Curious about Passing Audio Signal with a FET Switch (Simple - No Flip Flop)

[kimelopidaer]

Hello,

I'd like to control the flow of an audio (guitar) signal with a FET switch, like the Boss effects do - but i want to leave out the whole "flip-flop" portion and experiment to understand this one section.

I put something like this together on the breadboard; the choices of diode, capacitors and bias resistors informed by threads here on the forum and GEOFEX, and using what was handy for substituting.

the 1M resistors on Drain and Source went to 9v+
the capacitor to ground was 2nf i believe.
I had a 1n4001 diode, so i used it.

I tried to cut the audio signal by switching the Gate from v+ to ground,
though no luck. I gather it's about raising Vg to a threshold established by the resistors on Source and Drain.

Would anyone like to comment on what I may be overlooking?
If the gate is not being controlled by the flip-flop, how does the method below compare?

thankyou
K
 

[R.G.]

JFET signal switches change from essentially no conduction at all when they're off to a low-ish resistance when they're on. They are usually used as part of a degenerate voltage divider. There are a few changes to what you have show that I think you might want to try.

1. Change Vb to something less than 9V. This is so the switching voltage on the gate can get higher than the gate/source and reverse bias that gate diode. a 10K/10K/22uF Vbias source works well.
2. As it sits, there is 9Vdc on the output from Vb pulling the output pin up to 9V through a 1M resistor. I would put a capacitor in series with the output and another 1M to ground to remove the DC offset. It may be that the large DC is affecting whatever is after the audio signal out and keeping it from seeing any signal that you actually switch.

Other than that, the J201 is not all that great a switching JFET. It's optimized as a low-signal amplifier. It'll work, but a 2N5292, 2N5485, 2SK30A, 2SK117 and some others will work better. If it's all you have, it's OK to experiment with.

The drive scheme on the gate is OK. What you want to do is have a big enough signal to pull D1 to ground, thereby pulling the gate to nearly ground; the difference between the source at Vb and gate at nearly ground is the voltage which turns the JFET off. When the control signal is pulled high, it reverse biases D1, forcing the gate-source diode open, so the channel between drain and source is whatever resistance the raw silicon is, generally a few hundred ohms. All the resistor/cap on the drive signal does is slow down the transition speed of the control signal to eliminate pops and ticks.

[amptramp]

When the gate switch is taken positive, the gate voltage is indeterminate - it is set by the difference in leakage currents in the diode and the FET gate junction.  I have seen a similar circuit before where there was an op amp connected as a unity-gain buffer with its input at the FET source and the output taken to a resistor that connected to the gate.  Thus, the gate voltage was set by the resistor, not by leakage, but no current was flowing in the resistor since the voltages were the same on both ends.  The same diode connection was used to the gate, but there was no question of gate voltage - it was either dragged down by the diode or exactly at the source voltage.  I assume Boss uses a low-leakage diode and FET in their similar circuits with leakage characteristics that are guaranteed to let the gate approach the source voltage when the diode is reverse biased.  The leakage current would go through the 1 meg resistor to the switch - measure the voltage at the diode end of the resistor and see if you are really getting the diode to pull down in the off condition.

[Gurner]

Can I ask what purpose d1 is serving?

Now I'm no designer, but I''d have whipped the diode out (until someone at least educates me why it's there!) ...and also move the switching voltage from the present V+ to Vb. At least with such a scheme you can be assured what your fet gate voltage is with both switching scenarios.

[merlinb]

Can I ask what purpose d1 is serving?

It stops the FET's gate from going into conduction when the switch is pulled high.

Now I'm no designer,

Hmm...

[Gurner]

It stops the FET's gate from going into conduction when the switch is pulled high.

Oops, yep that'll do nicely!  But if the cct was modded to switch Vb to the gate  (instead of V+ ...i.e.like I said earlier), then you wouldn't need the diode cos the source & gate would be at the same potential....and the gate therefore wouldn't go into conduction.

[merlinb]

source & gate would be at the same potential....and the gate therefore wouldn't go into conduction.

True...until the audio signal comes along! Then the source will ride up and down above and below the gate, and you won't have a very good switch.... There's a reason it's done this way!

[mmaatt25]

I'm experimenting with relays and an arduino board to switch four guitars into one amp individually.  Could this circuit be used to substitute the relays?

Thanks

Matt

[Gurner]

source & gate would be at the same potential....and the gate therefore wouldn't go into conduction.

True...until the audio signal comes along! Then the source will ride up and down above and below the gate, and you won't have a very good switch.... There's a reason it's done this way!

I was clearly having a bad day yesterday! (I actually lay in bed last night & pretty much arrived at the conclusion that I've just seen in your reply - doh, I'll get me coat!)

[R.G.]

I'm experimenting with relays and an arduino board to switch four guitars into one amp individually.  Could this circuit be used to substitute the relays?

Probably. A JFET switch is a resistor that goes between a few hundred ohms and near-infinity. The input of a tube amp is often 1M. You don't have to get "near infinity" very wrong to make 1M significant. But if you get a good, solid "off" voltage on the gates, it'll probably work fine. The issue to watch for is signal leakage caused by signals which are a big fraction of the "off" voltage causing the source/drain channel to be moved nearer the gate voltage. This reduces the off bias on the gate and lets the JFET resistance drop. Well-designed JFET switching must make the gate much more "off" than the signal voltage.

