Is output common mode voltage DC quantity?

[circuitking]

Hello,

I am confused whether to use DC analysis or AC analysis to prove the below statements.

Case 1: Im3,4 >Iss
If the current in M3,4, I(M3,4), is high compared with Iss then that means the Vocm is high. To make Iss equal to Im3,4 we increase the Vp of M3,4 such that IM3,4 reduces.
Case 2: Im3,4<Iss
If the current in M3,4, I(M3,4), is low compared with Iss then that means the Vocm is low. To make Iss equal to Im3,4 we decrease the Vp of M3,4 such that IM3,4 increases.

The above logc makes sense to me If I follow the small signal analysis. However, my dilemma is that output common mode voltage (Vocm) is a DC quantity, in that case we should use something like Vop or Vom=VDD-Im3,4*R (resistance of M3,4). If I analyse in this view, the above statements are just opposite. Can someone confirm. Thanks

From: Design of Analog CMOS Integrated Circuits, Second Edition, Behzad Razavi

 

[danadakk]

Vcm is any voltage that is CM to a circuit.

One of the most Vcm voltages we deal with is 50/60 Hz as it is
"everywhere", coupling into circuits magnetically or electro statically.

Clock noise another common Vcm.


Regards, Dana.

 

[deep_sea]

Hello,

I am confused whether to use DC analysis or AC analysis to prove the below statements.

Case 1: Im3,4 >Iss
If the current in M3,4, I(M3,4), is high compared with Iss then that means the Vocm is high. To make Iss equal to Im3,4 we increase the Vp of M3,4 such that IM3,4 reduces.
Case 2: Im3,4<Iss
If the current in M3,4, I(M3,4), is low compared with Iss then that means the Vocm is low. To make Iss equal to Im3,4 we decrease the Vp of M3,4 such that IM3,4 increases.

The above logc makes sense to me If I follow the small signal analysis. However, my dilemma is that output common mode voltage (Vocm) is a DC quantity, in that case we should use something like Vop or Vom=VDD-Im3,4*R (resistance of M3,4). If I analyse in this view, the above statements are just opposite. Can someone confirm. Thanks

View attachment 167624
From: Design of Analog CMOS Integrated Circuits, Second Edition, Behzad Razavi

If I understand your question right, you want to know how the AC simulation accounts for the DC variations of the VOCM. The AC analysis is based on small signal analysis "around your DC operating point". So if the output CM is pulled towards either VDD or ground, then you will have transistors leave the saturation and consequently a small gain.

 

[sutapanaki]

Hello,

I am confused whether to use DC analysis or AC analysis to prove the below statements.

Case 1: Im3,4 >Iss
If the current in M3,4, I(M3,4), is high compared with Iss then that means the Vocm is high. To make Iss equal to Im3,4 we increase the Vp of M3,4 such that IM3,4 reduces.
Case 2: Im3,4<Iss
If the current in M3,4, I(M3,4), is low compared with Iss then that means the Vocm is low. To make Iss equal to Im3,4 we decrease the Vp of M3,4 such that IM3,4 increases.

The above logc makes sense to me If I follow the small signal analysis. However, my dilemma is that output common mode voltage (Vocm) is a DC quantity, in that case we should use something like Vop or Vom=VDD-Im3,4*R (resistance of M3,4). If I analyse in this view, the above statements are just opposite. Can someone confirm. Thanks

You don't consider M3,4 as resistors, because they are not. They are current sources. In very simplified terms your common-mode voltage is VDD-Vgs3,4.
If your output ac signals Vop and Vom are perfectly differential, then the common-mode voltage is pretty much DC. If, say, one side of vop or vom clips for example, then the common-mode voltage will vary, because outputs are not balanced anymore.
When you access the stability of the common-mode feedback loop then you apply ac analysis because you are interested in the small variations and look if they go crazy i.e. unstable.

 

[circuitking]

You don't consider M3,4 as resistors, because they are not. They are current sources. In very simplified terms your common-mode voltage is VDD-Vgs3,4.

To put my question differently "Why Vocm is high, If the current in M3,4, I(M3,4), is high compared with Iss"

 

[FvM]

You find the CMFB operation point by DC anaylsis.

 

[sutapanaki]

To put my question differently "Why Vocm is high, If the current in M3,4, I(M3,4), is high compared with Iss"

The circuit that you have attached doesn't suppose that the current in M3,4 is controlled differently than Iss. For common-mode transistors M3 and M4 are diode connected and thus the common-mode is defined by their Vgs when a current of Iss/2 goes through them. For differential mode those transistors act as current sources with high output resistance.