JFET guitar Pre-amp

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I've been looking for simple guitar pre-amp circuit build; Don Tillman's JFET circuit seems to be raved about: Don Tillman's discrete FET guitar preamp - Telecaster Guitar Forum

preamp.gif

It uses a J201 N-Channel FET which I'm sure I could get hold of, but I have a load of 2SK170 JFETS that need a home and I could do with learning a bit about circuit design. So I was wondering if I might be able to use one in this circuit? Is there a name for this type of circuit? I'd like to find some info on it and possibly modify for the 2SK170.

Any help getting started would be apriciated.
 
Excellent; thanks, that's ideal.

Just as you were posting that i found in a textbook:
There is a special parameter of FETs called transconductance factor (gm) and it is the slope of the transfer characteristics at the point of operation; it is represented in siemens which is equivalent to inverse R.
gm = 2xIDss/VP or {1-(VGS/VP)}


What would VP be?

Thanks again.

J
 
Pinch-off voltage. You really should read some introductionary text on the subject of JFETs, Google is your friend.

There is no point in calculating gain for your schematic though because it will vary bewteen transistors of same type with such a low supply voltage and arrangement where Vds is even lower due to self-biasing. Gain will be low either way (less than 10x, probably around 3x).
 
Pinch-off voltage.

So it would be the same as VGS (off) on the data sheet.

There is no point in calculating gain for your schematic though because it will vary bewteen transistors of same type with such a low supply voltage and arrangement where Vds is even lower due to self-biasing. Gain will be low either way (less than 10x, probably around 3x).

If you take the middleish values for Idss and Vp of the J201 it does indeed have a gain of 3.
I think a guitar has an output of up-to about 2v, so for a clean signal you wouldn't really want a gain of more than 3 on a nine volt battery. I guess that's why on Tillmans site he warns it might not work with all J201's
 
I've been playing with this circuit all evening and so far so Good I think; but I'm having some trouble with the biasing.
preamp.gif


I've been readin the JT Hawes design notes and following his process.

R1 = 4.4M (I don't have any 3M resistors; I dont think it makes any odds)
R3 = 680R (Reduce value of R3 for the higher current requirements of the 2SK170)
R4 = 220K
C1 = 4.7uF
R2 = I started with a value of 820R which attenuated the signal slightly and actually sounded quite good; which I can't understand because VD's Q-Point was only a few milivolts below the V+ And I would have thought it would clip.
I've been trying to tweek R2 to bias the amplifier so VD is about halfway between VS and V+; but I find I have to drop the value of R2 signifficantly to get a decent bias voltage; as low as 10R. But this pushes the gain right up; Ideally I'd like to be as close to unity as possible.
It does sound good though; just too much gain!

With R2 = 10R I get:
V+ = 8.3V (I need a new battery)
VD = 4.3V
VS = 50mv


Any thoughts on where I've gone wrong so far?
 
I think a guitar has an output of up-to about 2v, so for a clean signal you wouldn't really want a gain of more than 3 on a nine volt battery.

I have both a Strat and Telecaster. Playing hard they only put out 200mv. Normal playing is about 100mv. I also have an Epiphone SG with Humbuckers and that puts out about the same. I think the 2 volt figure is for active pickups.
 
I have both a Strat and Telecaster. Playing hard they only put out 200mv. Normal playing is about 100mv. I also have an Epiphone SG with Humbuckers and that puts out about the same. I think the 2 volt figure is for active pickups.

Peak to peak or RMS (AC) ? RMS reading is useless because it doesn't tell you how much headroom you need, it's peak to peak reading that you're after and based on what I've seen recently when tinkering with a scope there's no realiable way to calculate PP reading from RMS as pickup output signal is not a clean sinewave.
 
I've beeen looking around; But I can't find a definitive answer on max peak to peak voltage o/p of a guitar pickup; google turns up answers from 100mlV to 4v.
Has anyone tested on a scope before?

definately passive vs active

so the question is, does your head amp have passive or active input, or both

My head amp? My Amp is a tweaked amp-maker SE-5a

But the main reason i want a pre-amp is to sit infront of my effect pedals. which is also why I want low gain so i dont overdrive the first pedal.

I've been reading several docs on JFETS and common source-amps. But I could do with a decent text on designing a common source-amp and calculating the values.

I was trying to follow these examples FET Amp Designing but they are not very clear in places.

Does anyone have a link for a good design process?
 
I've beeen looking around; But I can't find a definitive answer on max peak to peak voltage o/p of a guitar pickup; google turns up answers from 100mlV to 4v.
Has anyone tested on a scope before?

50 mV PP with lightest touch of the strings using nylon plectrum, in excess of 2V PP when going as hard as I could. Nor all pickups are alike though. I'd say Google gave you a decent answer.
 
Nice one Arnulf

Please correct me if you think I'm wrong, I'm trying to learn; but after much messing and reading I'm begining to think the 2sk170 is unsuitible for a low voltage amp in Common Drain Config.
It seems because of it's high transconductance (22mS), that everytime I get IdQ and VdQ biased correctly I've pushed the gain through the roof and it clips badly.

I tried the circuit in Common Source instead with a potential divider at the gate setting the Q point.
This seems to work well as a buffer with a Gain of just over 1 (I think) I need to go over the circuit again with more attention to detail; but it seems OK;

Is there any dissadvantage of using CS configuration insted of CD?
 
Nice one Arnulf
Is there any disadvantage of using CS configuration insted of CD?

One potential disadvantage of common-source circuits is that they offer no voltage gain. In fact, CS circuits are slightly lossy. Usually the gain is almost 1, something like 0.95 to 0.99. In your case, the gain of 1 might turn out to be an advantage.

Another potential CS disadvantage is that the large source resistor produces a large negative bias. Your pickup must overcome this bias.

CS circuits offer two advantages: Isolation, due to high input impedance, and reasonably broad bandwidth. The isolation feature makes CS circuits great for buffering.

I examined the 2SK170 specs and you're right: This device has a rather high transconductance. (That is, at 10 volts and 1 kHz, which are the test parameters for that transconductance.)

Of course you can still vary the voltage gain. For example, with equal source and drain resistors, the gain drops to 1, just as with a CS. Plus, you now have a phase splitter. One advantage of the phase splitter over the CS is a lower negative bias. Plus, you can now choose either output phase. Or use both phases, but know that they cancel.

Your minimum Idss is some 2.6 milliamps (again at 10 volts). The Idss determines the recommended minimum current for stable operation. For example, let's assume that 2.6 milliamps will work as our average current with a 9-volt supply. (Probably a good assumption.) To make a phase splitter, use 820-ohm resistors for both the source and drain resistors.

I got 820 ohms by dividing and rounding to a standard resistor value...

(4.5 volts / 0.026) = 1,731
(1,731 / 2) = 866
866 is close to the standard value 820. We remain within the min / max
Idss specs.

If you want a little gain, divide the 1,731 unequally. With three-fourths as the drain resistor and one-fourth as source resistor, you achieve a gain of not quite three. Note that the FET's internal resistance and external loads reduce this gain.

When you're all through choosing resistors, you must still tweak the circuit. On the other hand, the phase splitter should work right away. My calculation method isn't perfect, but it's probably the easiest way.
 
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