Really, really silly voltage divider question...

[TomWaterman]

Ha! Yeah ignore the whole bloody thing! I also thought that 4 might have referenced a note somewhere. I searched for a while to find it!

I have another question this time concerning noise contribution from a simple divider.

It is my understanding that the noise is multiplied by the square of the attenuation gain.

That is R1 and R2 = 4kTR * A^2. where A = the overall voltage loss.

So for my example two 2500R as R1 and R2 and 50ohm source 20kohm load.

Each one contributes the following assuming 300K room temp:

1.66*10^-20 * 2500 * 0.4656^2

that is 8.99 x10^-18

Sum the two for 1.799 x10^-17 total power spectral density
Square root for voltage spectral density in V/sqrtHz

4nV/sqrtHz.

Is that even right?

I guess I need to consider noise bandwidth.

Can I do this?

R1 or R2 contribution = 4kTRfBW * A^2 where fBW = say 20-20k = 19980.

R1 noise = 1.66*10^-20 * 2500 * 19980 * 0.4656^2
= 1.797 * 10^-13

R1noise + R2noise = 3.595*10^-13

Total noise power = sqrt 3.595*10^-13 = 599 nV/sqrtHz.

Does that mean that if an audio fader changes position its noise contribution changes?

I'm trying to get things totally clear on simple things.

Thanks Tom

 

[JohnRoberts]

I don't see much clarity or sense from that math mumbo jumbo...

By noise contribution I assume you're talking about the "Johnson" or thermal noise? My understanding is that the noise contribution at the node is equal to the noise of a resistor equal to the actual impedance at that node, or the two or more legs in parallel. Assuming incoherent noise, this would combine with attenuated signal source noise floor as the SQRT of the sum of the squares.

Regarding noise contribution of a fader, yes indeed it will vary, rising to a max at equal resistance from wiper to top and bottom (-6dB), but even for a common value 10k, that's only 2.5k worst case.

So it might not be wise to use a 10k fader on a raw mic level signal, but this IMO is generally not a concern below line level signals. Keep in perspective that the noise floor will most likely be dominated by the signal you are passing. For the fader noise to even show up in a measurement it need to be closer than say 10dB below the attenuated source noise floor at that point, (and e.i.n. of following gain stage).

Perhaps of more concern, if that fader is connected to an opamp with significant input bias current, that changing source resistance with fader position will cause a changing DC component. If large enough this can give you a dreaded case of "scratchy pot". Some designers, cap couple to avoid this or use FET input opamps who have low enough bias current to not scratch. While this "scratching" doesn't in any way degrade performance in use (like a cheap blocking cap might), customers perceive noisy pots and clicky switches as evidence of inferior design, while the path may actually be superior to the more well behaved due to sundry blocking caps.

But the customer is always right, it's their money.

JR

 

[JohnRoberts]

[quote author="mediatechnology"]

Perhaps of more concern, if that fader is connected to an opamp with significant input bias current, that changing source resistance with fader position will cause a changing DC component. If large enough this can give you a dreaded case of "scratchy pot". Some designers, cap couple to avoid this or use FET input opamps who have low enough bias current to not scratch. While this "scratching" doesn't in any way degrade performance in use (like a cheap blocking cap might), customers perceive noisy pots and clicky switches as evidence of inferior design, while the path may actually be superior to the more well behaved due to sundry blocking caps.

Yes a pet peeve of mine. And even when using low bias op amps we need to remember that the wiper is not always in full contact with the element as it is being moved. So a high value R (relative to the pot) from wiper to ground also helps this.[/quote]

In general wipers are designed with 3, 4, or more, small spring finger contacts specifically to deal with contact bounce over microscopic surface irregularities. The design concept is one or two fingers may be airborne at any moment while there will still be one or more touching. If you think about it, a resistor loading the wiper, would also cause noise if fully floating as the signal pegs to ground (while this may be preferable to jumping to one rail or the other from bias current). Perhaps if you anticipate using a pot with lousy wiper, a small capacitor at that node might be less objectionable.

In my experience, pots from quality manufacturers will exhibit good wiper contact for a reasonable life if not abused.

JR

 

[TomWaterman]

[quote author="JohnRoberts"]I don't see much clarity or sense from that math mumbo jumbo...[/quote]Right OK - I'm trying...

[quote author="JohnRoberts"]By noise contribution I assume you're talking about the "Johnson" or thermal noise?[/quote] Yep.

[quote author="JohnRoberts"]My understanding is that the noise contribution at the node is equal to the noise of a resistor equal to the actual impedance at that node, or the two or more legs in parallel. Assuming incoherent noise, this would combine with attenuated signal source noise floor as the SQRT of the sum of the squares.[/quote]Ok so usual Zout equation for divider: (R1*R2)/(R1+R2) arriving at [quote author="JohnRoberts"]2.5k worst case[/quote] for a 10k pot set half way? Best case would then be down near 10 ohms ish for the pot at its ends stops either way?

