reached the stage where it talks about changing the 9v input resistance to get a reading of 4-5v on the transistor output
i tried a bunch of different resistors and was getting confusing readings so i hooked it up and tried it out by ear as im using a different transistor and not sure on what voltage it needs.
so question one is im assuming the higher the resistance the less power the transistor is getting?
1k gave me a fine signal 560R gave me a quiet signal and 10k gave me a loud signal
so assuming im right regarding the transistor power, the more power its getting the quieter the signal?
100k gave me nothing, so assuming there wasnt enough power for the transistor
18k gave me crackles and 15k gave me a struggling signal, so in those cases the transistor didnt have enough power?
wondering if im on the ball here or im missing something completely, the bit that doesnt make sense to me is that with more power the signal is getting quieter
hope all that makes sense, trying to work out whats going on really, what i really want to do is make a clean boost using some germanium diodes that i have, one that pushes them harder than a klon normally would
Just to be clear, @fluxpavilion, which resistor are you talking about? From the link it looks like you’re referring to the one connected to drain (the leftmost pin on the JFET). Also, do you mean [edited:] 2N5457? I can’t find anything about a BN5457.
No, an op amp is a circuit, usually on an IC. And that’s not the primary function of an op amp, but that’s another topic.
A resistor in and of itself does not change a voltage. However voltage can change across a resistor depending on what it connects to.
In that circuit the gate is not connected to ground. It’s connected to your input signal through a decoupling capacitor and a voltage divider (two resistors).
Are you sure it was R5 (the 15K one) you were changing? Not R1 or R4? Because I don’t understand the behavior you describe — then again, my grasp of transistor amplification is not that great.
Not quite, it’s a bit like a cross between a variable resistor and a switch, depending on how it’s wired up. There are opamps with JFET input stages, though (e.g. the TL07x series).
There are three things involved: voltage (V), resistance (R), and current (I). Ohm’s law:
V = I Ă— R
This works in all directions:
if you apply the voltage V across a resistor R, the current through the resistor will be I = V/R.
if you pull a current I through a resistor R, by adjusing the voltage across the circuit, the voltage across the individual resistor will be V = IĂ—R.
if you feed a voltage V through a circuit, and that results in the current I, the resistance is R = V/I.
(Yes, it’s pretty simple, at least if you pretend everything is DC. Simple enough that people thought Georg Ohm was an utter idiot when he proposed this.)
A lot of things follow from this, including e.g.
if you increase the resistance R in an existing circuit, you’ll lower the current – I = V/R so making R larger makes I smaller.
if you lower the current through a resistor, e.g. by adding another resistor in series, you’ll lower the voltage across it – V = I×R so making I smaller makes V smaller.
if you increase the voltage V across an existing resistor, you’ll increase the current – I = V/R so making V larger makes I larger.
if you put two resistors in series, the current has to go through both of them, so the total current is I = V/(R1+R2) and the voltage across each resistor will be V1 = I×R1 and V2 = I×R2 where V1 + V2 = V – this is called a voltage divider, and is a very common circuit.
(and so on)
I often find thinking in “what current goes through this thing” terms to be more helpful than “what’s the voltage here” (but you have to take the voltages into account when dealing with transistors and diodes, though, since their behaviour (“operational regions”) tends to depend on the voltage differences between the various pins).
The value of R4 can not be correct. The schematic you are referring to says 1 milli Ohm, and should probably be much higher. I think the used the wrong symbol. They probably ment 1 mega Ohm, or 1M Ohm, which is a 1000 million times higher than 1 milli Ohm.
A simple explanation of the working of the circuit is as follows:
the signal on the gate of the JFET will make a current flow through the JFET from the + to ground via the drain and source. This current depends on the value of the gate voltage and the resistors R5 and R1. The output voltage on the left pin of C4 is equal to 9 minus the voltage over R5. So if the current through R5 rises, the output voltage falls and if the current falls the voltage over R5 decreases so the output voltage rises. So one aspect of the circuit is that it inverts the signal you supply to it. If the input signal goes up, the output signal goes down and vice versa. That is most times not a bad thing because you ear is not sensitive to the phase of a signal ( as long as it is the sole signal you are listening to ). C3 and C4 make sure that if you connect this small circuit to other circuitry their DC voltages do not influence the working of this circuit. This is called DC decoupling. As a consequence on the right pin of C4 you will not measure a DC voltage but only the variation / the change in voltage that was caused by the current through R5, the JFET and R1. This will be the amplified signal (albeit inverted).
Aaah ok that makes sense! Or at least the way my circuit is behaving makes sense
I have used a 0.022uf cap for both instances and seems to work fine, how important are those capacitor values, would they affect the sound, or is it just R1 and R5 that would affect my sound in this case?
I’ve noticed with a 10k resistor on the drain the sound is overdriven, which I’m guessing is the desired effect with a circuit like this
The circuit is called an amplifier, not a distortion device. So I do not think this was meant to distort the signal.
