibo_mmx]

The book says Vx≈Vb-Vth2. Since Vx≤Vb-Vth2

Then why Vx cannot equal to a value much smaller than Vb-Vth2, such as 0 or 0.1mV? Why should be approximatly Vb-Vth2?(i.e. why must be at the edge of "on"?)

Thanks

James

 

[VVV]

That is because a sort of negative fedback takes place.

Assume for a moment that Vx was 0V, as you suggested. That would mean that M2 turns on, since its G-S voltage is Vb, which is greater than Vth2.
But if M2 turned on, then current would tend to flow through M1. Since M1 is off, its D-S resistance is very high (much higher perhaps than even the load resistor) and does not require significant current to allow Vx to rise and thus Vx rises to the point where M2 is just about at the threshold.
Why? Because if Vx rose higher, then the G-S voltage of M2 would be less than Vth2 and M2 would turn off, thus Vx would drop. But if Vx drops, then M2 turns on, forcing Vx to rise... and so on. And so Vx establishes itself at Vx≈Vb-Vth2.
It is approximately equal to Vb-Vth since when M1 is off, its resistance is not really infinite, thus some current is required to flow through it, and so the G-S voltage of M2 is slightly higher than Vth2.

Thus, M2 will always be at the threshold of turning on, if M1 is off.

If you replace M1 wih a resistance of relatively low value, then this requires some significant current through it to allow Vx to rise. To obtain that drain current through M2 would then require a G-S voltage greater than the threshold.
Then, Vx<Vb-Vth2.

 

[nibo_mmx]

I can undersand why Vx cannot be smaller than Vb-Vth2. But I am not sure I can fully undersand why Vx cannot be greater than Vb-Vth2. The following is my undersanding. Could you please see if I am correct? Thank you in advance:

When Vx is greater than Vb-Vth2. M2 truns off and the sum resistance of M1 and M2 is very high. That forces Vx drops to a rather low voltage, near ground. So, M2 cannot be off.

So since it cannot be on, also cannot be off. It should be at the edge of on and off but provides very little current to asure Vx wont be too high and wont be too low. Because if there is no current through M2 and M1, that will force Vx equals to ground. Since it has a little current, so Vgs2 should be a little bit higher than Vth2?

Is my understanding right?

 

[ambreesh]

Dear nibo_mmx,
In this discussion we are assuming no subthreshold conduction.
Vin<Vth1 so M1 is off.
Why should Vx= Vb-Vth2 ?
You have a capacitor at node X and this gets charged, this cap is the source to bulk cap of M2 and drain to bulk cap of M1.

And could u please elaborate your point about "When Vx is greater than Vb-Vth2. M2 truns off and the sum resistance of M1 and M2 is very high. That forces Vx drops to a rather low voltage, near ground. So, M2 cannot be off. "

 

[nibo_mmx]

Dear ambreesh,
many thanks for your idea, but I still have a question.

In your post:
"You have a capacitor at node X and this gets charged"

You mean when the capacitor is charged to Vb-Vth2, then M2 is off and there is no current to charge the capacitor anymore, so Vx will approximately equals to Vb-Vth2?

But you have an assumption here that initially Vx is below than Vb-Vth2 and M2 is on. How would it work if initially Vx is higher than Vb-Vth2, say V1? In that state, M2 and M1 are all off, and the charge at the capacitor has no way to follow, so Vx will be V1 all the time? If not, why?

 

[VVV]

We talk about Vx with respect to GND. When Vx is low, the source of M2 is almost at GND potential, VgsM2 is high (equal to Vb) and M2 turns on. As it turns on, Vx increases.
But if Vx increases, then VgsM2 decreases and at some point M2 would turn off. That point is Vth2.
M2 cannot really turn off, since then Vx would become low and that would turn on M2 (see first statement).
So M2 will be at the threshold.

 

[nibo_mmx]

VVV,
Thank you very much, I understand your explanation now.
Do you agree with ambreesh's point?

James

 

[ambreesh]

We talk about Vx with respect to GND. When Vx is low, the source of M2 is almost at GND potential, VgsM2 is high (equal to Vb) and M2 turns on. As it turns on, Vx increases.
But if Vx increases, then VgsM2 decreases and at some point M2 would turn off. That point is Vth2.
M2 cannot really turn off, since then Vx would become low and that would turn on M2 (see first statement).
So M2 will be at the threshold.

Dear VVV,
Correct me if i donot understand you. M2 turns off when VgsM2=Vth2. Or are u saying that M2 turns off when Vx=Vth2. If u are saying that M2 stops to conduct when Vx=Vth2, I beg to disagree with you.As this would only be decided by VgM2.
Now why should Vx once charged to (VgM2-Vth2), discharge. We have assumed that no lekage in form of subthreshold etc is happening. If we are not assuming this, then I agree that Vx would go down, but again we would also have subthreshold conduction through M2 so depending on sizes and various lekage currents equilibrium would be reached a dynamic equilibrium.

I hope I am clear

 

[jordan76]

Hi ambreesh,

Basically you and VVV are both right,but maybe you had some misunderstandings about VVV's statement:
But if Vx increases, then VgsM2 decreases and at some point M2 would turn off. That point is Vth2.

That point means the point where VgsM2 equals Vth2 instead of Vx equals Vth2.

Hope it helps you reach agreement on this question. Thanks.

regards,
jordan76

 

[nibo_mmx]

Dear ambreesh,
Thanks for your post. Your disscusion make the question clearer and clearer. As I asked in the former post:

How would it work if Vx initiallly is higher than Vb-Vth2, say V1? In that state, M2 and M1 are all off, and the charge at the capacitor has no way to follow, so Vx will be V1 all the time? If not, why?

 

[jordan76]

Hi nibo_mmx,

The analyses above have one assumption:no substhreshold current considered.
In fact, life is not perfect: substhreshold current always exists. So even if Vx initiallly is higher than Vb-Vth2(by any means),it will leak away to finally reach an equalibrium point depending on M1,M2 sizes.

Hope it answers your question.

regards,
jordan76

 

[nibo_mmx]

Thank you all, I understand it now.

 

[wanily1983]

I can undersand why Vx cannot be smaller than Vb-Vth2. But I am not sure I can fully undersand why Vx cannot be greater than Vb-Vth2. The following is my undersanding. Could you please see if I am correct? Thank you in advance:

When Vx is greater than Vb-Vth2. M2 truns off and the sum resistance of M1 and M2 is very high. That forces Vx drops to a rather low voltage, near ground.
quote]

is it right? why "That forces Vx drops to a rather low voltage, near ground.", i think in this condition it is a voltage divider.

 

[xdyrick]

VVV'S explaination is quite clear, but ......

 

[allan_guo]

From my point,the key point is subthreld current.then I want to rise a new question:in this circuit ,where is the most significant leakege path?parastic npn/pnp of the mos transitor?

 

[sabu]

hello friends,
i thnk for the above qustion, we can generalise the ans as " when power is swiched on m1 is off (as per question) then node Vx is starts charging (if Vb is > Vth) it can rise up to "Vb-Vth" after that M2 must be in OFF.And also Vx cannot discharge since M1 is in OFF state, ok.if iam wrong plz correct it.
regards
sabu