Op Amps, Single Supplies, and Virtual Grounds?

[barefoot]

Hi all,

I'm designing a circuit using integrated op amps and a single supply (no choice in the matter). I want to stick with the dual supply devices that I know and love, like the OPA2604. So, I'm looking for a high-end solution to the whole biasing and virtual ground issue.

Seems like there should be some really nice way to use active devices to create a virtual ground. It might even be an opportunity to achieve higher power supply noise rejection. Anyone know of any good examples I might work from? The circuit blocks will be primarily non-inverting gain stages.

Thanks,
Thomas

 

[clintrubber]

Two resistors in series between supply & ground, say 47k each. Put a cap across the lower one, say 47uF. Feed the tap to the non-inverting input of a unity-gain stable opamp. (you're adding this opamp to make a filtered and low-impedance middle ground to 'service' the other opamp-circuits.) Connect output of this opamp to the inverting input. This output is also your virtual ground.

It can be done, is simple & works fine. You may have it already seen in the various opamp-info you downloaded from the net - for which you found the links in one of the META's here. :wink:

 

[BradAvenson]

Check this out:
http://tangentsoft.net/elec/vgrounds.html

It seems to be a pretty good overview.

 

[PRR]

> the dual supply devices that I know and love

What dual-supply devices? Which ones have an explicit ground connection?

> solution to the whole biasing and virtual ground issue.

You do NOT want a virtual ground here. It just sneaks crap around the system.

You want a DC Bias source rooughly halfway between the rails. Return all the feedback loops and loads to real-ground with capacitors.

 

[barefoot]

[quote author="clintrubber"]Two resistors in series between supply & ground.... [/quote]
Yeah, I was just wondering if there might be something better, more stable or whatever.

Thanks!

[quote author="BradAvenson"]Check this out:
http://tangentsoft.net/elec/vgrounds.html [/quote]
Cool, thanks!

[quote author="PRR"]
What dual-supply devices? Which ones have an explicit ground connection? [/quote]
Not explicitly dual supply, of course. But typical implementations use dual supplies.


[quote author="PRR"]
You do NOT want a virtual ground here. It just sneaks crap around the system.

You want a DC Bias source rooughly halfway between the rails. Return all the feedback loops and loads to real-ground with capacitors.[/quote]
Can you show an example?

Thanks :grin:

 

[barefoot]

Are you talking about something like this?

 

[bcarso]

Sorta. You would do well to bypass the first voltage divider as well.

One of the problems with the buffered half-supply suggested earlier is the noise introduced by the buffer amp. However, there are ways to filter this at the output while preventing the amp from going ballistic. In many cases the little bit of extra noise is negligible.

 

[TedF]

If it's just one or two devices that you need to bias, then set up a couple of 150K resistors in series from rail to ground, by-pass the lower resistor with say a 1uF cap. Use the 'centre' point on the pos input of any lowish noise opamp. THEN connect a 680 ohm resistor from output to neg input.

NOW use the NEG INPUT of the opamp as the ground reference point for your circuit..... Now THAT'S smooth.

I'll let PRR explain the niceties.

 

[clintrubber]

virtual ground

Just realized that it was clear what the first question was after from the context, but to prevent (or add...) eventual confusion:

Isn't the term virtual ground better reserved for say what you get when looking directly into the inv-input of an opamp with feedback ?

To differentiate we could perhaps better call that point inbetween the supply rails a 'phantom ground'. I have little hope of this term becoming widely used, but still...

NOW use the NEG INPUT of the opamp as the ground reference point for your circuit..... Now THAT'S smooth.

That's interesting, and it actually answers Thomas' question exactly :wink: : he asked for a virtual ground and there he got it.

I'm fairly sure now this post will confuse, not unconfuse. :evil: :wink:

 

[barefoot]

[quote author="bcarso"]Sorta. You would do well to bypass the first voltage divider as well. [/quote]
I'm not sure what you mean? a capacitor to ground from the center point of the divider?

[quote author="TedF"]If it's just one or two devices that you need to bias, then set up a couple of 150K resistors in series from rail to ground, by-pass the lower resistor with say a 1uF cap. Use the 'centre' point on the pos input of any lowish noise opamp. THEN connect a 680 ohm resistor from output to neg input.

