Clean Op-amp Bias?

[ssashton]

I am building an audio circuit that powers the op-amps from a single 12V rail coming from an LM317. I therefore need a mid-rail bias of 6V. I am keen to make this as low-noise as possible as I am using RC4580 op-amp that have about 6nV/√ Hz noise at 1KHz.

I have thought of two options and I'm not sure which is best or if there is a superior alternative?

Option 1) LM4041 (or other?) V-ref.


Option 2) Resistor divider.


I know that the LM4041 has more voltage noise than the pure resistive divider Johnson noise, if driven by a perfect source. However, the source is not perfect and the resistive divider basically offers no rejection of the power rail noise. Is the 10uF capacitor enough to kill all audio band AC noise? I calculate it offers about -40dB attenuation at 1KHz and this will be subtracted from the voltage noise of the LM317 power rail. I'm really not sure how much voltage noise the LM317 outputs so I get a bit stuck here, especially trying to stay with the metric of nV/√ Hz for direct comparison to the op-amp self noise. Simply put, I'm afraid of ruining the low noise of the op-amps by directly injecting supply noise in to the signal!

[Zero999]

Why do you need to buffer it? How much current does it need to supply?

And here's a better way to connect two op-amps in parallel. The output impedance will be much lower than the two output resistors in parallel because of negative feedback.


Oh and is this a virtual ground circuit or not? If all of the voltages are relative to the bias point, then the capacitor doesn't make much difference. It reduces the noise on the negative side, at the expense of doubling it on the positive side.

[ssashton]

Thanks for your reply.
This Bias Out is feeding other opamps that carry the audio signal, either connected directly to the non- inverting input, or via resistors.

In my thinking, the reason to buffer is to have a high Input impedance for the filter cap. Also it doesn't seem it would do any harm having a low output impedance.

I find your updated circuit interesting! I suppose the first op-amp feedback loop will compensate the output impedance of both buffer resistors? Since the second opamp feedback loop is not connected around its buffer, but can not be or both feedback loops would be joined.

I am unsure the benefit of connecting the second opamp input to the output of the first, rather than parallel both opamp inputs. Can you give some insight on this choice?

The GND symbol is the negative output of the SMPS feeding the entire system, not a generated rail I'm calling GND, that's what my BiasOut will be, for the audio signal.

[ssashton]

So I made a little test circuit of the 'Resistive divider' schematic I shared in post #1.

It was powered like this: SMPS > LM317 > resistive divider circuit > AC coupled audio line input of computer.

In this way I can compare the noise level of the shorted line input and one fed from the bias supply. The audio input has a 2V rms input sensitivity.

First is the shorted input: -111dB noise floor.

(Click to enlarge)

Resistive divider with 10uF cap and op-amp buffer: -110dB noise floor. Slight mains harmonics.


I then tried connecting directly to the resistive divider without 10uF cap or op-amp buffer (i.e. LM317 simply followed by a divider): -96dB Noise floor. Wow, that is a lot worse!


This shows me the op-amp with the 10uF capacitor helps greatly to reduce audio band AC muck. It also shows me the cleaned up bias is at least good enough not to affect the noise floor of my 2V rms audio interface line input.

FFT of course averages away random noise, so it is difficult to say if the slight increase in noise floor with the bias signal is due to those 50Hz mains harmonics or random AC noise.

It seems to me I should focus on the filtering at the op-amp input to get something super clean, rather than worry about the source of that voltage (i.e. voltage reference, high performance regulator or just resistive divider). Perhaps I'll do a 2nd order filter to be confident about it.

[David Hess]

I am unsure the benefit of connecting the second opamp input to the output of the first, rather than parallel both opamp inputs. Can you give some insight on this choice?

In your example, the operational amplifiers in parallel with 22 ohm output series resistors provide an 11 ohm output impedance.

In the example shown by Zero999, U5.1 encloses the 22 ohm output series resistor within its feedback loop, so the output resistance is divided by the open loop gain for practically zero output resistance.  U5.2 then doubles the output current achieving the same output current as if the two operational amplifiers are in parallel.

[MasterT]

 Priority is "the noise", original circuits with 2 OPA in parallel would have  4nv / sqrt(2) or so.  Second circuits multiplyes both OPA's noise sources and likely to have 4nV x 2. 
 Better not to use second op-amp at all, or combine both outputs via 1 OHm resisors, since offset voltage a few mV only and likely to drift in the same direction (in dual package).

