Sigh. You understand that, once again, even if the loads are not "feather light" but are still identical, the balance is still perfect? Kirchoff's Law holds there, too.
Feedback schemes to the tail of an LTP may be useful for swing and distortion (take a look at the Curcio ST-70 mod for example, besides your own circuit), but they have nothing whatever to do with balance. If you removed that circuit entirely and your plate resistors (and load) were matched, the balance would still be perfect, independent of your tubes. And if the feedback resistors were tightly matched, again your balance would be perfect.
Geek,
I just wonder how come the difference is 16.5% at 85 Vp-p when you state in your schematic the tube section difference of +-10%, there must be some other tolerance mismatches on your components or between the scope channels?
Have you run the 85 Vp-p and connect BOTH scope probes to the SAME point to verify it's not your scope/probes?
Cheers Michael
I just wonder how come the difference is 16.5% at 85 Vp-p when you state in your schematic the tube section difference of +-10%, there must be some other tolerance mismatches on your components or between the scope channels?
Have you run the 85 Vp-p and connect BOTH scope probes to the SAME point to verify it's not your scope/probes?
Cheers Michael
I agree with SY, the outputs should be AC balanced. Some tolerance difference in the plate load resistors and feedback averaging resistors is likely and the scope probes maybe too. Also, Geek's CM Fdbk is operating at AC, there is no big cap in the CM feedback to keep response to DC only. What we are seeing is feedback correcting for the load resistor imbalance.
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This thread is very interesting to say the least.
The proof of the above hypotosis would be to swap the two load resistors and make sure the low AC gain side reverses with the high AC gain side. This should be easy to do with the higher amplitude drive signals where the gain difference is the largest.
Doug S. (aka Mickeystan)
The proof of the above hypotosis would be to swap the two load resistors and make sure the low AC gain side reverses with the high AC gain side. This should be easy to do with the higher amplitude drive signals where the gain difference is the largest.
Doug S. (aka Mickeystan)
Also, we should keep in mind that tubes are non-linear devices. Non-linearity of the right tube is reflected on left tube's anode swing, and vice verse.
Now, the question number 2: is it true that remote-cutoff tubes make more linear LTP than sharp cutoff ones?
I think Doug has the right idea on Geek's circuit. Should also check the two feedback resistors with an Ohm meter too. And try testing the scope channels/probes on the same test point to see if they show the same amplitude.
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"question number 2: is it true that remote-cutoff tubes make more linear LTP than sharp cutoff ones? "
I think tubes actually labelled as "remote-cutoff" (with weird grid winding spacing) will not be more linear in an LTP, they have logarithmic like stretched out curves. But tubes with curves like the 12AT7, 6LQ8(T) .... just might. These kind of tubes generally have a round cathode, a grid that looks shrink wrapped around it, and a flat plate. And G-- awfull Mu curves. Their gm curve versus current looks like a nearly flattened out ramp instead of strongly curved (flat Mu types). And ideally, have a flat linear ramp of gm versus grid voltage.
Ideal FETs have a square law curve of current versus gate voltage, and a linear ramp of gm versus gate voltage. gm versus current looks like a square root curve. The complementary/opposed linear FET gm ramps (versus input V) in an LTP, sum to a constant gm for the pair, giving linear gain.
Some tubes like 12AV7 go halfway between a flat Mu tube and 12AT7 like tube. And don't work right in any circuit.
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"question number 2: is it true that remote-cutoff tubes make more linear LTP than sharp cutoff ones? "
I think tubes actually labelled as "remote-cutoff" (with weird grid winding spacing) will not be more linear in an LTP, they have logarithmic like stretched out curves. But tubes with curves like the 12AT7, 6LQ8(T) .... just might. These kind of tubes generally have a round cathode, a grid that looks shrink wrapped around it, and a flat plate. And G-- awfull Mu curves. Their gm curve versus current looks like a nearly flattened out ramp instead of strongly curved (flat Mu types). And ideally, have a flat linear ramp of gm versus grid voltage.
