Well, there are some things that are fairly simple to understand (and thus explain) about that circuit... and some things that definitely go beyond my general understanding (yeah, I'm an EE, but a degree in digital design doesn't help me with active filters
). This discussion will make a lot more sense if you have the schematic in front of you; as I mentioned, a web search might turn this up for you but I'm not comfortable posting the document or a link to it myself because I'm not certain of the provenance and ownership.
This will get lengthy - I'll break it down and explain a little at a time, as clearly as I can.
First, a very few simple things to know. The Alembic Series preamp uses a "dual-rail" power supply, where "rail" means a connection to a power supply. One is positive (relative to ground) and the other is negative, the same voltage but opposite polarity. See? Simple.
Second, an operational amplifier (op amp) is a very high gain voltage amplifier. How high gain? Usually something around 100,000, or about 100 dB. Its symbol is a triangle with two inputs (+ and -) on the wide side and output on the point. The output voltage is the difference between the + and - inputs, multiplied by that enormous voltage gain. Why such a large gain is useful will become clear later. That's not hard either, is it?
Next, a junction field effect transistor (JFET) is a solid state device that behaves very much like a pentode vacuum tube. Its input is called the gate. The other two terminals are two ends of a semiconductor channel, they are called the drain and the source. The voltage on the gate relative to the voltage on the source controls the current from drain to source. For a given gate voltage the drain current is nearly constant, independent of the voltage difference between the drain and source. Again, it's a simple device. Even a child can understand that.
A child needs to be old enough to understand multiplication and division for the last thing: Ohm's Law. It says that if a current flows through a resistance, the voltage V across that resistance R is the current I times the resistance. V = I x R. By about third or fourth grade, that child has learned enough to be able to rearrange that as R = V/I and can calculate a resistance given voltage and current measurements, or I = V/R to calculate how much current will flow. (I hope they still teach this!)
And those simple things (plus a few more details to refine and clarify them) are enough to explain the input buffer and hum canceller parts of the Series preamp.
The "J" in JFET stands for junction. This is because the gate and the channel (to which the drain and source connect) are defined by a semiconductor junction. This has the characteristic that it only allows current to flow in one direction. In a JFET, current flowing from the gate to the channel is not useful for amplification, so a circuit is configured so that voltage between gate and source is such that this never happens. When the voltage takes the opposite polarity, negligible current flows between the gate and the channel. What happens (conceptually) is that the electric field created in the channel by the gate voltage "pushes" the current from drain to source to the far side of the channel; eventually, there's no room for any current to flow at all (this is called pinchoff). This is how the gate voltage controls the drain-to-source current.
A JFET has two very important characteristics defining the limits of its useful operation. One is the saturation drain current (I
DSS). This is the most current that will flow from drain to source when the gate-source voltage is zero, no matter what the voltage (up to some specified limit, beyond which damage may occur). The other is the gate-source cutoff voltage V
GS(off): if the voltage exceeds this, no drain current flows at all.
Now it's possible to explain the input buffer, consisting of nothing more than two JFETs in series. I had to stare at that a bit before I made sense (I think) of it; this is how I see it. The gate and source of one JFET are connected together to the - rail; that means the drain current must be I
DSS. The drain of the other one is connected to the + rail, the gate is the input signal from the pickup, and its source is connected to the other JFETs drain.
What I finally realized is that those two JFETs are a matched pair. (If you find a picture of the old PF-6 circuit board, they are the little black "water tower" things.) They are made at the same time on the same tiny chip of silicon, and have very nearly exactly the same characteristics - in particular I
DSS. That means that when the input voltage is zero (no signal from the pickup) the gate-source voltage of that JFET must also be zero. As the input voltage at the gate changes, the voltage at the source changes to match it. Because the gate-source voltage remains zero, Ohm's law tells us that no current flows - so the input buffer places no load on the output of the pickup and thus doesn't affect its frequency response. Contrast this with a passive bass, where the value of the volume pot (and in some circuits, the volume setting) change the sound of the pickup significantly.
One more thing: the Series instruments are usually used with an external power supply. They can run on batteries, but have a reputation for having quite an appetite for fresh ones. I think this is at least partly due to the JFETs operating at I
DSS all the time. (Exercise: how many high-quality 9V alkaline batteries can you buy for the price of a DS-5 power supply?)
Edits: fix a few typos visible in the light of morning.
Wow, that's a lot for tonight, and a lot for a child to absorb
. Sleep on that, ask me about anything that's unclear... then we'll be ready to move on to op amps and how they are used to control the gain and the hum cancellation.
