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In Reply to: Re: Motsenbacher and Fitchen "Low Noise Electronic Design" posted by Mahatma Kane Jeeves on August 21, 2006 at 18:54:08:
This does get a bit confusing :-). But basically, the noisy circuit is modeled as a noiseless one with a noise voltage source in series with the open-circuit input signal voltage all the way back at the input. That noise source is eni, and is the figure of merit for the noise performance. It's computed by taking the output noise voltage and dividing by the gain from the open-circuit input signal voltage to the output voltage.
If I understand your question correctly, you're referring to the output noise contribution from a noise source at a certain point in the circuit. But the way the equivalent input noise figure of merit is formulated, the effect when that contribution is reflected back to the input (by dividing by the amp's gain) is independent of whether there is feedback or not. Feedback reduces the output noise contribution, but the gain is correspondingly lower too. So it comes out the same when reflected back to the input.
No, that was exactly it. The answer was, "yes, but it's a trivial consequence of the gain reduction." I had only remembered the "yes" part. Many thanks!
Of course, negative feedback can help to reduce external noise from the power supply, but if you keep your forward gain low, and don't throw it away with high amounts of local feedback, then open loop design CAN be quieter than feedback design, because you don't have to add a feedback resistor in series (noisewise) with the input stage.
Your post makes me think about a common problem in many PA implementation: the feedback resistors (wired as a divider).
Noise coming from thess resistors is in serial with the feedback path, and so adds its noise without possible reduction by the FB loop.
So, it's a good idea to keep them low, in the 100 ohm range. Furthermore, it helps to maintain good high-frequency characteristics.
As a result, current flow and dissipation won't be insignifiant.
For a well-designed PA, 100 to 200mW at 100W PA power is common.
Here comes the issue:
To avoid thermally-induced distorsion at low frequencies, these resistors should be oversized types, at least 10 times the peak dissipation.
And on a real musical program, which most of its power in the bass range, you will have only peaks of dissipation.
So a resistor with high peak-to-average power tolerance is required.
Which bans metal-film resistors.
Since it has to be the less inductive possible, standard wirewound resistor is banned.
So, the FB resistor of choice should be:
- bulk metal resistors (manganin is the better alloy in this matter): the best choice or
- crenelled-wound wire resistor (much less inductive than standard wirewound,
- mounted on an heatsink, or it will be a high spot (best is TO220 shaped resistors)
Unhappily, I scarcely saw these refinements used in so-called high-end equipment.
Often, this resistor goes hot when the PA is used at full power. To avoid burning the PCB, the manufacturer often lifts the resistor a few centimeters above the PCB, on standoff if he's serious, or with the resistors' wires if he's clueless about mechanical vibration tolerance. But anyway the open area under the resistor's body creates a loop, in which any induced voltage will be ...in serial with the feedback...Jeeeezzz. Often, this loop is vertical, so that it can get flux from the Earth's magnetic field if the PA is "correctly" set. So, any vibration will translate into a signal.
(btw that's not the only issue that makes an equipment microphonic. Other issues exist too, but they are O.T on this thread. All these issues can be solved, and a SS equipment should not be microphonic at all if correctly designed )
Well, it makes the life easier for vibration control manufacturers, but, wouldn't it be easier and more respectful of the customer to have the equipment correctly designed from the beginning?
So, as you see (and John C knows), many issues are intermixed.
FB resistors can act as a noise generator, a distorsion generator at low frequencies, and can make the PA microphonic.
But you can avoid it. Just a question of knowledgeable and serious design.
Jacques, I think that I would rescale the resistors to be a higher value. Power amps are at the end of the amplifiying chain and the difference between a 100 ohm resistor and a 1K resistor would usually be unimportant. At a moving coil input stage, however, even 100 ohms would be too much, but then there is no signal output either.
For my power amps, I generally use a 47K 1/2W Holco (old) feedback resistor with perhaps 1.8K to ground. I might use a 1-2W Resista, if I had to sacale it down to 10K and 330 ohms or so. I would most likely never use a 100 ohm resistor, because of the problem of dynamic overheating, that could be fixed with a heatsinked power resistor, but then you develop other problems such as capacitive coupling to the chassis, etc.
Jacques, of course you meant the 100 ohm resistor as the resistor that goes to ground, but if you raised the value of that resistor to 1K or so, the noise in the power amp would still be very low, The main feedback resistor that swings almost all the voltage would then be about 25 times larger or more with modern power amps. This is the resistor that will overheat, if you are not careful.
...was the resistor to ground.
So, the FB resistor itself would be in the <1,000 ohms.
which would suck several watts up to tens of watts... For sure.
sorry for the misunderstanding.
Just to make it perfectly clear: You can use a somewhat higher series resistor in series with the input stage of a power amp, because it operates at a much higher voltage level than a preamp phono stage, or microphone input stage. A 1 Kohm feedback resistor to ground for a PA amp should be OK. Then the main feedback resistor would be 28.2K for a normal home theater power amp, and maybe more or less for some other amplifiers. If you use +/- 100 Volt supplies and presume a full square wave as a worst case output, then the main feedback resistor must have a rating of 100 squared =10000 divided by 28.2K. This is less than 1W, but it is always best to use the largest resistor that is practical in order to keep changes in temperature of the resistor to modulate the output voltage. Still, a 1W resistor can be compact enough to be board mounted and not require an external heatsink. Even a 1/2W resistor (what we use) is good enough with a good resistor and normal operation.
To avoid burning the PCB, the manufacturer often lifts the resistor a few centimeters above the PCB, on standoff if he's serious, or with the resistors' wires if he's clueless about mechanical vibration tolerance.""
Tis an unfortunate move. Most axial resistors are rated to dissipate power through the leads onto the pc traces. A typical rating for an axial diode, for example, has 3/8 inch long leads, with a specific pad size. It would be better to just dedicate a larger pad for the resistor connections.
