So... how does one measure noise?
I see preamps specs like S/N = 90db. Wow!
If you have a preamp with +/-15v rails the best it's going to do for a signal is about 10vrms.
The formula for Signal to Noise ratio= 10 log(signal/noise).
If you solve for noise, it equals 10nv!
Who in the world can measure that?
The lowest scale on my scope is 2mv - certainly can't say I can measure 10nv.
I suspect you need special gear.
Anyone?
I see preamps specs like S/N = 90db. Wow!
If you have a preamp with +/-15v rails the best it's going to do for a signal is about 10vrms.
The formula for Signal to Noise ratio= 10 log(signal/noise).
If you solve for noise, it equals 10nv!
Who in the world can measure that?
The lowest scale on my scope is 2mv - certainly can't say I can measure 10nv.
I suspect you need special gear.
Anyone?
...
The formula for Signal to Noise ratio= 10 log(signal/noise).
If you solve for noise, it equals 10nv!
No
Decibel - Wikipedia
Best wishes
David
Tx David. I checked out the link - I realize S/N ratio is expressed in DB, but, I'm trying to solve for the measured voltage values.
It stimulated thought though and I think I've answered my own question. From the link the ratio for 90db is 32,363. From this I assume only a highly resolved spectrum analyzer could measure with any certainty. 10v signal /32363 = 309uv of noise.
It stimulated thought though and I think I've answered my own question. From the link the ratio for 90db is 32,363. From this I assume only a highly resolved spectrum analyzer could measure with any certainty. 10v signal /32363 = 309uv of noise.
This number looks correct to me.Tx David. I checked out the link - I realize S/N ratio is expressed in DB, but, I'm trying to solve for the measured voltage values.
It stimulated thought though and I think I've answered my own question. From the link the ratio for 90db is 32,363. From this I assume only a highly resolved spectrum analyzer could measure with any certainty. 10v signal /32363 = 309uv of noise.
The resolution of your test equipment widely depends on the effective measuring bandwidth. Having said this any good soundcard with an FFT measuring software like ARTA has no problems resolving audio noise of several uVrms.
Btw, nowadays 90dB S/N is nothing superior. Look at the noise I measured at the output of my reference class D design.
The number is referred to 0dB=1Vrms.
Channel L = Equipment Under Test
Channel R = loopback of soundcard
Including the audio frontend with 6dB gain output noise is below 100uVrms here.
Considering a max output of approx 30dBVrms S/N is better than 110dB.
The number is referred to 0dB=1Vrms.
Channel L = Equipment Under Test
Channel R = loopback of soundcard
Including the audio frontend with 6dB gain output noise is below 100uVrms here.
Considering a max output of approx 30dBVrms S/N is better than 110dB.
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Excuse me, that should be 316µV, NOT nV!!!😡90 db below 10 volts rms is 316nV.
Now, can you HEAR noise that is 90 db down? Hmmmmm......
It depends.Excuse me, that should be 316µV, NOT nV!!!😡
Now, can you HEAR noise that is 90 db down? Hmmmmm......
When I was young I examined noise of the electric guitar signal chain and decided that 90dB would be the desirable S/N. A real challenge at that time😉
90dB S/N is hearable, but not at one point in time. Hearing can detect down to 0dB (SPL), and upto the threshold of pain (and above!), which is 140dB range. But play a loud sound (say 100dB) and you are temporarily deafened and won't be able to hear that 0dB level for some time. Play quiet sounds and listen up close one day, and loud things the next (from further back), and you are asking for quite a lot more S/N ratio than 90dB from your power amplifier (but not from the sound source or preamp, since you have adjusted the volume control between the two days).
6db of gain is one thing, even 30db. But, if you have a preamp with 65db of gain and you achieve 90db S/N, hats off.
I'have a discrete design intended for use with a Ribbon mic. It uses back to back diff amp sections and had concluded the PSRR was so good that regulation was likely not necessary. I was wrong. Small artifacts of ripple are showing up to yield a S/N that I approximate to be only 50db.
