Correct biasing of 2N5457 Booster-Buffer circuit

[PRR]

> the voltage at the source moves!

You mean, forever?

Or just a wobble? Audio etc?

For wobbly-audio amplifiers, we can know much if we know the average, which is usually the same as the idle (no-wobble) situation.

If you do want to study the wobble action, it is pretty much a stack of levers. There's "no" gate current. So drain and source currents are equal. Then Rd and Rs resistor voltage drops are proportional to resistance.

[ashcat_lt]

If you know the internal resistance of the fet, you can use ohms law to calculate the voltage drop across it. The same is true for the source and drain resistors. So now the fet amplifier stage looks like 3 resistors dropping voltage across each as the string goes to ground.

Yes.  Idle or wiggling, voltage at the source is dependent on the "resistance" of the transistor, which is dependent on the difference between that source voltage and the gate voltage, so essentially the source voltage is dependent on itself?

It's the same problem we have trying to calculate the actual instantaneous voltage in a diode clipper.  It depends on its own value and we can only get to it via iteration or W function or whatever.

[blackieNYC]

Someone once reminded me that the ROG Fetzer is a specific attempt to imitate a fender tube stage, and while it is certainly accurate and educational, it is not a full explanation of all aspects of fet gain circuit design.

[PRR]

> we can only get to it via iteration

For very large signals (including clipping), this is true. The device function is non-linear, not even a simple non-linear function.

For small signals it is sufficient to take the small-signal parameters of the device at the operating condition. Say Gm comes out 1,250 uMho. That's like 800 Ohms. Say Rs is 1K. Then the source moves about 1K/1.8K or about half of the gate voltage.

In low-Mu triodes there are additional effects. In most hi-Mu triodes, pentodes, BJTs, and FETs with significant drain-source voltage, these vanish in our overall haze of non-exactness.

We can predict amplifier small-signal gain to around 10% with just scratch-paper. 10% is usually good enough. When system gain must be tightly controlled, excess gain, trimmers, and testing will do it; or the feedback techniques taught by Black et al.

We can generally predict amplifier medium-signal distortion by assuming a zero-feedback BJT will rise to 25% THD, FETs including tubes to about 5%, just before clipping.

Start of clipping can usually be predicted on scratch paper. Where do the small-signal simplifications fall to pieces? Most simple amplifiers will just clip. (More complex amplifiers may develop sneak-paths and go wonky.)

Anyway, since the late 20th century, massive trial/error iteration has been super practical. Iteration is what computers do best. While I think it is wasteful to blow a million transistors for a second to approximate the action of a single transistor for a millisecond, such waste is now cheap and universal. I am of the generation which thinks that you can't understand (or accept) an answer until you understand the question. And that if you really understand the question, you can give a rough answer by inspection. Just throwing some parts in SPICE is too easy: you are likely to ask the wrong question and not even know you got the right answer for what you asked, not what you wanted.

[duck_arse]

I see, I will try small values. Should I just have a look at the cutoff frequency of the HPF it forms with the 10M to determine which cap is big enough?

there is a 10M (effectively) in parallel with another 10M, so calculate for 5M. 2n2 should be way plenty.

[iefes]

Hey again guys. I didn't want to open a completely new topic around this, so here we go:

I built the circuit discussed above and everything works nicely. I wanted to try to implement a JFET switch on the wiper of the gain pot instead of the mechanical switch, so I could switch on the boost and the LED with just one SPDT stomp.

I looked up a few schematics where this simple implementation is used, like the Abductor II from deadastronaut or the Glassblower from Valvewizard and put the same thing into my circuit.

The problem is that it wouldn't work. The FET seems like its conducting all the time. It doesn't matter if the serial resistor and the 1N914 diode on the gate are connected to +9V or not. The boost is always activated which means that there's always current going through the FET.
Why is that? How can I prevent this? Or is a FET Switch not suitable for this position?
I tried a J201 and one 2N5457 and none of them worked.

I would really appreciate some clarification around this :-)

[Groovenut]

edit: see better after coffee explanation below

[electrip]

J112 is a P channel device.

J112 is N-Channel!
2N5460/61 or 2n3820 are P-Channel.

electrip

[Groovenut]

J112 is a P channel device.

J112 is N-Channel!
2N5460/61 or 2n3820 are P-Channel.

electrip

Doh! You are correct!  for whatever reason I was thinking of the J17X series.

Apparently I should not answer posts before coffee......

