Beginnings in breadboarding

Using LED’s instead of the diodes?

If so then it’s the diodes that are the money winners for the sound, currently using germanium diodes and they are really lush

How would a double switch boost amplification?

I’m probably gonna use a 12 way switch in the end so I can have a stupid amount of diodes to switch between

I’m not sold on the JFET, but I’m pretty interested in guitar pedal building more than anything else so makes sense to get to know them

Otherwise maybe I’d use a TL072? But that’s a whole new kettle of fish where I’m not sure what the benefit would be as I’m gonna be using guitar as my input

1 Like

That being said, the circuit that I’m basing this off uses TL072’s

The Klon Centaur is my favourite pedal ever!!!

3 Likes

Using LED’s instead of the diodes?

Yep, or a few diodes in series to not reduce the voltage to 0.7 Volts that is lead to the gate of the 2nd JFET.

How would a double switch boost amplification?

In itself it wouldn’t but it could be used to change one of the resistors you have already learned that will change the amplification of the circuit.

Otherwise maybe I’d use a TL072?

It is of course your call to do whatever you like, but I would not jump from one schematic to the other if you are still trying to learn how stuff works.

2 Likes

There often is a lot of interaction between parts of a circuit and the rest of a circuit. So isolating some part and taking that to another circuit will not always be a success.

But I like your experimenting spirit :slight_smile:

3 Likes

Haha I literally just thought about maybe using a valve in there instead

3 Likes

What does series mean, one after the other?

Whereas at the moment they are in parallel?

I see those terms come up alot and dont actually k ow what they really mean

Btw I’m googling and reading stuff about all of this, but it really helps having humans to answer my basic questions, so thanks for being patient

2 Likes

Series means, one after the other.

Now they are anti parallel, which means they are connected with opposing poles (i.e. the arrows in the symbol point in opposite directions).

2 Likes

You want two sets of diodes in parallel, one in each direction, to handle each half of the AC wave.

But each set can be a number of diodes in series, to increase the voltage across the diode set before they start conducting (shorting out the 10k). Here’s an excerpt from the MS-20 filter schematics:

image

They have two sets of diodes in the feedback loop, each consisting of three diodes in series.

Sam’s filter uses two LEDs instead, which have a much higher forward voltage than your typical small-signal silicon diode (and has the nice side-effect that you see when they’re clipping). On the other hand, germanium diodes have a much lower voltage than silicon, so you need a more of them in series to get the same level – alternatively you can increase the gain in the next stage to compensate, or even add multiple output stages, one for each diode set, and switch between them instead of switching between the diodes…

2 Likes

yeah series does work a treat! but it changes the sound ever so slightly, will need to experiment!

yeah multiple output stages that might be a good way of doing it, potentially alot more complicated though!

but if im using a bunch of different diodes they are all gonna be jumping around in volume

2 Likes

ok so ive posted on some stompbox forums to get some more specific tonal advice as i appreciate this is a synth forum not a pedal forum, but you guys are awesome and i hope you dont mind indulging some more investigation

messing around with the resistance value on my drain pin some more to try and get more amplification pre diodes so they distort more and ive come accross something that makes no sense to me

checking the voltage coming through to drain

1k = 6.3v quiet
10k = 1.2v loud
4.7k = 300mv silent

i double checked too. the first two readings make sense as its an inverting transistor (or at least in an inverting circuit)

but that 4.7k doesnt compute in my simple understanding of maths, an inbetween resistance should be an inbetween voltage?

I assume you measure with no input? (i.e. gate at 0 V, via the 1M resistor). Did you triple-check the resistor value (with the multimeter, not by trying to decipher the color codes, nobody can do that reliably with these newfangled blue 1% resistors :slight_smile: ). Is the output connected to anything?

A simple explanation is that it’s actually a 47k, because even with the transistor (aka “variable resistor”) at a hypothetical 0 ohm you’d get 1.6 V out. So either your resistor isn’t what you think it is, or something’s leaking into the circuit, messing with your readings.

1 Like

yep bang on was a 47k resistor

i knew it didnt make sense, got a 7.5k one in there now and its working pretty nice

2 Likes

You’re developing a working intuition quickly :slight_smile: :+1:

I mentioned an expected 1.6 V earlier, and since this thread is as much about “beginnings” as this specific circuit, I thought I’d “show my work”:

The output part of this circuit is a voltage divider (yes, there it is again!) where the output voltage depends on the resistance below the output pin, in relation to the total resistance. If we pretend that the transistor is theoretically ideal and always fully on (0 ohm), the source, drain, and output voltages will all be the same, and the output is set by R5 and R1 only:

  • with R5 at 1k, source and drain and output will be 9V×1k/(1k+1k) = 4.5 V (if both resistors are the same, they split the voltage in half)
  • with 4.7k, they’ll be 9V×1k/(4.7k+1k) = 1.6 V
  • with 10k, they’ll be 9V×1k/(10k+1k) = 0.8 V
  • with 47k, they’ll be 9V×1k/(47k+1k) = 0.2 V

Your measurements are a bit higher than these values, which means that the JFET is adding some resistance to the circuit, shifting the output higher.

To get from 4.5 V to 6.3 V it needs to add 1.3k to the circuit (9V×2.3k/(1k+2.3k)=6.3V), to get from 0.8 V to 1.2 V it needs to add 540 ohm, etc.

Note how this resistance varies, even with a fixed input – the JFET’s drain-source resistance depends on the voltage difference between gate and source, i.e. the voltage across R1 – but that voltage depends not only on the values of R1 and R5, but also on what the transistor is doing. If the drain-source resistance goes up, the R1 voltage drops.

Yep, you read that right – what the transistor is doing depends on what the transistor is doing.

In practice, the transistor will find the right balance and stay there, and you can get some idea of where you’ll land from the datasheets, but things aren’t very exact. Here’s an excerpt from a 2N5457 datasheet:

I’ve highlighted two values that are important for this kind of amplifier; the gate-source voltage difference where the transistor cuts off completely, and the drain-source current at zero gate-source voltage difference.

Note the wide min-max range – these components are absurdly imprecise on their own. Given modern manufacturing standards, two transistors from the same production batch are probably rather close to each other, and often close to the “typical” value if there is one, but transistors from different batches/plants/manufacturers can vary a lot.

Relying on vaguely defined parameters is a very stompbox thing, of course, and mostly by tradition – many early designs were pretty haphazardly put together, for a variety of reasons, ranging from economics and component sourcing contraints, to people having absolutely no idea what they’re doing :smiley: Which is part of the fun, obviously, but can also be immensely frustrating, especially if you try to build two or more of something. And if you look at stompbox forums, it often leads to insane amounts of cargo culting.

(In JFET opamps, there’s additional circuitry to force the input JFETs to behave more consistently, and they use multiple amplifier stages to better control what’s going on instead of trying to get a single transistor to do everything. Also, opamp amplifier designs use negative feedback to further constrain the behaviour, bringing everything into much more well-defined regions. You can still do soft/hard clipping, but you do that by bringing the signal levels into the right range before the diode stage so you hit the diode’s “voltage-controlled resistor” knee at the right place, instead of trying to bring the diodes to the levels you have in the circuit for other reasons.)

4 Likes