Hi Folks
I'm planning an amp with a pre-amp similar to an early Orange-Matamp and a cascoded ECC82 as the third stage, this into a Split Load Inverter.
I'll have a triode half going spare so I thought I might try using it to buffer the anode output of the Concertina. like this:
http://s224.photobucket.com/user/bluebi ... sort=1&o=0
Anybody tried this? any reasons why its a bad idea?
Cheers
Shane
Split Load PI with added Buffer
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iknowjohnny
- Posts: 1070
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Re: Split Load PI with added Buffer
Hmmmm....i thought i'd thought of everything, but i never thought to do a split load on my PI. It's a LTP, but still i wonder how that would work because i have always felt the signal going to my PA on my amp is pretty hot and my pre PI master is super sensitive. (tried a few PPI masters and they worked hideously on my amp) Might have to try it. I never need nearly the volume my amp has, it's stupid loud. So it may just help with the sensitivity and maybe give a benefit tonally. Don't wanna hijack this thread, but anyone ever tried that on a LTP?
Re: Split Load PI with added Buffer
Seems it's not a new idea but is called the Kappler phase inverter there is some info about it out there!
Re: Split Load PI with added Buffer
Shane
So out of the cathodyne you buffer one phase, what does this get you? Or is it "...there's a spare triode and it might as well do something!!" I've just been looking into the early 70's orange design as well.
Any chance you post your preamp section schematic? Thanks....
So out of the cathodyne you buffer one phase, what does this get you? Or is it "...there's a spare triode and it might as well do something!!" I've just been looking into the early 70's orange design as well.
Any chance you post your preamp section schematic? Thanks....
In theory, theory is the same as practice. In practice it's different.
Re: Split Load PI with added Buffer
Guitar amp forums say that the outputs from a concertina/split load PI are inherently unbalanced. Audio forums say they are balanced and I see them more often on audio amps than LTPIs.
If it says "Vintage" on it, -it isn't.
Re: Split Load PI with added Buffer
And..
The Valve Wizard says "...if the loads are balanced it's a perfect inverter."
So 6 of this and a half a dozen of that. Need a scope to really see. But on the Orange PI there are equal 100K loads...so maybe!
The Valve Wizard says "...if the loads are balanced it's a perfect inverter."
So 6 of this and a half a dozen of that. Need a scope to really see. But on the Orange PI there are equal 100K loads...so maybe!
In theory, theory is the same as practice. In practice it's different.
Re: Split Load PI with added Buffer
The concertina has unequal voltage, but equal current. It is a current based device unlike the LTP which is a voltage based device.jjman wrote:Guitar amp forums say that the outputs from a concertina/split load PI are inherently unbalanced. Audio forums say they are balanced and I see them more often on audio amps than LTPIs.
Re: Split Load PI with added Buffer
I'd go to something like a switchable CF on the second half of the preamp, with a enhanced "compression effect" on one side of the signal.
Something like a FAT switch.
Something like a FAT switch.
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Tone Lover
- Posts: 261
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Re: Split Load PI with added Buffer
do you have a preferance in fat switch designs roberto.roberto wrote:I'd go to something like a switchable CF on the second half of the preamp, with a enhanced "compression effect" on one side of the signal.
Something like a FAT switch.
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gingertube
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- Joined: Mon Nov 14, 2011 2:29 am
- Location: Adelaide, South Oz
Re: Split Load PI with added Buffer
Some analysis:
The Output Impedance of the Cathodyne at the anode and cathode is identical so long as the loads on the anode and cathode are identical.
For equal loads on Anode and cathode on the Cathodyne
Zout = RL.ra/RL(u+2)+ra
The ra term on the bottom line is insignificant compared to RL(u+2) term so drop it. Then the RL terms top and bottom lines cancel leaving
Zout approx = ra/u+2
At typical values of u (>=20) u+2 approx = u, so simplify again
Zout approx = ra/u = 1/gm
Note that this value has nothing at all to do with the value of the Anode and Cathode Load resistors RL. In small amps we tend to see 56K driving a pair of 6V6 and in bigger amps we see 82K or 100K driving a pair or a quad of EL34, 6L6 or what ever.
The larger reistors on the larger amps are purely for ease of swinging the larger required grid voltage.
The problems start when we start to overdrive the output tubes
If driving output stage directly The equal loads on Anode and Cathode will NOT be guaranteed if:
1) Output stage strays out of Class A (When a tube cuts off it has no gain so Miller capictance will change, particularly with Triode Mode Output, less so with Ultralinear and less so again in Pentode Mode)
2) Output Stage strays into grid current (trying to overdrive the output tube). When this happend the load resistance presented by the output tube grid drops dramatically.
Calculating Overdrive Effects
From above - for equal loads at the anode and cathode the Zout at both is approximately 1/gm or about 650 Ohms for a 12AX7
Check the limiting conditions:
If the Anode load drops significantly then:
Zout cathode = RL+ra/(u+2) x ra/RL
The ra/RL term insignigicant so
Zout cathode approx = RL+ra/u+2
At usual values of u
Zout cathode approx = RL/u + ra/u = RL/u + 1/gm
That is it Zout cathode increases by RL/u
If the cathode load drops significantly then:
Zout anode = RLxRL(u+1)+RL.ra / RL(u+2)+ra
RL squared (u+1) is much larger than RL.ra and RL(u+2) is much larger than ra so
Zout anode approx = RLxRL(u+1)/RL(u+2)
and at reasonable values of u
Zout anode approx = RL
SUMMARY:
As the loads on Anode and cathode become unbalanced (as when overdriving the output tubes) then
Zout anode increases from 1/gm toward RL (from 650 Ohms toward 100 KOhms)
Zout cathode increase from 1/gm by maximum factor of RL/u (from 650 Ohms toward 650 + 100K/100 = 1650 Ohms)
The imbalance from an over driven output will therefore be worse with the 100K loads than with 56K loads but the difference between the anode and cathode output impedance in overdrive is so dramatic it doesn't really matter if its between 650 Ohms and 100K or between 650 ohms and 56K, Also the effect of this will be asymetrical, affecting one side of the push pull and therefore introducing large gobs of 2nd harmonic distortion.
