Hi,
we are discussing about OTL SE here:
OTL SE
with your tubes you can do something similar.
You can also see:
http://hem.spray.se/sound.brigade/images/head-otl-1.jpg
6C19P = 6S19P (I think).
Regards.
we are discussing about OTL SE here:
OTL SE
with your tubes you can do something similar.
You can also see:
http://hem.spray.se/sound.brigade/images/head-otl-1.jpg
6C19P = 6S19P (I think).
Regards.
Just not sure wither i can substitute 6S19P into that construktion whitout recalculating the whole thing.I imagine 6C19P and 6080 are quite different tubes. So you need a datasheet with curves to recalculate the values.
I saw it's not so easy to find curves in the web for 6C19P.
I think(but I am not so expert) you can trying empirically, using the schematic like it is, putting 5 tubes per channel, and changing only the cathode res(under the 5 tubes).
Start with high value of cathode res(500 ohm - 20W). Then measure the voltage across this resistor. Using ohm law you can easy find the current flowing for each tube. I=V/R*5
Then decrease the value of this cathode res gradually, until you don't have about 90mA for each tube(450mA tot.)
This is the value of max current admitted for your tube(I found this value in the web).
Remember that this res dissipate a lot of power. So you need big power type(For my schematic I used 5 x 20 Ohm - 20W in series to reach 80 Ohm - 80W)
Regards.
I saw it's not so easy to find curves in the web for 6C19P.
I think(but I am not so expert) you can trying empirically, using the schematic like it is, putting 5 tubes per channel, and changing only the cathode res(under the 5 tubes).
Start with high value of cathode res(500 ohm - 20W). Then measure the voltage across this resistor. Using ohm law you can easy find the current flowing for each tube. I=V/R*5
Then decrease the value of this cathode res gradually, until you don't have about 90mA for each tube(450mA tot.)
This is the value of max current admitted for your tube(I found this value in the web).
Remember that this res dissipate a lot of power. So you need big power type(For my schematic I used 5 x 20 Ohm - 20W in series to reach 80 Ohm - 80W)
Regards.
liberty said:I imagine 6C19P and 6080 are quite different tubes. So you need a datasheet with curves to recalculate the values.
That's right
I saw it's not so easy to find curves in the web for 6C19P.
No: http://frank.pocnet.net/sheets/113/6/6S19P.pdf
size of EL84, 7w P diss.
I think(but I am not so expert) you can trying empirically, using the schematic like it is, putting 5 tubes per channel, and changing only the cathode res(under the 5 tubes).
Since you have the specs, you can recalculate it. Empirically won't work, unless you make a huge amount of trials
Regards
Konstantinos
It depends on type of configuration. I imagine less than a couple of watts in single ended.
And about 20-30 Watts in pushpulls like these:
6C33C OTL
And about 20-30 Watts in pushpulls like these:
An externally hosted image should be here but it was not working when we last tested it.
6C33C OTL
Hi,
Not trying to sound negative but 6C19 is not an optimum valve for use in OTL's. The anode resistance is high and the transconductance is rather low so the output impedance even as cathode follower is high.
Typical data is Rp 400 ohm and Gm 7.5mA/V which gives a mu of ~3 and an output impedance as cathode follower of ~400/(1+3) = 100ohm, so you would need a lot of them in parallell.
There is another small russian valve that is much better, 6C41 which is similiar to a half 6C33C, that would be my choice if I wanted to build a small OTL.
Regards Hans
Not trying to sound negative but 6C19 is not an optimum valve for use in OTL's. The anode resistance is high and the transconductance is rather low so the output impedance even as cathode follower is high.
Typical data is Rp 400 ohm and Gm 7.5mA/V which gives a mu of ~3 and an output impedance as cathode follower of ~400/(1+3) = 100ohm, so you would need a lot of them in parallell.
There is another small russian valve that is much better, 6C41 which is similiar to a half 6C33C, that would be my choice if I wanted to build a small OTL.
Regards Hans
Not quite, the current of a 6C41 is half of a 6C33C so power will be 1/4, (P=R*I^2) that is one of the basic problems with building a small OTL, power falls quickly with current and effciency will also be lower, it is easier to make a high power OTL, my OTL with 4 6C33C give about 4 times more power than the one with 2 tubes.
Regards Hans
A 6C33C have a maximum allowed contionous cathode current of 600mA and that will give an output power of 1.44W in 8 ohm, I would not recommend to go much above 600mA so thats it.
In a class AB amplifier it is safe to run much higher peak currents, although max allowed peak current is not specified for the 6C33C experience show that there is no problem to run peak currents of 2.5A or more if the average current doesn't go much over 600mA. Also in a class AB amplifier while playing music the average current will be much lower than for peak conditions which give very long life of the ouput tubes.
I would recommend to use 6C33C in that case as it there will be no problems with tube matching, using parallell connected output tubes always give a whole range of new problems that it is better to avoid if possible, that is one reason why 6C33C is so an ideal tube for an OTL as it gives substantial power and low output impedance even for one pair.
Regards Hans
In a class AB amplifier it is safe to run much higher peak currents, although max allowed peak current is not specified for the 6C33C experience show that there is no problem to run peak currents of 2.5A or more if the average current doesn't go much over 600mA. Also in a class AB amplifier while playing music the average current will be much lower than for peak conditions which give very long life of the ouput tubes.
I would recommend to use 6C33C in that case as it there will be no problems with tube matching, using parallell connected output tubes always give a whole range of new problems that it is better to avoid if possible, that is one reason why 6C33C is so an ideal tube for an OTL as it gives substantial power and low output impedance even for one pair.
Regards Hans
Assume that we build a class B amplifier then the peak current is Pi * average current over one period. For 6C41C allowed max average current is 350mA so peak current can be ~1.1A which gives ~4.8W in 8 ohm, (Pout = Ipk^2*Rl/2). This calculation is valid also for a class AB amplifier but efficiency is lower due to higher idle current.
Given that it is an amplifier for music you can allow for a higher average current than specified as the current peks only will happen during music peaks which normally are 10 - 20 dB higher than the average output power when the ampliofier is close to clipping, so 6-7W seems reasonable and will give long life, an idle current of ~100mA should be OK.
Regards Hans
IMHO it is not needed, a fuse in the B+ line is enough in the unlikely event that a tube develop a catasthrophic failure. The speaker can withstand a DC voltage during the time it takes to blow the fuse, if one is anxious why not use ultrafast fuses.
If the fuse method is not deemed acceptable it is easy to build something around a window comparator that after LP filtering checks the DC voltage and switch of B+ in the event of failure.
Regards Hans
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