thermal noise in resistors - Baffled !

[frank]

Hello,

I am puzzled by the following equation:
Thermal noise in a resistor is as follows:

Vrms of noise = sqrt(4*k*T*R)*sqrt(B)

How is bandwith (B) supposed to be interpreted ?
Is bandwith suppose to be evaluated from the RC low pass filter that the
resistor forms with its own parasitic capacitance ?


-Frank

 

[Jim Thompson]

On 19 May 2004 17:23:32 -0700, frank8017@excite.com (frank) wrote:

Hello,

I am puzzled by the following equation:
Thermal noise in a resistor is as follows:

Vrms of noise = sqrt(4*k*T*R)*sqrt(B)

How is bandwith (B) supposed to be interpreted ?
Is bandwith suppose to be evaluated from the RC low pass filter that the
resistor forms with its own parasitic capacitance ?


-Frank

Look up "noise bandwidth", it's different from the actual 3dB
bandwidth... I can't remember the ratio right off the top of my head.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Reg Edwards]

I am puzzled by the following equation:
Thermal noise in a resistor is as follows:

Vrms of noise = sqrt(4*k*T*R)*sqrt(B)

How is bandwith (B) supposed to be interpreted ?

===============================

Can't see what your problem is. Bandwidth is simply a bandwidth. A number
of Hz. The band edges are not specified. B is the band between the band
edges.

The equation specifies the number of V volts of noise developed across a
resistor of R ohms, at a temperature T in ANY uniform bandwidth of B Hz.

Incidentally, V cannot be measured precisely because there is no such thing
as a bandpass filter with perfectly steep sides. But who cares? The 3dB,
1/2-power points, are good enough.
----
Reg.

 

[qrk]

On 19 May 2004 17:23:32 -0700, frank8017@excite.com (frank) wrote:

Hello,

I am puzzled by the following equation:
Thermal noise in a resistor is as follows:

Vrms of noise = sqrt(4*k*T*R)*sqrt(B)

How is bandwith (B) supposed to be interpreted ?
Is bandwith suppose to be evaluated from the RC low pass filter that the
resistor forms with its own parasitic capacitance ?


-Frank

Look for a book called "Low-Noise Electronic Design" by Motchenbacher
and Fitchen. The book has all sorts of wonderful information on noise
in electronics.

The bandwidth cited in the equation is the NOISE bandwidth of your
gadget. It is larger than the usual -3dB bandwidths that you are
accustomed to. For a simple RC lowpass, the noise bandwidth is 1.57
times wider than the -3dB bandwidth.

To answer your question, yes, the capacitance across the resistor
limits the bandwidth, thus limiting the total noise across your
resistor. If you didn't have capacitance across the resistor, you
would have infinite noise voltage across the resistor which would
cause it to smoke just floating in mid air. Thank Buddha for
capacitance! If you connect test equipment to the resistor, the
capacitance of your probe will change the noise bandwidth of said
resistor.

Don't get stuck looking at one lonely resistor. You must look at your
whole circuit as there are many things to consider when doing noise
analysis.

Mark

 

[John Larkin]

On 19 May 2004 17:23:32 -0700, frank8017@excite.com (frank) wrote:

Hello,

I am puzzled by the following equation:
Thermal noise in a resistor is as follows:

Vrms of noise = sqrt(4*k*T*R)*sqrt(B)

How is bandwith (B) supposed to be interpreted ?
Is bandwith suppose to be evaluated from the RC low pass filter that the
resistor forms with its own parasitic capacitance ?


-Frank

That's one way to look at it. B is just the bandwidth of the measuring
device, or of a filter that the noise is passed through. For audio,
for example, we mostly care about the noise in a 20 KHz bandwidth.

The equation implies that the open-circuit noise voltage is infinite.
The capacitance you mentioned limits the actual bandwidth, and at
extremely high frequencies the equation doesn't work any more anyhow.

Fun: if two resistors are connected, and one is at a higher
temperature, heat will flow from the hot one to the cold one, because
the hotter one generates more noise voltage.