So Total noise = SQRT (signal source^2 + 2k5^2)?

I'm just trying to prove it to myself that a 600 ohm fader is quieter than a 5kohm fader.

-Tom

 

[JohnRoberts]

Total noise with pot at 50% is

SQRT of ( ((input noise/2)^2)+ ((noise of 2.5k)^2) ).

Square root of, sum of, source noise reduced -6dB squared + 2.5k noise squared.

JR

 

[bcarso]

John has the right approach.

The thing about resistive attenuators: yes, they have thermal noise. But you can't make them arbitrarily low-Z because they load the source too much, so that sets a limit on the minimum noise contribution.

If there were no such limitation, then a voltage divider could be arbitrarily low noise. But the actual system situation is usually far more complex anyway---there is a noise contribution from the source to begin with, and there is a noise contribution from what's downstream from the attentuator. What's sometimes done is to analyze the signal-to-noise ratio in (at some point in the chain) and the signal-to-noise ratio out somewhere further on. This takes care of gain/attenuation and other variables.*

And the point about current noise is well-taken---it is always there, even with FETs (especially at higher frequencies).

Another thing to bear in mind: just because a node in a circuit has a certain impedance doesn't mean it perforce generates thermal noise. One can simulate Z's with a useful frequency range that look resistive but have far less noise than the equivalent resistor. However, it comes at a price, and usually there are other ways to get the same system noise performance that may be more convenient.


*this is done a lot in optoelectronic stuff, as streams of photons have some interesting statistics.

 

[TomWaterman]

Thanks guys - makes perfect sense that the source noise is attenuated by the divider and then summed with the divider noise.

My calculations say that max output impedance is halfway. However with the pot at min and max Zout is very small approaching zero...that can't be right?

What am I missing?

Also say I designed a bridged T-pad with 2k5 output impedance and a 6dB loss-could it be said that it also contributes the same noise as a 10kpot set to 50% even though it contains 2 more resistive elements?

The Handbook states that average noise for a T-pad is -100dB and constant, therefore varying with attenuation.

Thanks again.
Tom

 

[TomWaterman]

Ha thanks Wayne - just checking!

It all makes sense. Who thought the little divider could be so interesting?!

Cheers Tom

 

[bcarso]

[quote author="TomWaterman"]It all makes sense. Who thought the little divider could be so interesting?!
Cheers Tom[/quote]

I have a book somewhere that is truly arcane, with quite advanced maths: Paul Slepian, Mathematical Foundations of Network Analysis. At the end IIRC it says that it's complete, but that everything in it is applicable to strictly resistive circuits, and the extension to reactive passive networks remains to be done. :shock:

So it takes 195 pages just to handle resistors rigorously.

 

[PRR]

> I'm just trying to prove it to myself that a 600 ohm fader is quieter than a 5kohm fader.

Lower Voltage noise.

In Power terms, it does not matter. You go to your perfect transformer shop and transform the impedance.

Let's see if I can work a number or turn into a toad.

Assume the Perfect Transformer Shop is closed.

10K pot at -6dB. Impedance 2.5K. Noise voltage in audio band is like 7uV?

600 pot at -6dB. Impedance 150R. Noise voltage in audio band is like 2uV?

What signal?

Pick "10V" out of the air. 6dB down we have 5V.

5V/7uV= 117dB S/N

5V/2uV= 129dB S/N

Power required to slam 10V onto pot:

10V^2/10K = 10mW = +10dBm

10V^/600R = 167mW = +22dBm

It's not lower noise. It is higher signal Power.

When the Perfect Transformer Shop re-opens and sells you a 10K:600 and a 600:10K tranny, you can use 600R or 10K pot and get exact the same result.

The Power of a pot driver is fairly well set by Budget (cost, space, power).

The audio band Noise Power of a resistor is 2*10^-16Watts. Don't matter 1 ohm or 1.3Megs.

You agree with that, then wonder about parallel resistors. Is it double the power? Yes, but each resistor absorbs some of the power from its neighbors. As far as God and the little demon electrons are concerned, your several physical parts are one resistor. Slepian proves it.

You use pots which agree with your drivers. We like 10K because a chip will drive 2K and we may want to hang four pots on it.

We don't like super-high Z mixing because capacitance is everywhere.

We don't like super-low Z mixing because resistance and inductance are everywhere.

Overall, it should be moot. At the system input, low-noise is expensive. At the system output, power is expensive. "Line Level" is the place where we don't have to sweat either noise or power. It should be possible, even easy, for all real systems to hiss at the mike/head and clip at the speaker-amp, not in between. This does require some sense of gain-scale.

 

[TomWaterman]

Thanks the reply PRR.

Truly a joy to read and makes it all the more clear.