Ideal FETs have a square law curve of current versus gate voltage, and a linear ramp of gm versus gate voltage. gm versus current looks like a square root curve. The complementary/opposed linear FET gm ramps (versus input V) in an LTP, sum to a constant gm for the pair, giving linear gain.
Some tubes like 12AV7 go halfway between a flat Mu tube and 12AT7 like tube. And don't work right in any circuit.
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Don, which mu curve do you mean? Mu versus plate current or mu versus plate voltage?
I haven't tried taking distortion measurements of the differential output from my Hippogriff (I wish I had my Millett Interface working), but I suspect that even order would be extremely low and odd order extremely high.
I haven't tried taking distortion measurements of the differential output from my Hippogriff (I wish I had my Millett Interface working), but I suspect that even order would be extremely low and odd order extremely high.
Well, for constant Mu, it shouldn't matter. Flat single line for all plate Vs, like 12AX7. One would prefer constant Mu for SE.
In the LTP, the currents are constrained to sum to a constant, so one would like the gm's to sum to a constant also versus current sum (ie, complementary currents). Turns out that as the gm versus current curve flattens out toward a ramp, so does the gm versus Vg1 curve, although from different directions of curvature.
Here are datasheets for 6LQ8 and 6KR8. Same triode in them. GE gives curves versus current, RCA gives curves versus Vg1. The Mu doesn't look too pretty in either picture. But the gm's are straightening out toward ramps. Not totally straight, but much flatter than 5687 say.
http://scottbecker.net/tube/sheets/049/6/6LQ8.pdf
http://scottbecker.net/tube/sheets/135/6/6KR8A.pdf
http://scottbecker.net/tube/sheets/127/5/5687.pdf
In the LTP, the currents are constrained to sum to a constant, so one would like the gm's to sum to a constant also versus current sum (ie, complementary currents). Turns out that as the gm versus current curve flattens out toward a ramp, so does the gm versus Vg1 curve, although from different directions of curvature.
Here are datasheets for 6LQ8 and 6KR8. Same triode in them. GE gives curves versus current, RCA gives curves versus Vg1. The Mu doesn't look too pretty in either picture. But the gm's are straightening out toward ramps. Not totally straight, but much flatter than 5687 say.
http://scottbecker.net/tube/sheets/049/6/6LQ8.pdf
http://scottbecker.net/tube/sheets/135/6/6KR8A.pdf
http://scottbecker.net/tube/sheets/127/5/5687.pdf
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Maybe give it a measure. It's been a while.
I know my measuements, they are not equal.
You also have to consider with the concertina, it's the worst possible PI one can use because the differeing resistances and the way the tube is affected by the load. Big Cmiller loads will really do a number on it because it acts like a cathode bypass cap and the anode side begins to exhibit gain.
I have tried this with all 1% resistors. There's no such thing as a free lunch. It's better, but not perfect.
Scope is calibrated for matching channels before each test.
Yes.
Then explain why this happens with all 1% resistors too?
Here is what I suggest - do your own tests. Do them at good Vp-p and not low levels where any imbalance is moot anyway.
What is so hard to believe about a proven method of improving AC balance?
Cheers!
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Hi Wavebourn,
That's what I'm trying to say to folks...
It's used everywhere to correct little this-and-that. It works.
Cheers!
That's what I'm trying to say to folks...
It's used everywhere to correct little this-and-that. It works.
Cheers!
I've seen this technique applied in both solid state and tube based LTP where it improves CMRR at the input significantly.
Gregg makes an interesting comment about the concertina (split load) phase splitter, I have seen cases in vintage commercial amplifiers where the phase splitter amplitudes were becoming quite asymmetrical above 10kHz or so, and this clearly had to be the reason why. The effect seemed most pronounced with low transconductance/high rp types like the 12AX7... I have avoided the concertina phase splitter except in cost sensitive applications particularly with triode output stages for this reason - UL or pentode connected outputs have substantially lower miller capacitance and the concertina seems to do a better job here.
