I would like to use a JFET, like a J113, as an audio switch that is in series with the signal. The audio input signal is tied to the Drain, and the audio output signal is tied to Source. The audio signal fluctuates above and below 0V. The Gate is controlled by a control voltage, where 0V= switch is on, and -5V= switch is off. I have seen circuits where a diode is placed between the control voltage and the Gate. The anode of the diode is tied to the Gate and the cathode is tied to the control voltage.

Why is this put there? In describing this switch circuit, Kevin O'Connor in the book "The Ultimate Tone" writes on page 9-12, "A reverse-biased diode in series with the j-fet gate allows the j-fet to find its own "on-state." Does anybody know what he is talking about?

I understand that the Source-Gate voltage difference determines 'on resistance' of the jfet and the current that can pass through the jfet. Since the audio signal is going through the Source, there will be some distortion as the Source fluctuates in relation to the steady 0V at the Gate. Does Gate diode help with this? I also don't know why he calls it reverse-biased. Is that because it points in the opposite direction as the arrow on the jfet symbol?

It seems that a reverse-biased diode on the Gate lets the current flow to -5V when the control voltage is -5V, but when the control voltage is brought to 0V, it seems that the electrons at the Gate would be trapped there by the diode. My breadboard experiments show that this is not the case. The switch turns on and off. However, it is slower to turn on than off. Also, if I switch the diode around - i.e. cathode tied to the Gate - the switch turns on fast and off more slowly. Is there some diode leakage, in the micro or nano amp amounts, that is enough to change the voltage on the Gate?

How, does the diode's voltage drop fit into this, also? I have tried to measure this with my multi-meter, but I don't seem to be measuring any voltage drop between the anode and cathode. Only the cathode is tied to a voltage source, the anode/Gate side is floating. It seems to me that without the diode being tied to voltages on both sides, its behavior would be random (static electricity fluctuations).

I am sorry if I am not able to write more coherent questions. I just don't get what the diode does for the simple JFET audio switch circuit.

Thanks for any thoughts,
Dave

 

I should preface this by saying that most (all?) JFETs are symmetrical, so drain and gate are interchangeable. Furthermore, whichever one is lower in voltage than the gate assumes the role of the source.

Your input signal needs to be limited to ≈2.8V p-p (≈±1.4V). The control voltage needs to swing to at least +1.4V for turn-on,to ensure that the diode never conducts when the FET is on.
When the control is at -5V, the diode is forward biased through the drain-gate resistor, so the gate will be at about -4.4V, and the FET will be off as long as the input does not swing below -1.4V (the Vgs cutoff voltage is -3V).
When the control is at +1.4V (or higher), and the input is less than +1.4V, the the diode will basically stay off during the entire input cycle, so the gate voltage will follow the input (and the output will also follow). The path is through the drain-source resistor. Since the diode is not cunducting, there will be no voltage drop across the resistor, so Vgs and Vgd will be zero, ensuring maximum and constant conduction.
If your input signal never exceeds +1V, then the control voltage high level can be 0V so long as the load resistance is high, e.g., 100kΩ. Low load resistance could result in some harmonic distortion.
Switching time is influenced by the gate capacitance. When the control voltage goes low, it discharges the gate capacitance rapidly through the diode, which is very low impedance when conducting. When the control goes high, cutting off the diode, the gate has to recharge through the resistor, which is a much longer time constant.

 

Ron,

Thanks for the comments. This is making more sense!

You wrote:

I should preface this by saying that most (all?) JFETs are symmetrical, so drain and gate are interchangeable.

I think you mean that the drain and the source are interchangeable, correct?

And later in your post you wrote:

The path is through the drain-source resistor.

Do you mean drain-gate resistor?

