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Old 8th March 2003, 03:59 AM   #1
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Question Triode Amps

Having never designed a triode PP amp, I' looking for some help in finding a triode that will fit a 6.6K p-p xfmr, max 30W out. Is there any such animal (Non DHT,pse) out there? Help!
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Old 8th March 2003, 05:03 AM   #2
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Hmm...

At 6.6k, the Rp will be about 2k for each triode (rule of thumb: Rl = 3Rp). 30W into 6.6k represents 450V of swing, so that basically rounds it out to the power transmitter triodes, 845 and the like. You'll need a good 800V of supply for this; hope your OPT can handle it. I'm sure you can get by with some smaller triodes if you use less bias current, going into class AB1, maybe AB2.
I'm not doing any more math for you, my algebra homework is going slow enough as it is.

For a more realistic triode amp, drop your power or impedance requirements. There are many tetrodes/pentodes which are popular triode-strapped, and paralleling a few will greatly increase your choices.

Tim (might eventually make a sextet-of-6CL6-in-triode, approximate calcs show 0% 3rd HD at the class AB1 operating point )
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Old 8th March 2003, 07:03 AM   #3
316a is offline 316a  England
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Default triode-strapped

You're be pushing it for 30W . 12e1 , KT88 or paralleled 5894 triode strapped should do the trick and give you 20w on tap with about 350V HT. Retrovox of Australia has 12e1's going for peanuts at the moment (about $10 each) . The only 'real' triodes I can think of are 6C33 but these will not suit your transformers

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Old 8th March 2003, 07:05 AM   #4
316a is offline 316a  England
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Default 6528

...may also be suitable but requires a beefy driver

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Old 8th March 2003, 08:16 AM   #5
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KT88 (or even KT90) trioded will give you around 20W.
Using the 16 ohm tap for a nominal 3k3 load on an 8 ohm speaker will give a bit more power and distortion. Really depends which flavour you like.
I'd start looking at 350-400Va-k and 70 mA to get in the ballpark.
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Old 8th March 2003, 09:18 AM   #6
316a is offline 316a  England
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Default Not quite but almost !

With an 8ohm speaker on the 16 ohm tap the primary impedence will be less than 3.3k , more like 2k . Remember the turns ratio squared x secondary load = primary impedence

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Old 8th March 2003, 01:31 PM   #7
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6600R primary with 16R load on 16R tap gives us a turns ratio of 20.31.
Zpr=(Np/Ns)^2 x Rl
= (20.31)^2 x 8 = 3k3 (8R on 16R tap)

The secondary winding will have a higher R on the 16, as against the 8 ohm taps, which being in series with the load will actually make the Zpr a bit higher than 3k3. (Rl = Rsec + 8)

The transformer also has a given primary inductance, which is in parallel with the reflected secondary imperdance. Therefore the tube will see closer to Zpr in 3k3 mode than it will with the same primary L in parallel with a Zpr of 6k6.

If I was going the other way, looking to use the 4R tap to obtain a Zpr of 13k3, the secondary resistance would be lower (in series with Rl so less than expected Zpr), and the LF disortion would increase due to the decreasing load with decreasing frequency at a higher F than it will in either to designed implementation, or using a lower TR as in my example.
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Old 8th March 2003, 01:45 PM   #8
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Jack, a triode-strapped EL34 will work fine at 6600 p-p. And though you'll get about 22 W out for a plate voltage of 480V (that's what I've measured with my similarly-loaded amp), there's only about a 1.3 dB difference in max SPL between 22 and 30 W. Distortion will be low for that higher-than-optimal load. You could probably squeeze out a bit more power by raising the B+, at a cost of more heat and less reliability.

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Old 9th March 2003, 07:19 PM   #9
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If you are not concerned about efficiency it is ok to have some miss-match. You can use a 6.6k transformer on 5K tubes but the output will be slightly less. Another factor to consider is vacuum tubes produce less distortion when they are less loaded.

Another advantage to a higher impedance transformer is the resulting source impedance of the amp will be less. I see the lower source impedance presenting tighter bass and better hi frequency definition.
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