Here's the thing I keep designing the same amp. Need 100W per channel for the room. So my output stage is PP UL with 2x6550A on each half - a total of 8 6550A's and 12A of heater current for stereo.
So what a about a couple of KT120/KT150. Would look nice/something different. But I to get the same output power HT needs to go up (from 440 to 600V). I end up with the same issue. I need so much bias current to eliminate crossover distortion I will end up red plating the tubes - at least in simulation. There does not seem to be a better solution than two KT88/6550A in parallel. Am I missing something.
So what a about a couple of KT120/KT150. Would look nice/something different. But I to get the same output power HT needs to go up (from 440 to 600V). I end up with the same issue. I need so much bias current to eliminate crossover distortion I will end up red plating the tubes - at least in simulation. There does not seem to be a better solution than two KT88/6550A in parallel. Am I missing something.
In my experience, 6550’s don’t need stupid amounts of bias current to eliminate crossover distortion. It even gets down to tolerable levels with as low as 3mA per tube (More optimum at 10X that, but listenable nonetheless).
There may be more issues with gm doubling if the load impedance is too low (steep load line). With two tubes in parallel, the effective load impedance on each doubles and the load line shallows. My amp had 3 pairs - and during early tuning and testing phases, there was less crossover distortion running all 3 pairs when I had it biased down real low.
There may be more issues with gm doubling if the load impedance is too low (steep load line). With two tubes in parallel, the effective load impedance on each doubles and the load line shallows. My amp had 3 pairs - and during early tuning and testing phases, there was less crossover distortion running all 3 pairs when I had it biased down real low.
You can get 100W out of a single pair of 6550. Why do you need to use 8 tubes?
I built an amp using very similar voltages to this with Shuguang (cheapest set I could find) and they don't care about the voltage AKA they work fine, never arced. I hear a lot of new tubes won't handle it.
I built an amp using very similar voltages to this with Shuguang (cheapest set I could find) and they don't care about the voltage AKA they work fine, never arced. I hear a lot of new tubes won't handle it.
Last edited:
That's what I am not understanding the 6550A need very little bias to overcome crossover distortion whereas the KT120 you need to be 50W on the plate. Must be something to do with the curves.
B+ is 430V for the 6550A.
speakers are https://www.focal.com/en/home-audio/high-fidelity-speakers/aria-900/floorstanding-speakers/aria-948
maybe simulation model is inaccurate.
B+ is 430V for the 6550A.
speakers are https://www.focal.com/en/home-audio/high-fidelity-speakers/aria-900/floorstanding-speakers/aria-948
maybe simulation model is inaccurate.
I've always considered the 6550 the same as a KT88 but you're right. I'm running the tubes as shown in the sheet snippet l posted in the last post on 6550 in UL. It's worked fine for over a year now so I don't think it's an issue.
FWIW, I use a Hammond 278CX with cap input to power it. B+ is about 560V at the plates, and 50V at the cathodes for 510V @ 83mA, 42W Pd per tube. Fixed bias would get me 30% more power and the Pd would be closer to 25W per tube (at idle).
I think modern 6550A and KT88 may well be the same thing after all why have two different production runs when one will do. I am UL and servo bias and get 120W rms per channel with a 372LX transformer and 1650T OPT . Very pleased but wanted to try something else. Still not understanding why a KT120 needs so much bias when a 6550A does not. This is simulation for the KT120 - I don't have any.
Ah.. your transformer makes about 100V less than mine... That would also explain using 8 tubes instead of 4 🙂
It could be that the KT120 doesn't actually need more idle, just a bad model. I had a quad of them once. They wore out in 2 months while being used in a similar circuit I'm using the 6550A Shuguang tubes with that have worked for over a year now..
If only I could find someone who wanted to buy the amp, all of my amps would be using TV tubes instead - it's the only one using "audio" tubes for output.
I've also found that 2 parallel 6P3S or 6L6 (metal one) is electrically like a single KT88
In my apartment with all the others, but my most powerful is only 113WRMS@30Hz.
I have a Yorkville bridged to 950WRMS into 8R for my sub though.
I have a Yorkville bridged to 950WRMS into 8R for my sub though.
