Noiseless Biasing, 470k or 4.7M series resistor?

[matt239]

I've seen R.G.'s noiseless biasing described as using a 470k or a 4.7M series resistor after the 10k/10k voltage divider. Which is it?
How does the value of this resistor affect impedance? How is this calculated??
Thanks, all. 

[.Mike]

Hi Matt,

The answer changes depending on a lot of variables.

R.G. has an excellent explanation of what is involved calculating the answer to your question.

If it is beyond your understanding (not trying to insult, it was beyond mine for a long time), then post a schematic. Someone will probably be able to walk you through how to calculate a solution.

Mike

[merlinb]

How does the value of this resistor affect impedance?

The impedance of what?

[R.G.]

I've seen R.G.'s noiseless biasing described as using a 470k or a 4.7M series resistor after the 10k/10k voltage divider. Which is it?
How does the value of this resistor affect impedance? How is this calculated??
Thanks, all. 

I picked up that circuit from a textbook on low noise electronic techniques, and I doubt it was original in that textbook. The essence of the technique is to make the divider carry a current which is much larger than the current through the series resistor, but to bypass the bias voltage, whatever it is, to ground with a capacitor. This shunts any noise present at the bias voltage junction to ground. As a side effect, it also filters any noise that was present on the power supply that might have gotten into the bias voltage.

The series resistor was described as needing to be "high valued". In practice, the bias voltage end of the series resistor is at AC ground, held there by the capacitor bypassing the voltage divider. The device end of the series resistor then has an impedance effectively equal to the resistor. This appears in parallel with the input impedance of the device itself, and this determines the impedance at the device input.

Whether the series resistor is 4.7M, 470K, or 10K, or whatever, depends on the active device, how much current you need to bias it, and what impedance you have to have. If the active device happens to be a FET or FET-input opamp, or vacuum triode, the device impedance is effectively infinite at low frequencies, so the impedance you get is effectively just the series resistor. If the device is a bipolar, its input impedance depends on the device, its gain, any unbypassed emitter resistance, any overall feedback, and the phase of the moon.    So the overall input impedance at that point is the circuit input impedance in parallel with the biasing series resistor.

The series resistor should usually be as high as you can get away with for other reasons to get a high input impedance if that matters in the circuit application. In most guitar audio inputs, 1M - 2M is enough. There will usually be a pulldown resistor outside the capacitor which feeds audio to the device end of the biasing resistor. This pulldown appears in parallel to the series resistor and device input for determining input impedance to the outside world. All resistors have a thermal noise depending on their resistance and the absolute temperature. Getting high impedance with high resistance means more noise, so using too high a resistor causes other problems. This is usually not a big problem, and there is often not much you can do about it, but you ought to know if you are worried about calcuating things.

For a series biasing resistor feeding a bipolar input, the active device needs some actual current into the base. So the voltage which appears at the transistor base is lower (for NPNs) than the bias voltage string voltage on the other side of the series resistor by an amount equal to the input bias current times the series resistor. If the device has a very high gain and low input current, the voltage drop is minor. If it has lower gain and needs more current, the series resistor drops some significant voltage.

So to net it out: that series resistor value is set by (1) how much current you have to feed to the active device to bias it (2) what resistor string voltage you can allow for biasing (3) how much the series resistor value affects noise in your circuit and (4) what value of impedance you're trying to get. Usually one of these requirements predominates, and it determines a limit on the series resistor. Knowing which factor predominates requires knowing what you want your circuit to do.

[matt239]

R.G. has an excellent explanation of what is involved calculating the answer to your question.

If it is beyond your understanding (not trying to insult, it was beyond mine for a long time), then post a schematic. Someone will probably be able to walk you through how to calculate a solution.

No offense taken I don't understand fully yet, but I hope to. Thanks for the link!

