Hello all. I am halfway through Stan Gibilisco's book on "Teach Yourself Electricity and Electronics" and am curious about a particular amplifier circuit.

Q1: What exactly is the "input"? Can this be a microphone, for example? And if so, wouldn't a microphone generate "pulsating DC" rather than AC? I always thought that microphones, and similar devices (say, for example, a photodiode) simply allow more or less current to pass through them based on a given input. A sort of variable resistor, if you will. But, as explained in the text, it is evident that AC, not pulsating DC, is required, and I thought the two were different.

Q2: The voltage at the gate affects the current at the drain. But supposedly this current at the drain is DC, yes? Then why does the book mention that the capacitor at C2 allows AC current to pass through it? Where does the AC at the drain electrode come from? As you can see in the attached diagram, +12V is connected in close proximity, and no sign of AC current to be found.

I believe that is the real source of my confusion -- I do not know where the alternating current (AC) is coming from. I understand capacitors allow AC to pass, and inhibit DC, but if you've got a microphone at the source, then you're just getting pulsating DC, right?

And furthermore, even if there were AC around the input region, it seems to me that all a JFET can do is either increase or decrease the current from source to drain -- it should not generate AC. So why put a capacitor at the drain? Is the transistor really generating AC? Seems to me it should only allow DC to flow from source to drain, since the source is supposed to be negative, and the drain positive.

 

The input is anything that presents an alternating voltage to the input jack. It has to be alternating to get anything amplified because the capacitor will block DC and only one pulse will get through if you present a DC voltage to the input.

An old fashioned microphone is actually a generator based on magnets and a moving coil of wire.

As for the drain, J-fets always have some idle current when Vgs is zero. The idle current develops a DC voltage across R1, and that voltage reduces the drain current by raising the source terminal of the j-fet above zero. When that function has stabilized, the AC from the generator (microphone) makes the gate voltage vary, and thus the drain current varies.

As the circuit is, R1 resists the fast changes in current and limits the gain of the j-fet. A capacitor in parallel with R1 will pass the AC component and thus raise the gain of the j-fet.

 

Thank you for your post. I understand about the microphone input now.

However, if the JFET can only pass direct current (even though its amplitude may vary, its direction is constant), I am still not sure where the AC is coming from at the C2 capacitor.

I see that one of the output terminals is connected to chassis ground, which is also connected to one of the input terminals (from the left side of the diagram).

Since the input device generates AC, then this output node also varies in polarity. Yet at the same time, it is also connected to chassis ground, which in theory should mean that if there is a path from that node to the +12 V, then it will always flow in one direction -- from the chassis ground, to +12 V.

Again, I do not see how AC winds up on the output.

Could you further clarify? Thanks!

 

OK...The microphone creates an AC voltage signal because it is a wire and magnet generator being driven by the power of sound pressure waves traveling through the air. When the microphone voltage changes, it makes the current through the j-fet change quickly, more and less than the idle current. The voltage drop across R3 changes at the same speed but with more amplitude than the input voltage. These changes in voltage appearing at the connection of the J-fet and R3 are AC riding on a DC stream. C2 allows the changes in voltage to pass to the output jack, but does not let the DC stream leak to the output jack.

 

The second part is about "ground". Ground is a convenience. It is the place from which measurements are referred. One lead of your volt meter will always be connected to ground in this example, thus, ground does not vary in polarity or magnitude, everything else does.

When you think in those terms, this should become clear. If it does not, return and ask another question.

 

Maybe I misunderstand how they work, but I always thought that transistors allow current to flow in only one direction -- in this case, from source to drain.

In other words, only DC should be present at the drain.

Yes, I agree that the amplitude of that current may change according to the voltage applied at the gate, but its direction is constant.

In contrast, AC -- alternating current -- requires that the direction of the current change.

Since the current at the drain of the JFET is always one direction (from source to drain), then it stands to reason that only DC would appear at the drain.

That is the source of my confusion. You have helped me understand that indeed there is AC from the left side of the diagram (where the microphone is), however I am still not sure how the AC winds up on the right side of the diagram, since the JFET only passes DC.

Perhaps the JFET really does allow alternating current to pass through it? And so current can flow from drain to source, or even from drain to gate? Is this true?

 

We set the bias point in the middle of a Vcc (6V).
So change in input voltage change voltage on the drain in the "rhythm" of input ac-signal.
So on the drain we get "variable" current and we call this current AC-current.
For example drain voltage change from 6V to 9V in the negative half of a input voltage and then change from 6V to 3V in the positive half of a input voltage.
Read this:
http://forum.allaboutcircuits.com/showthread.php?t=36500&highlight=voltage+swing&page=2
http://forum.allaboutcircuits.com/showthread.php?p=149702#post149702

 

As long as the channel is "on", the JFET can handle current in either direction.

But I don't think that is what's going on here. The voltage at the drain can be thought of as the sum of an AC and DC signal, i.e. an AC with DC offset (caused by the 12V), which means that -- even though the signal contains a AC component -- the signal never changes direction.

The output cap then strips off the DC component, and you're left with the AC signal on the output.

That's my take on it. I hope it's right -- I've never built one of these, nor have I ever used a JFET so.. you get what you pay for on the internet

 

I'm going to go at this from the electron's point of view. The electrons flow up from ground, through the j-fet, through R3, and into the positive terminal of the battery. The current through the j-fet only goes one way, but it changes in amount. The changes in amount of current make the voltage across R3 become more and less.

As the voltage at the drain becomes less positive, it is because the j-fet is allowing more electrons (than the idle current) to flow from ground toward the + terminal. The increased concentration of electrons at the drain repel the electrons on the left plate of the capacitor and that repels the electrons on the right plate of the capacitor, IF (capital "IF") IF there is a path for electrons to get from the right plate of the capacitor to ground, electrons will flow out of the capacitor, through the load.

Shortly thereafter, the newly changed voltage on the gate of the j-fet will cause it to reduce the flow of electrons. At that point in time, the lesser concentration of electrons on the drain will be seen as a more positive voltage on the drain. That more positive voltage demonstrates a lesser concentration of electrons and the electrons on the left plate of the capacitor will rush in to fill the lack of electrons. At that point in time, electrons will flow from ground, through the load, and into the capacitor.

This demonstrates that electrons flow both ways through the capacitor, and that is called alternating current.

 

Continuing...The flow of electrons through the j-fet never reverses, it just changes in amount. The voltage at the drain becomes more and less. The concentration of electrons at the drain becomes more and less. IF electrons are available by way of passing through a resistance from the output terminal to ground, electrons are alternately repelled through the capacitor and then sucked back into the capacitor. The electrons reverse their direction of flow through the capacitor. That is the definition of alternating current.

 

Hi AsmCoder8088,

Don't get hung up on semantics. The original, and technically correct, meaning for AC is; an electric current that alternates from one polarity to the other.

However, many people will use this term to refer to an Alternating Voltage regardless of whether the current actually changes direction or not. The drain's voltage actually carries an pulsating DC voltage, which is approximately the same shape and frequency of the AC signal input, but with a larger peak-to-peak voltage value.

In the above description I could replace the phrase; "pulsating DC voltage" with; "AC voltage" and most people should still no what I mean even though it is not technically correct.

Read my signature line,
Ifixit

 

Okay, I believe I understand now.

Thank you Bychon and others for your generous help!

 

What I did was called electron theory. It's the process of accounting for every electron stream and the effects that stream has on the next part. Most people think in more abstract terms, and it works. It's like a higher level programming language. BUT...being able to do electron theory will get you answers that the abstract thinkers rarely consider and have even more difficulty explaining.