SImplest way to get a negative voltage to turn a JFET off?

[Marco]

I'm already using the 9V battery as a positive power supply for the rest of the circuit. My JFET is across my audio path as a "mute" with the Source at audio-ground, which is also power-ground for the +9V supply.

Then don't do that, couple the source to output with a coupling capacitor and use a resistor or current source to circuit ground.

[K3mHtH]

Alternative approach.

P ch jfet J175 or similar.

1M from control input to gate, B+ for channel open, take to ground to mute.

Wait, p-channel JFETs take a positive gate?

The J175 has an Rds(on) of 125 ohm, and I need something under 30.. but I'll start looking around for more options.

[K3mHtH]

Alternative approach.

P ch jfet J175 or similar.

1M from control input to gate, B+ for channel open, take to ground to mute.

Wait, p-channel JFETs take a positive gate?

The J175 has an Rds(on) of 125 ohm, and I need something under 30.. but I'll start looking around for more options.

I feel pretty dumb for not knowing p-channel JFETS take a positive gate, but a quick falstad circuiit simulator check and it seems to work. Thanks Xena E!

The selection of p-channel JFETS is way worse, must be harder to produce? The Rds(ON) spec is always higher as well, even on some $8 parts.

So.. is there anything stopping me from putting 4 or more J175 JFETS in parallel to get down to my desired Rds(ON)?

[Xena E]

For a one off, select your components and try parallel,  use separate gate drive resistors. I don't know enough about your application to qualify the suggestion.

For instance can you raise the source R to compensate for higher rds?

I suggested the J175 because they were cheap jellybean types and are pretty much symmetrical with interchangeable d/s.

Give it a try and good luck.

X

[Zero999]

What's the impedance of the signal?

How much attenuation is required?

The problem with shunting the signal is, a low on resistance switch is required to achieve a high level of attenuation.

A J-FET is a good idea. Just watch the voltage swing of the signal doesn't exceed the threshold voltage, otherwise it will start to turn on and cause distortion.

No gate resistor is required. A J-FET can also handle the gate being forward biased i.e. positive, for an N-channel device and negative, for a P-channel device, so long as the gate current rating is not exceeded. It's known as the enhancement region and reduces the on resistance.

P-channel devices do have a much higher on resistance. If it's practical you could just an N-channel P-JFET and AC couple to +V. The signal must be referenced to +V, but if your mute circuit is running off a separate battery, then this doesn't matter.

[BrianHG]

For a mute switch, you can just use a CD4053 for break and short.  Or, a CD4066 with all the switches wired in parallel for something like a sub 5ohm switch.

The cmos CD4000 series will draw virtually no current even powered at 9v.  The CD405x switched allow for a Vee negative supply allowing AC signals with a 0v GND bias and using the A/B switch of the CCD4053, A on source signal and B on GND with the Y to your output will generate an insane isolated mute which cannot be achieved with 1 J-fet.

The are also available in 5/6 pin SOT-23 single switch series...

[Zero999]

The problem is, the original poster can't break the signal link, for whatever reason.

How about just using a relay? A simple reed relay doesn't use much coil current. If it has to be solid state, use a photovoltaic opto-coupler, such as the TLP3905 and a couple of MOSFETs.

[BrianHG]

The problem is, the original poster can't break the signal link, for whatever reason.

Again, a CD4066 can be used without breaking the signal.  Signal on one side of the switch, and the other tied to the GND.

And again, we do not know it the OPs signal is DC biased or not, but with 4 analog muxes in 1 single CD4066, the on resistance will be around 10ohm with a 9v supply.

[Zero999]

The problem is, the original poster can't break the signal link, for whatever reason.

Again, a CD4066 can be used without breaking the signal.  Signal on one side of the switch, and the other tied to the GND.

And again, we do not know it the OPs signal is DC biased or not, but with 4 analog muxes in 1 single CD4066, the on resistance will be around 10ohm with a 9v supply.

I assumed the signal is AC coupled, in which case, I don't see how a CMOS analogue switch which isn't designed to a signal below the negative rail can work. I already suggested some analogue switches which can handle negative signals (presumably they have a built-in charge pump) a few posts ago, but it appeared to fall on deaf ears.

https://www.eevblog.com/forum/beginners/simplest-way-to-get-a-negative-voltage-to-turn-a-jfet-off/msg5359274/#msg5359274

[Xena E]

If this is just a mute on or off situation could it be done with a Reed relay? It would give low resistance and wouldn't matter about if the signal swung + or - around the ground rail.

