Fuzz Face output impedance - make it lower

[Lino22]

I was thinking about lowering the output impedance of a FF circuit.

If i am not mistaken, it can be counted as a Volume pot || Rc. When you lower the volume level, the "top" portion of the pot the signal goes through on the way out creates a significant additional resistance in series, right? Would enlarging the 10n output pot to, say 100n help? The cap's impedance should play a role.

[PRR]

Which one is Rc?



The 470r collector resistor is already a fine low impedance

The 500k pot is quite a high impedance, but the 470r collector resistor can easily drive 50X or 100X lower. Perhaps 0.5uFd into 10k pot? That's 2.6k worst-case Zout. I don't know what your goal is.

[Rob Strand]

When you lower the volume level, the "top" portion of the pot the signal goes through on the way out creates a significant additional resistance in series, right?

When you factor the pot into the output impedance the highest output impedance is when the pot is set to 250k + 250k.   The output impedance is then 250k/2 = 125k ohm (ie. pot value / 4)

If you had a high cable capacitance like 1nF then it would form a low pass filter with cut-off 1/(2*pi*Rout*C) = 1.27kHz.

If you decrease the pot to 100k and change the cap feeding it from 10n to 50n then the *high* pass filter will remain the same and the worst-case output impedance will be 100k/4 = 25k.   Which will push the low-pass filter upto 5x1.27kHz = 6.35kHz.

In many effect older fuzz pedals, the high volume pot impedance and the cable capacitance interact to produce a low-pass filter and that can shave-off a lot of nasty fizz.    It also means the sound depends on the set-up, which isn't great.

For an average set-up the cable capacitance won't be as high as 1nF,  nonetheless it is still part of the sound,
https://www.shootoutguitarcables.com/guitar-cables-explained/capacitance-chart.html

[Lino22]

Thank you guys. I am just trying to understand how is the Zout counted. The image below shows how i see it now, and i would like to know if it is correct.

From a simplistic AC point of view, caps are shorted and DC+ becomes ground. I am not completely sure what to do with the 8k2. Also, there is 500k volume pot set to the middle.

[Rob Strand]

Thank you guys. I am just trying to understand how is the Zout counted. The image below is most likely wrong, but this is how i see it now, and i would like to know what is wrong with it.

From a simplistic AC point of view, caps are shorted and DC+ becomes ground. I am not completely sure what to do with the 8k2. Also, there is 500k volume pot set to the middle.

You got it, spot on.   

For output impedance the 8k2 can be ignored.  It appears in series with the collector, which is a current source, and a current source has infinite impedance - so it hangs with no effect like your diagram.    (A real transistor isn't quite an ideal current source but since the 470R is so low it would be ridiculous to include the transistor's output impedance.)

[Lino22]

Oh mine, i can see it now (i misread my own diagram ). The "top" 250k portion is basically shorted via the 470 resitor, so the output resistance is

250k || (250k + 470) which is just what you said.

Is this correct?

[Rob Strand]

Oh mine, i can see it now (i misread my own diagram ). The "top" 250k portion is basically shorted via the 470 resitor, so the output resistance is

250k || (250k + 470) which is just what you said.

Is this correct?

Yes, as drawn that's it.

In this case the 470 ohm is so small compared to the 250k  it doesn't matter if you completely remove it from all calculations.

If you want to include the 470 ohm then technically the maximum impedance is actually at (250k + 470/2) // (250k + 470/2).  Here, the pot would be set with its resistance at 250k-470/2 at the top and 250+470/2 so the sum of the two *pot* resistances is still 500k.   What we are doing is making both the resistances the same from the *circuits* point of view.

If you are feeling keen, see what happens when the 470 ohm is 100k: The correct answer for maximum impedance is 150k.   If you set the pot midway you get (100k + 250k)//(250k) = 146k and if you completely ignore the 100k you get 250k/2 = 125k.   The answer gradually get more error as you take less into account.   The 146k means you haven't quite set the pot right for worst case and the 125k is just a rough approximation that you can eyeball directly from the circuit by doing calculations in your head.

[puretube]

IMHO, the 470R + 8k2 IS a 10k pot (within usual tolerances) ...
I (`d) use a 470R pot instead of the fixed resistor and route the wiper through a 10uF cap to the outputjack. (the big cap for equivalent rolloff-freq.)
Same max outputlevel.
And: yes, I use a big cap across the battery, so it IS AC-grounded on both poles.

