Reply by June 30, 20032003-06-30
Hi

Thank you for the helpfull answers

Jan


Reply by Peter Brackett June 27, 20032003-06-27
Jan:

For input referred noise density calculations the straightforward approach
is always to work from first principles and find/calculate the "equivalent"
source resistance Req.  The equivalent source resistance is the total source
resistance in series with the Op Amp inputs.

In the particulare case of your circuit this would be calculated as the
parallel connection of R1 and R2 in the non-inverting lead and the parallel
connection of R3 and R4 in the inverting lead.  Adding these two together
gives the equivalent single value of the total source resistance seen by the
amplifier inputs i.e.  Req = R1||R2 + R3||R4.

Then if en is the input referred Op Amp noise spot density is en expressed
in V/rtHz and in is the input referred Op Amp current spot density expressed
in A/rtHz, then the total input referred spot noise density will be:

vn = sqrt[ en^2 + in^2*Req + 4*k*T*Req]

There are several practical application things to watch out for with
resistor noise...

"In theory theory and practice are the same but in practice they are
different."  -- Yogi Berra.

Only *purely metallic* conductors exhibit the so-called Johnson noise
[quantum/thermal noise] calculated by the Nyquist celebrated formula i.e.
conductors in which the current is carried by free electrons above the
valence band and not bound by other forces in the conductor material.
Essentially only metals exhibit this kind of pure noise.  All other kinds of
conductors exhibit this quantum noise *plus* one or more other, usually
larger, noises which can only be determined empirically.  The celebrated
Nyquist formula for the Johnson [quantum thermal noise] is simply:

vrn^2 = k*T*R*B

[Nyquist's formula for the quantum/thermal "Johnson" noise power of a
resistance R in Volts squared at temperature T degrees Kelvin in bandwidth B
Hertz.  k is Boltzman's constant.]

Most commonly available resistors are *not* metallic conductors and usually
exhibit *considerably* more noise than that indicated by Nyquist's formula.
Thus you will be surprised by any measured results in low noise situations
if you are depending upon the Nyquist formula to predict your noise floor.
Your noise floor will be *way* off your Nyquist formula predictions by lot's
of dB's *unless* you are using metallic resistors!

In most resistors there are two other noises which are far larger than the
Johnson noise.  There is the generally larger so-called "shot noise" which
is proportional to the current through the resistor and which unlike the
white [flat] Johnson noise actually gets larger below a certain corner
frequency, i.e. it's a 1/f noise effect, plus there is also the larger
so-called current noise which is proportional to the voltage applied and is
usually rated in uVoltsrms/Volt which also rises below a corner frequency,
i.e. also a 1/f effect!

These latter two noises are not well characterized and as far as I know do
not have a simple physics based formula to predict their values, instead one
must rely on empirical measurements for these latter two 1/f noises.

So for most practical resistors which you purchase or have available in your
lab [e.g. carbon composition, carbon film, metal film, etc... ] there are
the three noises to consider, two of which are generally larger than the
Johnson noise and which also grow as 1/f below corner frequencies.

Metal film resistors are the most quiet of the three common resistor
technologies mentioned above, but... if you want really quiet resistors you
must use/buy bulk metal resistors!  Only bulk metal either of the wire wound
or metal foil type] resistors will exhibit the pure Johnson noise predicted
by the Nyquist formula!

Bulk metal resistors [Either metal foil or wire wound resistors] are
available but are more expensive than the common garden variety resistors.

See for instance the web site of Vishay which is a large passive component
manufacturer.  Vishay manufactures all  kinds of resistors, carbon comp,
carbon film, metal film, [in both thick  film and thin film styles] as well
as bulk metal resistors in both wire wound and metal foil formats.

If you are trying  to work down near the quantum noise levels without using
bulk metal resistors you will be sorely disappointed and will get noise
floor results far above that predicted by Nyquist's formula, unless...

you are using *bulk metal* resistors!

"In theory theory and practice are the same but in practice they are
different."  -- Yogi Berra.

