It just biasing to gate voltages of same polarity as the JFET diode, ie. positive Vgs for a N-Ch. JFET. Below the point where diode forward current comes into play this is a valid bias region, say below ~0.5V with sufficient low drive impedance. Corresponds 1:1 to the behaviour with tubes, and has the same gotchas: Beware of (medium/high-impedance) ac-coupling with that scheme, even a short overload pulse can de-bias the stage for a long time.AndrewT said:
- Klaus
uli said:
Agreed. I may be misremembering something.
Looks like ca. 3.75mA each JFET, maximum, and presumably less.
Grey
Nelson's Ono phono stage. Four 2SK170s in parallel with 22 Ohm Source resistors. The entire group is cascoded by a bipolar (something Japanese, can't remember which one, in the Ono, ZTX450 in the Pearl, which is a dumbed down version of the Ono). The rail at that point in the circuit is around 25V or so. It then goes to a 2SK170 in common Source mode. That's the head amp stage.
The schematic doesn't specify which grade of 2SK170 is used. I thought I remembered Nelson saying that it was BL, but I agree with Uli that the math doesn't look right, so it must be GR.
Grey
EDIT: I can send you the schematic if you don't already have it. Everyone else in the known universe has a copy, so I can't see any reason you shouldn't see it if you haven't already.
The schematic doesn't specify which grade of 2SK170 is used. I thought I remembered Nelson saying that it was BL, but I agree with Uli that the math doesn't look right, so it must be GR.
Grey
EDIT: I can send you the schematic if you don't already have it. Everyone else in the known universe has a copy, so I can't see any reason you shouldn't see it if you haven't already.
I asked earlier if you had any thoughts on the "vertical" H vs. the "horizontal" H. That could be interesting.
Grey
Grey
From the black box view (4-pole) they might be identical (when working within reasonable bounds), methinks. Too lazy to do the math proof, though.GRollins said:
- Klaus
Um, referring to CH's drawing (http://www.diyaudio.com/forums/attachment.php?postid=1413442&stamp=1201582814) it is a 4-pole, isn't it? It is not a 2-port (which is a specialized 4-pole, according to my literature and wikipedia entry, too)
And it's 3:30AM here, so I've got right to be lazy... should be already sleeping instead of lurking in forums 😀
So I'll leave it at that, for today...
And it's 3:30AM here, so I've got right to be lazy... should be already sleeping instead of lurking in forums 😀
So I'll leave it at that, for today...
From my point of view, anything that restricts communication between the two halves of a differential is going to reduce the differential's ability to produce an out of phase signal from the other side. Given that we're not talking about a lot of resistance below the differential (i.e. not a large value resistance or CCS), then if you intend to use the differential with a single-ended input to produce something close to a balanced output, you're going to want to limit the resistance between the two Sources of the same differential. Now, yes, you can point out that there's a sort of bootstrap effect in that the two opposing differentials tend to move together, but trust me, it ain't perfect. The non-driven side is down a bit from the driven side.
Okay, so how many circumstances come up where you want to use a single-ended input and a balanced output? Well, lots, actually. There's the case of a balanced system using a single-ended phono stage. I seem to recall that John favors single-ended phono stages, for instance. There's also the case of a single-ended signal driving a bridged amplifier--you're going to need to generate a signal to drive the other half of the amp (this being a topic very much on my mind at the moment since my all-FET JC-3 front end is ready to drive a balanced output stage--I've only got one output bank hooked up at the moment, so it's single-ended 25W instead of 100W).
You could change the second stage to be another differential instead of a folded cascode, then put sufficient impedance under its tail(s) to make it even out the signal...there's ways to correct for the imbalance, in other words...but assuming that you want to stay fairly close to this topology, you're going to have to keep an eye on how you lay out those resistors between the two differentials.
If you're using four ten Ohm resistors and a 200, then a horizontal H will give you 20 Ohms between the Sources of each differential. A vertical H will give you 220 Ohms. In my amp, I'm currently running either 33.2 or 47.5 Ohms (can't remember which) between the two differentials and no degeneration resistors at the Sources in the same configuration John used in the JC-3, i.e. a horizontal H with the legs equal to 0; crossbar equal to 47.5 Ohms or whatever it is.
Grey
Okay, so how many circumstances come up where you want to use a single-ended input and a balanced output? Well, lots, actually. There's the case of a balanced system using a single-ended phono stage. I seem to recall that John favors single-ended phono stages, for instance. There's also the case of a single-ended signal driving a bridged amplifier--you're going to need to generate a signal to drive the other half of the amp (this being a topic very much on my mind at the moment since my all-FET JC-3 front end is ready to drive a balanced output stage--I've only got one output bank hooked up at the moment, so it's single-ended 25W instead of 100W).