NOW use the NEG INPUT of the opamp as the ground reference point for your circuit..... Now THAT'S smooth.[/quote]
Don't you also need a resistor going from the negative input to ground? Otherwise the virtual ground impedance is always 680 Ohms, no?

And why is this better than just using the op amp output as the virtual ground?

 

[bcarso]

[quote author="barefoot"][quote author="bcarso"]Sorta. You would do well to bypass the first voltage divider as well. [/quote]
I'm not sure what you mean? a capacitor to ground from the center point of the divider?
[/quote]

Yes.

As far as Ted's suggestion: the action of negative feedback will make the impedance at the inverting input low, based on the frequency-dependent open loop gain of the op amp.

Many are the ways.

 

[TedF]

Answering Barefoot.... no, you don't need a resistor to ground from the neg.... that would just add noise.
If you use the opamp output as 'ground' you get the noise of the opamp appearing on the 'ground'.
The neg input is close to a perfect noiseless ground; absolutely zero impedance.... until you try to drag too much current from it.

 

[Johan]

[quote author="TedF"]If it's just one or two devices that you need to bias, then set up a couple of 150K resistors in series from rail to ground, by-pass the lower resistor with say a 1uF cap. Use the 'centre' point on the pos input of any lowish noise opamp. THEN connect a 680 ohm resistor from output to neg input.

NOW use the NEG INPUT of the opamp as the ground reference point for your circuit..... Now THAT'S smooth.

I'll let PRR explain the niceties.[/quote]

..stupid question perhaps ( and slightly different matter ), but I have to ask... how else can I learn?..

if I didnt used the neg-input for ground but for output and then fed a signal to the pos-input ( instead of the resistor/cap combo ), would that make this a noiseless 680ohm buffer?...
or did I get it all wrong?...

johan

 

[TedF]

Hmmmm.... Yes Johan, but it would have limited use: One way to think of an opamp is that when it's operating correctly the inputs will mirror one another exactly, so yes, you could feed a signal into the pos input and it would be a beautifully accurate unity gain buffer, but with an output current limited by ability of the opamp to drive the 680 ohm resistor.... If you make the resistor 0 ohms, then .... ha... suddenly it's a follower! :?
(but you've lost the noise buffering effect)

 

[Johan]

[quote author="TedF"]Hmmmm.... Yes Johan, but it would have limited use: One way to think of an opamp is that when it's operating correctly the inputs will mirror one another exactly, so yes, you could feed a signal into the pos input and it would be a beautifully accurate unity gain buffer, but with an output current limited by ability of the opamp to drive the 680 ohm resistor.... If you make the resistor 0 ohms, then .... ha... suddenly it's a follower! :?
(but you've lost the noise buffering effect)[/quote]

thanks Ted
the reason i ask is, I often have long cables going through hostile enviroment between guitar and hi-gain guitaramps so its high-z input..I'm not trying to drive heavy loads, just fighting the cables and the inviroment.. ( oh..that might not have come out right.. :? ) a regular follower improves things, but if it could also be in itself noiseless...that would be GREAT even if its not a perfect buffer as far a current is conserned..guess its time for the breadboard..

thank's again

johan

 

[barefoot]

[quote author="TedF"]Answering Barefoot.... no, you don't need a resistor to ground from the neg.... that would just add noise.
If you use the opamp output as 'ground' you get the noise of the opamp appearing on the 'ground'.
The neg input is close to a perfect noiseless ground; absolutely zero impedance.... until you try to drag too much current from it. [/quote]
I get it. :grin:

So, now the question becomes, which way is best?

PRR says NO to the virtual ground because "it just sneaks crap around the system".

Anyone have a counter argument? Any downsides to capacitive coupling with respect to linearity? I don't really need low frequency extension below about 100Hz for this application, so performance near DC isn't a consideration.

 

[PRR]

> set up a couple of 150K resistors in series from rail to ground, by-pass the lower resistor with say a 1uF cap. Use the 'centre' point on the pos input of any lowish noise opamp. THEN connect a 680 ohm resistor from output to neg input. NOW use the NEG INPUT of the opamp as the ground reference point for your circuit..... Now THAT'S smooth. I'll let PRR explain the niceties.

