# What is the purpose of this diode in the gate of a JFET

I am studying how the bypass circuit of the Tube Screamer pedal works. It is dependent on two JFET transistors that are used as switches. I am not great at understanding JFETs as switches, as I have allways used them as amplifiers. This is the circuit:

Image taken from https://www.electrosmash.com/tube-screamer-analysis

Below R16, thevoltage that controls the JFET switch is applied. It can either be 9V or 0V.

Can someone explain how this works? I cannott analyze the voltage Vgs because of D3. What would be the gate voltage Vg when 9V or 0v are applied below R16?

The purpose of D3 is to prevent the JFET junction from ever being forward biased, at least not to the point where significant current from the switching signal gets injected into the audio signal channel. At a guess, it permits the designers to use +9V as a "safe" switching signal potential, instead of having to derive a potential that's always "just a tiny bit lower" than the audio signal.

When the input signal at R16 falls to zero, D3 is initially forward biased, and gate potential will fall quickly to +0.5V or so. Following this, current eventually falls to an amount limited by the JFET's reverse biased junction, or C11 leakage. How far C11 continues charging towards zero, and how long it takes, depends on PCB, gate and C11 leakage. In any case you can expect gate potential to eventually fall to under +0.1V, but it might take a while to get there.

When the switch signal rises to +9V, D3's cathode rises at a rate limited by C11, but gate potential may not follow immediately, since D3 is has become reverse biased. The rate at which the gate rises in potential will be determined by the combined capacitance of D3 and the JFET junction, and leakage current through D3. While the gate will certainly, slowly, rise, it doesn't seem to be well defined behaviour. It seems reckless to rely on leakage current for timing. The designers probably knew more about D3 and Q2 than I was able to find on the web, so what looks like a design flaw to me might well have been deliberate.

Gate potential won't rise much further than the audio signal being switched (+4.5V), because that would forward bias the gate junction. The gate will begin to conduct D3's tiny leakage into the JFET's channel, at which point gate potential will stop rising. The gate will get stuck at just over +4.5V, or whatever the audio signal happens to be at the time.

It's worth noting that D3's reverse leakage current is likely to be orders of magnitude greater than the JFET's junction, and you can probably disregard JFET junction leakage. In other words, when both junctions are reversed biased, D3 has the (significantly) lower impedance. This difference is probably the only reason that gate potential is able to rise, and is why the circuit works at all.

Well, I can tell you, I wouldn't design an analog switch that way. But I would most likely be designing a proper analog switching circuit, not a effects pedal.

Just from this framing alone, we have reason to beware: beware that, ahead, demons lie; and that, what we are after, is not clean signal transmission, but a dirty mess, and they want it that way.

There's not really any accounting for taste, no analytical expression of audio effects; we can prepare various stock functions from expressions, from basic circuits or what have you, but what is ultimately picked, or how it's used in a system, is an artistic process.

So, I can't describe what this circuit will do in a song, how it will feel, for any particular instrument or other signal passed through it. I can only describe what the circuit does. Keep that in mind.

As for the circuit:

The basic structure is a JFET switch, with AC (R-C) coupling at the input and output. There is one extra output line which I have to assume can be ignored, and also have to assume a control voltage is applied from the bottom, 0/9V as given.

While the switch node is low, negative Vgs is applied to the JFET, and little or no signal will pass. How much, depends on the JFET capacitance, and node impedances. If impedance is maximum (the 510k resistors shown), this will give a cutoff of perhaps a MHz, above which attenuation is minimal and fairly flat, and below which, attenuation rises asymptotically (20dB/dec). For reasonable attenuation (audio muting, perhaps), you'd want maybe 80dB, or 4 decades below 1MHz, which is only 100Hz, which isn't very good. But a modest load impedance of say 10kΩ would bring that up to 5kHz or so, and maybe the quiet, tinny (high-passed) remainder isn't objectionable.

There isn't much data about this transistor, but I would assume it's a few-pF, ~1mA, 20V device, given what is available.

It's hard to determine maximum signal level without a cutoff voltage specification. Typical JFETs range from -1 to -10V, quite a spread. Note that, when signal peak drops below Vsw - Vgs(off), the JFET is turned on slightly, and signal passes -- but just the peaks, you get a clipped waveform.

The suggested replacement, 2SK304, has Vgs(off) from -1 to -4V, so could be a marginal choice given the bias levels, and typical audio levels (a couple volts).

When the line is high, the results are poorly defined.

The gate node has no driving source: only JFET and diode leakage act on it. MA150 is a switching diode with moderately low leakage, so it could take some milliseconds to stabilize. But it's impossible to tell what voltage it will stabilize at. I would hazard a guess, the diode's leakage dominates -- but diode leakage drops significantly towards 0V, so it won't stay there perfectly.

Note that the above mentioned clipping effect also pulls in, all the time while the gate voltage is rising. The result is a quiet, rough, heavily distorted signal, fading in to full amplitude as the distortion disappears -- granted, the fade might take place over just a few cycles of audio, so it might not be perceptible. But here too, we find a catch: the gate node is surrounded by PN junctions, one from the JFET, one from the diode. Any AC on the JFET channel, is sure to be rectified into offset voltage, and thus there is a relationship between Vg(on) and signal amplitude.

If I had to guess, this might be an intentional feature -- the distortion manifesting as a clipping or limiting (compression) effect, perhaps with interesting mixing effects given the time constants in play. This effect isn't at all reliable, as diode leakage depends exponentially on temperature, and it'll behave differently on a hot stage versus in the practice room.

It might even vary with light: if a glass-body diode is used, its leakage varies from ~nA to ~µA in daylight. Not very much, but, again, huge when you're comparing to basically nothing.

So yeah, madness it be. Whether that's consistent enough to be useful, and dirty enough to be interesting, I have no idea. But it's possible.

If one wants a proper signal switch:

Simply add a G-S resistor. This pulls the gate soundly up to the channel voltage when on, rejects diode or BJT collector leakage (an open-collector switch suffices to drive this, no diode needed), and switches off just fine. It might be desirable to cancel out the off-state bias current through the resistor, to avoid a "pop" sound on switching (as DC bias reestablishes itself), perhaps by driving an inverse signal into a resistor into the output, so there's always the same pull-down on it. It also helps to set a low impedance (say use a 10k or 1k load resistance), and maybe some buffer amplifiers would be used in service of that. That also improves high-frequency attenuation, as mentioned earlier.

• With the control signal at +9V, there's really no leakage to any potential less than +4.5V unless the circuit board is crap. So, the gate is going to end up around +4.5V. Commented Jul 25 at 16:36
• Hence the "stabilize within ms", but don't forget the rectification case. It depends! Commented Jul 25 at 21:57