--
Peter
Consultant
Indialantic By-the-Sea, FL


"Jan R&#4294967295;rg&#4294967295;rd Hansen" <jrh(IngenSpam)@(NoSpamTak)person.dk> wrote in message
news:bdfr20$1s6a$1@news.cybercity.dk...
> Hi > > Its been a while since last I did calculations on analog circuits, maybe
you
> can help with the following: > Consider the non-inverting amplifier on the drawing below (hope you can
see
> it). > > When calculating the resistors' thermal noise contribution to the total > noise voltage, En_total, I have seen application notes calculate each > resistors thermal noise and adding them (in the sum of squares way), like: > > En_thermal = SQRT[ 4kTdf( R1 + R2 + R3 + R4) ], k:Boltzmann's constant, > T:temperature, df:bandwidth > > Others calculate it by using the parallel resistances, R1||R2 and R3||R4: > En_thermal = SQRT[ 4kTdf( R1||R2 + R3||R4) ] > > Which method is correct? > > | \ > o-R1-------------| + \______out > | |---| - / | > R2 | | / | > | | | > GND |--R3------| > | > R4 > | > | > GND > > > If you feel you are moving to far from the digital world - feel free to > answer '01' for the first method, '10' for the second, '00': I got it all > wrong. > > Best regards > Jan > >
Reply by Jerry Avins June 27, 20032003-06-27
"Jan R&#4294967295;rg&#4294967295;rd Hansen" wrote:
> > Hi > > Its been a while since last I did calculations on analog circuits, maybe you > can help with the following: > Consider the non-inverting amplifier on the drawing below (hope you can see > it). > > When calculating the resistors' thermal noise contribution to the total > noise voltage, En_total, I have seen application notes calculate each > resistors thermal noise and adding them (in the sum of squares way), like: > > En_thermal = SQRT[ 4kTdf( R1 + R2 + R3 + R4) ], k:Boltzmann's constant, > T:temperature, df:bandwidth > > Others calculate it by using the parallel resistances, R1||R2 and R3||R4: > En_thermal = SQRT[ 4kTdf( R1||R2 + R3||R4) ] > > Which method is correct? > > | \ > o-R1-------------| + \______out > | |---| - / | > R2 | | / | > | | | > GND |--R3------| > | > R4 > | > | > GND > > If you feel you are moving to far from the digital world - feel free to > answer '01' for the first method, '10' for the second, '00': I got it all > wrong. > > Best regards > Jan
Usenet requires a fixed-width font and no tabs. It's the only standard there is. The Johnson noise power from a resistive source depends on the equivalent resistance (and Bolzmann's constant and the absolute temperature too, but you know what I mean), whether that be a series connection, a parallel connection, or some combination. Noise power from individual sources adds directly. If you keep the math straight, you will see that applying either basis leads to the same result. Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Reply by Al Clark June 27, 20032003-06-27
"Jan R&#4294967295;rg&#4294967295;rd Hansen" <jrh(IngenSpam)@(NoSpamTak)person.dk> wrote in
news:bdfr20$1s6a$1@news.cybercity.dk: 

> Hi > > Its been a while since last I did calculations on analog circuits, > maybe you can help with the following: > Consider the non-inverting amplifier on the drawing below (hope you > can see it). > > When calculating the resistors' thermal noise contribution to the > total noise voltage, En_total, I have seen application notes calculate > each resistors thermal noise and adding them (in the sum of squares > way), like: > > En_thermal = SQRT[ 4kTdf( R1 + R2 + R3 + R4) ], k:Boltzmann's > constant, T:temperature, df:bandwidth > > Others calculate it by using the parallel resistances, R1||R2 and > R3||R4: En_thermal = SQRT[ 4kTdf( R1||R2 + R3||R4) ] > > Which method is correct? > > | \ > o-R1-------------| + \______out > | |---| - / | > R2 | | / | > | | | > GND |--R3------| > | > R4 > | > | > GND > > > If you feel you are moving to far from the digital world - feel free > to answer '01' for the first method, '10' for the second, '00': I got > it all wrong. > > Best regards > Jan > >
The thermal noise of the Rs will be the second equation. -- Al Clark Danville Signal Processing, Inc. -------------------------------------------------------------------- Purveyors of Fine DSP Hardware and other Cool Stuff Available at http://www.danvillesignal.com
Reply by Paavo Jumppanen June 26, 20032003-06-26
"Jan R&#4294967295;rg&#4294967295;rd Hansen" <jrh(IngenSpam)@(NoSpamTak)person.dk> wrote in message news:<bdfr20$1s6a$1@news.cybercity.dk>...
> Hi > > Its been a while since last I did calculations on analog circuits, maybe you > can help with the following: > Consider the non-inverting amplifier on the drawing below (hope you can see > it). > > When calculating the resistors' thermal noise contribution to the total > noise voltage, En_total, I have seen application notes calculate each > resistors thermal noise and adding them (in the sum of squares way), like: > > En_thermal = SQRT[ 4kTdf( R1 + R2 + R3 + R4) ], k:Boltzmann's constant, > T:temperature, df:bandwidth > > Others calculate it by using the parallel resistances, R1||R2 and R3||R4: > En_thermal = SQRT[ 4kTdf( R1||R2 + R3||R4) ] > > Which method is correct? > > | \ > o-R1-------------| + \______out > | |---| - / | > R2 | | / | > | | | > GND |--R3------| > | > R4 > | > | > GND > > > If you feel you are moving to far from the digital world - feel free to > answer '01' for the first method, '10' for the second, '00': I got it all > wrong. > > Best regards > Jan
I can't seem to make out your schematic! Try using a monospaced font next time. But in any case the equivalent noise is just the equivalent resistance applied to the equation. En_thermal = SQRT[ 4kTdfR) ] Regards, Paavo Jumppanen
Reply by June 26, 20032003-06-26
Hi

Its been a while since last I did calculations on analog circuits, maybe you
can help with the following:
Consider the non-inverting amplifier on the drawing below (hope you can see
it).

When calculating the resistors' thermal noise contribution to the total
noise voltage, En_total, I have seen application notes calculate each
resistors thermal noise and adding them (in the sum of squares way), like:

En_thermal = SQRT[ 4kTdf( R1 + R2 + R3 + R4) ], k:Boltzmann's constant,
T:temperature, df:bandwidth

Others calculate it by using the parallel resistances, R1||R2 and R3||R4:
En_thermal = SQRT[ 4kTdf( R1||R2 + R3||R4) ]

Which method is correct?

                            | \
o-R1-------------| + \______out
              |        |---| - /      |
            R2      |     | /        |
              |        |                |
          GND    |--R3------|
                      |
                     R4
                      |
                      |
                   GND


If you feel you are moving to far from the digital world - feel free to
answer '01' for the first method, '10' for the second, '00': I got it all
wrong.

Best regards
Jan