I would if I could. I just don't see what you are getting at. This? (Ignore the dotted-box for now.)

My gut feeling is that the voltage at opamp "-" input is the voltage on the 1uFd cap, plus the differential noise of the opamp, unless the opamp is falling out of feedback (which won't happen). It is little different than taking the output from the output: the output adds the noise of the 680 resistor.

I don't trust my gut and I know this SPICE rig won't compute the noise without a ton of work. So I inject 1mV of 1KHz sine in series with an opamp input, where the noise appears. Noise won't be 1mV and won't be just 1KHz, but if this injection gets to the output, noise will too. It does, at unity gain. Noise at the opamp "-" input is 1uFd cap noise plus opamp noise. Output resistance is low (until the voltage drop in 680 ohms smacks the opamp to a rail), but it would be low at the output pin too.

How to kill that noise? Maybe we could hang a cap on that pin. But now the combined amp and R-C network is unity gain to ~1KHz, rises 32dB at ~3KHz, before falling off. The opamp is gain plus an R-C network, we add another R-C network, we get enough phase-shift to get a huge peak. This birdie tweets.

We can hang a passive R-C from the "-" input, or from the output, makes little difference. Now it gets audio-clean, but has real resistance.

Where am I missing a point???

 

[PRR]

> PRR says NO to the virtual ground

Depends on your definition of "virtual ground".

If your V/2 reference only carries small DC and miniscule AC current, OK.

But some folks dump heavy current, even headphone current, into a derived "ground". This often makes problems, and often where you don't expect trouble.

Look at today's posts in NYDave's preamp thread. He posted a plan that is, IMHO, nearly right. The derived V/2 reference supplies bias voltage to inputs. Outputs are cap-coupled, returned to ground. Feedback nets end in a cap to ground. The only place I think he has trouble is the Volume pot, returned to derived V/2 reference. Maximum bass attenuation is not high, and the 10K pot throws bass into the reference where it gets back into the input. He can upsize the reference bypass cap to huge size; or he could add two small coupling caps and hard-ground the Volume pot the way it should be.

So not a "virtual ground" that you tend to treat AS ground you can dump stuff into. No, a bias supply just used to set DC levels.

 

[TedF]

PRR, the problem with using the OUTPUT of an opamp as a reference point is that the impedance is not necessarily zero, and the 'ground' is the reference ground plus the noise generated in the chip.
If the open loop gain of the opamp is virtually infinite, then the -ve input will follow the +ve input precisely.... both in terms of voltage; AND noise.
This eliminates the chip noise... it is cancelled out by the topology.
ALSO, all the time the chip is within its normal operating parameters, the impedance at the -ve input is ZERO. (well, MIGHTY close to it!)

We used the concept for the first time in 1975 or thereabouts; I think it was a brainwave of Steve Dove. It was all about getting super-clean ground references on very large consoles where there was switching flying about... the very early days of logic control...... Ah, those were the days; masses of diode logic and not a processor in sight!

 

[bcarso]

You will still see the voltage noise of the op amp at the inputs (or between if you will), I believe.

See for example the discussion about the noise at feedback amplifier inputs in Arbel, Analog Signal Processing and Instrumentation, Cambridge UP, 1980, pp. 72-74.

Also, the impedance is not all that low as the frequency goes up. It can be made to look ~pure resistive over a fairly big range of frequencies with the proper choice of a feedback C across the feedback R.

EDIT: For example: take an LF347 that behaves like my circuitmaker model, with a 680 ohm feedback R. The inverting input looks like a resistor of about 7 milliohms to about the -3dB open-loop gain point of ~30Hz, after which the input looks inductive, with an inductive reactance of about 1 ohm at 4.6kHz, or a synthetic L of about 35uH.

Strap a C across the R of 7.8uF and the synthetic resistance stays about constant to about 460 Hz.

Put a perfect buffer between the real op amp output and the feedback network to get rid of the op amp's open-loop output Z, and you get flat R of the initial amount out to around the gain-bandwidth crossover frequency.