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bharat

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Question on small signal and Poles
Mar 12^th, 2007, 1:57am

I have following two questions:

In the small signal analysis, voltage sources are short ckted and current sources are open ckted. In conventional differential amplifier, once the bottom ideal current source is open ckted, why do I call source in NMOS and drain in PMOS of differntial pair as AC ground?
Is it because those nodes are not supposed to be changed?

The pole frequency is a frequency which makes the transfer function infinity. Considering, Bode's plot of single pole first order system, at pole frequency, the gain doesn't look like infinity.

Please explain.

Thanks
-Bharat

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monte78

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Re: Question on small signal and Poles
Reply #1 - Mar 12^th, 2007, 2:30am

In conventional differential amplifiers you can call the current source node as an AC ground because of the differential structure. If you have an increase current on one of the two transistors of the differential pair, in the other transistor you have a consequently reduction of current, so the current source node doesn't change its potential. This analysis doesn't take into account parasitics between the current source node and ground.
The pole frequency is a frequency which makes the transfer function infinity in the complex variable s. The bode plot represent the transfer function in the variable w. So if you consider w=1/RC this value doesn't make the transfer function have an infinity value.

Bye!

Monte

bharat wrote on Mar 12^th, 2007, 1:57am:

I have following two questions:

In the small signal analysis, voltage sources are short ckted and current sources are open ckted. In conventional differential amplifier, once the bottom ideal current source is open ckted, why do I call source in NMOS and drain in PMOS of differntial pair as AC ground?
Is it because those nodes are not supposed to be changed?

The pole frequency is a frequency which makes the transfer function infinity. Considering, Bode's plot of single pole first order system, at pole frequency, the gain doesn't look like infinity.

Please explain.

Thanks
-Bharat

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adesign

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Re: Question on small signal and Poles
Reply #2 - Mar 12^th, 2007, 2:35am

bharat wrote on Mar 12^th, 2007, 1:57am:

I have following two questions:

In the small signal analysis, voltage sources are short ckted and current sources are open ckted. In conventional differential amplifier, once the bottom ideal current source is open ckted, why do I call source in NMOS and drain in PMOS of differntial pair as AC ground?
Is it because those nodes are not supposed to be changed?

The pole frequency is a frequency which makes the transfer function infinity. Considering, Bode's plot of single pole first order system, at pole frequency, the gain doesn't look like infinity.

For the first part:
Yes at small signal analysis level, these voltages are not supposed to change and thus it is AC ground.

Second part:
Please refer the discussion at
http://www.designers-guide.org/Forum/YaBB.pl?num=1169098704

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avlsi

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singapore

Re: Question on small signal and Poles
Reply #3 - Mar 26^th, 2007, 1:30am

Hi,

Even though it a current source, we model it as a norton equivalent circuit. In differential mode, the source of the input pair doesnot change due to symmetrical behaviour of the two branches.

If current in one branch increases by delta I, current in other branch decreases by delta I. so, effectively, the voltage at the source node of the differential pair doesnot increase or decrease. This means that there is no AC voltage developed there, so we make it AC ground.

Please correct me if I am wrong.

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