OK you lot stop laughing. 

X

[dmills]

The challenge in a guitar pedal is to do it with sufficiently low current to not muller the 9V PP3 size battery (Which is where photomos usually fails) and to do the switching slowly enough to not cause charge injection clicks in what can be an annoyingly high Z circuit.

Suggestion: A couple of caps and some large resistors to move the audio reference up to the +9V rail, and just switch to ground to cut the fet off? -6V (Flat battery) will do it and -9V will do it easily.

Cost is probably three or four 1M resistors a couple of 100nF caps and the jfet, if you need better isolation then you can add a second jfet as a series/shunt sort of switch. 

[BrokenYugo]

Are modern coin cells really that good that after 10yrs even on the shelf, they are still ok ? IDK how bad they used to be, like in a digital wrist watch, but I'm guessing after 10yrs on the shelf, they would not be much good. IDK maybe they were tho.

I've seen Japanese brand CR2032s with 20+ year old date codes still holding data in a single SRAM chip, with the battery powering the chip probably 99+% of the board's life. A couple even measured above 3 volts, and that's nothing special, just consumer grade electronics, mostly Nintendo games. Note that when swapping those batteries I successfully used a ~10 year old salvaged bios battery from a laptop for the  temporary power supply to maintain the SRAM contents, because it already had wires on it and tested ok. You may be confusing the similar looking stainless canned 1.5V alkaline cells (e.g. LR44) that are usually more of a "button" profile with the CR series 3.0V lithium cells that are usually more coin shaped and have much longer shelf lives.

[BrianHG]

The challenge in a guitar pedal is to do it with sufficiently low current to not muller the 9V PP3 size battery (Which is where photomos usually fails) and to do the switching slowly enough to not cause charge injection clicks in what can be an annoyingly high Z circuit.

Suggestion: A couple of caps and some large resistors to move the audio reference up to the +9V rail, and just switch to ground to cut the fet off? -6V (Flat battery) will do it and -9V will do it easily.

Cost is probably three or four 1M resistors a couple of 100nF caps and the jfet, if you need better isolation then you can add a second jfet as a series/shunt sort of switch.

A 9v battery could power a CD4053 for years.
You can place the 9v on the VDD and VEE with the GND connected a 10meg between +&- of the battery for a +&- 4.5v dual supply system.

One side of the switch would be open and the other tied to GND, wiring the 3 switched in parallel for the low impedance short to GND when needed.

Instead of a 9v bat, 2 lithium coin cells could be used instead, offering 10 to 20 years battery life for this circuit, though a 9v to 2x9v batteries would get the 'ON' resistance short to GND down below 10 ohm as the CD4053's on resistance shrinks with larger supplies.


Anyways, if it is a problem with making a dumb guitar pedal mute, none of this is needed.  The J-fet would have the source at gnd, drain at the signal, and the 9v battery would have a 2x series 1meg from + to -, with the center of those resistors to GND.  Now, you have a +4.5v and -4.5v to drive the gate driven through a 10meg series resistor.   No extra batteries needed.

[Zero999]

The challenge in a guitar pedal is to do it with sufficiently low current to not muller the 9V PP3 size battery (Which is where photomos usually fails) and to do the switching slowly enough to not cause charge injection clicks in what can be an annoyingly high Z circuit.

Suggestion: A couple of caps and some large resistors to move the audio reference up to the +9V rail, and just switch to ground to cut the fet off? -6V (Flat battery) will do it and -9V will do it easily.

Cost is probably three or four 1M resistors a couple of 100nF caps and the jfet, if you need better isolation then you can add a second jfet as a series/shunt sort of switch.

A 9v battery could power a CD4053 for years.
You can place the 9v on the VDD and VEE with the GND connected a 10meg between +&- of the battery for a +&- 4.5v dual supply system.

One side of the switch would be open and the other tied to GND, wiring the 3 switched in parallel for the low impedance short to GND when needed.

Instead of a 9v bat, 2 lithium coin cells could be used instead, offering 10 to 20 years battery life for this circuit, though a 9v to 2x9v batteries would get the 'ON' resistance short to GND down below 10 ohm as the CD4053's on resistance shrinks with larger supplies.


Anyways, if it is a problem with making a dumb guitar pedal mute, none of this is needed.  The J-fet would have the source at gnd, drain at the signal, and the 9v battery would have a 2x series 1meg from + to -, with the center of those resistors to GND.  Now, you have a +4.5v and -4.5v to drive the gate driven through a 10meg series resistor.   No extra batteries needed.

