Chasm Reverb JFET problem

Started by uncle_rufus, May 03, 2018, 02:38:24 PM

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uncle_rufus

Hello all, welcome to my first post!  Long time lurker, first time poster

I've recently finished building a Chasm Reverb by deadastronaut (I considered putting this in the thread for it but it's very old now, and I think I have narrowed down the problem to be a slightly more general question).   I was emailing Rob back and forth but although he's been a great help I don't think he signed up to be my personal agony aunt for every little problem, so throwing it out to a wider audience...

The issue is that when the pedal is bypassed and I supply a large enough input signal (i.e. with my Strat it never happens, but switch to an LP or a metal guitar with higher output pickups and hit a boost or something earlier in the chain and it will) - I hear the effect... most noticeably if I play a single hard staccato chord the reverb comes through...

Based on my (very limited) electronics knowledge and the power of Google, I think what must be happening is the JFET (2N5457) which is gating the signal that goes through the effected part of the circuit isn't quite getting enough voltage to completely close (the datasheet seems to suggest 0V is fully open, and -0.5 to -6.0V is closed, presumably that means a full -6.0V will lock the signal completely)... I haven't busted out my multimeter yet to confirm but I'm expecting to see like -5.9V or something... The tiny amount it's letting through is normally so insignificant that the later parts of the effected path just eat it, but if the input signal to the pedal is high enough then it's making it through... Does that sound right?

Assuming so, what I'm not sure about is how to go about correcting it?  My first thought is that dropping the resistance of the 100k that is limiting the +9V coming in would see a higher amount hitting the JFET and hopefully reach -6.0V (or higher/lower - I must admit the fact that the quoted value there is negative confuses me a bit!) and completely shut it off... The datasheet sort of reads like the JFET can handle anything up to 25V before it is damaged so I can't see that this would break it or anything, but will messing with that resistor cause any other problems?  (I think the switching is done in this way to eliminate noise and I don't want to upset that)... Otherwise is there anything else I can try?

TL;DR - Reverb signal coming through in bypass if the input is loud enough, expect the voltage hitting the JFET is too low/high, but not really sure what I'm doing!

PRR

Welcome.

> I haven't busted out my multimeter yet

Electricity is invisible. Speculation is futile. METER!
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uncle_rufus

Thanks :)

So... err, any idea about my question?

PRR

I am hoping DeadAstro comes by; he knows much more about his chasm than I do.
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reddesert

Have a read through RGK's writeups on JFET switching:
http://www.geofex.com/Article_Folders/bosstech.pdf
https://www.premierguitar.com/articles/jfet-switching-1

Start by volt-metering the voltages on the JFET pins, and all the stuff that goes between the JFET gate and ground or +9V, in both the effect and bypass positions.

It helps to provide a link to the schematic. (I can google up a low-res copy, but ...)

For starters, it appears that the gate is connected to the footswitch through (in parallel) a 1K resistor, or a 100K resistor plus a diode. And the diode is oriented so that it conducts when the footswitch is high, not ground; high should be "JFET is on" state. So I don't think the 100K resistor is the cause of your problem.

uncle_rufus

I've read the articles, but not sure I completely understand it all!

Is the "high" really the "on" state?  I though zero volts corresponded to the JFET being fully open (i.e. "on" in terms of the signal passing through it from drain to source)... To me it looks as though (as you say) the Diode connected between the footswitch inputs and the 100K resistors in each case would only provide resistance when the footswitch connection is activated (+9V), with the other 1K resistor acting as a bleed resistor for the JFET when it turns off (and the capacitor is controlling how quickly the signal dissipates, or how quickly the JFET gate goes back to 0V and the signal path opens up)... that's what leads me to believe that the voltage across the JFET gate must be determined from (mostly) the 100K resistor...

The only link to the schematic I know of is deadastronaut's build document, but there may be a better one somewhere...

https://docs.wixstatic.com/ugd/ff4c6c_58f8a6b9d0db4a71b1e770234bb0db25.pdf

reddesert

RGK says "If you leave the Gate (G) of a JFET open, the Drain (D) and Source (S) will pass signal through, as the switch is connected.  If you pull the Gate to a voltage less than the Source, the JFET will no longer pass audio signal."  Remember that "open" means not connected - open like a switch, not open like a door.

When the switch is connected to ground, the capacitor will drain through the 1K resistor, and the JFET gate has a path to ground through a diode and 1K resistor. So the JFET gate should sink to one diode drop (0.7V) above ground.

There is no substitute for measurement. Remember, you have access to your as-built circuit, and no one else does. Measuring voltages on the various connections in both states of the switch will give a clue to what is going on.