I'have a discrete design intended for use with a Ribbon mic. It uses back to back diff amp sections and had concluded the PSRR was so good that regulation was likely not necessary. I was wrong. Small artifacts of ripple are showing up to yield a S/N that I approximate to be only 50db.
Got an idea. I'll use a bank of batteries to develop a quiet bipolar power supply to power the preamp with just to see if the output noise I see with my power supply goes away. I would assume if I can perceive a change from a flat line with my scope on the 2mv scale, the S/N is likely worse than 90db.
TI makes an interesting low noise dual rail linear regulator , the TPS7A3901DSCT.
Anyone embraced the new LDO regulators offered by TI or Linear Tech?
Perhaps the transformer/rectifier/filter ripple still won't be totally rejected. We'll see. I have an eval board coming.
TI makes an interesting low noise dual rail linear regulator , the TPS7A3901DSCT.
Anyone embraced the new LDO regulators offered by TI or Linear Tech?
Perhaps the transformer/rectifier/filter ripple still won't be totally rejected. We'll see. I have an eval board coming.
Mains induced hum and semiconductor noise are two different things.
Also the dB scale you cited is for power ratio. If you measure noise voltage referred to the nominal output, the correct formula is 20*log(Vout/Vin). If you refer to 1V output, -60dB is 1mV, -80dB is 100uV, -90dB is roughly 30uV. This can't be measure by a multimeter (partly because you'd need a True RMS meter), you have to use some specialized piece of equipment. Or just use your ears.
Also the dB scale you cited is for power ratio. If you measure noise voltage referred to the nominal output, the correct formula is 20*log(Vout/Vin). If you refer to 1V output, -60dB is 1mV, -80dB is 100uV, -90dB is roughly 30uV. This can't be measure by a multimeter (partly because you'd need a True RMS meter), you have to use some specialized piece of equipment. Or just use your ears.
ICsaszar, thanks for pointing that out. I did use the ratio listed for power.
I think you meant to say 10*log (vout/Vin) for volts.
Indeed semiconductor noise is another thing, but if I believe the specs for the transistors I've used, that will be down in the teens of nanovolts. Another source is resistor noise also quite minute.
I think you meant to say 10*log (vout/Vin) for volts.
Indeed semiconductor noise is another thing, but if I believe the specs for the transistors I've used, that will be down in the teens of nanovolts. Another source is resistor noise also quite minute.
ICsaszar, thanks for pointing that out. I did use the ratio listed for power.
I think you meant to say 10*log (vout/Vin) for volts.
Indeed semiconductor noise is another thing, but if I believe the specs for the transistors I've used, that will be down in the teens of nanovolts. Another source is resistor noise also quite minute.
No, 20*log(Vout/Vin) is correct.
You only use 10*Log for power ratios.
Yes, this is a frequent confusion. Another one is to speak about dB power and dB voltage as different things. dB is dB, however the related power ratio and voltage ratio are different (by a factor of 2).
Or put another way dB is always a power ratio (logarithmically expressed). The reference level might be denoted in various ways, such as 1W (dBW), 1mW (dBm), 1V (dBV), sqrt(0.6V) (dBu). If the reference is some sort of voltage it often means the load impedance isn't specified so that squared-volts is used as a stand-in for power. dB by itself represents a pure power ratio.
Apparently, the website How to Calculate Signal to Noise Ratio | Sciencing is wrong?
Here's an excerpt:
"SNR Calculation – Complicated
To calculate SNR, divide the value of the main signal by the value of the noise, and then take the common logarithm of the result: log(S ÷ N). There’s one more step: If your signal strength figures are units of power (watts), multiply by 20; if they are units of voltage, multiply by 10. For power, SNR = 20 log(S ÷ N); for voltage, SNR = 10 log(S ÷ N). The result of this calculation is the SNR in decibels. For example, your measured noise value (N) is 1 microvolt, and your signal (S) is 200 millivolts. The SNR is 10 log(.2 ÷ .000001) or 53 dB."
