I've been servicing guitar amps for awhile now but I still have a lot to learn, particularly when it comes to solid state amps. Recently I replaced a switch for the boost feature on a Traynor TS120B bass combo (schematic attached). I'm OK with understanding schematic symbols etc but I'm often stumped as to how various components interact with each other. So I was curious as to what is in fact happening with the 2N5245 JFET when the boost switch is engaged. To put it bluntly, how does the 2N5245 JFET (along with other components) increase the signal level? Is the 2N5245 JFET providing more current to the TL071 op amp by virtue of it being connected to ground when the switch is engaged? Any insight is much appreciated, thanks.

JS

 

The gain of a non-inverting opamp amplifier is Av = 1 + Rf/Rin. In this case the gain of the TL071 Av = 1 + [value of gain pot (Rf)]/[2.2K resistor (Rin)]. When active, the JFET acts as a switch and shunts the 2.2K resistor with a 1K resistor making it's equivalent value 688 ohms. A lower 'Rin' means higher gain.

 

The gain of a non-inverting opamp amplifier is Av = 1 + Rf/Rin. In this case the gain of the TL071 Av = 1 + [value of gain pot (Rf)]/[2.2K resistor (Rin)]. When active, the JFET acts as a switch and shunts the 2.2K resistor with a 1K resistor making it's equivalent value 688 ohms. A lower 'Rin' means higher gain.

OK, I think I follow to a degree. From https://en.wikipedia.org/wiki/JFET , " A JFET is usually on when there is no potential difference between its gate and source terminals." I take for the 1K resistor to be in parallel with the 2.2K the JFET needs to be "on" (conducting between source and drain). So turning the physical switch to the ON position should cause there to be the same voltage at source and gate. The source terminal is connected to ground (indicated by upside down triangle with number 2 inside referring to preamp ground) but there is a 100K resistor between chassis ground and the gate. So I don't understand how they could be at the same potential with that 100K resistor there.

Also, why is the 2.2K resistor considered Rin? The op amp is non-inverting, right? In the diagram on this page https://www.electronics-tutorials.ws/opamp/opamp_3.html the resistor between the inverting input and ground is not labelled Rin and in the diagram here https://www.electronics-tutorials.ws/opamp/opamp_2.html there is a resistor labelled Rin but the op amp in that case is inverting and the resistor is connected between the inverting input and the signal source and not ground.

 

A JFET is, as the name says, a "field effect" device - control is done with a voltage difference between the gate and the source. There is nominally "no" current flow in the gate circuit. In reality, there is a very tiny current, essentially the reverse bias leakage current of a diode formed between the gate and the channel of the FET. For the FET in question, the specified maximum current at 25°C is just one nanoampere (this is IGSS - "gate reverse current" in the datasheet). Because the current is so small, the voltage drop across the 100k resistor is just 100 µV (100k ohms x 1 nA; again, max. @ 25°C) which is inconsequential in this circuit. The resistor is probably there to help provide a measure of protection of the gate from spurious voltage such as static discharge.

The 2.2K resistor in this circuit really isn't "Rin" in the usual sense - just a part of the gain setting network.

 

OK, Rin is just nomenclature. The link you provide it is R2. What's in a name .

The specs on this JFET indicate it is 'ON' when the gate-source voltage is 0. It is 'OFF' when the gate-source voltage is -1 volt or lower.

The voltage divider of 50 volts/6.8k/1.5k produces a bias of about -9 volts at the cathode of the LED.

The JFET source is tied to ground. When the switch is off, that bias is connected directly to the gate and the JFET is 'OFF'. The 1K is out of the circuit.

When the switch is on, current from the voltage divider passes through the LED (causing it to illuminate), but because of the short on the control jack, there is no voltage on the gate and the JFET will be 'ON'. This connects the 1K in parallel with the 2.2K and increases the gain of the op amp.

If the pedal is plugged in and depressed, then the above will still be true. the Anode of the LED will be grounded, there will be no voltage at the JFET gate, the JFET will be on, and the gain will be higher.

When the pedal is released, the voltage from the voltage divide passes through the LED and is seen on the gate, turing the JFET off. JFET off, lower gain.

Actually pretty simple.

 

OK, Rin is just nomenclature. The link you provide it is R2. What's in a name .

The specs on this JFET indicate it is 'ON' when the gate-source voltage is 0. It is 'OFF' when the gate-source voltage is -1 volt or lower.

The voltage divider of 50 volts/6.8k/1.5k produces a bias of about -9 volts at the cathode of the LED.

The JFET source is tied to ground. When the switch is off, that bias is connected directly to the gate and the JFET is 'OFF'. The 1K is out of the circuit.

When the switch is on, current from the voltage divider passes through the LED (causing it to illuminate), but because of the short on the control jack, there is no voltage on the gate and the JFET will be 'ON'. This connects the 1K in parallel with the 2.2K and increases the gain of the op amp.

If the pedal is plugged in and depressed, then the above will still be true. the Anode of the LED will be grounded, there will be no voltage at the JFET gate, the JFET will be on, and the gain will be higher.

When the pedal is released, the voltage from the voltage divide passes through the LED and is seen on the gate, turing the JFET off. JFET off, lower gain.

Actually pretty simple.

Thanks Ylli and epb. I'm going to study your replies. I'm looking forward to a day when it is all simple to me as well.