The N channel fet switch will need the gate voltage to be lower than the source voltage by the Vgsoff amount in order to turn off. In your circuit you have the source at ground and the control voltage low at ground. This doesn't allow the fet to turn off.

If you have a look at Merlin's Glassblower schematic, you can see the source voltage is at Vbias. When the mechanical switch is connected to ground it pulls the gate below the source by ~4.5VDC turning the fet off.

[iefes]

Ah, I see! Thank you!

So is there a way to improve my circuit to get the result I am looking for? Could I put a resistor between GND and the Gate just like on the Input of the Buffer stage (R3)?

Or would I need to put a resistor from source to GND in order to raise the voltage on the source? I want as less resistance from the wiper to GND as possible so this wouldn't be ideal.

Any more suggestions? :-)

I know it would be easier to just use a DPDT switch, but I would really like to learn how this could work. Just helps me understand things better and gives me some ideas :-) Besides I just have a SDPT stomp-switch on hand atm

[Groovenut]

You could use a P channel fet like a J175 in place of the J201 in your schematic. P channel will turn off when a positive voltage higher than Vgsoff is applied to the gate. This will work in your circuit but you would need to wire the LED differently to coincide with the on state of the boost. I would think connecting CLR (the LED current limiting resistor) to 9V -> LED anode -> pin 2 of the mechanical switch would work.

[iefes]

Thanks a lot! I will have a look if I have a P Channel FET laying around :-)

[Groovenut]

I think you could also use a MosFet in the same application. Something like a 2N7000 would work I believe. Easier to find than a P channel Jfet. Worth a try.

[iefes]

I put in a BS170 and it seems like it works :-) Thank you very much! But this is a N-Channel device as well...?! Now I just need to go through this again and understand why it works with this one.
I didn't change anything about the setup, just took care of the pinout.

[Groovenut]

Do a little bit of research on the differences between enhancement and depletion devices. That's a good start and should explain quite a bit as to why the bs170 works in this application. As well as why the N channel Jfet did not.

[iefes]

Thanks a lot up to now but I came across another question Still in the process of learning :-)

Even after reading a huge amount of articles about impedance I still don't really understand how to determine the output impedance of my circuit. I know how to tune the input impedance but about the output impedance I just know that BJTs have an even lower output impedance than FETs. Which resistor or which component determines the output impedance of my circuit posted above?

Building a buffer I would want to have a very small output impedance to easily drive following pedals but how could I achieve this in my example?
I thought about adding a BJT stage after the FET but then I still don't know how to determine the impedance on the output. I thought about adding another stage anyway to include a simple LPF (internal trimpot) in case it sounded too sparkly.

Thanks guys! I'm getting closer to fully understand what I'm doing :-)

[ashcat_lt]

It's a complex question.  Really it changes as the signal swings.  I think for most things you can call it the larger of the drain or source resistor, though when the source is bypassed by a cap, it's basically shorted for audio, so call it the drain resistor.

[Groovenut]

It also very much depends on the load it's connected to. For most guitar amps, it will be the smaller of R_D or R_S unless the R_L for the circuit is smaller, then it will be R_L.

When I was studying tube circuitry, the output impedance for common cathode amplifiers was R_P//R_L. However, most of the literature that I have studied for jfets uses the the smaller of R_D or R_S

[iefes]

Hello guys, I've now been using this little Buffer/Booster for half a year and thought about getting some pcb's of it made. But when I tested it with the guitar of a friend of mine, which has quite high output humbuckers I discovered a lot of distortion going on when the boost is cranked.
Why is that? Is the JFET not capable of being driven so hard? Should I increase the value of R4 (470R) to limit the amount of boost?
Could I do anything else on this circuit to prevent distortion of the FET? It only distorts when the boost is engaged and cranked and I am quite sure that it's not distorting the pedals/amp coming after it.

I will try to run it on 18V and see how this helps, but for the PCBs I made it would be hard to also squeeze a voltage doubler circuit inside the small 1590A enclosure.

Any other suggestions would be great! Do you think it wouldn't distort with a lower input impedance (1M) ? Should I lower the values of Rd and Rs? Please help :-)

Thanks a lot in advance and sorry for picking up on this topic again!

[midwayfair]

High output pickups could clip the FET at the gate. This is a device-level restriction, and the the circuit supply voltage won't affect it. It's actually possible to create more distortion. Or you might be clipping the amp on the other end.