Pale Horse Jfet question

[jmasciswannabe]

Asked over at MBP with no luck. Thought I would try here:

So I decided to use a socket due to all the comments about jfet causing issues. I tried a j113 which was terrible (could be counterfit, ordered from mouser to do a comparison in a few days.) I also tried j201 which worked a lot better but still is a bit fizzy compared to not having one in there at all. Is that the sound (what it sounds like without one in there) that I should be shooting for in the first place?

UPDATE: j11 3from mouser still fizzing out

V+ = 9.5  and v-9.3 without jfet 0 on drain - with j113 in there v- drops to -9.0 and drain reads -8-1. In the build doc drain is supposed to = 0. Could someone shed some insight on what this jfet should be doing and what might cause that voltage on drain. Triple checked components and did continuity tests. Trid different opamp. If one of the bs170s were bad, could it cause an issue?

AS always any help appreciated and thanks!

http://vfepedals.com/schematics/palehorse.pdf

[Rob Strand]

From what I can work out the JFET is wired as a constant current source,
https://www.vishay.com/docs/70596/70596.pdf

The 2N5484 has a *typical* IDSS of around 2mA to 3mA so that would mean there should be a voltage drop of 2 to 3V across the 1k resistor.    The J113 has *minimum* idss of 2mA.  The typical could be 4mA more which will cause more drop across the 1k resistor.

From what I can see the voltage at the drain should be 0V, which agrees with the build DOC.  The opamp output would sit at about 2V.  If the JFET was pulling a lot of current then then opamp would not be able to swing to a high enough to the feedback loop to keep the drain at zero volts.

In you case I think things are worse than that.  I really think there is a pinout issue and the JFET looks like a diode and it's pinning the source at 0.7V above the negative rail - you could measure that to confirm the theory.

Beyond that, I'd pull the JFET temporarily and see if the opamp output biases at 0V.  Then if that's OK perhaps put a 2k2 in place of the DS terminals then check the "drain" point is at 0V.

[jmasciswannabe]

Rob! Thank for chiming in and including the article. So, with no jfet in the output on the opamp is at -0. Everything sounds great. Once i put that sucker in i get voltages from -0.4mv to -8.2 at the drain. This varies with the drive. It also swings the opamp output from 3.2 to 2.3 volts respectively. Triple checked orientation of JFET, It's orientated correctly. With the 2.2k in the terminals I get -1.58 at D and 1.8 at opamp output. The fizziness is better but still there on the extremes of the compression control and low gain settings.

Obviously, I want to get to the bottom of why this is doing what it does and have it work as intended, so let me know if you have any other suggestions on how to debug. Additionally, what is the benefit of the constant current source and how does that translate audibly?

[bool]

it will include a DC offset to tilt the clipper duty cycle. (it's just a CCS to DC bias the 1k resistor in the nfb loop).

if you want to play, replace the fet with a 100k trimpot and set it for the "best tone".

or keep the fet and replace the 1k resistor with a suitable trimmer, and trim it for the best tone.

[jmasciswannabe]

Tried both. Settled on leaving the j113 in there and a 51ohm resistor. Close enough for rock and roll.

[Rob Strand]

Obviously, I want to get to the bottom of why this is doing what it does and have it work as intended, so let me know if you have any other suggestions on how to debug.

Bool's post covered it.

So the key requirement is to have a *stable* voltage across the 1K.  Stable in that when the battery goes flat the voltage stays constant.  However there's more to consider. When the makes large swings, say near the positive and negative rail, the constant current source ensures the voltage drop across the 1K is fairly constant at all time.

Additionally, what is the benefit of the constant current source and how does that translate audibly?

I actually don't know.   It *may* contribute something.  When the output is close to the -ve rail the current source gives up.  You would have to AB test against alternatives to be 100% sure.     An obvious alternative is a transistor based current source,

https://i.stack.imgur.com/PEm2I.png

Rather than attempt to get the voltage drop indirectly using V=Iconstant * R (=1K) we could simply put a voltage source in place of the 1K resistor.   For example we can replace the 1K with a string of diodes, or a VBE multiplier.  In this case we need to put in biasing resistor from the old drain connection to the -Vcc rail.

https://upload.wikimedia.org/wikipedia/commons/0/03/Basic_Vbe_multiplier.png


There's many ways to do it but you would have to try all of them to know which ones sounded best, or different, or in fact the same!

ried both. Settled on leaving the j113 in there and a 51ohm resistor. Close enough for rock and roll.

That solution is pulling probably pulling a lot of current.   Perhaps a better solution is to leave the 1K then place a resistor in the like Figure 1 of this paper posted,
https://www.vishay.com/docs/70596/70596.pdf

You can trim the resistor for 2 to 3V drop across the 1k.    In fact figure 10 has a graph.  The resistor will probably end up around 330 ohms.   You could argue it's not 100% the same because of the drop across the 330 ohms.  That would depend on how far the opamp swings.

[jmasciswannabe]

....just when I thought the back was screwed on forever. You know, I noticed that the first time I looked at that graph that there was a resistor there but not in the circuit design. You are correct that sticking a resistor between the gate and source achieves desirable results.