John

 

[Don Pearce]

On 19 May 2004 17:23:32 -0700, frank8017@excite.com (frank) wrote:

Hello,

I am puzzled by the following equation:
Thermal noise in a resistor is as follows:

Vrms of noise = sqrt(4*k*T*R)*sqrt(B)

How is bandwith (B) supposed to be interpreted ?
Is bandwith suppose to be evaluated from the RC low pass filter that the
resistor forms with its own parasitic capacitance ?


-Frank

If you are dealing with radio, there are normally filters that
determine bandwidth. In audio you can either use the audible 20,000Hz,
or use a psophometric "A" weighting to give you about one third of
that. Every system ultimately has a bandwidth.

d
Pearce Consulting
http://www.pearce.uk.com

 

[Kevin Aylward]

news:k46oa0litli61tkvvbljiso3b4snu2qgqe@4ax.com...

On 19 May 2004 17:23:32 -0700, frank8017@excite.com (frank) wrote:

Hello,

I am puzzled by the following equation:
Thermal noise in a resistor is as follows:

Vrms of noise = sqrt(4*k*T*R)*sqrt(B)

How is bandwith (B) supposed to be interpreted ?
Is bandwith suppose to be evaluated from the RC low pass filter that
the
resistor forms with its own parasitic capacitance ?


-Frank

Look for a book called "Low-Noise Electronic Design" by Motchenbacher
and Fitchen. The book has all sorts of wonderful information on noise
in electronics.

The bandwidth cited in the equation is the NOISE bandwidth of your
gadget. It is larger than the usual -3dB bandwidths that you are
accustomed to. For a simple RC lowpass, the noise bandwidth is 1.57
times wider than the -3dB bandwidth.

To answer your question, yes, the capacitance across the resistor
limits the bandwidth, thus limiting the total noise across your
resistor. If you didn't have capacitance across the resistor, you
would have infinite noise voltage across the resistor which would
cause it to smoke just floating in mid air. Thank Buddha for
capacitance!

Even with no capacitance there would still not be infinite noise. The
thermal noise formula is only the classical approximation. Quantum
Mechanics corrects the formula, KT-> hf/(exp(hf/KT)-1). This issue is
one of the actual foundations of QM. How come no infinite energy? Look
up blackbody radiation e.g.
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html


Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.
"qrk" <SpamTrap@reson.com> wrote in message

 

[Ric]

qrk <SpamTrap@reson.com> wrote:

To answer your question, yes, the capacitance across the resistor
limits the bandwidth, thus limiting the total noise across your
resistor. If you didn't have capacitance across the resistor, you
would have infinite noise voltage across the resistor which would
cause it to smoke just floating in mid air. Thank Buddha for
capacitance! If you connect test equipment to the resistor, the
capacitance of your probe will change the noise bandwidth of said
resistor.

KT is just an approximation to:
hf/(exp(hf/KT)-1)
where h is Planck's constant.

At frequencies of interest to electronic engineers, hf << KT,
so the exponential term can be approximated by just two terms
of series expansion:
exp(hf/KT) ~ 1 + hf/KT

So the expression for the thermal energy becomes:
hf/(1 + hf/KT -1)
= kT

At temperatures close to absolute zero, or at frequencies
above about 10^12Hz, the approximation breaks down. So as
f->infinity, the engergy->0, saving the universe.

Note that k*T is expressed in Joules, so if you're interested
in power (Joules/second), you need to integrate kT with
respect to delta-frequency (which is a bandwidth) giving you
kTB, which has units of J/s, i.e. power.


--
Rick

 

[Jim Meyer]

qrk <SpamTrap@reson.com> wrote in message news:<k46oa0litli61tkvvbljiso3b4snu2qgqe@4ax.com>...

To answer your question, yes, the capacitance across the resistor
limits the bandwidth, thus limiting the total noise across your
resistor. If you didn't have capacitance across the resistor, you
would have infinite noise voltage across the resistor which would
cause it to smoke just floating in mid air.