Heya Kevin,
That's something I completely neglected to mention I found in my tests.... you can get away with a lesser power supply to have decent performance. Also that the problems of decoupling are reduced in high gain circuits (like guitar amps).
My hopes here is people experiment and get the circuit bugs worked out and perfect values tweaked-to-the-nines and the benefits to all will be great!
I made my concertina worst-case tests with a low mu triode and that's where it got ugly in the HF spectrum.
Pentodes should be more immune. I will explore this further.
Cheers!
That's something I completely neglected to mention I found in my tests.... you can get away with a lesser power supply to have decent performance. Also that the problems of decoupling are reduced in high gain circuits (like guitar amps).
My hopes here is people experiment and get the circuit bugs worked out and perfect values tweaked-to-the-nines and the benefits to all will be great!
I made my concertina worst-case tests with a low mu triode and that's where it got ugly in the HF spectrum.
Pentodes should be more immune. I will explore this further.
Cheers!
Then you have either made an unsuspecting error in the measurement (and don't feel bad, many people have done so with this circuit) or the loads are not matched, or both. Or Kirchoff's Law and the First Law of Thermodynamics are wrong, but that's not where I'd put my money.
If there's an unbalance hiding somewhere in my Hippogriff, it's hiding pretty damned well- with matched probes at the plates and a 3581A's AC voltmeter, AC output at the previous test conditions was within 0.2%, which is as well as I matched the resistors and as well as my scope will indicate with the "add" function. The square waves I showed before indicate the HF relative balance. And remember, this is with wildly different tubes, wildly different DC voltages, and wildly different plate currents.
Hi SY,
Oh, that is entirely possible.
One reason I want to revisit this.
I actually have a MK-III driver w/6AN8 sitting on the bench waiting me to test-jig and output stage... I'll use that
In a great circuit, it shouldn't be noticeable until you come up on high voltage swings.
Kevin's KTA-ST-70 driver is one such circuit for example...
Cheers!
Oh, that is entirely possible.
One reason I want to revisit this.
I actually have a MK-III driver w/6AN8 sitting on the bench waiting me to test-jig and output stage... I'll use that
In a great circuit, it shouldn't be noticeable until you come up on high voltage swings.
Kevin's KTA-ST-70 driver is one such circuit for example...
Cheers!
Sy, its same resusts if you replace one triode with a Pentode, BJT, or MOSFET.
The weaker device determines the current delta for both. AC currents take the
U shaped path, they can't get in or out through the CCS. Its a folded cascode.
AC drops across equal resistrs will be equal and opposite. Current in, current out.
If the device curves are wildly different, you get SEPP curve only of the weaker
device. Regardless which device you actually drive, and which you may ground...
There's plenty O' interesting ways to abuse this effect to good purpose.
If transformer coupling, rather than a pair of resistors... The Trode plate "feels"
what happens at the Pentode plate through inductive coupling. Plate impedance
of the pentode drops to match the Triode, cause the Triode is actually in control
both branch currents.. Assuming the Triode was the weaker (lesser Gm) device.
The weaker device determines the current delta for both. AC currents take the
U shaped path, they can't get in or out through the CCS. Its a folded cascode.
AC drops across equal resistrs will be equal and opposite. Current in, current out.
If the device curves are wildly different, you get SEPP curve only of the weaker
device. Regardless which device you actually drive, and which you may ground...
There's plenty O' interesting ways to abuse this effect to good purpose.
If transformer coupling, rather than a pair of resistors... The Trode plate "feels"
what happens at the Pentode plate through inductive coupling. Plate impedance
of the pentode drops to match the Triode, cause the Triode is actually in control
both branch currents.. Assuming the Triode was the weaker (lesser Gm) device.
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