But anyway the open area under the resistor's body creates a loop, in which any induced voltage will be ...in serial with the feedback...Jeeeezzz. Often, this loop is vertical, so that it can get flux from the Earth's magnetic field if the PA is "correctly" set. So, any vibration will translate into a signal. ""
Loop, whatsa loop? :-)
I would also be concerned with the magnetic fields that are being generated internally within the chassis, by the xfmr, the power leads, and all the high current leads around the zistors, the output binding posts, everything. I do not see good poweramp wiring design practices much, and the magnetic hash within the chassis will indeed be coupling to the FB loop, especially where it will do the most damage. The half gauss of the earth's magnetic field is small taters.
The half gauss of the earth's magnetic field is small taters.Yes, but appreciable anyway: Lorentz induction through an area of several cm2 varying about one thousandth by mechanical vibration at 1,000Hz in a field of 0.5gauss would give something in the tens of microvolts, so less than -100dB.
OK, just appreciable ;-)
Yes, but appreciable anyway: Lorentz induction through an area of several cm2 varying about one thousandth by mechanical vibration at 1,000Hz in a field of 0.5gauss would give something in the tens of microvolts, so less than -100dB.
OK, just appreciable ;-) ""
500 Wrms into 4 ohms, sqr(500/4) = sqr(125) = 11.18 ampsRMS.
Peak I 11.18 * 1.414 = 15.8 amperes peak.
B(tesla) = (μ 0 I)/(2 pi r)
μ 0 = 4 pi 10 -7
1 Tesla = 10000 gauss = 10 4 gauss
B = 2 10 -7 10 4 I / r
B = 2 10 -3 I / r Gauss
at 500 wrms into 4 ohms,
B = 2 10 -3 15.8 / r Gauss
at the wire surface (r =1mm, or .001 meters 10 -3 meters ):
B = 31.6 10 -3 / 10 -3
B = 31.6 Gauss.
at 1cm, B = 3.16 gauss.
at 10 cm, B =.316 gauss.
Mind you, this field it time varying, and proportional to the output current. AND, if the pos and neg rails or output devices are not equidistant, they will couple differently..meaning, feedback error coupling will be dependent on the quadrant of operation of the amplifier/load.
And, also? The haversine charging of the supply caps. The output current I used? The charging currents through the bridge are a lower duty cycle..at 10% duty cycle, your talking 3 Gauss fields at 10 cm...and they will also depend on which cap is being pulled.
Haversine = ripple in this case.""
Two points of concern..the first, the primary currents of the xfmr, odd order harmonics which are indeed haversines..the second, the FWR current pulses, even harmonics in the secondary loop.
What is never worried about is the fact that the ripple currents within a capacitor, externally, are just a wire. External to the cap case, the difference between a wire carrying the ripple current, and the capacitor itself doing the same, is zero..If the cap is carrying 15 amps of 1khz ripple, the equations I gave here model correctly.
It should be possible to spot the difference between primary and secondary coupling, but that requires test methods I use here but have not seen externally..
meaning, feedback error coupling will be dependent on the quadrant of operation of the amplifier/load.which brings us back to the previous thread about how the amp behaves under the voice coil back-emf induction.
And, also? The haversine charging of the supply caps.../...
and also the eddy currents induced in the chassis by the above, and their own induction. (should act to mitigate all those sneaky inductions).
Anyway, all these issues are manageable. Twelve years ago, I designed an airborne equipment half analog in the audio band, half digital, which is set in helicopters just beneath the magnetic compass with its exquisite sensitivity, and just above the radar pulses generator...
It worked the first time with those neighbours, and is now certified.
No heavy shields, just a light aluminium enclosure.
But careful layout.
Now, take that 500 w amp, and go a bit more:
EMF = -N δΦ/δt N is turns..
Φ = BA = magnetic flux...A is area in meters.
At 10 cm from the wire, BA per cm 2 is:
.316 gauss times 10 -4 T/gauss = 3.1610 -5 Tesla
As a 1 Khz signal 10 cm away, B = 3.16 * 10 -5 sin(ωt) Tesla
Rate of change of B is it's derivative.
ω = 2 pi f, or 6.283 10 3
so B = 6.283 10 3 * 3.16 * 10 -5 sin(ωt) Tesla
= 19.84 * 10 -2 sin(ωt)
If the feedback loop encompasses 1 cm 2 , 10 cm away from a current lead, then..1cm = 10 -2 meters, 1 cm 2 = 10 -4 meters 2
EMF = - 1 * 19.84 * 10 -2 sin(ωt) * 10 -4
= 19.84 * 10 -6 sin(ωt) volts....20 μvolts per square cm area.
Now, keep in mind star grounding techniques. They do not consider a time varying magnetic flux environment..It is not inconceivable to imagine 10 square centimeters of loop in a poweramp chassis in the grounding system. So, perhaps 200 μvolts??
And god forbid, some kind of inductor in the output..what, 10 turns? 2 millivolts?
Is that a worry??
That calculation is of the error voltage after the divider, not before.
If it's before, then the error is divided down also. If it is before, it is pure error introduced into the front end.
If it were caused by the secondary haversines, then...it can be somewhat masked by the power output.
If one tries to measure this output current to input error, what would one expect? If it's the output lines, it'll just be a phase shift. If it were one of the rails, it gets more complicated, based on quadrant of operation.
Many good points here; Although I think the resistor effects could well be swamped by the compression and distortion of the loudspeaker.
Low noise is NOT an attribute that high end wants as far as I can tell, and this includes both measurements and listening. It appears that they desire high levels of noise for masking, and the "appearance" of greater dispersion in the loudspeaker. That's my impression of what's going on.
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