Mark

That's an interesting story, Uncle Mark. But it doesn't quite
make sense. Regardless of the capacitance, noise voltage, noise
current, bandwidth, or resistance, no resistor will ever have a
temperature due to noise power (current times voltage) alone that is
even a microdegree above (or below) ambient.

Jim

 

[Phil Hobbs]

John Larkin wrote:

On 19 May 2004 17:23:32 -0700, frank8017@excite.com (frank) wrote:



Hello,

I am puzzled by the following equation:
Thermal noise in a resistor is as follows:

Vrms of noise = sqrt(4*k*T*R)*sqrt(B)

How is bandwith (B) supposed to be interpreted ?
Is bandwith suppose to be evaluated from the RC low pass filter that the
resistor forms with its own parasitic capacitance ?


-Frank




That's one way to look at it. B is just the bandwidth of the measuring
device, or of a filter that the noise is passed through. For audio,
for example, we mostly care about the noise in a 20 KHz bandwidth.

The equation implies that the open-circuit noise voltage is infinite.
The capacitance you mentioned limits the actual bandwidth, and at
extremely high frequencies the equation doesn't work any more anyhow.

Fun: if two resistors are connected, and one is at a higher
temperature, heat will flow from the hot one to the cold one, because
the hotter one generates more noise voltage.

John




One good way of deriving the Johnson noise formula (the sqrt(4kT) thing)

is from classical equipartition of energy. The stored energy in a
capacitor is a single classical degree of freedom, and hence (when
connected to a thermal reservoir, e.g. connected in parallel with a
resistor at temperature T) has a mean energy of kT/2, and since the
energy is CV**2/2, its rms noise voltage is sqrt(kT).

The noise bandwidth of a one-pole RC lowpass is (pi/2)*(3 dB BW) =
1/(4RC). Therefore, the noise power spectral density in the flatband is

p_N=(kT/2)*(4RC) per hertz,

so setting p_N=C(e_N)**2/2, we get

(e_N)**2 = kT*4R

and

e_N = sqrt(4kTR) per root hertz.

This is the same noise that correlated double sampling in CCDs was
designed to deal with. The advantage of this way of looking at it is
that the resistor doesn't have to be linear--CMOS reset switches behave
the same way.


Cheers,

Phil Hobbs


PS: My current research involves building crystal radios for 1.5 um
light (200 THz), which is well above the frequency at which the room
temperature Johnson noise rolls off due to the Planck function. I'm
probably the only one on this NG for whom that rolloff is an actual
engineering issue--or is there anyone else?

 

[Peter O. Brackett]

Jim:

[snip]

Look up "noise bandwidth", it's different from the actual 3dB
bandwidth... I can't remember the ratio right off the top of my head.

...Jim Thompson
[snip]

It sure is different from the 3dB bandwidth... and it is not a constant
ratio
it varies depending upon the transfer function under consideration.

If a transfer function [of a filter] is given by:

T(p) = N(p)/D(p)

Where p = sig + jw is the Laplace transform variable[Heavisides operator
"p"] then
the "total square integral" is defined by:

I = 1/2*pi * Integral [-00 - +00] {N(p)N(-p)/D(p)D(-p)}dp

Where the integral is taken along the jw axis.

i.e. I is the total area along the jw axis under the function...

1/2*pi * |T(jw)|^2

Such an integral is evaluated using Cauchy's Residue Theorem from the theory
of complex variable
calculus. This integral, i.e. the number I, is the total power delivered
through the transfer function
T(p) by a unit value white noise source applied to the input of the
filter/transfer function.

The noise bandwidth of the system with transfer function T(p) is then the
bandwidth of a brick wall
filter of unity gain that has the same integral value, i.e. the value of I.
If the brick wall filter has bandwidth
B then B = I, etc...

So to find the equivalent noise bandwidth you gotta first evaluate and get a
number for I for your transfer
function or filter and then the equivalent noise bandwidth is revealed as
the value of that total squared integral.

