Hallo Chow,
I would also be interested to build something like that as I have 8 of those 6c19pi tubes lying about idly.
Should not be too hard. I suppose it would just be 4 * 6c19pi in parallel..configured as a cathode follower, with a big fat output cap. Just add a driver stage (6H30pi) and voila.
The only "tricky bit" is to ensure the same amount of current is flowing through each tube...but you can always do that later...
I'll try to come up with something one of these days.
Regards,
Bas
I would also be interested to build something like that as I have 8 of those 6c19pi tubes lying about idly.
Should not be too hard. I suppose it would just be 4 * 6c19pi in parallel..configured as a cathode follower, with a big fat output cap. Just add a driver stage (6H30pi) and voila.
The only "tricky bit" is to ensure the same amount of current is flowing through each tube...but you can always do that later...
I'll try to come up with something one of these days.
Regards,
Bas
Not sure..isn't the current set by the cathode resistors?Where is the current coming from? Resistor? Inductor?
Would a cathode follower not be the way to do it?
That shows you how little I knowThink you don`t get 1,5 Watt from a resistor.
Ok...I give up..how is it done
Hi Reinhard...I won't ask.Have the schematic, but don´t ask, think it is intelectuell property of Bruce Rozenblit.
The info you gave me should be enough for me to rig something up that will work.
Funny..just found your thread where you mentioned the chokes! I also have 6C33C and 6C41C...but I'll try the 6C19pi's first..I had plans for something like that a while ago, I even made me a pair of chokes. One PL519 could replace 4 x 6C19, however I never got around to build the circuit. Maybe some day...
Hmm, I finished a pair of speakers with Fostex FE167E last weekend and I think this kind of amp would be a great match for those.
Unfortunately I don´t have the chokes anymore, but since it has to be cap coupled to the speaker I guess a constant current source would work just as good if not better.
With a little luck that might even be a good way to get a decent autobias point for a bunch of 6C19.
I think I have everything except the tube sockets.
Since idle current is the limiting factor for output power I´d probably increase the number of tubes to 5 or 6 per channel, I have quite a few of those 6C19 IIRC.
Dang, I just remembered that the filament transformer I have only can supply 10A.
Unfortunately I don´t have the chokes anymore, but since it has to be cap coupled to the speaker I guess a constant current source would work just as good if not better.
With a little luck that might even be a good way to get a decent autobias point for a bunch of 6C19.
I think I have everything except the tube sockets.
Since idle current is the limiting factor for output power I´d probably increase the number of tubes to 5 or 6 per channel, I have quite a few of those 6C19 IIRC.
Dang, I just remembered that the filament transformer I have only can supply 10A.
Have you built one of those 1.5W OTL's? Does it sound good?
Hi Bas,
Very Good! I have had mine since they came out.
I would love to see a higher power version.
Andrew
Hi Andrew,Very Good! I have had mine since they came out.
Worth pursuing then... How does it compare with say an EL84 single ended amp?
Regards,
Bas
P.S. Long time no talk
SE Otl
Hi Bas,
you don´t get the same with a single 519, because the 19 has a peak current of 1 A, 4 A for a quad.
The amp has very much detail and great bass, with the right speaker, fantastic.
For more power you can use 8 pcs. 6C19 and two inductors to get
4 Watt.
Reinhard
Hi Bas,
you don´t get the same with a single 519, because the 19 has a peak current of 1 A, 4 A for a quad.
The amp has very much detail and great bass, with the right speaker, fantastic.
For more power you can use 8 pcs. 6C19 and two inductors to get
4 Watt.
Reinhard
Re: SE Otl
Won't you get 5/6W with a single EL519? I mean... P=I*I*R = 1 * 8 = 8, let's say that the impedance could drop but... where do my calculations fail?
reinhard said:Hi Bas,
you don´t get the same with a single 519, because the 19 has a peak current of 1 A, 4 A for a quad.
Won't you get 5/6W with a single EL519? I mean... P=I*I*R = 1 * 8 = 8, let's say that the impedance could drop but... where do my calculations fail?
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