JFET switch bleed in off state

[agent_sprinkles]

Hi - I'm trying to work with a p-type JFET switch as shown in the attachment. The JFET is an FCS P175, and  the diode is an FCS 2N5962. The switch works great when it's on, but in the off state, there is still a bit of bleed through I can hear if I turn my amp all the way up. Using my computer audio out as a function generator and my multimeter Vrms to measure the output, I'm measuring about -35dB of attenuation in the off state. Is there anything I can do to get the attenuation higher?



Many thanks!

--Alex

[R.G.]

If it lets signal through when it's "off", then it's not "off" enough.

JFETs vary in the amount of reverse bias needed to make them turn fully off. You have, at most, 4.5V at the source and 0V on the gate, but you're losing 0.5-0.6V of that in the reverse diode.

If the JFET Vgsoff voltage is near 4V, then it will bleed. The cure is to either (1) make the Vgsoff voltage bigger by raising the bias voltage on drain and source, or (2) pick a JFET with a smaller Vgsoff.

Try this: your bias to 4.5V might as well be to +7V or so. Or even +9V, although that will let in any power supply noise. Try raising the "4.5V" to maybe 6-7V and see if it's not more "off".

Note that there are other ways signal can bleed through and I'm assuming you've stoppered them up, too.

[PRR]

> in the off state

It never goes utterly OPEN off.

The "off"-ness has to be considered relative to the *other* resistor.

Because such switching is ALWAYS a Voltage-Divider.

You do not show any load resistor except the 1meg. If the JFET went to 55Megs (pretty dang "off"), that's 55Meg:1Meg or 56:1 which is pretty near 35dB.

Your amplifier may be another Meg. Still have a need for EXTREME off-ness in the FET.

Also your "off" condition leaves the amplifier input un-connected. It is wide-open for hum, radio, even signals leaking from cords before the switch.

Also the assumed 55Meg is shunted by capacitors, parasitic in the FET and in the wiring. Just 30pFd working against 1meg will give hardly-any attenuation (3dB) at 5KHz.

You need to consider it as a Voltage Divider, meaning you need to give the FET something to work against.

A 10K load on the output will give pretty good attenuation. But consider if the signal source can drive that. (Many pedals can, some won't, and guitar will be loaded bad.) 10K is also about the least load I'd want on a "disconnected" amplifier input.

For serious switching, you use a second switch to ground. If Q1 is 55meg series and Q2 is 1K shunt you have 90+dB attenuation. OTOH when Q1 is 1K and Q2 is 55Meg with another 1meg load you have less than 0.2dB loss in the on condition. This may complicate your drive circuit and layout, true.

The Ed Oxner book on Designing with FET has a couple dozen pages on FET switching. It's a lot about complexity versus performance. How much will you pay for how much/little signal?

[amptramp]

You also have a fight between the leakage current through the diode connected to the gate and the gate junction itself (although you show a MOSFET, you are asking about a JFET so I assume the FET symbol is wrong).  Some people go as far as to have a voltage follower drive a 1 or 2 Megohm resistor connected from the source to the junction of the gate and the diode.  When the diode anode goes high, the FET switches off and the gate voltage is at a diode drop below the gate drive signal.  When the diode anode goes low, what sets the gate drive?  It is set by the relative leakage current of the diode and the FET gate.  If the FET leaks more, the gate may stay near 4.5 V.  As Paul mentioned, you have to reduce the signal source and output load impedances to make sure you can get the attenuation you want.

For fuzz and clipper uses, diode switching works better than FET switching and there is no need to select components.  You may want to use antiphase diode bridge switching to reduce the switching noise.  As with the FET switch, proper control of impedances allows you to get a good on/off ratio.

The other possibility is to use the CD4009 / MC14007 in the panning circuit shown in the Motorola app note in the datasheet:

http://html.alldatasheet.com/html-pdf/11942/ONSEMI/MC14007/367/2/MC14007.html

It is easy to set this up for very low switch pop by driving pins 3 and 10 from opposite polarities through an RC lowpass.

[PRR]

> When the diode anode goes high, the FET switches off and the gate voltage is at a diode drop below the gate drive signal.  When the diode anode goes low, what sets the gate drive?  It is set by the relative leakage current of the diode and the FET gate.

In fact, *for JFETs*, this plan is reliable. (JFET gate currents are order-of-magnitude smaller than generic diode leakage.)

It always looked "wrong" to me, true.

[Seljer]

A 10K load on the output will give pretty good attenuation. But consider if the signal source can drive that. (Many pedals can, some won't, and guitar will be loaded bad.) 10K is also about the least load I'd want on a "disconnected" amplifier input.

For serious switching, you use a second switch to ground. If Q1 is 55meg series and Q2 is 1K shunt you have 90+dB attenuation. OTOH when Q1 is 1K and Q2 is 55Meg with another 1meg load you have less than 0.2dB loss in the on condition. This may complicate your drive circuit and layout, true.

The Ed Oxner book on Designing with FET has a couple dozen pages on FET switching. It's a lot about complexity versus performance. How much will you pay for how much/little signal?

ah, hence the buffers in the Boss/Ibanez pedals

[R.G.]

We're confusing the original poster with subtleties, guys.

His JFET doesn't have enough off drive to turn it entirely off, especially with some signal variation. That same circuit works reliably in zillions of Boss and Ibanez pedals - and a very few of them have this same issue. More off-bias will almost certainly fix it.

[PRR]

.

[R.G.]

Yes, but notice he's only included "in" and "out". Whether he recognizes it or not, there are impedances with each of those. He'd have to load it with an "out" load of more than 10M to make the 1M bias resistor be the only thing counted. I sincerely doubt he's doing that.

[tubegeek]

Get a room, you two.