The value of total squared integrals [The "I" that I mentioned.] is actually
tabulated in at least a couple of
textbooks for many transfer functions in terms of the coefficients (e.g.
N(p) = n0 + n1*p + n2*p^2 + ... nn*p^n)
of the rational polynomials of the transfer functions N(p) and D(p) up to
the order 10 or so in several textbooks.

cfr: Kaiser et all "Stochastic Control Systems".

For simple filters/transfer functions, like say first or second order then
the integral is easily evaluated by
anyone with a smattering of first year integral calculus. In fact for
simple single pole low pass filters which
is what Jim Thomson was referring to, the ratio of the total square integral
I to the "actual" 3dB bandwidth
is about 1.35, i.e. the equivalent noise bandwidth is about 35% wider than
the 3dB bandwidth. But you
can work it out for yourself. Of course the result will be different for a
second order low pass filter, and
in the general case of arbitrary N(p)/D(p) you would have to work it out
yourself or get the value of I from
the available tables.

--
Peter

 

[Jim Thompson]

On Thu, 20 May 2004 23:38:26 GMT, "Peter O. Brackett"
<none@no-such-domain.nul> wrote:

Jim:

[snip]
Look up "noise bandwidth", it's different from the actual 3dB
bandwidth... I can't remember the ratio right off the top of my head.

...Jim Thompson
[snip]

It sure is different from the 3dB bandwidth... and it is not a constant
ratio
it varies depending upon the transfer function under consideration.

If a transfer function [of a filter] is given by:

T(p) = N(p)/D(p)

Where p = sig + jw is the Laplace transform variable[Heavisides operator
"p"] then
the "total square integral" is defined by:

I = 1/2*pi * Integral [-00 - +00] {N(p)N(-p)/D(p)D(-p)}dp

Where the integral is taken along the jw axis.

i.e. I is the total area along the jw axis under the function...

1/2*pi * |T(jw)|^2

Such an integral is evaluated using Cauchy's Residue Theorem from the theory
of complex variable
calculus. This integral, i.e. the number I, is the total power delivered
through the transfer function
T(p) by a unit value white noise source applied to the input of the
filter/transfer function.

The noise bandwidth of the system with transfer function T(p) is then the
bandwidth of a brick wall
filter of unity gain that has the same integral value, i.e. the value of I.
If the brick wall filter has bandwidth
B then B = I, etc...

So to find the equivalent noise bandwidth you gotta first evaluate and get a
number for I for your transfer
function or filter and then the equivalent noise bandwidth is revealed as
the value of that total squared integral.

The value of total squared integrals [The "I" that I mentioned.] is actually
tabulated in at least a couple of
textbooks for many transfer functions in terms of the coefficients (e.g.
N(p) = n0 + n1*p + n2*p^2 + ... nn*p^n)
of the rational polynomials of the transfer functions N(p) and D(p) up to
the order 10 or so in several textbooks.

cfr: Kaiser et all "Stochastic Control Systems".

For simple filters/transfer functions, like say first or second order then
the integral is easily evaluated by
anyone with a smattering of first year integral calculus. In fact for
simple single pole low pass filters which
is what Jim Thomson was referring to, the ratio of the total square integral
I to the "actual" 3dB bandwidth
is about 1.35, i.e. the equivalent noise bandwidth is about 35% wider than
the 3dB bandwidth. But you
can work it out for yourself. Of course the result will be different for a
second order low pass filter, and
in the general case of arbitrary N(p)/D(p) you would have to work it out
yourself or get the value of I from
the available tables.

Showoff ;-)

I should have gotten out my copy of Motchenbacher and Fitchen.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Kevin Aylward]

Peter O. Brackett wrote:

Jim:

[snip]
Look up "noise bandwidth", it's different from the actual 3dB
bandwidth... I can't remember the ratio right off the top of my head.

...Jim Thompson
[snip]

It sure is different from the 3dB bandwidth... and it is not a
constant ratio
it varies depending upon the transfer function under consideration.

If a transfer function [of a filter] is given by:

T(p) = N(p)/D(p)

Where p = sig + jw is the Laplace transform variable[Heavisides
operator "p"] then
the "total square integral" is defined by:

I = 1/2*pi * Integral [-00 - +00] {N(p)N(-p)/D(p)D(-p)}dp

Where the integral is taken along the jw axis.

i.e. I is the total area along the jw axis under the function...

