# Johnson noise potential divider

A quick question: If I have a potential divider with two 10kΩ resistors, is the noise that you would measure between the two resistors simply the the Johnson noise of a 20kΩ resistor?

On a similar note, if I have two 10kΩ resistors joined together in series, and measure the noise across just one of them, i.e. the second resistor is unconnected at one end, is the noise just the Johnson noise from a single 10kΩ resistor, or does the floating 10kΩ resistor add any Johnson noise? (I suspect it would somehow act as an antenna or something.)

-
I think this question can get good answers if you explain why you are wanting to know this. The thermal noise in a resistor at room temperature would be very difficult to measure and not have much value. Even if you can, as soon as you apply current, the temp will change which will cause the noise to change. I can potentially see you modeling Rf systems or front ends as a circuit with resistors at different temperatures, but you don't say that is what you are doing. – Kellenjb Mar 5 '12 at 12:20
Unfortunately a 20k system does not have negligible Johnson noise - At room temp it provides 12nV/sqrt(hz) which can be a large component of a lower noise op-amp (input noise 1nv/sqrt(hz)). I'm just interested in if the opamp is measuring the noise across one of the 10K resisters, does the other floating resister contribute at all? – Ross W Mar 5 '12 at 16:23
I am by no means an expert on this stuff, but since no one has answered I am just trying to see if I can get you an answer. If you have 5MHz bandwidth, 100x gain, you are still only talking about a few milivots of noise on the output. I can't imagine a system that is so sensitive to that noise that you would consider putting a resistor just floating in the system. It could act like an antenna and add noise to the system, but this would be separate from thermal noise. – Kellenjb Mar 5 '12 at 16:41
@Kellenjb, I can imagine it. Also, he may be using those resistors for a circuit he must have and is wondering what kind of noise characteristic to expect. – Kortuk May 29 '12 at 5:37
Some flame about the topic – clabacchio May 30 '12 at 8:40

It's the noise of a 20k$\Omega$ resistor.

edit
The first paragraph is a bit ambiguous to me. The 20k$\Omega$ is when you measure across both resistors (that's what I thought you meant, since you mention the 20k$\Omega$). If, however, you would measure across one of the resistors you'd have to see both in parallel, and then it would become the noise of a 5k$\Omega$ resistor.

But noise can't be added arithmetically because of its stochastic character. Noise voltage is defined as

$v = \sqrt{4kT R\Delta f}$

so it's proportional to the square root of the resistance. Therefore the noise of a resistor of 20k$\Omega$ is $\sqrt{2}$ higher than that of a 10k$\Omega$ resistor, not twice as high. Placing the resistors in parallel will cause the noise voltage to be $\sqrt{2}$ lower.

I've seen a couple of designs where this property was used to improve the noise of an opamp by placing two of them in parallel. But this will only improve noise figures by -3dB. Barry also mentions this. Probably not worth it.
A few years ago Elektor published a power amplifier design based on 32 parallel NE5532s. Then it becomes interesting. The noise is reduced by 15 dB, and the specs for the amplifier show a S/N ratio of 110dBA.

edit
For the two resistors in series with the other end of the second resistor open: that resistor's noise voltage doesn't count. The open end would show some Johnson noise with respect to the common node, but the other resistor won't see it. (The noise it would pick up as an antenna is not the resistor's Johnson noise.)

-
 Sorry, to understand better: if you measure the noise in the middle of the divider, the resistor will appear in parallel, as for small signals model, right? In that case, the noise adds or splits? – clabacchio♦ May 30 '12 at 8:12 I feel ashamed because I've upvoted both you and Rocket Surgeon, but you say opposite things, basically. Really, the model suggests that each resistor has its otn noise voltage source, but at the same time a resistor may be the abstraction for a series or a parallel. – clabacchio♦ May 30 '12 at 8:29 Well, he also says that it's the noise of a 20k resistor if they're in series. That's what I say as well. But he doesn't say that this isn't twice that of a 10k. I explain that it's $\sqrt{2}$ higher, by the equation and the stochastic character. His answer just isn't complete, but we don't contradict each other. – stevenvh May 30 '12 at 8:35 The fact is that I'm pretty sure that they behave as a parallel, but the point is if the noise sums or divides. IMO :) – clabacchio♦ May 30 '12 at 8:37 Indeed, it will most likely pick the 50/60 Hz and other nasty noise – clabacchio♦ May 30 '12 at 11:14

If I have a potential divider with say two 10k resisters - Is the noise that you would measure between the two resisters simply the the Johnson noise of a 20k resister?

No. The noise measured for potential divider will be noise of equivalent resistor with value measured by AC ohmmeter with defined badnwidth. In theoretical case with noiseless signal source with zero impedance (consider top resistor grounded as well). So it will be noise of 5K resistor when you probe the output of divider.

The noise at the far ends of 2x10K resistors will be the noise of 20K resistor.

Noise of each resistor when connected in series and floating will be same as single disconnected resistor, because AC is disconnected.

Longer story: To understand it easier, start with studying of concept of noise, othrogonality and correlation.

Noise fundamentally has no amplitude (there is a probability to find a million volt on unpowered resistor).

Noise caused by thermal effects (natural) is orthogonal to every signal or other noise.

-
 Like I just explained to clabacchio (comment to my answer) I think your answer is incomplete. You don't say how high the noise level of the 5k is, users may get the idea that it's half that of a 10k, which isn't the case. – stevenvh May 30 '12 at 8:39 Thank you. Just tried to provide quick answer when person presented a "quick" question. The orthogonality is just part of what can be omitted in haste. There is more to miss there: bandwidth, concept of AC, does a cable or free space with equivalent impedance have johnson noise properties, etc. – Rocket Surgeon May 30 '12 at 10:37 Ross asks more about topology of 2 parts of divider than about properties of single part. Are they serial or parallel or is it behaving only to degree of single half?. Property is well identified already, as question involves the correct term (Johnson noise). – Rocket Surgeon May 30 '12 at 11:01

Each 10k resistor has a noise voltage associated with it whose magnitude is given by the Johnson noise equation. If you put two 10k resistors in series, then the equivalent resistor is 20k but the equivalent noise voltage is only the square root of 2 times the noise voltage of a 10k resistor. This is because the noise voltage of each 10k resistor is independent of the other so the voltages do not simply add but combine as the square root of the sum of their squares. Thus if the noise voltage of each resistor were 1nv/sqrt(Hz), then the series noise voltage would only be 1.414nv/sqrt(Hz). This principle is sometimes used when designing very low noise amplifiers. The output of two separate but identical amplifiers is summed. This results in the signal level being twice that of either amplifier but the noise level is only sqrt(2) times the noise level of either one (assuming the noise level of either one is the same). Thus the overall signal-to-noise ratio has been improved by a factor of sqrt(2) or 3 dB. This can be extended to more than 2 amplifiers but the low of diminishing returns sets in. The main point is that noise cannot be treated the same as a deterministic quantity such as ohms of resistor or volts of signal.

-