AMZ Super Buffer Question

[Agung Kurniawan]

I have built AMZ Super Buffer but only 2 opamp section. I place this one inside my guitar cause buffer sounds best near the signal source. It tottaly working, it bring the highs on, but i cant figure the value of the output impedance. Anyone know the output impedance on AMZ Super Buffer?

[antonis]

About 50 Ohms...
(two 100 Ohms isolating resistors in parallel)

Actually, you have to count the 2M2 resistor in parallel with the impedance of what comes next..

[PRR]

Yes, 50 Ohms +plus+ 10uFd.

Why do you need an exact value?

[Agung Kurniawan]

Im just corious with the output impedance, what is the formula?

My amp input impedance its about 1M, so the tottal impedance would be near 47,6Kohm right?

[antonis]

Im just corious with the output impedance, what is the formula?

In this case (actually, in most of the cases) is a simple* voltage divider..
(*) It becames simple after you've calculated the total impendaces for the upper & lower divider resistors..


Any total resistance (impedance) is placed in series between source's out and final out..

My amp input impedance its about 1M, so the tottal impedance would be near 47,6Kohm right?

Can't get the meaning of "total" impedance..
(neither your calculations for the 47k6..)

As Paul said, Buffer's greatest Output impedance is 50R + 100nF +10μF, leading to 252 Ohms at 80Hz..

This value is theoretical, because is calculated under "no load" - when the buffer is connected to somewhere, the 2M2 resistor from Out to GND must be considered in parallel with whatever is the somewhere's impedance..
(You may omit the anti-pop 2M2 resistor and take in mind only the 252R & your amp's 1M to form the voltage divider..)

With the upper calculations, you have a divider of 252/687500 at the lowest frequency of interest.
(when freq raises, impendances of caps are getting lower -> less output impedance -> less "loss" on divider..)

[Agung Kurniawan]

Ok i get it now, but there is still  somethink disturb me about impedance.
I put my buffer inside the guitar, and i plug the guitar straight to the amp and then i use send return jack for my fx chain. There is a switch for engage the buffer on my guitar. If i engage the buffer, i can hear the highs. But when i disengage the buffer, the highs is gone.
Why did this happen when i have 1M amp input impedance?

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[R.G.]

Ok i get it now, but there is still  somethink disturb me about impedance.
I put my buffer inside the guitar, and i plug the guitar straight to the amp and then i use send return jack for my fx chain. There is a switch for engage the buffer on my guitar. If i engage the buffer, i can hear the highs. But when i disengage the buffer, the highs is gone.
Why did this happen when i have 1M amp input impedance?

It's a matter of details.

The reason highs get cut from guitar inputs is that a guitar pickup is a high-value inductor. Single coil inductors are in the 1/2 to 1 >>>Henry<<< range. Humbuckers can be 2 to 4 henries. That's huge. And an inductor has an impedance that rises with frequency. So at low frequencies, the guitar signal is only going through about the 4K to 20K of resistance that the wires have. But the impedance of the inductance rises with frequency. The impedance of a 1H inductor is Zl = 2*pi*F* L = 2*pi*10,000*1 = 62.8K at 10khz and 125.6K at 20kHz.

If an input is not more than about ten times that, there will be noticeable treble loss, more at the high end than the low end. 1M is about the >>minimum<<< for not losing a lot of highs. Much greater than 1M reduces the losses further.

So it depends on whether your buffer is much higher than 1M when switched in, and whether the buffer switching circuit adds more capacitive loading to the pickups than connecting directly without the buffer circuit and its bypass.

And remember that your guitar is further loaded by about 20pf to 100pf of capacitance per foot for each foot of cable between the guitar and the amplifier. The amplifier impedance could be infinite and the cable capacitance would still form a treble-cut filter with the inductance of the pickups. That's another thing that buffers do, they make that capacitance vanish as far as the guitar pickup inductance is concerned.

[ashcat_lt]

How is that switch "engage"-ing the buffer?  It wants to be at least two poles for a True Bypass arrangement.  If you're leaving the low-Z output of the buffer connected, and just switching the input between the buffer and the jack, you'll definitely lose all kinds of treble if not also overall volume.

Otherwise, the big thing that this buffer is doing is cutting out the cable capacitance like RG said.  That will make a somewhere between a subtle and a rather noticeable difference.

Where are the V and T pots with respect to this buffer?  Before or after?  What value are they?  That could make a difference also.

[PRR]

> i have 1M amp input impedance?

AND a cable.

There is capacitance everywhere. Capacitance sucks highs. Cables have capacitance. Cables suck highs. Shorter is better.

Figure 30pFd per foot (100pFd/meter).

A 30 foot (10m) cable is a 1,000pFd capacitor.

1,000pFd "looks like" about _10K_ impedance at the top of the audio band.

Guitar impedance may be under 10K in bass but nearer 100K in treble. (R.G. explains this in more detail.)

1,000pFd working against 100K is a 1.6KHz high-cut. Most of the guitar overtones.

1,000pFd working against just 50 Ohms is (on paper) a 3,000KHz high-cut. Far far beyond the audible range. (In practice the chip's accuracy will be poor several/many Octaves above the audio band.)

You really only "need" a buffer output under 1K Ohms to cover all the guitar range plus. But modern chips make it dead-easy to get extremely low source impedances.
____________________________

> what is the formula?

You have to know stuff. Each opamp has an open-loop output impedance and a closed-loop gain excess. TL072 output may be 300 Ohms naked. But it has OL gain of over 1,000 even at the top of the guitar band. 300r/1000 is 0.3 Ohms.

Plus the resistor, 100 Ohms each opamp. Because we use 5%(2%) resistors, "100" may be 95-105 (98-102), so we don't worry too much about the estimated "0.3r" at the op-amp. We usually pencil an opamp output as "zero".

So 0.3+100 is 100 Ohms for our practical purposes.

Two in parallel should follow the Usual Rules. 50 Ohms.

If you are fighting hum/buzz, the 10uFd becomes the effective output impedance. Below 333Hz the 10uFd acts like more than 50 Ohms reactance. At 60Hz this becomes over 250 Ohms. Still "low", but I have seen cases where a solid LOW impedance damps a lot of hum/buzz induced from too-close power cables. (A better fix is don't run precious signal next to wall-power cables, keep them apart and cross at right-angles.)

[Agung Kurniawan]

How is that switch "engage"-ing the buffer?  It wants to be at least two poles for a True Bypass arrangement.

Where are the V and T pots with respect to this buffer?  Before or after?  What value are they?  That could make a difference also.

Of course, its true bypass with DPDT toggle switch.

The pot value is 500k
The route without the buffer:
Pickup-pickup selector-volume-out jack

With buffer:
Pickup-pickup selector-buffer-volume-out jack
Anythink wrong?
I just remove my tone pot, it useless for me.

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[antonis]

IMHO you should place the buffer after the Vol pot..