What is the JFET switching in this pedal doing?

[OwenM]

Hi there

I've breadboarded this pedal and it works fine. I get a lot of the passive electronic stuff that's going on, not sure about the PT2399's but that's for much later. What I can't get my head around is how the switching works. With only the 9v applied voltage to the switching circuit (top right of schem) I can't see what it can do to alter the Q1, J112 gate voltage to make it switch. It DOES switch, and I see generally about half a volt more when the IC/reverb part of the circuit is on. The on-off gate voltages wander from (off; -0.5v, on; 0v) to around (off; -1.5v on; -1v). It wanders around a fair bit, but generally always works once I've fixed all the wires that come loose. It could be that my variance is caused by relatively-complex-circuit-on-a-breadboard, but either way I don't get how, with that reverse biased diode D3, the gate could be changed in voltage via the switching circuit. Or is there some kind of small voltage allowance of a reverse biased diode, before it blocks current, that I haven't heard about yet?

Generally, I get confused when theres a reverse biased diode FACING the working voltage but something later on responds to the earlier voltage change, I've seen it in a few circuits and found a few explanations but none that illuminate me to this particular circuit!

I've read all manner of tutorials going into JFET usage that for far into the detail of the different zones you get at different operational points but nothing that has that diode (d3)...

Thanks for any help



PS - not looking for an in depth lesson, one sentence that I can research further off would be very welcome

[g1]

Think backwards as if there were a positive voltage at the point 'sw'. One way you want to allow leakage to ground, one way you want to guarantee there can be none (reverse biased diode).
The Fet is normally 'short' from Source to Drain with the gate floating. FSa connected to FSb, LED on, diode reverse biased.

Fet's like this need a voltage to turn them off. When you connect FSb to FSc, you offer the gate a path to ground so it can be lower voltage than the Source (because this is an N-channel type Fet). Now you have turned off the Fet. The LED is also off because the positive voltage from FSa is removed.

[Mick Bailey]

That.

The FET junction is easily destroyed when forward biased and D3 ensures the junction can only be reverse biased. If you only had the FET gate either connected to ground (off) or floating (on) then D3 would not be necessary unless the signal at the source went negative with respect to the gate (which it can't in this circuit). Because you have a switching circuit that gives a smooth switching transition and gives a visual indication via the LED that the circuit is on, you need D3 to isolate the gate from the positive voltage at the junction of R22/C30

[diydidi]

Ive built these circuits numerous times. The main function of D3 is to ensure that the Jfet stays off with varying input signal on the Drain source pins. Even if the DS pins are biased at 1/2Vcc they could still turn on due to large signal transients. The diode is needed for both Pch and Nch devices, whichever you choose to use.

[OwenM]

Thanks, you guys are great! I'm reading the three descriptions and trying to get my head around it. I think I get it in the main. Am I right in thinking in this circuit that the JFET doesn't create a broadly mechanical 'off' when the pedal switch is set to off, when no input signal is present the jfet is just sitting there waiting, but as soon as a chord is played at the input the source voltage is higher than the gate's because the gate has a path to ground, so the higher source voltage causes the fet to stop conducting, causing the effect to be off? (I guess this is the voltage that is turning it off, from G1's answer)

I get get the diode now, regarding protecting from the 9v and allowing the path to ground from the gate's point of view!

Cheers for the help!

[Mick Bailey]

With no signal input FET source sits at 1/2 the supply voltage (4.5v). This is because the circuit has a single 9v supply rail, so the opamps are biased at 4.5v to give a possible +/- voltage swing for the signal. So your signal can go from 4.5v down to 0v, or 4.5v to 9v. Well, not quite; the opamps in this circuit cannot swing the voltage that close to the supply (0v or 9v), so the maximum signal is a fair bit less than the theoretical maximun for a perfect opamp. There are rail-to-rail opamps but the TL072 isn't one of them. Anyhow, with a guitar signal this is rarely a consideration.

The FET has a voltage relationship between the gate and source that determines whether it's on or off. To prevent the signal peaks from switching the FET, it has to be selected with the correct Vgs(off) voltage for the particular circuit. This is given as a range in the spec sheets because the parameters of a FET are not precise. Sometimes with a switching circuit you have to try a few devices to get the best performance, depending where it sits in the range of possible allowed values.

[OwenM]

With no signal input FET source...

Ahhhh, ok cool, thanks very much, I see what I was not getting. I've been thinking of the op amp power as just being generic to make the device switch work but not as a biasing voltage, but that makes sense. I've also been confused about how they ask for positive and negative rails but also seemingly work positive and ground, etc. This is starting to join the dots for me.

Thanks again!

[OwenM]

Having read up as much as I could find and looking at the circuit in question for a while could I ask one more question?? Of course, pass this thread by if the answer is no for you

So if we want to put 4.5v on the U1:B positive pin so, working as a buffer that is just concerned with keeping it's + and - pins the same voltage, it's output has 4.5v on it before adding guitar signal, giving potential for an ac signal to swing to near ish 9v and near ish 0v. Why then do we need R3, a 1M resistor? Is it so the guitar signal can't get to ground through only the 100k R21, which would lower input impedance considerably?

This may be can-o-worms opening; I can see why R3 would help input impedance, but wouldn't most of the 4.5v destined for the OpAmps positive input be dropped over it? I have a sneaking feeling I'm taking basic ohms law lessons on multiple resistors and missing a fundamental thing about voltage, or probably just miss-applying anything I've learned on ohms law in practice. Or maybe the 4.5v reference doesn't come from the lower side of R3 and I've barked up the wrong tree. I'm trying to understand all this stuff from a mixture of online tutorials and this forum, and so far it's been working, but I understand completely if only those with an enthusiasm to teach get the urge to post

[g1]

Why then do we need R3, a 1M resistor? Is it so the guitar signal can't get to ground through only the 100k R21, which would lower input impedance considerably?

Not through R21, but through C26, which is basically ground for AC. Same as most caps in power supplies, you can think of them as AC (signal) ground.

[Mick Bailey]

....wouldn't most of the 4.5v destined for the OpAmps positive input be dropped over it?

The theoretical idealised model for an opamp has an infinite input impedance, therefore no current flows in the inputs. Because C1 blocks DC, there is only one path for DC via R3 & R2 to the input. No current flow means no voltage drop according to Ohms law, so the 4.5v is not dropped by either resistor. Whilst a real-world opamp doesn't exactly match the ideal characteristics, the real voltage drop is so small as to be inconsequential in an audio circuit.

[OwenM]

Not through R21, but through C26, which is basically ground for AC. Same as most caps in power supplies, you can think of them as AC (signal) ground.

Haa, yes, of course, I think I did the thing where I got so excited that I had figured something out I missed the much larger obvious thing sitting right next to it! Thanks

[OwenM]

The theoretical idealised model for an opamp has an infinite input impedance, therefore no current flows in the inputs. Because C1 blocks DC, there is only one path for DC via R3 & R2 to the input. No current flow means no voltage drop according to Ohms law, so the 4.5v is not dropped by either resistor. Whilst a real-world opamp doesn't exactly match the ideal characteristics, the real voltage drop is so small as to be inconsequential in an audio circuit.

Ah brilliant, great explanation and makes total sense thanks!