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Old 28th June 2003, 03:31 PM   #1
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Question Recommended primary output impedance for triode strapped EL34?

I have two choices for output transformers for my SE EL34 amp. I can use One Electron transformers with a primary impedance of 3K or Hammond 1627SEs with a primary impedance of 2.5K. I've looked at a mullard EL34 datasheet but they don't specify an output impedance for triode strapped EL34s. Any opinions or advice. BTW The Ra for a triode strapped EL34 is 910 ohms according to the data sheet. I'm not sure if they where taking into account bypassing the cathode resistors with caps or not. Any help is welcome. Thanks.

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Old 28th June 2003, 06:38 PM   #2
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Id say 3k or higher.
A friend did some listening tests at 2,5k and 5k and according to him 5k sounded way better, and I believe him without doubt.
He also claimed that 6L6s sounded a lot better than EL34s in the same circuit (triode SE), although they give much less power.
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Old 28th June 2003, 07:24 PM   #3
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Hi Fuling,
I guess that I could connect my 8 ohm speakers to the 4 ohm taps of the Hammonds, which would reflect about 5K to the plates of the EL34s. What do you think?

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Old 28th June 2003, 07:32 PM   #4
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Hi

If you already own those 1627SEs, then that is one option.
Works perfectly well according to my friend who did it.
If you havent bought the iron yet you may have a look at the 1628SE instead, that one is 5k as-is.
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Old 28th June 2003, 07:52 PM   #5
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Did your friend try using the UL taps and compare how it sounded in contrast to triode strapped?

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Old 28th June 2003, 08:23 PM   #6
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Nope.
When he tried the EL34 with different loads he used 125ESE iron which has no UL tap.

When he did the same thing with the 1627s he used some other tube (QQE06/40 triode strapped if my memory serves)
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Old 10th August 2005, 07:51 AM   #7
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Hi,
I don't why this is true, but this is what I read from magzine
Quote:
The primary impedance of the output transformer is more closely matched to the plate impedance of the output tubes for more efficient current transfer to the speaker
It donesn't make any sense to me.
We should consider the load line from the tube characteristic curve first in any case!
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Old 10th August 2005, 11:15 AM   #8
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The moral of this story is, "Stop reading magazines." Or if you insist on reading them, assume that 90% of what's written is total nonsense.
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Old 10th August 2005, 03:48 PM   #9
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> Ra for a triode strapped EL34 is 910 ohms according to the data sheet. I'm not sure if they where taking into account bypassing the cathode resistors with caps or not.

They are selling the tube, not the circuit or the cap. All tube-chart data, unless otherwise note, is taken with perfect bypassing, perfect transformers, perfect drivers, etc.

> One Electron transformers with a primary impedance of 3K or Hammond 1627SEs with a primary impedance of 2.5K.

Ignoring the difference in quality: these are the same thing, 2.5K or 3K. Your speakers vary from 7Ω to 50Ω, your tube varies +/-20% from batch to batch, and as we'll see later the "optimum" load is very broad.

> I don't why this is true, but this is what I read from magzine: "The primary impedance of the output transformer is more closely matched to the plate impedance of the output tubes for more efficient current transfer to the speaker" ... It donesn't make any sense to me.

Basic electric calculation. To get the maximum flow of load power from a given source, the load should match the source.

Example: There is a 100 Volt source in town. I'm on a farm 10 miles away. I buy a 10-mile power cord which has 100 ohms of resistance.

What is the maximum power I can get at the farm? I'm a farmer so I just try different loads.

infinite: 100V, 0 amps, zero power
1,000Ω: 91V, 0.091 amps, 8 watts
200Ω: 67V, 0.33 amps, 22 watts
100Ω: 50V, 0.5 amps, 25 watts
50Ω: 33V, 0.67 amps, 22 watts
10Ω: 9.1V, 0.91 amps, 8 watts
short: 100V, 0 amps, zero watts

Maximum power in the simple case happens when load equals source.

Note however that for "best" result, I get 25 watts on the farm and 25 watts lost in the cable. The man with the power source has to put 50 watts into the line, and will make me pay for 50 watts, even though I only get 25 watts. For the 10Ω case, 100V*0.91A= 91 watts into the line, 8 watts out, while for the 1,000Ω case I get my 8 watts with only 9.1 watts into the line.

So maximum power means matching, but matching is often very wasteful.

The "maximum" is broad: note that 50, 100, and 200 ohms all make 22-25 watts, or 23 watts +/-7% for a 4:1 range of loads.

In this case, I would consider the continuing cost of the power lost in the line against the one-time cost of a fatter power line. For many utility-power situations, the economics suggest buying wire big enough to keep losses down to a few percent. But if I had a very short-term need for big power, and was able to use any voltage I got, a very thin lossy line might be cheapest over all.

So in this case, the magazine is not wrong. Just incomplete.

And as you point out, tubes are not simple resistors, and getting audio out of a tube is a different problem.

Because an SE tube only makes DC, and we want AC, there is a factor of 2. Best power happens around 2Rp, for specific and reasonable operating condition.

BUT you have to stay inside the ratings. And tube ratings are expensive.

For EL34 you have a choice of up to 800V B+. Within the 25 watt plate dissipation rating, this suggests biasing to 800V, 31mA. When RL is much higher than Rp, and distortion is not a big concern, a "good power" load will be about 800V/31mA= 25K. If Rp were zero, the tube would swing from 800V 31mA to 0V 62mA and to 1,600V 0mA. Rp is 0.9K in the book, probably more like 1K for this extreme condition: actual swing will be 1K/(1K+25K) less, or about 770V each way.

This is RL= 25*Rp, far away from "matching" or "RL=2*Rp" rules, yet it works very well on paper.

In practice: 800V is nasty stuff, and 25K is hard to do in iron with high-fidelity. (Also I forgot that for EL34, only the plate is rated 800V, the G2 has a lower rating and for triode operation that is the limit.)

400V supply suggests operating near 25W/400V= 60mA (round down, be cool). For a first approximation to a maximum power load, try 400V/60mA= 6,700 ohms. This is higher than the ~1K Rp, so we get most of the 400V as plate swing. Swing will be from 50V 112mA to 750V 8mA.

The huge swing from 112mA to 8mA will cause huge distortion. It is better to keep the minimum current to about 1/4 to 1/3 of the maximum current. Staying with 400V and 60mA, that means swinging +/-30mA, or 90mA to 30mA. With 30mA peak at a little less than 400V peak, the load might be 12K.

This leads to a general rule: higher impedance gives lower distortion. It gives less power, but the change from 2*Rp to 5*Rp is often "better on the ear".

> One Electron transformers with a primary impedance of 3K or Hammond 1627SEs with a primary impedance of 2.5K.

For EL34, I think you have to go down past 250V 100mA for 2K-3K loading to make sense. Even then, 5K probably sounds better. EL34 is not a low-impedance tube to begin with. It seems to attempt to push the efficiency limit of tubes. Efficiency improves as plate voltage rises: same heater power, more voltage, more power (assuming a hi-Z load). The zero-bias triod ecurve is asymptotic to about 80V, which means terrible efficiency as you get below 400V supply.
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Old 10th August 2005, 04:11 PM   #10
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The Telefunken data sheet specifies the following operating point for a single EL34 in triode mode:

Uag2 375 v
Rk 370 ohm
Ia + Ig2 no signal 70 mA
Ia + Ig2 max OP 73 mA
Ra 3k ohm
Power @8% thd 6 w RMS

So, according to Telefunken, you need 3k ohm load.
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