How do I calculate the current draw on a bias network/ virtual ground?

[Max999]

I have been reading R.G's excellent article on Geofex called "Designing Bias Supply (Vbias or Vb) Networks for Effects", found here http://www.geofex.com/circuits/Biasnet.htm

Another article I have read about this topic is Tangensofts "Virtual Ground Circuits", found here http://tangentsoft.net/elec/vgrounds.html

I thought I understood what the current draw on the biasing network was ( while biasing an opamp) after reading R.G's article: the input bias current spec found in the opamps datasheet.
But then I went to the Tangensoft article, and it has a diagram and text that explains that the load also draws current from the voltage divider, using this diagram and text:




"The 1 mV battery (Vos) simulates the op-amp's input offset voltage. This is a reasonable value for an OPA132, though it does vary between chips in practice.

This offset forces 1 mV across R3. Because op-amps always force their input voltages to be equal, this in turn forces 10 mV across R4. As you can see, this puts 11 mV of DC across the load. If the load is 32 Ω at DC (such as a pair of Grado SR-60s), 0.34 mA is forced through the load. This current can only come from the rail splitter, which looks like two parallel resistors to the load. Ohm's law tells us that since the current is 0.34 mA and the resistance is 2.35 kΩ (two 4.7 kΩ resistors in parallel), the voltage at the midpoint of the divider is forced ~0.8 V away from the ideal midpoint."

But we in the stompbox world use a capacitor before the load, and we don't reference the load to the virtual ground but to the "real" ground.

I wish I knew more about physics, but this means that the Tangensoft problem is not in effect while using our topology right?
Did I understood well that R.G. explains in his article that the only current draw on the bias network is the input bias current?

[PRR]

> the Tangensoft article, ... the load also draws current from the voltage divider...

Yes; but there was a REAL problem on the Cmoy amp used with low-Z headphones and a floating battery. While the issue appeared in VTVMs it was usually mild. Some Cmoys, some chips and loads, just fell-over.

I guess the difference is: the Cmoy inadvertently returned *load current* through the bias divider. You probably are not doing that.

Not claiming any great smarts, but as the bottom of the page says it was I who drew-it-up for general understanding.

[Max999]

Wauw this is such a great forum. I missed that PRR, "The DC-perspective CMoy pocket amplifier schematic above and the original explanation of it is due to PRR of Headwize"! I liked these two articles and what do you know: one written and one inspired by a diystompbox member.

Let's see, am I returning load current through the bias divider .. Well I use an output coupling cap, so I would think that would stop any load current from reaching the bias divider. But I am not sure.

What I actually want to accomplish is create a perfect bias for an opamp, because when operated on 9v there is not much headroom anyway.
I understood R.G's explanation, but regarding the tangensoft article your explaination goes way over my head PRR. I hope to someday solve this puzzle you laid out for me in that diagram   

[PRR]

> What I actually want to accomplish is create a perfect bias for an opamp

Perfection is impossible.

What practical engineers and carpenters aim for is "good enough". Dan and I are building a roof. I think the ridge is 3/8" off center. So what? Nobody will notice. Fudging it over 3/16" would give 0.047" more attic headroom, which is nothing.

And if your truck scrapes the garage ceiling, you don't want another inch, you want a foot or two more overhead.

If your bias point is within 0.25V of "ideal", your headroom is within 1dB of maximum. In audio, that is no difference.

Also, it is not a skyscraper or moon-rocket. You can build it in your lap (bench) and poke it, nobody dies if it "falls over".

[Max999]

I envy your relaxed attitude towards designing these little boxes PRR. Maybe when if I ever come to a point where I have seen it all like you I will be just as relaxed

So with opamps the current drawn is the input bias current, correct?