I was a bit quick to dismiss old fashioned analogue switches.

You're on the right track, but the AC signal needs a low impedance path. The problem is, with an AC signal, the current flows in both directions and connecting one side of the switch to 0V could cause problems.

I would recommend the newer CD74H4066.
https://www.ti.com/lit/ds/symlink/cd74hc4066.pdf

Remember the analogue switches need all of the voltages connected to them to be within supply rails.

Connect the both sides of the analogue switches to either supply rail with a potential divider. One side needs to go to either supply rail, depending on which one is the least noisy, via a large capacitor. The other needs to go to the signal, again via a large capacitor, which unfortunately needs to be non-polarised.



In the above schematic C2 could be two 22µF electrolytic capacitors connected back-to-back, to make a 11µF capacitor. Here's an example of two polarised 10µF capacitors used to make a 5µF non-polarised capacitor.


You might need to use 1M resistors, rather than 10M, because it'll take ages (several minutes) to charge the capacitors, which might have too higher leakage to ever fully charge. Tantalum might be the best choice for this circuit.

[BrianHG]

The challenge in a guitar pedal is to do it with sufficiently low current to not muller the 9V PP3 size battery (Which is where photomos usually fails) and to do the switching slowly enough to not cause charge injection clicks in what can be an annoyingly high Z circuit.

Suggestion: A couple of caps and some large resistors to move the audio reference up to the +9V rail, and just switch to ground to cut the fet off? -6V (Flat battery) will do it and -9V will do it easily.

Cost is probably three or four 1M resistors a couple of 100nF caps and the jfet, if you need better isolation then you can add a second jfet as a series/shunt sort of switch.

A 9v battery could power a CD4053 for years.
You can place the 9v on the VDD and VEE with the GND connected a 10meg between +&- of the battery for a +&- 4.5v dual supply system.

One side of the switch would be open and the other tied to GND, wiring the 3 switched in parallel for the low impedance short to GND when needed.

Instead of a 9v bat, 2 lithium coin cells could be used instead, offering 10 to 20 years battery life for this circuit, though a 9v to 2x9v batteries would get the 'ON' resistance short to GND down below 10 ohm as the CD4053's on resistance shrinks with larger supplies.


Anyways, if it is a problem with making a dumb guitar pedal mute, none of this is needed.  The J-fet would have the source at gnd, drain at the signal, and the 9v battery would have a 2x series 1meg from + to -, with the center of those resistors to GND.  Now, you have a +4.5v and -4.5v to drive the gate driven through a 10meg series resistor.   No extra batteries needed.

I was a bit quick to dismiss old fashioned analogue switches.

You're on the right track, but the AC signal needs a low impedance path. The problem is, with an AC signal, the current flows in both directions and connecting one side of the switch to 0V could cause problems.

I would recommend the newer CD74H4066.
https://www.ti.com/lit/ds/symlink/cd74hc4066.pdf

Remember the analogue switches need all of the voltages connected to them to be within supply rails.

Connect the both sides of the analogue switches to either supply rail with a potential divider. One side needs to go to either supply rail, depending on which one is the least noisy, via a large capacitor. The other needs to go to the signal, again via a large capacitor, which unfortunately needs to be non-polarised.

(Attachment Link)

In the above schematic C2 could be two 22µF electrolytic capacitors connected back-to-back, to make a 11µF capacitor. Here's an example of two polarised 10µF capacitors used to make a 5µF non-polarised capacitor.


You might need to use 1M resistors, rather than 10M, because it'll take ages (several minutes) to charge the capacitors, which might have too higher leakage.

Use CD4066, it will operate on a 9v rail and less supply current than the 74HC4066.
Your idea is fine.  The difference with the 4053 with it's Vee pin is that you can wire the GND signal with your 22uf to the actual GND to the A or B side of the MUX.  And the signal side 'Z' goes directly to the audio signal being muted.

This is how the OP keeps his source signal directly routed without any biasing.

Remember, the CD4053 has a Vdd, GND and Vee pins.

It is the 9v battery supplying the +4.5 and -4.5 to the CD4053.
And now the OP may choose active high or active low logic by just choosing which mux input, A or B gets shorted to GND while the other gets left open.

Better yet, short all A&B inputs to GND, all logic inputs to GND except the 'Inhibit' pin.  Tie the audio signal to all mux 3 'Z' pins.  This will make a noise free circuit and antistatic design as everything is grounded on the IC.  1 input, 'inhibit' will either open all 3 mux 'Z' ports to hi-z, or undoing the 'inhibit' wil short all 3 'Z' mux IOs to GND.

