regarding psrr analysis of error amplifer

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ashvani88

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hello friends
i am using pmos input two stage error amplifier (0.18um technology). i am using tanner tool can any one tell how to perform psrr analysis , currently i am following these syntax if i am wrong tell me i am not getiing any sensible value of psrr by this analysis. syntax are as follows...

vSource_v_ac_2 N_4 in 0 AC 1.0 90.0>>>>.syntax for ac source connected to supply voltage Vin for dc analysis supply voltage have max value of 1.8 volt
Vin in Gnd dc 1.8>>>supply voltage
Vref ref gnd dc 0.67>>>reference voltage
.ac DEC 5 1 10000MEG
.print V(in) V(out)
.print vdb(out) vp(out)>>output port of error amplifer
.op
.end
 

Can you give the schematic? I am assuming that you have an ac source of 0V DC sitting on top of the DC source of 1.8V. Does Tanner Tools have single source to give both together?

From what I can understand, you have Vin between the node 'in' and 'Gnd' and vsource_c_ac_2 between 'N_4' and 'in'. In this case if 'in' is at dc = 1.8V with ac=0 and N_4 has ac = 1 with dc = 1.8V.

Which is the supply for the amp?
Is it 'in' or is it 'N_4'?

Is the plot of the output would be something like -400dB throughout the frequency range?
If it is so then you have connected the 'in' node to the supply which has no ac component and would give absurd results like -400dB.
 

yes it is....i will post soon with schematic and more detial
- - - Updated - - -

i connect an ac source along with dc of source equal to Vdd=1.8v.
Vin is node corresponding to differential structure where reference voltage and feedback signal will be applied
N-4 did not signify NMOS its correspond to an internal node errorrr.png
 

Hi,
The Gate of the PMOS of the second stage is connected to ground effectively turning it off.

For a PSRR analysis, keep the inputs of the amp at ac=0. You want to see the effect of a signal at supply only for PSRR analysis, so an input at the 'in' node should not be there. Put the fixed bias voltage there (usually Vdd/2 or what you have designed it for).
At the supply put 1.8V DC with an AC = 1 source in series. This would effectively put 1.8VDC + 1V AC at the supply for analysis.
Then run the simulation.
 

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