Bias Network questions

[robkeeler]

I read through (many times) R.G. Keen's Bias Network article http://geofex.com/circuits/Biasnet.htm and I am still a bit clueless.  I was wondering if someone could point out exactly what should look on a datasheet (for example http://focus.ti.com/lit/ds/symlink/rc4559.pdf ).  Then how would I translate that info. 
Thanks and I really appreciate the help

[R.G.]

OK.  What we're trying to do is figure out how much current is pulled out or pushed into that bias current junction and keep it below 10% of the total current in the two bias resistors.

I pulled up the RC4559 datasheet. Looking for things that may let current into or out of the opamps, I find:

Iib - input bias current - at 25C, "typically" 40nA, but guaranteed to be less than 250nA and less than 500nA at all temps from 0C to +70C.
Rin - input resistance - at 25C, minimally 300K, "typically" 1M, no maximum given.

That's it. The most telling one is that input bias current, which tells you what you're looking for directly. With no signal at all, you need to supply something between 40nA and 500nA to the inputs to get them to stay in the linear region, which is what the datasheet means when they say the output is at 0V on a +/-15V supply.

I put quotes around the word "typical" because that's a lot like "improved" on a box of soap or toothpaste - it has no meaning that you can count on whatsoever.  Designing to "typical" on a datasheet is a beginner's mistake.

So you need to assume that the + input of a RC4559 will have to be supplied with half a microamp from somewhere to keep it at a stable bias point. And that's for each amplifier, so you need to provide that twice if you have two of them. Still, that's only one uA.

You need to add up the DC currents into and out of the bias network for EVERY SINGLE THING THAT CONNECTS TO THE BIAS POINT. The worst are bipolar transistors, which tend to be current hogs. If you read my stuff in the recent "that does each component do" posts, you know how to compute how much current the base of a BJT need to suck to hold its emitter up. (that is, it needs emitter current divided by a guess at hfe). That means you have to actually know how much curren't going through the emitter, and that means you have to know how to bias them, but that's reasonably easy. BJTs in small signal circuits typically use one to ten mA of emitter current, so they use 100uA of base current for a 10ma bias point and a hfe=100 down to a few uA for 1ma and high gain.

But go through the circuit and all them all up. PNP BJTs must have current sucked OUT of their bases, and this may offset the currents used by NPN BJTs. Opamps may have to have current supplied into their inputs or pulled out, depending on the input structure.

[robkeeler]

Yet again R.G. comes through.  Thank You!!!

sidenote did you get that picture I sent you of you, me and ben from NAMM?