Am I Having a Saturday Morning Stupid Moment?

Am I Having a Saturday Morning Stupid Moment?

Is the noise of two 1Meg resistors in parallel less than a single 500K resistor?

Or do I need to go have another cup of coffee?

...Jim Thompson

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| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson
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Probably :)

No!

Think about a 500k "block" of resistive material and just imagine a line through it dividing it into two 1M ones.

That'll be it! :)

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John Devereux
Reply to
John Devereux

Are the two halves correlated ?:-)

...Jim Thompson

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| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Well you connected them together so the voltages are the same :)

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John Devereux
Reply to
John Devereux

And!

If it was less everyone would be using hundreds of resistors in parallel in their low-noise amps. You could buy "pre-wired" super-low-noise resistor networks.

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John Devereux
Reply to
John Devereux

Johnson and other excess noise aside, the noise _power_ available from any resistance, no matter how it is arrived at, is solely dependent on (and proportional to) temperature. It is -174dBm at room temperature if _my_ head is screwed on straight this morning.

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http://www.wescottdesign.com
Reply to
Tim Wescott

I was hoping for divine intervention ;-)

...Jim Thompson

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| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Maybe something a bit stronger than coffee. The noises should be equivalent or about 91 nV/rtHz for 500k ohms or two 1meg resistors in parallel. The two resistors are considered uncorrelated noise sources.

All bets are off if your resistor is in a state variable filter feedback loop since you get something loosely correlated to a Q multiplication factor in noise.

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Mark
Reply to
qrk

No.

Johnson noise of a resistor is Vn^2 = 4*k*T*R , with source impedance R.

So the voltage noise^2 of the first 1M resistor is going to be k*T*R (divide by two because it's shunted by the other 1M resistor)

as is the second 1M resistor

Uncorrelated noise sources add as vx = sqrt(vn1^2 + vn2^2)

So Vn^2 = 2 * k * T *R = 4 * k * T * (R/2), which is the same as the Johnson noise of a 500K resistor.

Active loads can do better than resistors.

Best regards, Spehro Pefhany

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"it\'s the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
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Reply to
Spehro Pefhany

You could "uncorrelate" them if you'd dump some remaining ice from yesterday's champagne cooler on one of them 8-D

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SCNR, Joerg

http://www.analogconsultants.com/
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Reply to
Joerg

With a shot or two of Jack Daniels I can probably get totally uncorrelated ;-)

...Jim Thompson

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| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

Marketing analysis provides an answer to a purely technical question. No engineering knowledge needed.

--
Dirk

http://www.transcendence.me.uk/ - Transcendence UK
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Reply to
Dirk Bruere at NeoPax

Two 1M resistors in parallel is by all accounts supposed to be a 500K resistor.

If the parallel pair managed less noise than the single, then I would suspect that the 500K version was noisier than it should be.

First move on my part would be to switch to a different manufacturer of

500K resistors (should I need 500K resistors - those are not "standard values" to extent of being on the usual 24-per-decade lists of ones of 5% tolerance - the closest value on such lists is 510K, and 470K is on the 12-per-decade 10% tolerance lists, and even the 6-per-decade 20% tolerance lists.) (My experience is that 20% tolerance resistors mostly date back *at least* to when most TV receivers sold "new" in USA and most components therein were manufactured in USA, and when vacuum tubes in those outnumbered semiconductor components - maybe even dating back mostly to pre-TV days! It must have been 15-20 years since I trashpicked anything with 20% resistors!)

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

Doggone it, I now remember that this is also proportional to bandwidth

(in power terms, proportional to square root of bandwidth in voltage/current terms).

By any chance did two 500K resistors in parallel have enough capacitance to ground increased from one 1-meg one enough to decrease the bandwidth significantly?

Not that I expect this to reduce noise more than signal should both occur at affected frequencies or to occur in any circuit "not needing more work"...

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

The two load each other down in terms of voltage of noise generation to extent of equating a parallel pair to a single. Otherwise a parallel pair of 1-meg resistors at 20 degrees C would have a net transfer of power to a

510K resistor whose temperature is 20.01 degrees C, transferring heat from the cooler mass to the warmer one naturally. That is a "thermodynamic impossibility" worth a Nobel Prize.

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

Some cheap high-volume toys use resistors made out of a resistive paste silksreened directly on the PWB, generally with tolerances of 30% or so. Makes the analog portion of the design interesting...

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Guy Macon
Reply to
Guy Macon

I'm sure I meant to say "power spectral density". But I didn't, did I?

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http://www.wescottdesign.com
Reply to
Tim Wescott

No. It would violate conservation of energy.

Considering COE solves a lot of problems.

John

Reply to
John Larkin

There are many people who don't believe in that kind of logic. Such as hedge fund managers, mortgage bankers, politicians (especially when it comes to budgets), etc.

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SCNR, Joerg

http://www.analogconsultants.com/
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Reply to
Joerg

I've designed stuff like that. When it had to be more accurate you made the paste a bit "rich" and lasered away at it until within required tolerance.

Chip designers like Jim have to live with 30% or more pretty much all the time. But they can rely on very accurate ratios, much better than the 1% we discrete guys usually get to enjoy.

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Regards, Joerg

http://www.analogconsultants.com/
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Reply to
Joerg

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