CD4049UBE....is it inherently noisy?

[Peter Snowberg]

http://www.physics.ucdavis.edu/Classes/Physics122/Phys122_Johnson_Noise.pdf

[Stephen]

MAN Every single circuit I see with a 4049 inverter is way over on the values in the feedback loops....beleive me this is TRUE FOR SURE...Look at the circuit man you are really driving that buffer twice as hard as you need too..In the first inverter remove the 570K I think it was and use a 100K and than us a 500k Pot...In the other three MANNN cool it it on those 270K and make them like 100k ...I have had trouble with this long time ago and became aware that I cant turn the volume up because of this!!!  I WAS DRIVING THE BUFFER WAY TOOOOOOOOOO hard!!!!  You will notice in a Llama it works plenty good half throttle and out of control over that...Why have so much resistance in the feedback loops ..You DONT need them!!! This will work!!!!!!

[B Tremblay]

This circuit works as drawn, trust me.  Changing resistor values will affect much more than gain.

[powerplayj]

I'll try to enclose it as suggested and then try the small cap if necessary.

[R.G.]

Nota Bene:

There are two ways to limit signal current to low levels. One is with a series resistor of high value. The other is with a T network. The series resistor is cheap and simple, but you have the problem of thermal noise. T networks are more complicated, but can use much smaller resistors and therefore you get less thermal noise and less parasitic capacitance effect.

The trick is to think in terms of current, not voltage. F'rinstance: at the input of the Mr.EQ schemo, there is a series 1M input resistor. This resistor deposits charge at pin 3 that has to be pulled off pin 3 by the 560K/1M pot feedback resistor. If you had 1/10th the input voltage through 1/10 the input resistor, the same current would be deposited at pin 3. So let's change the 1M resistor to a 100K. Now we have to somehow reduce the incoming signal voltage by 10:1.

How about - a resistive divider! We put a 100K/10K resistor divider on the front end, dividing the input voltage down by 10:1, and hook the newly lowered 100K series input resistor to the divider junction. Now the same amount of charge gets deposited on pin 3 as it did for a 1M input resistor, but there is no resistor over 100K being used.

The careful electronicist will be saying "But wait - the divider network and series resistor are close but not exactly the right numbers." and that's true. To get an exact 10:1 drop and exactly equivalent of 1M current, I have to juggle the values of the divider and series resistor by using Thevenin-Norton transformations (or just some math) to get the source impedance to be exactly 100K and not 109K as it is now, and to get the divider to be exactly 10:1, not 9:1 as it is now since I ignored the 100K input resistor's loading. But those are secondary effects. This discussion sets up the understanding. The fine points come later.

You can do the same thing on the feedback network. You can replace the 560K/1M pot with a 56K/100K pot driven by a low impedance divider network on the output. Now both input and feedback network have had their impedances lowered by 10:1, therefore parasitic capacitance effects are 10:1 down, and the resistors all have much lower thermal noise. Where there are DC problems, like with where do I connect the lower end of the divider network resistor, you can block DC with a capacitor. So the input 10K divider resistor goes to a capacitor to ground, as does the divider to the feedback network. The DC conditions are left unchanged and the dividers work for signal.

As a side note, this is how you get a high input impedance inverting opamp stage. You use a large resistor at Rin, perhaps the 1M that seems to be the standard, then use a T network feedback "resistor" to cut the fed back signal.

That's a place that you really want T networks. If you have a 1M Rin, and you want a gain of 10, a single feedback resistor has to be 10M. That's out at the limit of what you can get. And what do you do if you want a gain of 30? 30M resistor? There aren't any commonly available. The T network makes this work with stock resistor values.

[A.S.P.]

http://www.diystompboxes.com/smfforum/index.php?topic=47736.0

[R.G.]

...as long as you take care of the appropriate input-/output impedances of the "opamps"
involved (especially in the so-called linear "crossover-region", that is of interest here with the CMOS - inverters...)

Yep, that's right. Fortunately, CMOS inverters have an input impedance that's equal to 20 Volts worth of glass at DC. I goes down some with increasing frequency as the parasitic capacitances take over, but it's pretty good for audio, especially low audio. Also fortunately, the output impedance of a CMOS inverter is generally suitably low, especially compared to the 100K input resistors I mentioned for t he divider networks.

I don't know anything that works for every situation, so there is always a trail of special cases and things you have to watch for and look out for.

But you have to know the generalities before you can effectively cope with the special cases, yes?

[A.S.P.]

...as long as you take care of the appropriate input-/output impedances of the "opamps"
involved (especially in the so-called linear "crossover-region", that is of interest here with the CMOS - inverters...)

Yep, that's right. Fortunately, CMOS inverters have an input impedance that's equal to 20 Volts worth of glass at DC.
I goes down some with increasing frequency

I don't know anything that works for every situation, so there is always a trail of special cases and things you have to watch for and look out for. ...

just for the records. 

[R.G.]

Was there any question? I'm confused.