You could change the second stage to be another differential instead of a folded cascode, then put sufficient impedance under its tail(s) to make it even out the signal...there's ways to correct for the imbalance, in other words...but assuming that you want to stay fairly close to this topology, you're going to have to keep an eye on how you lay out those resistors between the two differentials.
If you're using four ten Ohm resistors and a 200, then a horizontal H will give you 20 Ohms between the Sources of each differential. A vertical H will give you 220 Ohms. In my amp, I'm currently running either 33.2 or 47.5 Ohms (can't remember which) between the two differentials and no degeneration resistors at the Sources in the same configuration John used in the JC-3, i.e. a horizontal H with the legs equal to 0; crossbar equal to 47.5 Ohms or whatever it is.
Grey
Remember Grey, it is the change in CURRENT that makes the stage amplify. Resistance is not that important.
Hi John,
I have a topic that is more related to your design,
and that is about local open loop regulators.
If I have understood it correctly, you use OL local regulators after the series regulators (LM3-regs?). Do you have some technical ideas (not looking for a proof) that explains your using OL regulators?
What sonical advantages have you found?
Sigurd
I have a topic that is more related to your design,
and that is about local open loop regulators.
If I have understood it correctly, you use OL local regulators after the series regulators (LM3-regs?). Do you have some technical ideas (not looking for a proof) that explains your using OL regulators?
What sonical advantages have you found?
Sigurd
You get excellent HF noise rejection. The only assumption is you have quite stable current consumption.
Standard feedback regulators will not behave well with high frequency transients. You can try to make faster regulators, but open loop followers are fastest.
john curl said:
True, but my line of reasoning leads elsewhere.
The current between the two differentials (the N-ch pair up top and the P-ch on the bottom) will follow the path of least resistance, so to speak. Given 4 x 10 Ohms and 200 Ohms as the values, if the predominant current flow to/from the N-ch JFET in the upper left quadrant is to/from the P-ch in the lower left quadrant, then it's not inducing a signal in the JFETs on the right. The predominant signal flow will be between vertical pairs of JFETs, not horizontal pairs.
Lower distortion? Well, maybe...but maybe not...
Assuming that you have a single-ended signal entering the complementary differential from the left, there's a vertical pair of JFETs (one N-ch and one P-ch) that are functioning as both common Source and common Drain. The common Source gives a source at the Drain, which goes on off to the second stage in one way or another. The common Drain mode gives a signal at the Source that would, in a single differential, enter the right side of the differential and drive that side of the differential (in common Gate mode) giving a signal at that device's Drain that is in phase with the original signal and out of phase with the common Source signal on the right.
But if the path of least resistance leads the predominant current flow down into the left-hand P-ch device in a complementary differential, then the right-hand vertical pair (one N, one P) is not driven as hard and the available output from that side will be correspondingly weaker. Yes, you get a signal...I'm not saying that there's no signal at all...it's just that it's down a bit. In a single-ended application this wouldn't be a problem, but if you're seeking as balanced an output signal as possible, then the horizontal H (at 4 x 10 Ohms and 200 Ohms) will give you better balance than a vertical H. Assuming that you are using a complementary differential front end to drive two banks of devices in a bridged output amplifier, unequal drive to the two sides would result in increased 2nd harmonic distortion, yes?
Taken to extremes, as the value of the crossbar resistor in a vertical H approaches infinity you'd lose all coupling between the two sides of the differential(s) and have no signal at all on the right. At that point, you might as well discard the right half, as it does nothing. You'd have your choice of a common Drain signal from the remaining N-ch/Pch pair (complementary follower--potentially useful, but not a differential) or common Source, which would give you two signals separated by some arbitrary DC value (also potentially useful, but still not a differential).
Other resistance values could change things, but I don't see that there's such a black and white, clear superiority regarding horizontal vs. vertical Hs.
Grey
Sorry Grey, but your reasoning is wrong.
'Predominant' current flow is from one N-ch JFET to another N-ch JFET, your current path will be with severe overload, when one N-ch JFET and one P-ch JFET are cut-off. Take a pencil and check with the help of the Ohm law.
I agree Dimitri, I just worked it out when in the shower. 😉 The voltage drop across the feedback resistor (I ac) forces the other side to generate the same current into all four drains, some with reverse polarity.
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