In pedal/guitar switching with only 9V power to work with, JFETs work well for signals up to about a volt, maybe a little more. Then you start running out of overall voltage this way: The gate really needs to be held 'off' by Vgsoff plus the peak signal voltage so the JFET doesn't leak on signal peaks on one polarity. The gate diode needs to be driven 'on' by at least a fraction of a volt on D1 (this circuit) to keep the gate open circuited. The signal voltage peaks in one polarity also subtract from that gate+diode voltage when 'on'. In a 9V circuit this works well up to a couple of volts peak, and then you may start getting leakage. If the signal is raw guitar level, 100-500mV, you're fine. If the signal is a boosted, gained up drive from a pedal, you may be getting much more than that. The company I work for makes a pedal that will put out maybe 12-15V peak to peak, intended for punishing the input triode in an amp. That output could easily overwhelm a 9V powered JFET switcher.

A better technical choice might be a CD4066 or CD4051. The CMOS switches have internal circuits which make them much more tolerant of large signal swings (within their power supply limits!) while still acting like a clean on-off switch. Especially where you're only switching between guitars and not footswitching in a song, this would work well and ease up some of the issues with CMOS switches. 

As a side issue, an Arduino is an expensive way to do the switching. If all you want is a radio-button selector, look at the 74C373 circuit on geofex. The 373 is about $1.

[amptramp]

This topic came up in another thread ( http://www.diystompboxes.com/smfforum/index.php?topic=95733.0 ) where I suggested a few other possibilities:

You could use a low slew-rate switcher such as the Boss / Ibanez circuit shown here:

http://geofex.com/Article_Folders/bosstech.pdf

which suppresses switch popping by gradually transferring from one source to another.  You can add resistor / capacitor / diode networks at the flip-flop outputs to obtain the correct priority for switch actuation.  No need for programming.

You can also use vactrol (LED / photoresistive) switching, diode switching and CD4007 low slew-rate switching.  For the latter, check the following datasheet figure 1:

http://pdf1.alldatasheet.com/datasheet-pdf/view/11942/ONSEMI/MC14007.html

and eliminate the inverter connections to pins 6, 8 and 13 and drive pins 3 and 10 directly from a flip-flop as presented in the geofex article above.  Note that this site has some Boss pedal schematics scattered all over in various locations but a more complete compendium is here:

http://www.generalguitargadgets.com/tech-pages/45-schematics/55-boss-schematics

so you even have the parts values.  Almost any small-signal NPN transistors will do for the flip-flop.  There is nothing critical in these circuits except that for the diode / JFET circuit used in the Boss pedals, the JFET must be low gate-to-source cutoff and the diode must have lower leakage current than the JFET gate.  (One more reason to like the CD4007 / MC14007 circuit.)

[kimelopidaer]

Hi,
It's been a real help reading this thread, though at the moment my replies can't be as quick as i'd like -
anyhow... I enjoy the other forum members' tangential discussions.

With reference to the simplified switch schematic above, I took the advice of Mr. Keen and
added the voltage divider to reduce Vb to 4.5volts. I also added the series output cap and resistor.
Notes on what happened when i fired it up:

First - With the Gate drive voltage switched to ground, the signal still passed, however if I touched the 1M resistor going the gate, the signal would be cut!

Second - I added a 4.7uf capacitor to ground and I was suddenly able to switch the signal on and off!
Well, nearly. A tiny amount of signal is passing through in the off state. I can tell by turning up the reverb on my amp and strumming hard
(For the switch, I'm using an SPDT switch that connects the 1M resistor from the Gate to either +9v or Ground)
Please see below:



Third - I will order the recommended FETs that might be more suitable for what i'm doing - but right now I wonder
is the unsuitability of the j201 i'm using the reason that I can't completely block the signal when the JFET is in its 'off' state?
I have tested a number of them as well as 2N5457 JFETs, which don't work very well at all with the component values I used.


Cautiously optimistic that I'll be able to cut out the signal leakage entirely after reading Mr Keen's comment regarding leakage:
"If the signal is raw guitar level, 100-500mV, you're fine. If the signal is a boosted, gained up drive from a pedal, you may be getting much more than that."

alright, if anyone cares to remark
cheers
K

[kimelopidaer]

Right now i'm only switching a clean guitar signal, and there's a small amount of signal leakage coming through - at anything other than mild volume. What happens if I want to switch something on and off like a fuzz or high gain distortion?


I was thinking, perhaps I could limit the voltage of the input signal coming into the switch with a voltage divider -
as insurance against leakage,
and then route the output signal through a 'gain recovery' stage if things were too quiet.

I may be overcomplicating
when considering how the input signal voltage issue would be addressed to
accommodate various signal strengths


K

[PRR]

> there's a small amount of signal leakage

In guitar switching we usually have A _and_ B. When A is open, B closes the circuit, and also absorbs leakage from the imperfectly open A switch.

> I may be overcomplicating

Yes. It isn't as hard as you are making it, and you are following wrong paths.

> I added a 4.7uf capacitor to ground and I was suddenly able to switch

Something is wrong in wiring or parts. The gate network goes +9V and DC ground. With that cap, the lower voltage is undefined (except that a real 4uFd cap will leak-down).

[gritz]

If you have a look at e.g. Boss distortion schematics there will be a fet switch before the distortion circuitry as well as after it, so when the effect is bypassed the distortion is muted at it's input as well as it's output.

Consider also that the fet forms the top half of a voltage divider, so the amount of attenuation is proportional to the resistance of the bottom half of the divider. In your diagram this is 1M in parallel with the input impedance of the circuit it's feeding. Try reducing the value of the 1M resistor (the one to the right of the fet). The tradeoff will be that you will eventually start to lose a bit of signal with the fet in the "on" state (and we're probably all familiar with commercial pedals that do this!).

Hope this helps.  

edit: PRR beat me to it... He's right about switching the gate to ground - I missed that.