OpAmp: Better to actively attenuate, or drive full range into a divider?

I want to get a +-15V differential signal (60Vp-p) into an ADC that has a single 3.3V supply. When I asked a friend to proofread it, he was concerned about the noise performance of R5-R6 vs. using R1-R4 to make U2A output the desired level to start with: simulate this circuit – Schematic created using CircuitLab

(input protection not shown for clarity)

My understanding is that an opamp's noise level does not depend on signal level, therefore a hotter signal is better. So I only attenuated U2A enough to match the full input range and left the rest to R5-R6.

(Another benefit is that if something goes wrong and U2A becomes a comparator, then the ADC input is still within the limits. Probably not likely, but possible.)

My friend thinks I would actually have less noise by putting all the attenuation in R1-R4, deleting R5-R6, and driving C1 directly. Is this true?

• You can't use an LM833 with a 60V input. Peek at the data sheet. – Scott Seidman Dec 3 '14 at 0:22
• @Scott: It's 60V peak-to-peak. +in=+15v and -in=-15v gives 30v between inputs. Do that the other way, and you get -30v. The difference between the two extremes is 60v, hence 60Vp-p. – AaronD Dec 3 '14 at 0:29
• Is the Johnson noise of your divider the dominant noise source in the circuit? (I think the noise from a divider, driven by a noiseless voltage source is the parallel combo of the R's) If not you can mostly likely forget it.. otherwise cut down the R5,6 values. (What's the 6k on the ADC input about?) And I don't understand your answer to the 60 Vp-p question. – George Herold Dec 3 '14 at 2:07
• @GeorgeHerold: The 6k comes from the ADC's datasheet; it's a minimum spec with no maximum. My answer to the 60Vp-p question was just to clarify how I got it. I'm not sure how to figure Johnson (thermal) noise, but that was the concern compared to the opamp's 4.5nV/rtHz equivalent input noise and the 24-bit ADC's 103dB dynamic range. – AaronD Dec 3 '14 at 2:29

I assume that 6K is intended to represent the input impedance of the ADC, so it doesn't contribute noise.

The R5/R6 combination contributes 2.0nV/sqrt(Hz) at the ADC input. The output of the op-amp has noise of about 18nV/sqrt(Hz) (mostly due to Johnson noise in the resistors), so about 1.9nV/sqrt(Hz) at the divider output, or a total of 2.8nV/sqrt(Hz) at the ADC input. In other words the two contributions are similar (divider and amplifier).

I encourage you to do a full noise analysis and look at the various contributions individually. Since they add in quadrature, the magnitudes of the squared values are important. This Analog Devices paper gives you the information you need. I may have made an error, but when I do the analysis with an attenuator made with 10K/546.5 ohm resistors rather than 10K/5K and no divider (gain is 0.0546 in both cases), I get a noise for your circuit of 51nV/sqrt(Hz) referred to the input of the diffamp vs. 119nV/sqrt(Hz) for the no-divider circuit, corresponding to 2.8nV/sqrt(Hz) vs. 6.49nV/sqrt(Hz) at the ADC input.

All this stuff refers to the region above the voltage and current noise corners, of course, which should be valid for most of the audio spectrum (but it never seems to be for the stuff I do).

• Thanks for the reference. I'll probably do through that several times before I'm done with this project. If your math is right, it looks like the hot opamp into a divider is better in this case, but they're both well below the ADC's noise floor anyway. As long as I don't do something stupid with the layout or power supply, it should be fine. Right? – AaronD Dec 3 '14 at 3:42
• Yes, layout and power supply (and reference noise) are probably of a lot more concern given the dynamic range of your ADC. There are 24-bit ADCs where it would be noticeable. – Spehro Pefhany Dec 3 '14 at 3:48
• As far as I can tell the 6k is in series with the signal..(as input protection??) and so that dominates the noise. But nice link. S. Franco's opamp book is good for noise analysis. – George Herold Dec 3 '14 at 17:23

Why in the world are you worried about Johnson noise?

Consider that the LM833 has a power bandwidth less than 1 MHz. Assuming your ADC has a 3 V input and 16 bit resolution, 1 lsb is ~46 µV. A 1 kΩ resistor with a 1 MHz bandwidth has a Johnson noise voltage of ~4 µV, and noise scales as the square root of the resistance. It would take a 529 kΩ resistor to give you 1 lsb of noise.

You don't say what your desired input impedance is. Let's say that it's 1 MΩ. You could use a 475kΩ/50kΩ/475kΩ divider network to get your input down to 3 Vp-p, followed by a simple unity-gain differential receiver, and still have ⅓ lsb of noise. And, of course, any lower value of input resistance will directly cause a lowering of resistor noise.

I really don't see the problem.

• Part of the problem is that it's a 24-bit ADC with 103dB of dynamic range. With an input range of 70% supply at 3.3V (this all comes from the datasheet), it can measure ~16uV with some semblance of accuracy. ~4uV is ~12dB down from there, so probably not noticeable on its own. – AaronD Dec 3 '14 at 2:58
• @AaronD - So, what's your sample rate? – WhatRoughBeast Dec 3 '14 at 3:02
• Variable between 44kHz and 96kHz. – AaronD Dec 3 '14 at 3:03
• @AaronD - OK, so 96 kHz says maximum analog bandwidth of 200 kHz (or maybe a bit more if you don't want to invest a lot on input filter. 1kohn at 200 kHz is 1.8 uV rms. That's just about the input noise (4 nV / Root Hz) of the LM833. So any input attenuator with a total resistance less than 20 kohm will produce less noise than the op amps. – WhatRoughBeast Dec 3 '14 at 3:54