Its good to see you still posting around here.
I really appreciate it.

Cheers Tom :guinness:

 

[TomWaterman]

One thing - I just ran your numbers and didn't get the following answers.

[quote author="PRR"]10K pot at -6dB. Impedance 2.5K. Noise voltage in audio band is like 7uV?

What signal?

Pick "10V" out of the air. 6dB down we have 5V.

5V/7uV= 117dB S/N[/quote]

I get 10k @ 50%, Zout = 2k5, Noise voltage 20-20kHz = 0.91uV

5v/0.91uV = 134.8dB S/N

[quote author="PRR"]The audio band Noise Power of a resistor is 2*10^-16Watts. Don't matter 1 ohm or 1.3Megs.[/quote]Sorry, where does that come from?

Cheers Tom

 

[JohnRoberts]

[quote author="TomWaterman"]One thing - I just ran your numbers and didn't get the following answers.


I get 10k @ 50%, Zout = 2k5, Noise voltage 20-20kHz = 0.91uV

5v/0.91uV = 134.8dB S/N


Cheers Tom[/quote]

It been quite a while since I did this math but the way I used to calculate thermal noise contribution of a resistance was to memorize the value for 600 ohms and scale that up or down for whatever R I am interested in.

noise = SQRT ( (2,500/600) * (9.94x10^-18) * (20,000) )

= SQRT ( (4.16) * (9.94 x 10^-18) * (20,000) )

= SQRT ( 8.2833 *10^-13 )

= 0.00000091

My calculation agrees with yours and is pretty much in the ballpark of what I would expect.

Regarding S/N, faders are typically followed by 10dB gain stage so output noise will likely be dominated by that gain stage. If bipolar the noise current will also need to be factored into the 2.5k worst case source impedance (plus feedback resistor's contributions.

Still not much noise compared to typical line level sources you will ever encounter.

JR

 

[TomWaterman]

Good one John. Thanks.

Yeah I have noted the point the about the following amp.

This forum is mad - when you join you can easily bounce forward a few steps as you get excited about building this and that...especially if you come from a recording background and not an electronics one. Its easy to get bedazzled by the stunning projects here and skip some of the really useful basic understanding.

I'm going through a back to basics approach now as I realise where the holes in my knowledge are.

This has been a great thread for me.
Thanks to everyone.

T

 

[bcarso]

[quote author="TomWaterman"]
Sorry, where does that come from?
Cheers Tom[/quote]

The power per unit bandwidth is 4kT, e sub n squared over R, or i sub n squared times R, which at 300K is about 1.656 * 10^-20 W (the R terms drop out). Then multiply by the bandwidth of about 20kHz to get 3.31 * 10^-16 W.

 

[CJ]

there are a bunch of these things online if you are lazy:

http://my.athenet.net/~multiplx/cgi-bin/att_pad.main.cgi

 

[bcarso]

That particular one is only for equal input source and output loading impedances.

But it is free...

 

[PRR]

me> Let's see if I can work a number or turn into a toad.
me> Impedance 2.5K. Noise voltage in audio band is like 7uV?
me> Impedance 150R. Noise voltage in audio band is like 2uV?

TOAD!!!

It is very humid here. The decimal-point on my slide-rule is sticking.

Divide my wrong answers by "10". 0.7uV, 0.2uV. Do another 20dB on the other calculations.

It looked wrong at the time but my brain is braised by heat and humidity.

> I get 10k @ 50%, Zout = 2k5, Noise voltage 20-20kHz = 0.91uV

0.7uV, 0.9uV, what-ever. Maybe the sliderule swelled that much. Maybe I rounded-up when I shudda rounded down. Maybe I still think in 15KC bandwidth.

> 134.8dB S/N

What's in your studio? A 35dB SPL air vent and a 105dB SPL singer? You got 70dB S/N at the source. Aim it at the middle of your >130dB S/N in the mix stage. Room rumble/hiss is 30dB below mix-hiss, mix-amp clipping is 30dB above the singer's shout.

Yeah, you may have super mufflers on the studio ventilation and get near 0dB SPL midrange. You may be close-miking a trumpet or Fender Twin, 115dB SPL. You can still get 5dB-10dB away from both hiss and clipping.

And in the end, you probably have to dump to 96dB S/N CD.

And play OK on home systems which rarely beat 80dB S/N between noise coming through the walls and limited amplifier power.

(Not to mention car-audio.)

If the music is good, you can use a Wollensak and sell it.

If the music is bad, it hardly matters.

 

[JohnRoberts]

Yup, I never found the decimal points on my slip stick.. working with them you pretty much had to do the math in your head, and just let the slide rule provide some precision, as it was.

If the meat computer is overheating, stuff happens. The DR prescribes chilled liquid refreshments in copious quantities, administered orally until relief is experienced. :sam: :guinness:

JR