Wow, in an apartment? I'm impressed, I can't even use my 30W amp at full power in my house, too loud!
I've been going through these pages:
http://www.normankoren.com/Audio/Tubemodspice_article.html
The model provided is not accurate for very negative grid voltages. I've redone this as best I can and the KT120 does look better with crossover distortion gone at 80ma bias. I think without a alumi model that's the best guess.
.SUBCKT PENTODE1 1 2 3 4 ; A G2 G1 C
RE1 7 0 1MEG ; DUMMY SO NODE 7 HAS 2 CONNECTIONS
E1 7 0 VALUE={V(2,4)/KP*LOG(1+EXP((1/MU+V(3,4)/V(2,4))*KP))} ; E1 BREAKS UP LONG EQUATION FOR G1.
G1 1 4 VALUE={(PWR(V(7),EX)+PWRS(V(7),EX))/KG1*ATAN(V(1,4)/KVB)}
G2 2 4 VALUE={(EXP(EX*(LOG((V(2,4)/MU)+V(3,4)))))/KG2}
*G2 2 4 VALUE={PWR(if( V(2,4)/MU+V(3,4) < 0 , V(2,4)/MU+V(3,4), 0 ) ,EX )/KG2}
RCP 1 4 1G ; FOR CONVERGENCE A - C
C1 3 4 {CCG} ; CATHODE-GRID 1 C - G1
C2 1 3 {CPG1} ; GRID 1-PLATE G1 - A
C3 1 4 {CCP} ; CATHODE-PLATE A - C
R1 3 5 {RGI} ; FOR GRID CURRENT G1 - 5
D3 5 4 DX ; FOR GRID CURRENT 5 - C
.MODEL DX D(IS=1N RS=1 CJO=10PF TT=1N) ;
.ENDS PENTODE1
.SUBCKT KT120 1 2 3 4 ; A G2 G1 C (Beam Tetrode)
X1 1 2 3 4 PENTODE1 MU=10.1 EX=1.35 KG1=480 KG2=4200 KP=36 KVB=30 CCG=29P CPG1=1.8P CCP=10P RGI=1K ;
.ENDS
http://www.normankoren.com/Audio/Tubemodspice_article.html
The model provided is not accurate for very negative grid voltages. I've redone this as best I can and the KT120 does look better with crossover distortion gone at 80ma bias. I think without a alumi model that's the best guess.
.SUBCKT PENTODE1 1 2 3 4 ; A G2 G1 C
RE1 7 0 1MEG ; DUMMY SO NODE 7 HAS 2 CONNECTIONS
E1 7 0 VALUE={V(2,4)/KP*LOG(1+EXP((1/MU+V(3,4)/V(2,4))*KP))} ; E1 BREAKS UP LONG EQUATION FOR G1.
G1 1 4 VALUE={(PWR(V(7),EX)+PWRS(V(7),EX))/KG1*ATAN(V(1,4)/KVB)}
G2 2 4 VALUE={(EXP(EX*(LOG((V(2,4)/MU)+V(3,4)))))/KG2}
*G2 2 4 VALUE={PWR(if( V(2,4)/MU+V(3,4) < 0 , V(2,4)/MU+V(3,4), 0 ) ,EX )/KG2}
RCP 1 4 1G ; FOR CONVERGENCE A - C
C1 3 4 {CCG} ; CATHODE-GRID 1 C - G1
C2 1 3 {CPG1} ; GRID 1-PLATE G1 - A
C3 1 4 {CCP} ; CATHODE-PLATE A - C
R1 3 5 {RGI} ; FOR GRID CURRENT G1 - 5
D3 5 4 DX ; FOR GRID CURRENT 5 - C
.MODEL DX D(IS=1N RS=1 CJO=10PF TT=1N) ;
.ENDS PENTODE1
.SUBCKT KT120 1 2 3 4 ; A G2 G1 C (Beam Tetrode)
X1 1 2 3 4 PENTODE1 MU=10.1 EX=1.35 KG1=480 KG2=4200 KP=36 KVB=30 CCG=29P CPG1=1.8P CCP=10P RGI=1K ;
.ENDS