Step 4: Designing a JFET Amplifier Using the Load Line Technique

In this section, I’m going to tell you how to design a JFET amplifier using the load line technique. This was used many years ago to design original valve equipment and the same techniques can be applied to JFETs provided you have access to the drain characteristics chart.

How this works is you must be able to edit the chart in some way (for example using Microsoft paint or printing off the chart and doing it with a pencil.) and you must draw on the load line of your chosen values.

Since my chart only shows up to 1.2mA of drain current, I’m going to use a current value of 0.8mA at 9v so I can utilize my chart.

So the first step in using the load line technique is to believe it or not, draw a load line! A load line is literally a line that is drawn from the Vcc (drain to source voltage on this chart) to how much current will be flowing, so in my case from 9v to 0.8mA. This line will be used to calculate the bias point and the resistor value used for Rd and consequently Rs.

Calculating Rd is very easy in this case, it literally consists of:

Rd = 9/0.0008 = 11250 Ohm

To calculate Rs, we must now look at the load line and decide where we want our bias point. To bias the amplifier so it can have a full swing, we must choose the bias point in the middle of the load line. As you can see in my picture, I have put a little blue dot where the bias point lies. If you look at the chart, you can see the thick black lines, they are the characteristics of the JFET. If you see to the right of the chart, there are voltages (VGS values). As you can also see, my bias point lies just below the -0.4v point (between -0.4v and -0.6v) it is about -0.44v. All that this means is that if the gate is 0.44v LOWER than the source, it will follow this curve. Since we are pulling the gate to ground, we can achieve the same thing by putting the source to +0.44v. Therefore the gate is -0.44v relative to the source.
We can now calculate Rs:

Rs = V(Bias) / Id = 0.44 / 0.0008 = 550 Ohm

Now, the only problem with this method is if you look at the curves of the JFET, once you get past the linear region and into the saturation region, not much happens in terms of transconductance, you increase Vds and the Id hardly increases. This has a profound effect on the output waveform by squashing one end of the waveform. This can be very musical on guitar and adds a fair amount of second harmonic content (something lots of guitarists crave!). This can be changed by moving the bias point much more towards or away form the linear region, all of which can be done by replacing the source resistor for a potentiometer or variable resistor.

Once again, since these aren't standard value resistors, the values:
Rd = 10k Ohm
Rs = 560 Ohm
Could be used.
<p>If it's class A it has to have 1/2 Ucc on the drain right? If Iwant to calculate Rd why do not use 4,5V? </p>
<p>I'm not sure also why to use the graph with Vgs = -1.2V for the load line while there is other graphs in the specs. What do they mean? (I'm terribly sorry for my ignorance!...) </p>
<p>Hi,</p><p>why am I reading Vgs(off) as -0.5V to -6.0V on the datasheet? Where comes from the -2.5V in your calculation? thanks</p>
<p>can I plug this directly to computer line in? </p>
<p>I mean like this guitar&gt;preamp&gt;computer soundcard or guitar&gt;preamp(w/powersupply or battery)&gt;computer?</p>
<p>Actually:</p><p>Rs = Vgs / Idq = 0.44V / 0.45mA = 977 Ohm</p><p>If we use the &quot;tube&quot; approach...</p><p>Here is how I do it with a single 2SK170. Since I want a low noise I chose Rd of 1k and supply(Vdd) of 10V. The load line is calculated by Vdd/Rd = 10mA.</p><p>I chose a bias point at -0.3V/2mA for low power and small signals(MC cartridge input). </p><p>Rs = Vgs/Idq = 0.3V/2mA = 150 Ohm</p><p>Regards, </p><p>Venci</p>
Its been a long time since I wrote this so I apologise for any wrong parts! Thank you for clearing up the load line equation :)<br><br>Harris
im an electronics dummy. how would this look like in an actual circuit board?
Really nice job. Thanks for putting in all the effort to develop this instructable so completely.
I have two question. <br> <br>1. Why Vsg(off) is -2.5V? A catalog says it's -6V, but also says Vgs(not cut off) is -2.5V. <br>2. In equation: Rs = Rs + Rd x ( -Vgs(off) / 10 ), what kind of approximation is used? I'd like to know if its value is available for any kind of jfet. <br> <br>Thank you! <br> <br>
Hi, the source resistor is used to bias the whole amplifier. all the Vgs(off) rating means is that when the source is 2.5v HIGHER than the gate, the jfet will not be conducting (at all) so if you divide this value by a preset number (I used 10 as a simple round value) you will be biasing the source between full conduction and in the off state, this will allow for amplification. If you look at a datasheet of any jfet, there are generally two paramaters, Vgs and Vgs off. Vgs is the parameter which says the maximum voltage you can have between the gate and the source, Vgs off is the useful parameter used for this equation. <br> <br>Hope that helps!
..Amateur audio electronics hobbyist here.. so forgive the silly question. But how to I turn this into a fully functioning amp, do I just replace the signal symbol with a mono 1/4&quot; jack and how do I connect it to an output/speaker? <br> <br>cheers <br>mdog
Hey, to turn it into a fully working amplifier, replace the signal symbol for the jack and connect the output to the 10k resistor furthest right of the schematic diagram, you'll need to connect this to something such as a computer sound card or a power amplifier as it is not capable of driving low impedance sources (such as a speaker)<br><br>Cheers,
would it work through my bass guitar practice amp (combination amp) if i plugged it into the input?
Yeh if you plug your bass into where the signal symbol is on the schematic diagram and plug the amp input into where the 10k resistor is (If that makes sense!).
Yeah so I'd neead an output jack which one side would go to the resistor and One side would go to ground?
How well do you think this would work as a microphone preamplifier? I was trying out a few microphones I have the other day plugged into a PC mic input and one is much lower volume than the rest. So I am thinking maybe a little preamplification might help it out some.
Hi pfred2, it should work fine for any form of microphone :) infact, many JFET input amplifiers (such as the TL072) are used in microphone amplifiers. If the microphone is dynamic, plug it in and it should be fine :) you may want to change the input and output filter to reduce extraneous noise such as low frequency rumble. If the microphone is a condensor, you can add a 6.8k resistor between Vcc and the input of the input capacitor to provide power. Hope this helps! :)
I looked at that microphone again after I posted to try to get a little more information about it and found out it needs a battery. Funny it worked at all when I tried it. So I think the first thing I'm going to do is get it a fresh battery. Yeah I feel pretty dumb now.
Haha yeh true, most probably a condensor mic you used then if it required power to function, but that is quite strange how it worked! I'm going to have to put my wits on saying that some of the signal was leaking through capacitances in the preamplifier circuit, causing a slight bit of signal to pass through. Hope it is all working though!
If the computer, laptop for instance, has a built in mic is it possible it could have been recording through that because the mic had no battery?
Yup, if it wasn't set to recording through the soundcard and was recording through the internal mic, it could've picked up some sound as mdog93 has stated :)
This is a really great tutorial on JFET amplifiers! I plan to refer back to it again and again. Thank you!
I'm glad you have found it useful and thank you for the feedback! :)

About This Instructable




Bio: I'm a student from Cheshire, currently studying at University of Nottingham!
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