Jfet switching.. sharing gate voltage, can do?

[dschwartz]

Hi there folks.. i have a small question..

standard jfet switching needs a high ohm resistor, a cap to ground, and a reverse diode to drive the jfet Gate..

what if i need to close (or open) 2 fets simultaneously, can i share the voltage coming from 1 diode to both gates so i dont need to duplicate the control circuit? there´ll be crosstalk if so?

just a cheap lazy guy question
cheers!!!

[R.G.]

standard jfet switching needs a high ohm resistor, a cap to ground, and a reverse diode to drive the jfet Gate..

what if i need to close (or open) 2 fets simultaneously, can i share the voltage coming from 1 diode to both gates so i dont need to duplicate the control circuit? there´ll be crosstalk if so?

Mother Nature insists that every cotton-picking thing be correct, or it won't work. Well, actually, it will always work - but it will work the way Mother Nature's Laws say it's going to work, not how you thought it ought to work.

In this case, the answer is "it depends". It depends on what the DC and signal voltages on the drains and sources of the two JFETs are doing. If the sources are at the same DC voltage, and the signal is "small" compared to the DC voltage, you may be able to make it work, although using two diodes has a much better chance.

A JFET is a resistor made with doped semiconductor material. An N-channel one is a slab of very, very, very pure silicon with a controlled amount of N-type impurities that lower the nearly-infinite resistance of the pure silicon down to some lower resistance. Switching-type JFETs are often in the 10 ohm to 500 ohm range. This resistance is the lowest resistance that the drain-source can ever be.

The gate is a region of P-type (in an N-channel device) silicon on top of and overlaying the channel. If you pull the gate positive with respect to the channel, it starts conducting when it hits the silicon diode turn-on voltage.  If you pull it negative compared to the channel, the voltages force all the charge carriers to be depleted from a certain thickness on each side of the junction area. "Charge carriers depleted" means "I ain't gonna conduct NOTHING!" for the area with charge carriers depleted. The bigger the reverse voltage, the bigger the thickness of depleted charge carriers, so the higher the resistance of the channel, as more and more of the conductive area is being squeezed shut. I call this the Garden Hose effect; you get less flow by squeezing the garden hose closed.

The reason I'm blathering about this is that if you have two JFETs with their sources tied to the same DC voltage and "small" signals, then yes, you can turn them both off by pulling the gates way below the source voltage. If the source voltages are different, then the "off-ness" is different for the same gate voltage on each one.

The series diodes change things. If you have a series diode that only conducts OUT OF the gate, and you pull it the non-conducting way for the gate diode, then the gate is effectively open circuited, and the drain-source channel acts like the N-type resistor it is without the gate being involved, as the gate is open. If you pull the diode the conducting way, negative for the setup for N-channel devices, then the diode is forward biased, and it pulls the gate to one diode-drop-less-negative than the diode is being pulled.

If you use two diodes and one R-C slowdown network, and the pull DOWN is big enough for both devices at the same time, they both turn off. If the voltage on the two diodes gets positive enough to turn them both off, yes, the JFETs turn on independently.

As you can tell by the description, there are several ways to get this wrong.     But yes, iin theory, it's possible.

If you put a voltage on it that lets

[dschwartz]

Wow, thanks RG, for your amazing response, as usual.

I'm trying to get 2 switchable gain and volume pots from a single dist+ type distortion, like a dual pedal..

For gain switching, a pair of 2sk30a in parallel with each gain pot at the feedback loop, and for volume switching, a jfet spdt after the volume pots..

All the fets will be referenced to 4.5v (thats the voltage at the opamp feedback, and the volume pots will be referenced to 4,5v)..

BUT, after reading your response I'm  doubtful..since the voltage swing at the opamp feedback will be large (up to opamp clipping), it is fairly possible that i will get fet distortion instead of opamp clipping..

Should i use a cd4066, or 4053 instead? Or use a relay?

[R.G.]

If you're feeding this a signal as big as the opamp's clipping and referencing it to the opamp output, you're going to have problems turning the JFETs off.

JFETs are off when the gate is more than Vgsoff negative (for N-channels) with respect to the source. If the source is at 4.5V, the signal is as bit as -3V, then you have to pull the gate negative by the sum of -3V (the signal negative peak) PLUS Vgsoff. For the 2SK30A,Vgsoff can be 3-4V, so you'll need a gate turn-off voltage or about -7 to -8V to keep the signal voltage from turning the JFET partially on.

Using CMOS switches would be my next step. They work really well and are not complicated to drive. Relays are more expensive, more complex to drive, and wear out.

[dschwartz]

For bypass I'm using a tq2 5v latching relay, driven by a 4049 flip flop, with 5.1v zeners at the coil to limit the pulse voltage and a 22uf cap from coil to gnd..it's very simple and at least it worked very consistenly on breadboard..
I dont plan to use a relay for channel switch, since I'm afraid of popping and costs... The option would be a 4066 since it is low noise and a very low On resistance

[R.G.]

I dont plan to use a relay for channel switch, since I'm afraid of popping and costs... The option would be a 4066 since it is low noise and a very low On resistance

I'm not sure what you mean by 'low noise' for the 4066. CMOS transmission gates are very much the same within a series.

Low ON resistance makes a difference only when the other resistances in the circuit are low. If, for instance, the pots are in the 10K range and up, a few hundred ohms for a switch won't be noticeable - and probably not measurable.

But if you already have 4066s, that trumps buying other chips.

[dschwartz]

I also have 4053 available..it's just that 4066, according to the data sheet is more suitable for analog signal switching..

I just want to avoid pops when switching the channel, that's why i like jfets as you can make a slow turn on switch..cmos are instantaneous (or almost instantaneous) when switching, so i expect a click or pop when switching..

Maybe i can get away by using 2 parallel drive channels , and switch the in and out signals, which are far from the rails, and easily handled by jfets with seamless switching..

I'll try with cmos nevertheless..time to breadboard!!

[R.G.]

I used the 4053 in - literally - many thousands of pedals built to my designs. They do not pop when used properly.

The secret to "properly" is at geofex.com in the article on CMOS bypassing. Shhhh... don't tell anyone else. It's only been posted there since 2000. 

[dschwartz]

I know RG, I've read your articles a thousand times, and probably will a thousand  more. Since my first approach to music electronics, geofex has been my go-to site..you, Forrest Mims, Tim escobedo, Teemu Ktala, muzique and ROG are my "mentors"..

My insecurity about cmos switching on the feedback loop, is that on simulation, it shows sharp peaks when switched...

Ive seen the guts of visual sound pedals and as i saw the 4049/4053 chips, aha!! RG was here, i though. But I'm weird and i like doing stuff my own way at first, thats why i refuse to make clones, and designed some neat (at least to me) circuits..

I'm very happy having this conversation with you..and I thank you for all the inspiration and knowledge you share so uninterestedly..,

[R.G.]

My insecurity about cmos switching on the feedback loop, is that on simulation, it shows sharp peaks when switched...

I believe that may be because of the break-before-make nature of the CD4053. I'll have to look at the switching timing, but I think it might leave a feedback loop open for a hundred nanoseconds or so. In that case, you just arrange the feedback parts so the 4053 is not opening the loop, but switching in things in parallel. But there are always special cases. I'd have to look at your schematic.

I'm very happy having this conversation with you..and I thank you for all the inspiration and knowledge you share so uninterestedly..,

You're welcome to any help I can be. I just didn't know if you'd read it.