But you’ve discovered how distortion is often implemented: by messing around with the resistor values you are changing the amplification factor or the circuit. When sending a signal to a circuit that is amplified with an extremely high gain factor its output voltage swing ( difference between lowest and highest value ) should actually exceeds the supply voltage of the circuit. Obviously this can’t be done because the maximum voltage a circuit like this can output is the supply voltage. As a result e.g. a sine wave will not have a nice curved top (and bottom) anymore but will be flattened. The sine is topped off, so to speak and start to resemble a square wave. This way of distortion introduces harmonics to the signal, and thus it sounds differently.
The capacitors are chosen in such a way that for the frequency content of the signal you are sending through the circuit they act as a DC blocker and they have little influence on the signal. If you deviate too much from the suggested value, they can behave as filters, blocking the signal to some extend.
R1 and R5 shifts the signal around (see the voltage divider I mentioned earlier), and if it gets too close to the rails (i.e. within some distance from 9V and 0V respectively) the “variable resistor” can get into non-linear territory and you get various kinds of distortion, including outright clipping if you hit the limits hard.
There’s some extra complexities here in how a JFET works; the resistance depends on the gate voltage (other transistor types have different control mechanisms), or specifically on the voltage difference between the gate pin and the source pin, and that needs to be in a specific negative range (i.e. the gate voltage needs to be below the voltage across R1), otherwise you hit the minimum resistance (and things go non-linear again) or the JFET cuts off completely. I started writing something about input biasing, but got lost in the datasheets. Maybe later.
As for the capacitors, you can see them as “variable resistors” too (if you squint enough, everything is a resistor) – but in that case, the “resistance” (*) depends on the frequency. The higher the frequency, the lower the “resistance” – the formula is 1/(2×pi×F×C) where F is the frequency (in Hz) and C is the capacitance (in F) – you can use programs to plot nice curves of the frequency response and other properties, but you can see that since you divide by both F and C, the higher the frequency and/or the larger the capacitor, the lower the “resistance”. The output capacitor forms a voltage divider with the 500k output potentiometer (yes, yet another voltage divider) so the higher the frequency, the less voltage is dropped over the capacitor. 1 uF is a pretty common value here, but if 0.022 uF sounds good to you, there’s nothing wrong with that
There’s one more “resistance” that affects the circuit, and that’s the input impedance of the thing you plug this into; if that’s low enough, non-trivial amounts of current will sneak out that way and mess with the calculations.
*) resistance in quotes because the correct name is “reactance” which is a form of “impedance” which is kind of like resistance but for AC signals, but we’re simplifying everything here.
Still, quite an omission because what people will return to is the schematic, or at least I do. But I hope that similar to you other people will read the posts here from head to tail, so that is is not a waste of time to comment.
its the schematic originally linked, with the values i currently have in the breadboard
as you can see ive added some diodes, i looked at a klon schematic and saw that formation so i chucked it on my output and it sounded alright, it was quiet so i added another amplifier on the end
the idea for the pedal i want to build is have multiple sets of diodes of all different flavours and just be able to flick between them, its gonna be called Diodysus
now i literally started looking into this today so it all is a bit confusing to me, but it works!
it sounds . . . . ok, i would quite like a gain knob to dial in the effect of the diodes, i checked it against a klone i have and its sitting at about 50% of what the Klone can do in terms of gain, and doesnt sound anywhere near as warm
wondered if anyone might take a look at it and give me a hand visualising how to make it nice and pretty, need to get some kind of bypass function in there too
oh and if i bypass the diodes it is LOUD, then with the diodes its a little underwhelming
Well, diodes are variable resistors too, this time controlled by the voltage across them (see here)
(this is what causes clipping – once the output voltage gets high enough, typically somewhere between 0.3 and 1.0 V depending on diode type, the resistance quickly drops to near zero, and the excess voltage is shunted away)
Awesome experimentation there! I’d probably go for opamps for the later stages, though; easier to reason around, and more people here who can design for them. JFETs are more a guitar pedal thing (both by tradition and because guitar pickups have very high impedance, while synth gear tends to have low output impedance which makes it much easier to build input stages).
Electronics is not something you pick up in a day. It is a form of applied physics which requires quite a bit of training and experimenting to be mastered. But don’t worry, if things start to work that can be very satisfying.
oh and if i bypass the diodes it is LOUD, then with the diodes its a little underwhelming
Well, that is because the diodes lead quite a bit the signal from the first JFET to GND, so there is less left for the second JFET to amplify. In fact they will reduce the voltage of the signals to about 0.7 Volts (depending on the type of the diode), because that is the maximum voltage drop a diode has when you send a current through it (its so called forward voltage). What you could try is limit this effect by putting a few diodes in series or using an LED because that has a higher forward voltage (this voltage depends on the material the LED is made off, so if the LED has a different color, the Voltage will be different). So you could experiment with a few LEDs here.
On the other hand if you want to stick with the single diodes, if your switch were a double switch it could maybe compensate for the loss in signal by boosting the amplification of the 2nd JFET resulting in less of a signal drop.
