vari-Z buffer ...link inside..

[Johan]

..dont know how usefull this would be to others, but here is something I came up with..
http://aronnelson.com/gallery/johan/vari_z_buffer

the output is taken from the negative input, not the output...I came up with this after reading a few threads by Ted Fletcher. I dont fully understand exactly HOW it works, but it cancels the noise from the opamp and becomes VERY low noise. with the pot being 20k, you could vary the output-Z from 0-20k and still have the guitar see only high Z input

..so when would it be usefull?...imagen you want to put a buffered pedal before a FF ( for example..)..it just messes things up...put this inbetween, and you can dail in the desired Z. or you could play that same FF without loading you pickups, and still have the FF see high Z..
..so, ok..uses might be limited, but hey.I came up with it..so why not share..?..

johan

[Coriolis]

Hi Johan
Looks interesting - a nice and simple building block to experiment with.
Thanks for that. 
C

[R.G.]

Wouldn't you get a very similar thing by connecting the (-) input to the output and then using the 20K pot as a variable resistor after the opamp?

[davebungo]

It is subtley different as it will maintain unity gain irrespective of the load presented on the output but I don't understand how any noise could be "cancelled".  It would be interesting to hear the theory behind this.  Noise should not be a problem with this circuit in isolation anyway as it is low (unity) gain.

[Johan]

I got the idea from ( and in) this thread about virtual grounds, a few months back
http://www.prodigy-pro.com/forum/viewtopic.php?t=13478&postdays=0&postorder=asc&start=15

look for TedF's replies...I still dont fully get it, but it works..

here is from Ted's last reply..
"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!) "

..so yes, it assumes a perfect world, and assumption is the mother of all fxxk up's( a 741 is NOT a perfect world). but for buffering a guitarsignal, it's close enough..

johan

[davebungo]

I got the idea from ( and in) this thread about virtual grounds, a few months back
http://www.prodigy-pro.com/forum/viewtopic.php?t=13478&postdays=0&postorder=asc&start=15

look for TedF's replies...I still dont fully get it, but it works..

here is from Ted's last reply..
"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!) "

..so yes, it assumes a perfect world, and assumption is the mother of all fxxk up's( a 741 is NOT a perfect world). but for buffering a guitarsignal, it's close enough..

johan

The following is all meant to be constructive... The same could be said for a simple follower i.e. output tied to inverting input (or with the variable resistor set to 0) so I don't see how this is any different.  Have you compared the noise output with the variable resistor set at minimum and maximum?  And yes, the output impedance will appear to be very small because the op-amp will do whatever necessary (within drive, slew and bandwidth limits) to maintain very little difference between the inputs but then again this rather negates your original aim to vary the output impedance.  As R.G. mentioned you may as well just use a unity gain follower and stick the variable resistor on the output.

[Johan]

.well, it would be kind of silly of me trying to debate something i dont fully understand and perhaps I'm just fooling myself and tricking my own brain..
... and obviously, when the pot is turned down(shorted) it  is just a follower, but when it's turned up a little, the noise does get a little quieter...so something is going on...hmmm..more thinking to do...

johan

[christian]

.well, it would be kind of silly of me trying to debate something i dont fully understand and perhaps I'm just fooling myself and tricking my own brain..
... and obviously, when the pot is turned down(shorted) it  is just a follower, but when it's turned up a little, the noise does get a little quieter...so something is going on...hmmm..more thinking to do...

johan

It doesnt remove the noise from the buffer itself, but from the resistor. If you'd do this as R.G. suggested, use a resistor after normal buffer, the resistor would create some noise, but doing it this way cancels that.

Theres just a small problem if you use this before very low impedance circuit. Since the opamp doesnt care where you take the output, it may create some "internal" gain and produce clipping, or slew rate problems, etc..

ch.

[R.G.]

OK, so the idea is to eliminate the noise of a 20K resistor by folding it inside the opamp feedback loop?

A quick bit of calculation indicates that the thermal noise of a 20K resistor is 2.11nV/sqrtHz.  If we decide we're worried about 82Hz to 20kHz, that's a bandwidth of 19,918Hz, and that leads to the voltage noise being 298nV. That's 0.3uV of noise.

This adds in an RMS fashion to the noise from the opamp and the input resistor. If you use a 1M input resistor, that resistor has a thermal noise of 126nV/sqrtHz, or 17.782uV of noise for the same bandwidth. The addition of the 20K noise causes the RMS noise from the 1M resistor at the input to be increased from 17.782uV to 17.784uV (I did this by taking the square root of the sum of the squares). The noise increases by 0.01%.

Of course, that's the noise you get if you don't connect a pickup. If you connect a guitar pickup to it, the 1M is effectively paralleled by the guitar's pickup and controls, so you could get as low as the pickup impedance, which would vary from the winding resistance at DC up to well over 100K at the top end where the pickup inductance and capacitance dominates things.

LT gives a good paper on selecting opamps for low noise at http://www.linear.com/pc/downloadDocument.do?navId=H0,C1,C1154,C1009,C1026,D6539 where they show a lot of their opamps being between 1 and 10nV/sqrtHz, or 0.141uV to 1.411uV of input noise depending on which one. That's a range of 20K matters some to matters little.

The conclusion I take from this is that what you connect on the input is more important than whether the 20K output resistance is inside or outside the feedack loop. But it might be fun to play with.