The circuit you are describing has a resistor that is tied between the drain and the gate. I have seen circuits where a 1M or a 10M resistor are used. A source-gate resistor would also work, correct?. This circuit make much more sense than the one described in O'Connor's book. He has the jfet in series with the audio signal and a diode between the control voltage and the gate, with no resistor. So, the gate side of the diode is floating and not tied to a definite voltage. Curiously, the switch still works without the resistor. I don't understand how it should work. I have seen other jfet switch circuits, where there is no drain-gate (or source-gate) resistor. See the first circuit in this article:

http://www.premierguitar.com/Magazine/Issue/2007/Jun/JFET_Switching.aspx

Your design with the resistor makes more sense to me. When the control voltage is -5V on the gate, some dc current is being added to the audio signal through the source-gate resistor, correct? Is this a problem? Can this effect the sound?

Thanks,
Dave

 

You wrote:

Quote:
I should preface this by saying that most (all?) JFETs are symmetrical, so drain and gate are interchangeable.
I think you mean that the drain and the source are interchangeable, correct?

Oops! Yes, of course.

And later in your post you wrote:
Quote:
The path is through the drain-source resistor.
Do you mean drain-gate resistor?

Oops again!

The circuit you are describing has a resistor that is tied between the drain and the gate. I have seen circuits where a 1M or a 10M resistor are used. A source-gate resistor would also work, correct?. This circuit make much more sense than the one described in O'Connor's book. He has the jfet in series with the audio signal and a diode between the control voltage and the gate, with no resistor. So, the gate side of the diode is floating and not tied to a definite voltage. Curiously, the switch still works without the resistor. I don't understand how it should work. I have seen other jfet switch circuits, where there is no drain-gate (or source-gate) resistor. See the first circuit in this article:

http://www.premierguitar.com/Magazin...Switching.aspx

Your design with the resistor makes more sense to me. When the control voltage is -5V on the gate, some dc current is being added to the audio signal through the source-gate resistor, correct? Is this a problem? Can this effect the sound?

I don't see how the circuit can work without a resistor, at least not reliably. It doesn't work in simulation.
If you connect the resistor to the output, some of the control signal will show up there, attenuated by the voltage divider consisting of the gate-to-output resistor and the load resistance. The resistor should be from gate to input. You need to drive the input side with a very low impedance to prevent this sort of coupling. If you don't already have this, you should add an operational voltage follower or some other sort of amplifier which has essentially zero output impedance.

 

Ron,

It is very interesting that the circuit without the resistor does not work in simulation. Maybe some kind of electron leakage through the diode is enough to change the state of the gate, since there is so little capacitance (2 pF) on the gate. I have really been scratching my head on this for a while. I will stick with your design, because it makes sense.

Also, I realize that when dc current is being drawn through the source-gate resistor, the jfet is off and no audio signal is being allowed to the output. So, I don't think dc current is a problem. There is some dc current in the audio signal when the jfet is transitioning from on-to-off or off-to-on, though.

Thanks,
Dave

 

What type of diode are you using? A high-capacitance diode can allow the transistor to switch. In simulation, it works with a 1N4001, but not with a 1N4148.

 

I am using a 1N4148 and the switch works without the source-gate resistor. There is a noticeable slow turn on time, and a fast turn off.

How does a high capacitance diode make the simulation work?

Thanks.

 

I am using a 1N4148 and the switch works without the source-gate resistor. There is a noticeable slow turn on time, and a fast turn off.

How does a high capacitance diode make the simulation work?

Thanks.

I was running a relatively high frequency on the control signal. I slowed it down, and it is apparent that diode leakage eventually charges the gate capacitance, turning on the JFET. The 1N4001 has a larger junction and therefore higher leakage, so the turn-on delay is less than with the 1N4148. Both diodes turn the JFET off rapidly, of course, because of their low forward resistances.
SO - as you observed, you don't need the resistor, if you can tolerate an indeterminate delay. The slow turn-on may reduce popping when switching, but I think I would prefer use the resistor so I have predictability, and then control the switching times and slew rates by shaping the control voltage.

The higher capacitance of the 1N4001 couples enough of the control signal to the gate if the control voltage is 10V p-p (there is a voltage divider formed by the diode capacitance and the gate capacitance) to turn the JFET on immediately. With a 5V p-p signal, there was still a delayed turn-on because leakage was required to finish getting the gate voltage above the threshold.

