AoE noise calculation - bias chain noise? - Page 1
Forum section: electronics / beginners

author: aneevuser
author_rank: Frequent Contributor
timestamp: March 12, 2020, 10:10:46 am
content: In AoE, on page 486, they present a noise calculation for a common emitter amplifier with a 750K/150K bias chain on the base. They seem to be claiming on page 487, in the final sentence of the noise calculation section, that the noise from this is negligible.This strikes me as odd - I would have thought that the base would be seeing the uncorrelated Johnson noise from both resistors and given that they are by far the largest resistors in the circuit, that this noise would massively dominate all other sources. What am I misunderstanding?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 12, 2020, 10:18:05 am
content: It's in parallel with a stonking great speaker, much lower Z.To expand on that: the Johnson noise at a given node is the Thevenin equivalent resistance at that point. There's probably a lot more qualification on that for how resistors (or other impedances) are connected, but that works for passives.Tim

author: aneevuser
author_rank: Frequent Contributor
timestamp: March 12, 2020, 01:55:45 pm
content: Quote from: T3sl4co1l on March 12, 2020, 10:18:05 amOK, this struck me as very unlikely but I think that I've managed to derive the result for parallel resistors so I'm going to take your word for it. So that means then, that if I have a big noisy resistor, I can quieten it down by putting it in parallel with a small quiet one? Seems weird to me.(Derivation details: Model parallel \$R_1, R_2\$ as ideal resistors in series with a noise source with \$v_n^2=4kTR_n\$, then \$R_{Th}=R_1 || R_2\$ and we'll have Thevenin voltage contributions \$v_{1rms}=\sqrt{4kTR_1} \dfrac{R_2}{R_1+R_2}, v_{2rms}=\sqrt{4kTR_2} \dfrac{R_1}{R_1+R_2}\$ but these are uncorrelated so total Thevenin noise power \$v^2_{Th} = 4kTR_1 (\dfrac{R_2}{R_1+R_2})^2 + 4kTR_2 (\dfrac{R_1}{R_1+R_2})^2 = 4kT R_1||R_2\$)

author: aneevuser
author_rank: Frequent Contributor
timestamp: March 12, 2020, 02:01:40 pm
content: Quote from: T3sl4co1l on March 12, 2020, 10:18:05 amHmm, does this not also mean that if the speaker is disconnected, then the audible noise at the o/p will rise significantly?

author: RandallMcRee
author_rank: Frequent Contributor
timestamp: March 12, 2020, 03:54:39 pm
content: Quote from: aneevuser on March 12, 2020, 02:01:40 pmIt won't be audible because you disconnected the speaker!This discussion reminds me of another point made in Low Noise Electronic Design, a thought problem which asks:This discussion might lead one to consider using a large-valued resistorand a wide bandwidth (whlch can produce several volts).in series with a diodein an attempt to power a load such as a transistor radio. It should be obviousthat this will not work, but can you explain the flaw in the reasoning?

author: aneevuser
author_rank: Frequent Contributor
timestamp: March 12, 2020, 04:20:23 pm
content: Quote from: RandallMcRee on March 12, 2020, 03:54:39 pmIt's probably worth pointing out at this juncture that the speaker is being used as the mic.QuoteHmm. It *should* be obvious...My only thought at the moment is that I'm not sure what determines the b/w here. You'll have some capacitance involved somewhere so the transferred energy will start to roll off at some point. Not sure if that resolves the puzzle.Quite apart from that it's worth noting that the 4kTR thing is only really true at low frequencies - it's not truly white noise all the way up to infinity.