[BrianHG]

Ok, you made me do it, here is the schem: (Change the CD4054 to a 74HC4053)



Almost every pin is tied to GND!, No static or EMI noise or floating pins.
Device Quiescent current for the 9v bat will be around 16ua + your resistor divider's current.
Device Quiescent current for the 2x3v bat will be around 8ua + no resistor divider's current.
On resistance = 45ohm / 3 = 15 ohm short to GND.

No biasing on the audio through line, just open or a strong short to GND with a +/- 4.5v range if you are using a 9v battery.  +/-3v range if you are using 2x lithium coin cells which will last around 20years in this circuit.


CD4053 datasheet: https://www.digikey.com/en/products/detail/texas-instruments/CD74HC4053E/376767

Still available in dip or SMD.

[K3mHtH]

Ok, you made me do it, here is the schem:



Almost every pin is tied to GND!, No static or EMI noise or floating pins.

No biasing on the audio through line, just open or a strong short to GND with a +/- 4.5v range if you are using a 9v battery.  +/-3v range if you are using 2x lithium coin cells which will last around 20years in this circuit.


CD4053 datasheet: https://www.digikey.com/en/products/detail/texas-instruments/CD4053BE/67309

Still available in dip or SMD.

Thanks BrianHG!

Is there any advantage to this over JFETS though? I only ask because I already have some p-channel JFETS in the mail for prototyping.

[BrianHG]

74HC4053 is still under mass production by at least 6 different manufacturers, all with the same specs.
And it has been for available for over 50 years because of it's great utility in analog audio mixer/pre amp switchers.

It is under 1$.

It is available in dip and SMD as all those tied GND pins makes it easy to dead-bug style hand wire.

When 'open', the pins 4,14,15 will truly float unloaded in voltage anywhere between the Vdd and Vee supply rail pins.  No variable resistance as you approach the on threshold like a J-fet give you and no gate current when forward biased.

Do the math with your battery chart to fine the battery life of your circuit.
EG: a 250mah lithium coin cell like those in computer motherboards for their real-time clock will last around:
250000/8 = 31250 hours, or 1302 days, or 3.5 years.
(Note that 500mah and 1000mah lithium coin cells exist, IE 7 years or 14years life.)

A duracell 9v battery is 580mah. at 9v the IC takes 18ua plus you need to power your 2 meg resistor divider, that's 4ua for a total of 22ua.
580000/22 = 26363 hours, or 1098 days, or 3 years.

Lithium battery CR2032 = ~250mah.  No leaks like Alkaline batteries.  2 of them will cost around 1$.

[BrianHG]

I got the logic on the INH pin backwards.

Tie to Vdd for open audio.
Tie to GND to short the Audio and pins 4,14,15 to GND with a 15 ohm resistor.

The 2 lithium coin cell solution, if the power supply wires are short, you might not need any caps, but I still recommend a 0.1uf between Vdd & GND and another between GND and Vee.

[Zero999]

Ok, you made me do it, here is the schem:

Almost every pin is tied to GND!, No static or EMI noise or floating pins.

No biasing on the audio through line, just open or a strong short to GND with a +/- 4.5v range if you are using a 9v battery.  +/-3v range if you are using 2x lithium coin cells which will last around 20years in this circuit.


CD4053 datasheet: https://www.digikey.com/en/products/detail/texas-instruments/CD4053BE/67309

Still available in dip or SMD.

What if the audio signal goes below -V or above +V because it's not biased around the middle of the power supply (imagine there's a resistor going from the audio signal, to either +V or -V)? To prevent this, Y & Z need to be biased to ground and AC coupled.


Other than that, it's fine. The plain old 4053 has a higher on resistance, whilst the 74HC4053 has a much lower on resistance and assuming you're right (I haven't checked the data sheet), a higher quiescent current at 9V.

Thanks BrianHG!

Is there any advantage to this over JFETS though? I only ask because I already have some p-channel JFETS in the mail for prototyping.

None that I can think of, other than perhaps simplicity? P-channel J-FETs will work, but have a much higher on resistance, than an analogue switch.

[John Coloccia]

Hey everyone, I have a 9V battery powered circuit for a guitar effect.
I'd like to use a JFET to mute something in the audio path, but I'm already using the 9V battery as a positive power supply for the rest of the circuit. My JFET is across my audio path as a "mute" with the Source at audio-ground, which is also power-ground for the +9V supply. So I'm wondering how I can get at least -6.0V at the JFET's gate.