 

Thanks Ron!

It is the leakage current in the diode that allows the gate to return to 0V. That finally makes sense to me!!

Even though there is a delay, I am impressed that the jfet switches on as fast as it does.

I was looking at a 1N4148 datasheet, which states that a reverse voltage of 20V causes a leakage current of 25 nA, and a reverse voltage of 75V causes a leakage current of 5 uA. They do not give a rating for 5V.

I think it would be difficult to calculate the turn on time, since the leakage current of the diode decreases as the reverse voltage decreases. Sounds like a calculus problem to me.

I think I would prefer use the resistor so I have predictability, and then control the switching times and slew rates by shaping the control voltage.

I agree with you! A simple RC network on the control voltage worked perfectly to slow the on/off switching time. That is, I put a resistor in series between the control voltage and the gate diode, and a capacitor connected at the gate diode's cathode and to ground. I realized that if I made the RC network's resistor too large in relation to the source-gate resistor, I got a voltage divider that kept the jfet from turning off fully. So, I made the source-gate resistor 10M and the RC nework resistor 1M. That seemed to work.

By placing another diode in parallel with the RC network resistor, I found I could have fast-on/slow-off or slow-on/fast-off depending on which way it was facing. This second diode in the circuit lets me bypass the RC network's resistor in one of the two directions that the voltage can be oriented. What do you think?

The higher capacitance of the 1N4001 couples enough of the control signal to the gate if the control voltage is 10V p-p (there is a voltage divider formed by the diode capacitance and the gate capacitance) to turn the JFET on immediately. With a 5V p-p signal, there was still a delayed turn-on because leakage was required to finish getting the gate voltage above the threshold.

Could you rephrase what you said here? I am not quite following what you are saying.

 

By placing another diode in parallel with the RC network resistor, I found I could have fast-on/slow-off or slow-on/fast-off depending on which way it was facing. This second diode in the circuit lets me bypass the RC network's resistor in one of the two directions that the voltage can be oriented. What do you think?

Yep, that's a common way of tailoring rise and fall times.

The higher capacitance of the 1N4001 couples enough of the control signal to the gate if the control voltage is 10V p-p (there is a voltage divider formed by the diode capacitance and the gate capacitance) to turn the JFET on immediately. With a 5V p-p signal, there was still a delayed turn-on because leakage was required to finish getting the gate voltage above the threshold.

Could you rephrase what you said here? I am not quite following what you are saying.

I'll try.
Let's say the diode has Cd=12pF of capacitance, and the J113 has Cg=6pF of gate capacitance. Also, assume the control signal swings from -5V to 0V. The gate starts out at about -5V. When the control signal transitions to 0V, the diode will cut off, but due to the capacitive divider made up of Cd and Cg, the transition will couple 5V*12pF/(12pF+6pF) = 3.33V to the gate, raising the gate voltage to (-5+3.33) = -1.67V. This may turn the JFET on partially, but diode leakage will have finish charging it to 0V gradually. I think you can see what will happen if the control voltage goes from -5V to +5V, or if you use a diode with less capacitance.

 

Ron,

I am not familiar with a capacitive voltage divider. I found a description here, with the formula you used:

http://en.wikipedia.org/wiki/Voltage_divider

But, I am still not getting how it is working. Are you saying that when the diode's cathode is brought to 0V, immediately a voltage of 3.333 volts is seen across the diode and a voltage of 1.67 is seen across the gate and the source of the jfet? How does that work? Does the diode's capacitance draw electrons away from the gate's capacitance?

I will keep trying to figure this out. I also see a capacitive divider being described in Chapter 12 of the AC book.

Here is another example of a soft transition jfet mute switch that uses a diode and various resistor voltage dividers to adjust on/off times and shunts the audio signal to ground. It seems a bit overly complicated, in my opinion. I like the addition of the red and green diodes.

http://www.edn.com/article/CA468422.html?spacedesc=DesignIdeas&taxid=10536

Thanks for the help,
Dave