Mini-boost transistor question

[spi]

On the AMZ site, it says that "The Q1 transistor is the one that determines gain (mostly)".  In AMZ schematics, I think Q1 is the one on top.

In my notes from a few months ago, I have "The bottom transistor in the mini-booster config is where the gain comes from...   the top transistor makes very little difference".   Since I left this in quotes in my notes, it was probably taken verbatim from somewhere, although I didn't save a link to the reference. 

However, I did find this article http://www.geofex.com/article_folders/foolwfets/foolwfets.htm, which backs up my notes:   "The entire trick to understanding this circuit is in noticing that the gain comes only from the bottom JFET. The upper JFET is merely acting as a source follower...."

So either my notes contradict the AMZ site, or Q1 and Q2 are labeled differently on different schematics.  Maybe on the AMZ circuit boards they labeled Q1 and Q2 differently than they do in their schematic?

I was hoping someone can clear this up for me:  which is the transistor responsible for the gain?  The one on top or the one on bottom.

[Fancy Lime]

Bottom transistor determines gain.

Andy

[amz-fx]

On the AMZ site, it says that "The Q1 transistor is the one that determines gain (mostly)".  In AMZ schematics, I think Q1 is the one on top.

That is a quote from http://www.muzique.com/pcb.htm#mini

Q1 is the bottom transistor on the pcb and the assembled module. It is the other way around on the old schematic.

I did a lot of tests with different transistors in the top and bottom positions, and I can confidently say that gain is impacted by the transistors in both positions. Maybe more by the bottom one, but I would need to look at my notes. Q1 and Q2 do not need to be matched, or even the same part number.

regards, Jack

[antonis]

It's always good to have a schematic reference.. 



As Jack clearly stated, both Q1 & Q2 contribute in Gain..
Q1 as Q2's Drain resistor and Q2 with its individual transcoductance..

[spi]

Thanks everyone for clearing it up!

[Steve.mg]

Hey Jack,
I have always wondered why the PCB's had "3" positions for Transistors? I have searched and searched but can't find
any Schematic that shows "Q3"...

Kind Regards,
Steve

[Clint Eastwood]

I would describe the top device as a controlled current source, controlled by the signal through C2. The top device can also be a bipolar transistor or a mosfet, with adjusted R2 and R3. I have tried with a bipolar and it works fine.

[antonis]

I would describe the top device as a controlled current source, controlled by the signal through C2.

Actually, the top device is a Common Gate configuration where the lower device is a classic Common Source one, effectively acting as a push-pull combo via feedback capacitor..

[merlinb]

Ugh, there are so many bad explanations of this circuit. The problem is that its operation changes depending on whether a single resistor is added.
In your version the upper JFET acts as a simple impedance, basically like a dumb resistor. Depending on the size of that impedance, and the gm of the lower JFET, you get some gain as usual:
Gain = gm * R
So both devices 'are where the gain comes from'. (Where does the area of a field come from, length or width?)

[antonis]

You're right Merlin..

I had in mind BJT analogous configuration, where upper transistor intrinsic Emitter resistor plays the role of small value R..

[merlinb]

You're right Merlin..
I had in mind BJT analogous configuration, where upper transistor intrinsic Emitter resistor plays the role of small value R..

You need an actual resistor to make it push-pull, intrinsic resistance is trapped inside the transistor...

[Clint Eastwood]

In your version the upper JFET acts as a simple impedance, basically like a dumb resistor.

So why then does the circuit with the JFET have much more gain than the same circuit with an ordinary dumb drain resistor? I don't get what you are saying... clearly C2 makes the JFET do more than a dumb resistor can?

[antonis]

So why then does the circuit with the JFET have much more gain than the same circuit with an ordinary dumb drain resistor?

'Cause ordinary dumb resistor value is limited by voltage drop across it (determined by working current and supply voltage) where upper JFET apparent value (as seen from lower JFET Drain viewpoint) can be several orders of magnitude higher..

[Clint Eastwood]

So why then does the circuit with the JFET have much more gain than the same circuit with an ordinary dumb drain resistor?

'Cause ordinary dumb resistor value is limited by voltage drop across it (determined by working current and supply voltage) where upper JFET apparent value (as seen from lower JFET Drain viewpoint) can be several orders of magnitude higher..

Yes.. and this high apparent resistance is caused by C2 and is not constant, right?

[merlinb]

So why then does the circuit with the JFET have much more gain than the same circuit with an ordinary dumb drain resistor? I don't get what you are saying... clearly C2 makes the JFET do more than a dumb resistor can?

Because the upper JET 'looks like' a really big resistance, bigger than you could practically use an actual resistor. But if you did use a resistor of the same size (you would need a much higher supply voltage) you would get the same gain.

[merlinb]

Yes.. and this high apparent resistance is caused by C2 and is not constant, right?

Yes at low frequencies the cap's reactance rises, so it does not pass all the AC to the grid of the upper JFET, so the 'impedance magnification' effect will be reduced.

[Ben N]

Sorry about my head-scratching interjection: Maybe we need to ask the question differently for it to be clear to dolts like me. It seems fairly obvious that Q-top (see what I did there?) contributes to gain, because the gain of the overall circuit is greater than a simple common source amplifier. Is the real question perhaps whether the *gain* of Q-top contributes to the overall gain?

[merlinb]

Is the real question perhaps whether the *gain* of Q-top contributes to the overall gain?

Q-top looks like a resistor equal to its internal resistance, i.e. the inverse of the slop of its IV curves.
So it depends where the operating point of Q-top happens to be. In the saturation region the 'gain' of Q-top doesn't really affect the outcome, but in the triode region it does.

[Ben N]

Thanks, Merlin, that makes sense to me. But in practical, real life terms (like in Jack's variation of the circuit used as a guitar booster), in what region does it spend its time?

[ElectricDruid]

Merlin, one query - is "10mF" on your schematic (set of three schematics) a typo?

I've never seen "mF" before. 10mF would be 10,000uF, right? Quite a big cap for an audio circuit!