 Some examples:

[David Hess]

Priority is "the noise", original circuits with 2 OPA in parallel would have  4nv / sqrt(2) or so.  Second circuits multiplyes both OPA's noise sources and likely to have 4nV x 2.

Since the second operational amplifier is within the feedback loop of the first one, part of its noise is removed.  Reference noise or noise from the input filter will dominate anyway except at high frequencies, and at high frequencies low pass filtering on the output will remove high frequency noise.
 

Better not to use second op-amp at all, or combine both outputs via 1 OHm resisors, since offset voltage a few mV only and likely to drift in the same direction (in dual package).

I would not use that configuration at all.

An operational amplifier can usually operate with a large enough capacitance connected directly across its output, relying on dominant mode compensation because of its output resistance.  This provides low AC impedance from the capacitance, low DC impedance from the operational amplifier, and a minimum of added noise.  Some experimentation will be required to find the output capacitance.

If more current is required, then I would add a current buffer inside of the operational amplifier's feedback loop, and still directly drive a larger output capacitance.  4-quadrant power supplies work this way, and closely resemble class-ab audio power amplifiers.  The buffer could be a single transistor, a 2 or 4 transistor diamond buffer, an LT1010, or a current feedback operational amplifier.

I wonder what the noise performance the below configuration is.  It would be simplified for this application since no gain is required.

[Zero999]

I don't understand why you want to buffer the bias point. Normally the bias point only passes the tiny DC input bias currents from the op-amps. I normally select resistor values to make the impedance the same as the other input, but it's not really necessary.


The main application for a buffered bias point is a as a split supply/virtual ground/earth, in which case all of the inputs and outputs are referenced to the output of the buffer. This can be useful as it eliminates the need for AC coupling capacitors. Here's an example showing a theoretical power op-amp driving a low impedance load into a virtual ground. This can be used in a headphone amplifier circuit.


EDIT:
I missed another reason for a buffered reference: an instrumentation amplifier's refeence pin!.

[ssashton]

I suppose you are selecting bias resistor values that equal the other resistors around the op-amp in order to maintain CMRR?

In my case this Bias Out is feeding about 9 op-amp sections, so I can not match the bias network output Z to a particular value as far as I see.

This is an example of how the bias is being used.




P.s. The bias opamp is not happy with 10uF directly on the output, without a buffer resistor.

[Zero999]

Why not use an instrumenation amplifier? A differential amplifier needs closely matched resistors for a high CMRR.

As per the edit to my previous post, a buffered reference can be used to drive the reference pin of an instrumentation amplifier IC, in a single supply application.

[ssashton]

I've not seen them used in audio applications TBH. I suppose there must be a good reason such as noise or distortion levels? There is the SSM2143, which has excellent CMRR but noise and THD are unremarkable. Cost is also significantly higher than 'normal' opamps like the RC4580 or even LM4562.

[Zero999]

That's not a propper instrumentation amplifier.

Instrumenation amplifier ICs are very common in balanced microphone amplifiers, when high common mode rejection ratio is essential to reduce pick-up from external noise sources such as mains hum. If low noise is important, then the INA217 might be suitable.
https://www.ti.com/jp/lit/ds/symlink/ina217.pdf

The op-amp can have very low noise, but it's no good if the mismatch in resistors result in a crappy CMRR, causing it to pick-up noise.

Here's an applicaion note which deals with the effect of resistor mismatch on diffeential amplifier CMRR.
https://www.ti.com/lit/an/sboa582/sboa582.pdf

[ssashton]

That looks like a nice part, I'll remeber it. For this project it is a bit expensive for the cmrr benefit.

[Zero999]

I suppose you are selecting bias resistor values that equal the other resistors around the op-amp in order to maintain CMRR?

It's not really important for AC signals, because the capacitors bypasses the resistors to ground at AC.

In my case this Bias Out is feeding about 9 op-amp sections, so I can not match the bias network output Z to a particular value as far as I see.

You could use one potential divider for all of them, just make the capacitor large enough for the cut-off frequency to be well below 20Hz, with all the bias resistors in parallel.

The normal way to do this is to use a separate bias network for each amplifier. It's more parts but reduces the crosstalk.

Alternatively, share a bias network for amplifiers where crosstalk isn't a big deal.

P.s. The bias opamp is not happy with 10uF directly on the output, without a buffer resistor.

That's hardly surprising, but why would you connect the output to ground via a such a large capacitor anyway?