1/2*pi * |T(jw)|^2

Such an integral is evaluated using Cauchy's Residue Theorem from the
theory of complex variable
calculus.

Nitpick here on the phrasing. Integrals of this type are often evaluated
by the residue theorem, not "is evaluated". There are, of course, many
ways to evaluate such integrals, the residue theorem is but one method.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Rene Tschaggelar]

Phil Hobbs wrote:

PS: My current research involves building crystal radios for 1.5 um
light (200 THz), which is well above the frequency at which the room
temperature Johnson noise rolls off due to the Planck function. I'm
probably the only one on this NG for whom that rolloff is an actual
engineering issue--or is there anyone else?

You apparently use a light sensor and they have different noise
models than resistors.

Rene
--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
& commercial newsgroups - http://www.talkto.net

 

[Phil Hobbs]

Rene Tschaggelar wrote:

Phil Hobbs wrote:



PS: My current research involves building crystal radios for 1.5 um
light (200 THz), which is well above the frequency at which the room
temperature Johnson noise rolls off due to the Planck function. I'm
probably the only one on this NG for whom that rolloff is an actual
engineering issue--or is there anyone else?


You apparently use a light sensor and they have different noise
models than resistors.

Rene

No, actually. I'm building devices based on metal-insulator-metal tunnel
junctions, which are fast enough to rectify light. The tunnelling time has
never been measured accurately, but most estimates come in at slightly under
1 fs, which is fast enough for my purposes. We normally think of optical
processes as relying on transitions between single-electron states, e.g. from
the valence band to the conduction band in a photodiode, but anywhere below
the plasma frequency of the metal (somewhere in the UV), we can get normal
metallic conduction, which is a collective motion of many electrons. Thus
it's still possible to think in circuits terms about 200-THz currents, and
the Johnson noise rolloff is a genuine design fact, as is Coulomb-blockade
behaviour due to the fact that a single electron moves the bias on the MIM
junction by about 20 mV.


Cheers,

Phil Hobbs

 

[John Larkin]

On Fri, 21 May 2004 20:08:19 -0400, Phil Hobbs
<pcdhSpamMeSenseless@us.ibm.com> wrote:

Rene Tschaggelar wrote:
Phil Hobbs wrote:



PS: My current research involves building crystal radios for 1.5 um
light (200 THz), which is well above the frequency at which the room
temperature Johnson noise rolls off due to the Planck function. I'm
probably the only one on this NG for whom that rolloff is an actual
engineering issue--or is there anyone else?


You apparently use a light sensor and they have different noise
models than resistors.

Rene

No, actually. I'm building devices based on metal-insulator-metal tunnel
junctions, which are fast enough to rectify light. The tunnelling time has
never been measured accurately, but most estimates come in at slightly under
1 fs, which is fast enough for my purposes. We normally think of optical
processes as relying on transitions between single-electron states, e.g. from
the valence band to the conduction band in a photodiode, but anywhere below
the plasma frequency of the metal (somewhere in the UV), we can get normal
metallic conduction, which is a collective motion of many electrons. Thus
it's still possible to think in circuits terms about 200-THz currents, and
the Johnson noise rolloff is a genuine design fact, as is Coulomb-blockade
behaviour due to the fact that a single electron moves the bias on the MIM
junction by about 20 mV.


Cheers,

Phil Hobbs

So, will we one day get thermal focal-plane imagers that are diode
arrays, as opposed to thermistors? That would be very slick... a
cheap, fast, megapixel CCD-like thermal imager.

How about a superhet version, if the diodes won't work high enough?
Then you could do pseudo-color thermal imaging, swept spectral
imaging, all sorts of fun stuff.

John

 

[Peter O. Brackett]

Kevin:

[snip]

Nitpick here on the phrasing. Integrals of this type are often evaluated
by the residue theorem, not "is evaluated". There are, of course, many
ways to evaluate such integrals, the residue theorem is but one method.