The trick here is to use large grid stops on the output tubes to reduce the loading effects on the cathodyne. Take those output tube grid stop resistors way up to say 47K. This is effectively tuning the amount of 2nd harmonic distortion you get in overdrive. This is what Merlin says too.
OR
Buffer the anode output as per your suggested circuit.
Cheers,
Ian
The Output Impedance of the Cathodyne at the anode and cathode is identical so long as the loads on the anode and cathode are identical.
For equal loads on Anode and cathode on the Cathodyne
Zout = RL.ra/RL(u+2)+ra
The ra term on the bottom line is insignificant compared to RL(u+2) term so drop it. Then the RL terms top and bottom lines cancel leaving
Zout approx = ra/u+2
At typical values of u (>=20) u+2 approx = u, so simplify again
Zout approx = ra/u = 1/gm
Note that this value has nothing at all to do with the value of the Anode and Cathode Load resistors RL. In small amps we tend to see 56K driving a pair of 6V6 and in bigger amps we see 82K or 100K driving a pair or a quad of EL34, 6L6 or what ever.
The larger reistors on the larger amps are purely for ease of swinging the larger required grid voltage.
The problems start when we start to overdrive the output tubes
If driving output stage directly The equal loads on Anode and Cathode will NOT be guaranteed if:
1) Output stage strays out of Class A (When a tube cuts off it has no gain so Miller capictance will change, particularly with Triode Mode Output, less so with Ultralinear and less so again in Pentode Mode)
2) Output Stage strays into grid current (trying to overdrive the output tube). When this happend the load resistance presented by the output tube grid drops dramatically.
Calculating Overdrive Effects
From above - for equal loads at the anode and cathode the Zout at both is approximately 1/gm or about 650 Ohms for a 12AX7
Check the limiting conditions:
If the Anode load drops significantly then:
Zout cathode = RL+ra/(u+2) x ra/RL
The ra/RL term insignigicant so
Zout cathode approx = RL+ra/u+2
At usual values of u
Zout cathode approx = RL/u + ra/u = RL/u + 1/gm
That is it Zout cathode increases by RL/u
If the cathode load drops significantly then:
Zout anode = RLxRL(u+1)+RL.ra / RL(u+2)+ra
RL squared (u+1) is much larger than RL.ra and RL(u+2) is much larger than ra so
Zout anode approx = RLxRL(u+1)/RL(u+2)
and at reasonable values of u
Zout anode approx = RL
SUMMARY:
As the loads on Anode and cathode become unbalanced (as when overdriving the output tubes) then
Zout anode increases from 1/gm toward RL (from 650 Ohms toward 100 KOhms)
Zout cathode increase from 1/gm by maximum factor of RL/u (from 650 Ohms toward 650 + 100K/100 = 1650 Ohms)
The imbalance from an over driven output will therefore be worse with the 100K loads than with 56K loads but the difference between the anode and cathode output impedance in overdrive is so dramatic it doesn't really matter if its between 650 Ohms and 100K or between 650 ohms and 56K, Also the effect of this will be asymetrical, affecting one side of the push pull and therefore introducing large gobs of 2nd harmonic distortion.
The trick here is to use large grid stops on the output tubes to reduce the loading effects on the cathodyne. Take those output tube grid stop resistors way up to say 47K. This is effectively tuning the amount of 2nd harmonic distortion you get in overdrive. This is what Merlin says too.
OR
Buffer the anode output as per your suggested circuit.
Cheers,
Ian
Re: Split Load PI with added Buffer
Thanks Gingertube.
Yes I thought it may be of some benefit during PT OD.
I've had various opinions expressed ranging from 'fine' to 'complete waste of time!'"
Even Merlin says in a post somewhere that the Kappler circuit "fixes a problem that doesnt exist" and that it "looks like it was designed by someone who doesn't know how a cathodyne works"!!
I may still try it However.
I'm also investigating the Aikido split load inverter from the TubeCAD site that looks interesting too {basically a split load with the bottom resistor replaced with a current source).
I'd prefer to keep it simple though
Cheers
Shane
Yes I thought it may be of some benefit during PT OD.
I've had various opinions expressed ranging from 'fine' to 'complete waste of time!'"
Even Merlin says in a post somewhere that the Kappler circuit "fixes a problem that doesnt exist" and that it "looks like it was designed by someone who doesn't know how a cathodyne works"!!
I may still try it However.
I'm also investigating the Aikido split load inverter from the TubeCAD site that looks interesting too {basically a split load with the bottom resistor replaced with a current source).
I'd prefer to keep it simple though
Cheers
Shane
Re: Split Load PI with added Buffer
ahahah! Not so much personally, but many people likes so much this kind of mod on their amp. In such an amp can be a nice option.Tone Lover wrote:do you have a preferance in fat switch designs roberto.