[PRR]

...not completely sure what to do with the 8k2

The 8k2 goes to the collector. Collector impedance of a BJT is typically 1,000 times the emitter impedance, which we know from Shockley's Law.

Rounding to half-mA and 60 ohms we pencil 60k Ohms. Which with the 8k2 and the 470r is 467r. Which is within 1% of the 470r resistor. So forget it.

As you say, 99.8% of the at-jack Zout can be the pot. Which "can" be smaller. Arbiter won't rap your knuckles.

[iainpunk]

if you really want to lower Zout of a fuzz face, add a buffer onto the end after the volume control.

cheers

[antonis]

And: yes, I use a big cap across the battery, so it IS AC-grounded on both poles.

Which, IMHO, is an overkill due to 1 to 2 Ohms internal resistance of a 9V alkaline battery..
(2 Ohms is the impedance of a 1000μF cap at 80 Hz..)

[Lino22]

Also the cap across the battery filters out the DC riddles that happen when the battery us getting old and you play hard notes.

[Timelord]

So if you remove the volume pot the Zout is 470R?

Referring to schematic earlier in thread.

If the 470R was changed to 2K then Zout would be 2k and so on?

[fryingpan]

So if you remove the volume pot the Zout is 470R?

Referring to schematic earlier in thread.

If the 470R was changed to 2K then Zout would be 2k and so on?

Not "so on". So on as long as R1 (eg. 470R) << RC.

Anyway, I have never understood the obsession with Fuzz Faces. They sound nice, but they come with lots of drawbacks.

[antonis]

Not "so on". So on as long as R1 (eg. 470R) << RC.

Actually, no..

CE amp output impedance is the parallel equivalent of Collector load and Collector output impedance (r_c or r_o)..
The later ([V_A + V_CE] / I_C ) being much higher than the former, results into the resistance going to AC (or DC) ground, 'cause R_C is just in series with r_c (or r_o)..

You see, output is taken from 470R/8k2 junction and not directly from Q2 Collector..

More precisely, for the above circuit, output impedance is 470R//(8k2+r_c)//500k (considering VOL pot at max and 10nF capacitive reactance negligible)..
(r_c can be calculated by paricular device Early voltage and Collector current..)

[fryingpan]

Not "so on". So on as long as R1 (eg. 470R) << RC.

Actually, no..

CE amp output impedance is the parallel equivalent of Collector load and Collector output impedance (r_c or r_o)..
The later ([V_A + V_CE] / I_C ) being much higher than the former, results into the resistance going to AC (or DC) ground, 'cause R_C is just in series with r_c (or r_o)..

You see, output is taken from 470R/8k2 junction and not directly from Q2 Collector..

More precisely, for the above circuit, output impedance is 470R//(8k2+r_c)//500k (considering VOL pot at max and 10nF capacitive reactance negligible)..
(r_c can be calculated by paricular device Early voltage and Collector current..)

It's basically the same result. If 470R were increased by 20x it would be at least  comparable to RC + rc (small) and the simplification to 470R would no longer apply. Of course, there is the volume pot to consider too.

[antonis]

It's basically the same result.

Sorry but no (again).. 

Any value of (here) 8k2 resistor should be ignored..
What counts is only the value of (here) 470R..

P.S.
For 470R inceased by X20, Collector current should be decreased by the same proportion so r_c can't be small..
(actually, r_c should be increased by X20..)

[fryingpan]

If Early Voltage is 40V (a sensible value) and collector current is 500uA,and also considering Vce, we're talking about 30 + 8k, and if 470R is increased by 20x it gets close to 10k, so it's not much smaller.

[antonis]

If Early Voltage is 40V (a sensible value) and collector current is 500uA,and also considering Vce, we're talking about 30 + 8k, and if 470R is increased by 20x it gets close to 10k, so it's not much smaller.

I think you lost me..

For V_A = 40V and I_C = 500μA, r_c = 80k (neglecting  V_CE)..
As for 470 increased by X20, see my P.S. above..
(you can't increase resistor without decreasing current or else you'll result into mis-bias..)

[fryingpan]

I'm drunk, never mind. (Literally).