Thanks for your detailed answer R.G. I understand some of what you're saying, (but it hurts my head   )

So, lets say I want to build an input stage for a guitar pedal, using 1/2 an LM833. Pull down resistor is probably 1M?, series input resistor?, cap? unknown, single 9V supply, so need to bias to 4.5V
Gain yet to be determined, do not wish to allow op-amp to clip to rails.. x5? x10?

[matt239]

So just using the 10k/10k voltage divider, & a 1M series resistor for the bias network would be a pretty safe bet for a guitar input right?

[R.G.]

I would start there, yes.

[amptramp]

Don't forget, the input resistor is in parallel with the source resistance above frequencies where the input coupling capacitor permits, so I usually use a large coupling capacitor to push the turnover point well into the subaudible region.  I hear no audio below 35 Hz, so I usually use something at most one tenth of that frequency if I can.  The effective noise resistance is the parallel resistance of the output resistance of the previous stage in parallel with the input resistance.  If you are using a guitar input, you may have a 500K volume pot on the guitar fed by a ~7K pickup so the parallel resistance at the highest point (just above half resistance) is 126.75K.  A 1 M resistor at the input would give 112.49167K.  A 10 M resistor would bring the value closer to 126K.  If the guitar volume control was set to the maximum, the input resistance would be in parallel with the 500K level control and the 7K pickup impedance, which would be just under 7K.

Moral of the story:
1. Don't obsess over the input resistor about noise
2. Set the system up so the guitar volume control is set at or near maximum unless there is a reason to make it variable at the guitar.

[R.G.]

1. Don't obsess over the input resistor about noise

Correct. The reason to worry about input resistance is treble loss. You want as high an input impedance as is practical, considering all the various issues, so that the inductive nature of the pickup doesn't cause loss of more treble than bass. Or at least any more than you can stand.

We can't change this for historical reasons, but this would all be a lot simpler if the guitar pickup could be buffered at the pickup, before any tone controls or cabling. A lot of hum, noise, and tone issues would vanish.

[PRR]

> I doubt it was original in that textbook.

I'm helping-out on a similar circuit: a "Battery Eliminator" from the 1920s.

Early radio-sets "had" to run on batteries. Filtering AC to low-hum DC was very costly, radio power demands were low, 22V 67V 90V and even 135V batts were common items.

Of course battery hassle gets annoying, part prices fell, so someone built a lamp-socket powered DC supply.

Power transformer, tube rectifier, choke, capacitor.

Then since battery radios come in many voltages, a "bias divider" to supply the various voltages which might be required. This was actually one big resistor with several taps:



> make the divider carry a current which is much larger than the current through...

We want divider current "much larger", but we don't want to burn the house down, run-up the electric bill, or pay more for the battery eliminator than we were paying for batteries.

Current to the radio could be 0.1mA or 10mA.

In this case: power from the lamp-socket is "cheap", but DC power from a rectifier is $1 for 30mA (type '01 tube) or $10 for 100mA (type '80 tube).

Therefore this was designed for 25mA-30mA.

Long-long before your lo-noise textbook.

[matt239]

OK how does this look?

http://www.aronnelson.com/gallery/main.php/v/Schematics-etc/Input+Section.JPG.html?g2_imageViewsIndex=1

[sault]

Thank you, RG! Reading that link gave me a lot of good information!

Now, because the voltage reference is connected to an op-amp, at most its going to pull a few nano-amps ("input bias current" according to TL072 datasheet), so that's not a concern, and since there's a serial 680k resistor the input impedance shouldn't be an issue.... so no problems, almost any value of resistor over, say, 10k ohms would work, right?

The question that I have is that of the cap in relation to the divider. As is, it looks like it forms a high-pass around a few hz.... does this frequency cut-off matter?

Oh yeah, second question - does that serial 1k resistor (R3, I mean) do anything significant?

[R.G.]

Sorry - which link?