author: TimFox
author_rank: Super Contributor
timestamp: March 12, 2020, 05:31:29 pm
content: Depending on the circuit, it is sometimes better to treat Johnson noise from a resistor as a current rather than a voltage. It’s the same noise power, so use only one equivalent circuit at a time. Usually, when the resistor is in series with the source, the voltage is relevant, but when the resistor is in parallel with the input, the current is relevant. For the two equivalent circuits, the noise voltage increases for higher resistance, but the noise current decreases for higher resistance.With a low source resistance, the high resistance of the bias chain makes a low noise current that makes a low contribution to the input voltage.The available noise power is independent of the resistance. The quantum effects on noise power are not important at audio frequencies, but do keep the power from becoming infinite over an infinite frequency range.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 12, 2020, 08:41:13 pm
content: Quote from: TimFox on March 12, 2020, 05:31:29 pmIndeed, a terminated transmission line is just a low frequency TEM mode of black body radiation, and obeys the same cutoff; it just so happens that, at room temperature, the wavelength near cutoff is many times smaller than the width of that TEM cavity, so it doesn't propagate in any meaningful way as a TEM wave anymore. Not to mention the typical dielectric being strongly absorptive up there.I suppose at low temperatures, the cutoff would be in the microwave range where it is measurable (in a TEM line); I wonder if anyone's done this? The obvious difficulty being, a resistor at say 1 K needs to be sensed with an amplifier having a noise temperature significantly lower than 1 K, no small feat.Quote from: aneevuser on March 12, 2020, 01:55:45 pmYup! Consequence of electrical signals only having two degrees of freedom (in whatever way you wish to express that, whether AC steady-state I and Q (Fourier domain), or Hilbert domain, or orthogonal time-domain functions, or..). N noise sources in parallel and series, still only has 2 degrees of freedom. The noise voltages and currents add as they do, and the Thevenin equivalent pops out.Perhaps even more perplexing, but quite understandable once reasoned out -- capacitors have a fixed noise voltage, period! The derivation assumes some noise resistance, which therefore has some cutoff with the capacitance. The noise and cutoff are inverse, so the integrated noise over that bandwidth is constant.Which will also affect the circuit in question. With no speaker connected, the transistor is left to its own devices; I suppose base input resistance still dominates, plus whatever excess noise the transistor has. Hmm, actually it's close, isn't it? If r_e ~ 250Ω and h_fe ~ 500, r_pi ~ 500*250 = 125k, comparable to the bias resistors. So the Thevenin combination of all of them will be the answer.Whereas with the speaker in, the presumably low impedance shunts the other noise sources, and R_bb' probably dominates.(Which also perhaps means the minimum noise point is the geometric average of these, about 4kΩ? Probably different once excess noise is factored in.)Or for the one remaining case, with the speaker in, but the ground connection broken instead... BUZZZZZTim

author: aneevuser
author_rank: Frequent Contributor
timestamp: March 14, 2020, 03:44:09 pm
content: Quote from: T3sl4co1l on March 12, 2020, 08:41:13 pmNot sure I see how the degrees of freedom determines this aspect of noise - do you have a pointer to something that discusses this?QuoteBy "cutoff" here, do you mean some kind of LPF rolloff effect?Anyway I shall have to study noise in some more depth - looks like there's a fair bit of interesting physics going on.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 14, 2020, 10:17:26 pm
content: Quote from: aneevuser on March 14, 2020, 03:44:09 pmI probably shouldn't have said that...The equipartition theorem states that a system in thermal equilibrium has that energy distributed evenly over all available states. States meaning, a particle can be in this or that energy level, and the differences between levels, and the number of levels accessible at a given mean energy (~= temperature), determines the thermodynamics. Discrete levels are great for quantum mechanics; for classical waves, we can use continuum energy and simply (hah) integrate over the available states (bandwidth and amplitude).So, something tiny like a molecule, doesn't have many states available -- give or take symmetry, a few degrees of freedom per atom. The dimensionless specific heat capacity of a monatomic gas is 3/2: 3 because of the three dimensions of movement and half because half; diatomic 5/2, because two added rotational modes; polyatomic approaching 3 (6/2) because the number of modes added (bending and sometimes stretching) counts with the number of atoms.Electrical signals have a factor of 1 (2/2), for those two degrees of freedom. You can access more thermal power by having multiple electrical signals together -- but there's no way to rectify all of them together, without having to deal with one overall signal at a time, and thus you can't get more than unit noise out of it.Another way to think of it: two degrees of freedom represents a pitiful amount of energy compared to the ~Avogadro's number degrees of freedom a solid has. While we can connect a resistor to a low-noise amplifier, in effect making an electronic refrigerator -- the rate of cooling is essentially negligible. We could wire an array of nano-scale resistors and amplifiers, and presumably get perceptible cooling, but, ahhh, what a PITA!Or we could "wire" the surface of a solid to fields in space, which still doesn't transfer much energy at room temperature, but it's pretty good at higher temperatures. This being radiative heat transfer, of course.So, that's a terrible and brief explanation, but it makes a very fundamental point. It just takes years of thermo and stat mech to understand that point. ...Much of which I've since forgotten, so at this point I merelythinkI understand.QuoteYes- there's a resistance somewhere, that's your noise; and there's a capacitance in question. Together, you get an LPF.Tim

author: aneevuser
author_rank: Frequent Contributor
timestamp: March 15, 2020, 08:57:12 am
content: Quote from: T3sl4co1l on March 14, 2020, 10:17:26 pmOK I see what you're getting at now.I'm familiar with the equipartition theorem, but I'm not sure I see that it applies to a resistor - for equipartition to hold in the way that you suggest, we need a classical Hamiltonian with two independent quadratic terms, and I don't see what they are in this case. Anyway this is probably off topic for this site, so I will investigate it elsewhere.QuoteOK, I shall investigate this further elsewhere too - surprising result to me though. I need to see a derivation, I think.