[SOLVED] LDO Output resistance.

[saqib.shah06]

Hi everyone,
I have a question regarding the output impedance of a LDO, which I am trying to simulate in HSPICE.

It is a basic design with an error amplifier and a PMOS pass transistor. I am having the following doubt.

For any given LDO, the dominant pole lies at P= 1/RC (R = rds of the pass transistor and C=output cap ). My HPSICE simualtions confirm this. I am injecting the AC signal by breaking the feedback loop (this ahs been perfectly design - i have checked it multiple times). However when I do a .tf analysis in HSPICE, the .lis file says that my output impedance is 100milliohm. (the rds of the pass transistor is around 75 ohm). Can someone please explain this discrepancy?

btw I am using very large feedback resistors (100k each), and a current source load. The output current is around 100mA

Thanks a lot!

 

[erikl]

... when I do a .tf analysis in HSPICE, the .lis file says that my output impedance is 100milliohm. (the rds of the pass transistor is around 75 ohm). Can someone please explain this discrepancy?

rds is a static (DC) parameter, the output impedance is a dynamic (small signal) parameter. Due to the LDO regulation mechanism with nearly full feedback, i.e. gain≈1, the (low frequency) output impedance -- in a 1^st (coarse) approximation -- is rds/A_ol -- A_ol being the open loop gain of the LDO. So your .tf analysis result seems quite correct.

 

[saqib.shah06]

erikl , thanks for you reply. but I still have one query. When I simulate the LDO, I am breaking the loop gain by inserting a large (1meg) inductor and using a cap to inject the input signal, effectively making turning it into an open loop - in such a case the outpur R shouldnt be affected by the loop gain, should it?

rds is a static (DC) parameter, the output impedance is a dynamic (small signal) parameter. Due to the LDO regulation mechanism with nearly full feedback, i.e. gain≈1, the (low frequency) output impedance -- in a 1^st (coarse) approximation -- is rds/A_ol -- A_ol being the open loop gain of the LDO. So your .tf analysis result seems quite correct.

 

[erikl]

I am breaking the loop gain by inserting a large (1meg) inductor and using a cap to inject the input signal, effectively making turning it into an open loop - in such a case the outpur R shouldnt be affected by the loop gain, should it?

Also with 1MH in the feedback path you still have a DC gain of ≈1 , isn't it? At higher frequencies the output impedance will increase of course, due to decreasing A_ol and feedBack.

 

[saqib.shah06]

Also with 1MH in the feedback path you still have a DC gain of ≈1 , isn't it? At higher frequencies the output impedance will increase of course, due to decreasing A_ol and feedBack.

@erikl - perfect! I talked to my professor today and he gave the exact same explanation, I understand now :smile:. Thanks a lot! The whole .tf thing being about DC caused the confusion.