noise circuit reduction

[prestonee]

I am seeking to better understand the reduction of circuitry to simplify noise calculations.

for example vnoise of resistor is sqrt(4ktr), and total noise of a resistor with a cap to ground is sqrt(kt/c) all good and dandy with the basics.( i hate relying on memorizing such answers since this can get you in trouble so i like to be able to hand check. in this case I do vin^2=4ktr, and vo^/vi^2= ((1/sc)/(r+1/sc))^2= 1/(scr+1)^2, to get vo^2=4ktr/(scr+1)^2 = 4ktr* 1/(1+j*2*pi*f*c*r)^2, take the abs to remove the i gets you to 4ktr* 1/(sqrt(1+4*pi^2*f^2*c^2*r^2))^2 which finally ends in 4ktr*1/(1+4*pi^2*f^2*c^2*r^2), integrate over freq and you get 4*k*t*r * pi/(2*2*pi*c*r) = k*t/c .

as you see solving a simple rc is a bit of work, but then if you add a resistor on the other side of the cap, rcr. it gets nasty quick(solving the system twice using superposition for the two res noise sources, and the integral is much nastier and hard to do). But I have seen where people take a rcr, and magically flip the bottom r to the top and solve as a rc with r =sum of two r's. and the circuits i try to solve are much more complex with multiple r's and c's and while simulation can give me a value I wish more to be able to see the relationship of noise sources to the overall noise sum. how can rc noise sources be reduced? if you have a res in series with a cap in series with a res to ground, how can you simplify this to a single r, where one noise source has a rc filter impact and the other source must travel through the c, effectively making one noise source low pass filtered and the 2nd noise source highpass filtered? someone in person spoke to me about in a switch cap circuit its the noise being sampled on the cap that matters, but converting the switch to its on res and solving as above i believe is equivalent. however when the noise is being transferred to the next circuit that same res impacts the noise still, so it must be included in both calculations, calculating the voltage noise stored on the cap, and while at it the non ideal switch on the other end of the cap is connected to ground making the rcr i was referring to, this r contributes as well because it effectively is randomizing an offset ontop of the ideal ground that the voltage is in respect to. then when the voltage on the cap is used in the next phase the resistor on the ground side of the cap is still impacting the voltage while the new switch in r for phase 2 is also being added.....so to me even a simple switch cap when you account for the nonideal switching r's and both phases it gets very complex to calculate one simple circuit...

please help me out of the black cave ...
-Preston

 

[crutschow]

You explanation is a little hard to follow. If you post a schematic you would be more likely to get a good response.

 

[chuckey]

A RCR circit can be turned into a RC circuit by using Thevenims formula. Where the new R is the parallel combination of the original two Rs, and the new output voltage is Vin/2 (in this case)
Frank

 

[prestonee]

Sorry I realize describing circuits is difficult with text, thought i could get away with it with a rcr. here is what I am referring to.


the inclusion of the 2nd r between the cap and ground is throwing me off. i know many just lump it into the 1st r with a simple sum, but i cant convince myself its equiv due to the superposition results of the 2 sources at out in phase 1.
Also when adding the 2nd R between the cap and ground, it makes the integral of vo^2...nasty ....am i wrong?

- - - Updated - - -

Sorry you are right on the two res with a cap between, i meant all in series with the cap located between them, i mis-described earlier.

 

[crutschow]

For a switched-cap circuit, the ON resistance of the MOSFETs is typically fairly low and so are the capacitance values, so the kT/C noise dominates. You can do a quick calculation of the two to see if that's true. If so, then for engineering purposes, you can ignore the contribution of the ON resistances and just use the kT/C noise in your calculations.

 

[prestonee]

as I attempted to explain in the 1st post, the kt/c is the result of the switch r and cap. in text books they typically refer switch cap kt/c to series resistors with a cap in the middle, in phase 1 the 2nd switch(r2) is open, so it is equal to kt/c. but in the case of charge transfer where they are all in series and there is always two switches shut, i do not believe kt/c is still accurate. my math suggest it is not accurate. I am stating this in the hopes you will tell me i am wrong because of point(a) and show me how I am wrong, because i do not think the 2nd switch's resistance is ignorable.
-Preston

ps- i recently saw a wally cartoon, of him describing his work ethic to the performance evaluation team as not being a morning person , and finally becoming efficient at 12, then after lunch he feels too tired from eating to do productive work, which explains why hes such a productive eater
(next scene showed the evaluator selling his company shares)

 

[crutschow]

I don't see that whether you have two switches or one affects the kT/C noise. That noise is independent of the circuit resistance. As I stated, if it is significantly larger than the transistor ON resistance thermal (kTR) noise, then the resistor noise can be ignored. You haven't said whether that's true or not.

 

[prestonee]

I partly agree. for a single R and single C making a rc filter( which is the case for a typical sc sample and hold passing a stored voltage), the R value doesnt matter for the total integrated noise but i disagree thats its because the thermal noise from the switch is so small, because the kt/c noise is the result of integrating the resistive noise over its bandwidth and as showing in the first post, the terms all cancel except the k t and c value.

however my difficulty is when you add another r on the back side of the cap( such as in the case of a sc integrator where there is always 2 r's, one on each side of the cap for the charge storing and the charge transfer phases. if i try to work out the the equation for integral(vout^2* transfer function, 0,inf) its nearly impossible for me to solve. and i dont see a lumping of the resistors in noise to an equiv rc circuit. so for me the question comes down to if you have something like in the schematic before, how do you simply this to something solvable but accurate? if say both resistors have similar values, then what i see is a voltage noise source for r1 having some kind of lowpass filtering on it, while the 2nd r seems to have a high pass filter on it since that noise must travel through the cap to reach the output node.

 

[crutschow]

If you can answer my question about the relative noise between the resistor and capacitor, that will help clarify the discussion. If you are strictly interested in a theoretical discussion about the math, then I can't help you there. I'm allergic to complex math formulas. :wink:

 

[chuckey]

One thing that seems to be missing is the source and terminating impedances. These must be included in any calculations in the noise figures. As the load impedances increases so does the noise. Integrate the bandwidth of the filter, then use this in the thermal noise calculations. At a first approximation for a single RC circuit, its gain is 1 up to the 3db point at F1, it then decreases at a rate of 6dB/octave, so the gain is .5 at 2F1, .25 at 4F1. . . (it never gets to 0!!, try a limit of 5%?) or the limit of the frequency response of succeeding stages.
Frank