Thermal noise & input impedance

[decc]

On a non-inverting op-amp (or a something else with a high-impedance JFET input) the input impedance seen by the guitar is the value of the bias resistor to Vref, correct? Does the issue of thermal noise with high-value resistors apply here as well? If so is there any advantage to the non-inverting config over the inverting one (other than maintaining correct phase) since you have the same problem of balancing high input impedance with low resistors for noise reduction?

[Cliff Schecht]

The noise is there. That's why it's a designers job to choose the correct resistor value for their purposes, it not only sets the noise through the op amp but also sets the current that gets used to bias the op amp (which is negligible here), the cutoff frequency if you have a DC blocking cap, the effect on your AC gain, etc.. Voltage drop isn't a concern here because the bias currents are so low (nA to pA) that no power gets dissipated in the resistor.

As far as the actual input impedance, that is set by the parallel combination of your op amps input impedance in parallel with the VREF pull-up resistor. Since the input impedance is so high for a FET op amp, you end up with 99.99999999999999% of the resistance you place in for your VREF resistor (remember the formula for two parallel resistors? (R1*R2)/(R1+R2)). But in this case, your VREF point better be feeding a low impedance DC voltage source (like an op amp voltage follower) or you'll run into the problem of your VREF not being quite 4.5 V.

[decc]

So for the simple case of an AC coupled unity gain input buffer stage with a low-impedance source of Vref:

Non-inverting: 1M pull-up resistor to Vref sets input impedance to 1M and adds thermal noise of one 1M resistor.

Inverting: 1M series resistor to the "-" input sets input impedance to 1M and adds thermal noise of one 1M resistor. Or is there 2x the noise as there is also the 1M feedback resistor needed to set the gain to 1?

[alanlan]

You also have to factor in the source resistance because this is also in parallel with the bias resistor and any resistive input impedance.  This is usually much less than 1Meg and so will dominate proceedings.  If you imagine a 1K resistor across the input, then the noise produced by the 1Meg is effectively subject to a 1000:1 attenuation.  The effective noise in this case would be that of the 1K input.

[Cliff Schecht]

So for the simple case of an AC coupled unity gain input buffer stage with a low-impedance source of Vref:

Non-inverting: 1M pull-up resistor to Vref sets input impedance to 1M and adds thermal noise of one 1M resistor.

Inverting: 1M series resistor to the "-" input sets input impedance to 1M and adds thermal noise of one 1M resistor. Or is there 2x the noise as there is also the 1M feedback resistor needed to set the gain to 1?

The series and feedback resistor of an inverting op amp are accounted for in parallel, not in series. It isn't necessarily half the noise however, because you have a noisy active device somewhere in the mix. Your noise is also depended on your gain and bandwidth, both of which are completely interlaced (look up gain bandwidth product for more info).

[decc]

You also have to factor in the source resistance because this is also in parallel with the bias resistor and any resistive input impedance.  This is usually much less than 1Meg and so will dominate proceedings.  If you imagine a 1K resistor across the input, then the noise produced by the 1Meg is effectively subject to a 1000:1 attenuation.  The effective noise in this case would be that of the 1K input.

Ah, OK. So we would end up with the 1M input impedance but the effective noise of, say, the wiper position of the (typically) 250K volume pot on the guitar or the 100K volume pot on the preceding pedal.

The series and feedback resistor of an inverting op amp are accounted for in parallel, not in series. It isn't necessarily half the noise however, because you have a noisy active device somewhere in the mix. Your noise is also depended on your gain and bandwidth, both of which are completely interlaced (look up gain bandwidth product for more info).

When taking into account alanlan's comment, would we then have:

Non-inverting: 1M input impedance with 1M||R_source effective noise (+ op-amp active noise of 1x gain @ 10kHz bw)
Inverting: 1M input impedance with 1M||1M||R_source effective noise (+ op-amp active noise of 1x gain @ 10kHz bw)

Making the inverting configuration slightly better, but insignificantly so since the noise is dominated by the much lower R_source?

Where all of this is going is I have a design that, in order to maintain correct phase, currently has an inverting input buffer (unity gain.) Do I get any benefits of adding a non-inverting buffer before it and lowering the resistors on the inverting buffer?

[R O Tiree]

Better to stay with the single, inverting stage and keep the resistances low. The higher the resistance, the more thermal noise you get, but the lower the current consumption is, so it's a trade-off. There is also Schottky Noise (aka shot noise, because it sounds like lead shot hitting a concrete wall) present at very low levels in all conductors, but really starts to be noticeable in any reverse-biased semi-conductor junction.

Next it is wise to look at the signal voltage in relation to the supply voltage - an opamp may have a published signal-to-noise (S/N) ratio of, say, 80dB, but if you are running it at +/- 9V and the output signal is only +/- 1V, then the S/N ratio is going to be worse... someone check my maths, but I think it might be as low as approx 60dB.

Another problem is flicker noise, particularly prevalent in carbon comp resistors, much less so in metal-film.

There are a few other sources that may or may not be relevant to your application.

Source:- "Opamps For Everyone, Chapter 10 - SLOD006B, Texas Instruments, Aug 2002, Ed: Ron Mancini - This is an invaluable resource. The maths gets a bit hairy at times...

[decc]

Source:- "Opamps For Everyone, Chapter 10 - SLOD006B, Texas Instruments, Aug 2002, Ed: Ron Mancini - This is an invaluable resource. The maths gets a bit hairy at times...

Thanks for the link. That does explain a lot.

I'm not really concerned with current consumption as I rarely use batteries, so it comes down to the noise/impedance issue for me. I would think that a circuit designed to go first, such as a compressor, would benefit from larger resistors in order to not lose treble, while something like a delay would be OK with lower resistors as it's likely being driven by a low-impedance pedal output.

No single simple answer, but I think I know what the trade-offs are now.

Thanks!