Kevin Aylward
[snip]

Calculus is nice, but...

I like the table look up method myself!

Kaiser's [Stochastic Control Theory] book has those integrals tabulated in
terms
of the coefficients of the polynomials N and D up to somewhere around10th
order.

Handy.

--
Peter

 

[Phil Hobbs]

John Larkin wrote:

On Fri, 21 May 2004 20:08:19 -0400, Phil Hobbs
pcdhSpamMeSenseless@us.ibm.com> wrote:


Rene Tschaggelar wrote:

Phil Hobbs wrote:



PS: My current research involves building crystal radios for 1.5 um
light (200 THz), which is well above the frequency at which the room
temperature Johnson noise rolls off due to the Planck function. I'm
probably the only one on this NG for whom that rolloff is an actual
engineering issue--or is there anyone else?


You apparently use a light sensor and they have different noise
models than resistors.

Rene

No, actually. I'm building devices based on metal-insulator-metal tunnel
junctions, which are fast enough to rectify light. The tunnelling time has
never been measured accurately, but most estimates come in at slightly under
1 fs, which is fast enough for my purposes. We normally think of optical
processes as relying on transitions between single-electron states, e.g. from
the valence band to the conduction band in a photodiode, but anywhere below
the plasma frequency of the metal (somewhere in the UV), we can get normal
metallic conduction, which is a collective motion of many electrons. Thus
it's still possible to think in circuits terms about 200-THz currents, and
the Johnson noise rolloff is a genuine design fact, as is Coulomb-blockade
behaviour due to the fact that a single electron moves the bias on the MIM
junction by about 20 mV.


Cheers,

Phil Hobbs



So, will we one day get thermal focal-plane imagers that are diode
arrays, as opposed to thermistors? That would be very slick... a
cheap, fast, megapixel CCD-like thermal imager.

How about a superhet version, if the diodes won't work high enough?
Then you could do pseudo-color thermal imaging, swept spectral
imaging, all sorts of fun stuff.

There are several groups working on just those things--there's a DARPA BAA
that came out recently on uncooled focal planes based on antennas. My
interest has more to do with single-element detectors and modulators, but
there are some interesting things you can do by combining heterodyne
techniques with antenna-style detectors. Many of those things you can do
with ordinary heterodyne optical detection, of course, which has been done
for 30 years. The difference now is that you can potentially put 10**6
separate heterodyne detectors on a chip, with lots of DSP attached, which
could be, ahem, interesting.

Cheers,

Phil Hobbs

Advanced Optical Interconnection
IBM T. J. Watson Research Center
Yorktown Heights NY

 

[Mikko Kiviranta]

PS: My current research involves building crystal radios for 1.5 um
light (200 THz), which is well above the frequency at which the room
temperature Johnson noise rolls off due to the Planck function. I'm

Don't forget the (1/2 hbar omega) quantum noise contribution
which replaces (and exceeds) the missing hi-freq part due to the
Planck function.

probably the only one on this NG for whom that rolloff is an actual
engineering issue--or is there anyone else?

I'm working on cryogenic circuits: the rollof takes place at
100GHz on 4.2K and at 2GHz on 100mK. It has consequences for certain
Josephson circuits and for sub-mm detectors. But that's not "room
temperature Johnson noise".

Anyway, the range you're working on is interesting. Direct conversion
of solar power into dc by an antenna and rectifier was suggested in
Kraus' classical book on antennas, and I think NIST did actual work
on such themes at some time in the past.

Regards,
Mikko

 

[Mikko Kiviranta]

How about a superhet version, if the diodes won't work high enough?
Then you could do pseudo-color thermal imaging, swept spectral
imaging, all sorts of fun stuff.

I think a superhet imaging array gadget has been demonstrated by
TRW (at SPIE Aerosense conference a few years ago), but that operated
somewhere in the sub-THz range. Still, even at such long wavelengths
there appears to be enough blackbody radiation tail available
to perform passive thermal imaging.

Regards,
Mikko