I see that there are charge-pump IC's like LT1054 that can obtain a negative voltage from a positive one.. awesome. But I really don't need any current here, just enough to turn the the JFET off... is there an easier way that maybe I'm missing? Are there 1:1 transformers maybe where the primary can go across my +9V supply and the secondary winding be used in the reverse direction for a negative voltage to the JFET's gate?

Any chance you could post the circuit? FWIW, you need to be careful where you ground the audio path, if that's what you end up doing, or you could get loud pops when you turn the thing off. It's pretty common for battery operated effects to be referenced at 1/2 Vdd. Just kind of curious what you're up to. Could be easier to figure it out with the actual circuit.

[Zero999]

Hey everyone, I have a 9V battery powered circuit for a guitar effect.
I'd like to use a JFET to mute something in the audio path, but I'm already using the 9V battery as a positive power supply for the rest of the circuit. My JFET is across my audio path as a "mute" with the Source at audio-ground, which is also power-ground for the +9V supply. So I'm wondering how I can get at least -6.0V at the JFET's gate.

I see that there are charge-pump IC's like LT1054 that can obtain a negative voltage from a positive one.. awesome. But I really don't need any current here, just enough to turn the the JFET off... is there an easier way that maybe I'm missing? Are there 1:1 transformers maybe where the primary can go across my +9V supply and the secondary winding be used in the reverse direction for a negative voltage to the JFET's gate?

Any chance you could post the circuit? FWIW, you need to be careful where you ground the audio path, if that's what you end up doing, or you could get loud pops when you turn the thing off. It's pretty common for battery operated effects to be referenced at 1/2 Vdd. Just kind of curious what you're up to. Could be easier to figure it out with the actual circuit.

I agree. That's the rationale behind me strongly recommending biasing resistors and AC coupling capacitors on the analogue switches. We don't know what DC voltage the audio signal might be floating on. They might not be needed. It's likely there's another part of the circuit it can be biased to.

[BrianHG]

What if the audio signal goes below -V or above +V because it's not biased around the middle of the power supply (imagine there's a resistor going from the audio signal, to either +V or -V)? To prevent this, Y & Z need to be biased to ground and AC coupled.

Unlike the CD4053, the 74HC4053 does have IO protection clamping diodes.  But they are weak as I believe the data sheet says something like continuous 100ma max.  (3 in parallel means 300ma protection!)

This is what the OP asked for.  No break in the audio line circuit.
I guess you could add a 47uf non polarized cap between pins 1,2,3 and the 'Audio' line meaning this circuit will have 1 cap in it.

The protection diodes in the IC will keep that cap floating.
However, a 'Jfet' might not fair as well.  You won get that low 15ohm on switch, nor will you have static or high voltage protection built into the IC.
You also wont get that 'Digital' transition high impedance digital grade control input with a guaranteed 1/2Vdd transition voltage.

Other than that, it's fine. The plain old 4053 has a higher on resistance, whilst the 74HC4053 has a much lower on resistance and assuming you're right (I haven't checked the data sheet), a higher quiescent current at 9V.

The battery current consumption calculations I did were from the 74HC4053 data sheet.  So we are good.

[BrianHG]

Just for Zero999.  Now safe for even studio XLR with 48v phantom power on the audio line: (maybe make the cap 63v or 100v just to be safe...)

[Zero999]

What if the audio signal goes below -V or above +V because it's not biased around the middle of the power supply (imagine there's a resistor going from the audio signal, to either +V or -V)? To prevent this, Y & Z need to be biased to ground and AC coupled.

Unlike the CD4053, the 74HC4053 does have IO protection clamping diodes.  But they are weak as I believe the data sheet says something like continuous 100ma max.  (3 in parallel means 300ma protection!)

This is what the OP asked for.  No break in the audio line circuit.
I guess you could add a 47uf non polarized cap between pins 1,2,3 and the 'Audio' line meaning this circuit will have 1 cap in it.

The protection diodes in the IC will keep that cap floating.
However, a 'Jfet' might not fair as well.  You won get that low 15ohm on switch, nor will you have static or high voltage protection.

I thought the diodes were a parasitic element of the CMOS process. Either way, DC coupling can still cause trouble.

Imagine the audio signal has a fairly low impedance path to negative rail, meaning it goes outside of the supply rails. Then there's the big DC transient when the switch connects it to the half the power supply voltage. It's better to keep both sides of the switch at the same DC voltage, which means the signal needs to be capacitively coupled.