Here's an excerpt:
"SNR Calculation – Complicated
To calculate SNR, divide the value of the main signal by the value of the noise, and then take the common logarithm of the result: log(S ÷ N). There’s one more step: If your signal strength figures are units of power (watts), multiply by 20; if they are units of voltage, multiply by 10. For power, SNR = 20 log(S ÷ N); for voltage, SNR = 10 log(S ÷ N). The result of this calculation is the SNR in decibels. For example, your measured noise value (N) is 1 microvolt, and your signal (S) is 200 millivolts. The SNR is 10 log(.2 ÷ .000001) or 53 dB."
definitely wrong!Apparently, the website How to Calculate Signal to Noise Ratio | Sciencing is wrong?
Here's an excerpt:
"SNR Calculation – Complicated
To calculate SNR, divide the value of the main signal by the value of the noise, and then take the common logarithm of the result: log(S ÷ N). There’s one more step: If your signal strength figures are units of power (watts), multiply by 20; if they are units of voltage, multiply by 10. For power, SNR = 20 log(S ÷ N); for voltage, SNR = 10 log(S ÷ N). The result of this calculation is the SNR in decibels. For example, your measured noise value (N) is 1 microvolt, and your signal (S) is 200 millivolts. The SNR is 10 log(.2 ÷ .000001) or 53 dB."
Well yeah, the level of PSRR required will definitely depend on the signal levels and ripple levels you are dealing with. Also, it is possible to ruin real-life PSRR of an otherwise high-performance circuit by implrmentation issues like unlucky PCB layout, e.g. having relatively high-impedance shared signal and power ground returns with large rail decoupling capacitors and a low-impedance power supply connection. (That's why you find series resistors in the power lines in some designs.) Fully differential circuits tend to help with problems like that due to their inherent PSRR. Even so, even high PSRR is ultimately finite.6db of gain is one thing, even 30db. But, if you have a preamp with 65db of gain and you achieve 90db S/N, hats off.
I'have a discrete design intended for use with a Ribbon mic. It uses back to back diff amp sections and had concluded the PSRR was so good that regulation was likely not necessary. I was wrong. Small artifacts of ripple are showing up to yield a S/N that I approximate to be only 50db.
Sounds like you could use another 40 dB of PSRR, which shouldn't be a real challenge for any decent pair of low-noise regs (including the trusty LM317/337 with ADJ pin capacitors) given a good layout. Depending on the current needs of your preamp, you may even get away with a passive (2nd-order) filter + opamp buffer approach (a trick commonly used for low current draw, but very noise sensitive circuits, like oscillators or D/A converter Vref).
Is this amp supposed to be for ribbon mics with or without an integrated transformer?
"Naked" ribbons tend to be so spectacularly low in impedance (read <10 ohms or so) that you generally not only have to parallel a whole bunch of input devices but also cable and contact resistance become an issue... seems tricky to do anywhere except inside the mic itself, as a prepre. (In this case you may also prefer an unbalanced input to avoid incurring a noise penalty.)
By contrast, ribbons with transformers electrically aren't an awful lot different from dynamic mics, really. 65 dB of gain would seem to be the absolute maximum ever needed in this scenario, and that is assuming pro (nominal +4 dBu) output levels.
As far as the original question goes, you don't directly measure SNR. You amplify the noise to the point where you can easily compare it to a known signal level, e.g. 20+ dB above the noise floor of a soundcard input of known (calibrated) input level. Then you can do the math to infer input noise level. Using an A/D input has the additional advantage of potentially showing what your problems are.
Apparently, the website How to Calculate Signal to Noise Ratio | Sciencing is wrong?
Here's an excerpt:
"SNR Calculation – Complicated
To calculate SNR, divide the value of the main signal by the value of the noise, and then take the common logarithm of the result: log(S ÷ N). There’s one more step: If your signal strength figures are units of power (watts), multiply by 20; if they are units of voltage, multiply by 10. For power, SNR = 20 log(S ÷ N); for voltage, SNR = 10 log(S ÷ N). The result of this calculation is the SNR in decibels. For example, your measured noise value (N) is 1 microvolt, and your signal (S) is 200 millivolts. The SNR is 10 log(.2 ÷ .000001) or 53 dB."
They got it reversed!
Jan
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