[DavenPaget]

OK how does this look?

http://www.aronnelson.com/gallery/main.php/v/Schematics-etc/Input+Section.JPG.html?g2_imageViewsIndex=1

Go harder on the input resistors , it's a JFET-input opamp .
1M isn't of much worry anyway .

[sault]

RG - this one http://www.geofex.com/circuits/biasnet.htm. Effing awesome. Just when I thought I "got" it, I learn something new!



My two questions remain...

Any specific reason to use the R3 resistor? And does it matter what the cap value on the voltage divider is?

[R.G.]

Any specific reason to use the R3 resistor?

If you mean the 1K in series at the input, yes. It's there to help limit current into the opamp in fault conditions.

See http://geofex.com/circuits/what_are_all_those_parts_for.htm

And does it matter what the cap value on the voltage divider is?

Yes. The cap to ground at the voltage divider has to be large enough to make the resistive voltage divider look like ground at all signal frequencies of interest. It  needs to be a small impedance compared to the bias resistor, which is easy, and also compared to the bias string, which is much harder. Make it big enough to prevent power supply noise from being felt at the middle of the bias string.

[matt239]

OK how does this look? 

http://www.aronnelson.com/gallery/main.php/v/Schematics-etc/Input+Section.JPG.html?g2_imageViewsIndex=1

[sault]

If you mean the 1K in series at the input, yes. It's there to help limit current into the opamp in fault conditions.

Makes sense. Kinda like a resistor into a tube grid. The capacitance makes sense, too.

And thank you for another article! I don't know how I managed to miss that one, I thought I'd already swept through most of the Geofex articles... awesome. Some of that I knew, some I definitely did not (the diodes at the end, for instance).

[R.G.]

OK how does this look? 

I would
(1) Change R4 to about 2M
(2) Move R4 outside C1, to the signal source side. This prevents R4 and R5 from fighting to try to move the input to ground and Vr at the same time, and ending up halfway between.
(3) Change R5 to 1M to 2M. Exact value probably not important.
(4) Think about changing C1 to maybe 0.22 or even 0.1uF.
(5) Add a new capacitor in front of R8 to break the DC path from the opamp.
(6) Add a new capacitor between R2 and ground to break the DC path for amplification, so the opamp doesn't try to amplify the Vr dc level. Make this cap bigger than C = 1/(2*pi*1K*F) where F is the lowest frequency you want to pass through relatively un-attenuated. For guitar, F is about 82Hz, so the cap has to be bigger than 1.9uF; for bass at 41Hz, bigger than 3.9uF. 10uF is a good place to start, unless you WANT to attenuate bass response.
(7) Think about and experiment with removing D1, D2, and R9. When the opamp is running as a linear amplifier, there will be only millivolts between pins 2 and 3, held that way by feedback. The diodes won't conduct until something goes way outside normal linear operation, and maybe not then.

[matt239]

OK now it looks like this:
http://www.aronnelson.com/gallery/main.php/v/Schematics-etc/Input+v3.JPG.html

I couldn't get it working at all until I added C6 from R2 to ground as you suggested. I had thought this was an optional part.. Thought I'd seen schems without it.
Maybe they had bi-polar supplies instead of biasing? Do you need this cap just for the circuit to function?

Do I need to lower the value of the resistors in the voltage divider? (R6,R7) - If I use op-amps with bi-polar inputs later in the circuit, would I need more bias current than these can provide?
(Let's say 4 op-amp sections, from 2 dual devices, a TL072, & an LM833.)
I could make R6 & R7 100k each? (Or should I make a reference voltage using an op-amp??)

I don't expect current to flow in D1,D2 under normal conditions, it's just a protection clamp. I just guessed at a value for R9 though.
I was thinking: if someone plugs the output of another pedal into this input, rather than just a guitar, there could be a few volts of signal. - I don't want these diodes to clip/clamp audibly with such a signal, nor the op-amp, so I added R9 to raise the threshold, & soften any clipping. What value should I make R9?
Maybe I should use LEDs for the clamp?