[Moved]: a question for dc psrr for a 2 stage circuit ??

Status
Not open for further replies.
C

cmos_ajay

Full Member level 2
Joined
Jan 30, 2007
Messages
126
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,296
Visit site
Activity points
2,247
What is the DC PSRR (power supply rejection) for this attached circuit ?
 

Attachments

  • dc psrr.JPG
    74.5 KB · Views: 64

In this configuration the PSRR will be terribly bad, even > 0dB *), as the interference source lies inside the input control loop.

*) if PSRR is defined like gain: PSRR = 20log(v_out_i / v_ps_i) ; i = interference or noise
 
Last edited:

Hello Erikl,

I was only looking for an equation for PSRR in terms of R1, R2, gm1 etc....
Basically an expression ?
 

I'd think in this configuration (power supply interference source in the input control loop),
PSRR is the same as gain (no rejection at all), i.e. - in first order approximation, neglecting the output impedances of the transistors -
PSRR = gain ≈ gm1*R1 * gm2*R2

If you move the power supply interference source outside of the input control loop, PSRR ≈ gm2*R2 ,
- s. the right side of the PDF: View attachment PSRR.pdf
 

I can derieve it as follows ->
psrr = 20*log(vdd/vout) and [1/psrr] = 20*log(vout/vdd)

id1 = gm1*vsg1 = gm1(vs1-vg1) = gm1(vdd - 0) = gm1*vdd

id2 = gm2*vsg2 = gm2(vs2-vg2) = gm2[vdd - (id1*R1)]

But, id1 = gm1*vdd as found earlier

and so id2 = gm2[vdd - (gm1*vdd*R1)]

id2 = gm2[vdd{1 - gm1R1}] = gm2*gm1*R1*vdd ....approximately

vout = id2*R2 = [gm2*gm1*R1*vdd]*R2

Vout/vdd = gm2*gm1*R1*R2

Any comments on this ??
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…