Effect Dry/Wet circuit

[knutolai]

So I came by this circuit the other day and got curious about the position of the polarized 10uF capasitor. I'm wondering why this circuit isn't producing positive feedback. Is it because current wount travel through the polarized capasitor from negative to positive? Would this circuit still work if a ceramic capasitor was inserted in place of the 10uF? Im relatively new to electronics, but I know that capasitors block DC and work as a frequency dependent resistor. 


Thanks a lot for helping a guy out

[Bill Mountain]

So I came by this circuit the other day and got curious about the position of the polarized 10uF capasitor. I'm wondering why this circuit isn't producing positive feedback. Is it because current wount travel through the polarized capasitor from negative to positive? Would this circuit still work if a ceramic capasitor was inserted in place of the 10uF? Im relatively new to electronics, but I know that capasitors block DC and work as a frequency dependent resistor. 


Thanks a lot for helping a guy out

I think the cap is simply a coupling cap for the buffer.  If you turn the knob completely "dry" there needs to be a dc blocking cap after the buffer.

[knutolai]

hm that makes sense. Then I should rephrase my question. What is preventing signal to go from "effect output" to "effect input". I can totally see that going to "output" is the "easiest" rout for the signal from "effect output", but is this enough to prevent there to be any signal feed back into the "effect input"?
This is just really confusing

[Bill Mountain]

hm that makes sense. Then I should rephrase my question. What is preventing signal to go from "effect output" to "effect input". I can totally see that going to "output" is the "easiest" rout for the signal from "effect output", but is this enough to prevent there to be any signal feed back into the "effect input"?
This is just really confusing

Because the pot is blending the two signals.  Look up passive blending circtuits.  All it is is 2 resistors.  There may be some bleeding of signals.  It won't be perfect.  I prefer active (op amp) blending circuits because they avoid cross talk.

Edit:  I feel I should expand slightlty.  The pot is creating a voltage divider between the wet and dry signal.  Think of it like this: a regular volume pot is blending the effect with "gound".  This set up is just replacing "ground" for the "dry" signal.

[knutolai]

thanks for clarifying. I thought there was some magic here I was missing out on

[PRR]

> current wount travel through the polarized capasitor

Nothing to do with feedback.

The cap sure will pass audio from FX Out back to FX In.

Some FX have very little gain so we could ignore that path. (Some invert and it becomes NFB.)

The real reason it is not prone to howl: FX Out must flow through 100K (pot) to JFET Source. The JFET Source is about a 1K impedance. FX would have to have gain over 100K/1K or 100 to cause PFB. Most don't. (Some do.)

But that circuit as-drawn is _BAD_. The FX input gets 5V DC!! Most FX do have input blocking caps. Some don't. It is simple courtesy to _not_ have DC on audio jacks. There should be a blocking cap from JFET Source to FX Out.

While considering "what could go wrong?", consider these:

* Strong radio signals or LOUDspeaker outputs could upset the JFET Gate. 33K in series with the Gate does no harm to the audio but reduces radio-signal and limits excess-voltage current so the Gate don't blow.

* If your FX box/cable is shorted, the direct path is also shorted. 1K-10K from JFET Source (and blocking cap) to FX Out allows direct signal to keep happening even when FX is shorted, so that simply turning to Dry lets the show go on without embarrassing re-patching and de-shorting.

[knutolai]

Is there a formula for this calculation?
Looking at your calculation it follows that the lower the impedance of a path the more "eager" a signal is to travel through it. Am I right? How would I calculate a approximate impedance of a buffer?

I'm currently reading Forest Mims "Getting started with electronics" book, but have not yet found a satisfying explanation on the matter of impedance. Is there a fast and oversimplified explanation that can help me wrap my head around it? 

[chptunes]

I'm currently reading Forest Mims "Getting started with electronics" book, but have not yet found a satisfying explanation on the matter of impedance. Is there a fast and oversimplified explanation that can help me wrap my head around it? 

Go here:  http://www.muzique.com/lab/imp.htm


-Corey

[knutolai]

Simply put:
frequency-dependent resistance! Thanks a lot for the link! this is great!

[knutolai]

About the article by Jack Orman: http://www.muzique.com/lab/imp.htm
In the forth circuit shown a extra pull-down resistor is needed to prevent "pops" when the circuit is being switched. Why is this a issue for this circuit and not for the third circuit in the article?

[PRR]

> In the forth circuit ... Why is this a issue for this circuit and not for the third circuit

It is a tutorial about Design. Step by step. The 3rd design is incomplete, it lacks the bleed resistor. He adds a "reasonable" bleed to make design step 4, then tells why it compromises the goals of the 3rd design (no point using 10Meg bias for HIGH impedance with 1Meg bleed in front of it).

He doesn't really finish the story.

In general, you want a bleed resistor. You don't always need it, but when you do you DO.

In this case (10Meg bias) I'd be thinking 10Meg bleed. Guitar now sees 5Meg, up from the 500K of many guitar inputs.

It is also possible to omit the input capacitor and the bleed resistor. Thousands of tube-amps do, and this JFET is biased like a tube. Whether JFET or bottle, you also want 10K-50K in series with the Gate/Grid to deter radio signals and high (accidental) voltages.

[knutolai]

Would this modification of the circuit affect the transistor source impedance?