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AC performance of LDO

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tia_design

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Hi,

For all those LDO papers, they all give figures showing the AC performance with pole location. I don't understand this figure. To get this figure, I have to do AC simulation. But in AC simulation, what is the input and what is the output so that I can get this curve? Thanks
 

You must break the loop at some point, inject a AC signal at the right and check the output at the left.
 

and you must add some caps and res between this two cutted points and vcom for dc option point work.
 

For example, you can break the loop, then insert a transformer, so DC option point is built and AC signal can be coupled from primary side to second side by transformer.
 

It is better to use STB analysis.
 

Yes, but with stb analysis you can not see the AC signals in every node, while with AC you can. You just have to open the loop and put a dummy cap when open with a value as close to reality as possible
 

break the loop anywhere in the LDO. apply test signal and observe the output.Vo/Vi will give the loop gain and phase. from the u can find PM,UGB.
 

WTF ?

DO NOT BREAK THE LOOP ANYWHERE !!!
THAT'S BRUTE FORCE

If u r using typical topologies (error amp, pass transitor, feedback net), u must break the loop in the feedback path (which connects with the gate of one of the differential transistors), and introduce a large inductor in series (e.g. 10 H). Then, u should place ur ac signal with a huge decoupling capacitor (e.g. 10F) in series, an introduce it into the feedback gate. then run ur ac simulation. Let me now if u need further assistance.

ez, don't u think ?
 

break the loop after where you apply the feedback to the Error amplifier. You can do stability analysis using the component "iprobe". you dint hav to put big cap or inductor like before. this works very fine and gives you the loop gain and phase reqd for the stability analysis of the LDO
 

why the use of the inductor and the capacitor?
why don't simply break the loop and apply a AC source betwen the positive input of the ampli and the the feedback?
it is easier i think
 
imar said:
why the use of the inductor and the capacitor?
why don't simply break the loop and apply a AC source betwen the positive input of the ampli and the the feedback?
it is easier i think
Click to expand...

YES that´s true !
If you can find a node within the loop which can be opened with only a minor and neglectable change in load conditions (very small output impedance and/or very high input impedance) this simple method works very well.

To tia_design:

For you it is not only important to know how to perform an ac simulation of the loop gain.
More important is to know why you are doing this and how to evaluate the results !
 

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