Data sheets for EL-34’s that I’ve seen recommend Ra-a for various PP operating conditions (pentode, UL, various Va, etc.), anywhere from 2800-6600. Why would there be so much variation? Is there a sweet spot for UL with 400 Va?
Second point of confusion: some transformer manufactures (like Hammond) list their impedance with the ct (center tap) indication. What is the actual Ra-a impedance? I.e. does “3400 ct” = 3400 or 1700 or 6800 Ra-a?
Finally, the notion of reflected impedance. If I run an OPT with ½ the recommended primary impedance, can I make up for it by using the 16 ohm secondary tap (assuming 8 ohm speakers)?
Thanks in advance for your help.
Dave
Second point of confusion: some transformer manufactures (like Hammond) list their impedance with the ct (center tap) indication. What is the actual Ra-a impedance? I.e. does “3400 ct” = 3400 or 1700 or 6800 Ra-a?
Finally, the notion of reflected impedance. If I run an OPT with ½ the recommended primary impedance, can I make up for it by using the 16 ohm secondary tap (assuming 8 ohm speakers)?
Thanks in advance for your help.
Dave
6k Ra-a = 6kct.
For all intents and purposes, a 5k:8r transformer is also a 10k:16r or 2.5k:4r transformer. The load is reflected. This datasheet advises 6k Ra-a with 43% UL taping at 430V: https://frank.pocnet.net/sheets/129/e/EL34.pdf
For all intents and purposes, a 5k:8r transformer is also a 10k:16r or 2.5k:4r transformer. The load is reflected. This datasheet advises 6k Ra-a with 43% UL taping at 430V: https://frank.pocnet.net/sheets/129/e/EL34.pdf
You'd put your 8R speakers on a 4R tap to double the primary Z, if it had a 4R.
Operating conditions such as amplifier feedback level, speakers impedance curve, bias of tubes and voltage.
Higher tube voltage increases maximum power at the cost of higher impedance.
El-34 can be operated anywhere from 250V to 500V with great results as long as it has appropriate feedback, voltage, OT, and current bias.
Higher tube voltage increases maximum power at the cost of higher impedance.
El-34 can be operated anywhere from 250V to 500V with great results as long as it has appropriate feedback, voltage, OT, and current bias.
Notice that the 6.6K a-a example is for K biasing using 470R on each tube, the other is for fixed bias.
How would I do that ?
example 1: PP, 3.2k:8, ratio 0.0504, 50Hz...12.5kHz, L= ???
example 2: SE, 5.2k:8, ratio 0.0366, 40Hz...20kHz, L= ???
Simply: you can run the same tubes at 300V 300mA or 600V 150mA. Results are similar but the implied impedance changes by a factor of 4.
Same as if I needed 1,000 Watts of lamps. We don't buy lamps by Ohms but let's see if we did. In 120V land I would rig 8.3 Amps of lamps, and 120V/8.3A is 14.4 Ohms. But in 230V world I would rig 4.35 Amps of lamps, 230V/4.35A is 53 Ohms. Same power, different voltage, different V/I ratio, different impedance.
Some tubes do not have much choice. You need a certain voltage to get enough current to get the max power possible, and the max voltage is not much higher than this. The original 6V6 did not have a lot of current and you had to work ~250V to get enough to justify a 12W plate. But the max plate voltage rating (on paper) was not much over 300V. Full power loads are all 4K-6K. The EL34 has huge current (can run at low voltages) *and* insane plate voltage rating. 400V @ 3.4K makes sense, 600V @ 8K makes sense.
Inductance is important, but most good full-bass push-pull OTs have way more than needed for "frequency response". They need inductance to get a good THD for high power at low frequencies. It is often easy to hit 8Hz at 1W even without NFB. However 50 Watts at 50Hz may come out bent. There's not a dead-stupid (buyer friendly) way to specify this. Jensen publishes curves of power vs THD vs freq; but these are not Big-Watt transformers.
well you cannot tell from that 3.2K:8 because this has nothing to do with it.
You have to trace your Ra and multiply by 2. This is the tube in PP in class A.
When the tube goes into Class AB this value will be higher.
However for the transformer for which we wish to solve L, we have to assume/guess the manufacturer intended voltage/tube/UL/ = Ra.
-->not our calculation from our design!!!
L of transformer is then = Ra of tube (Z) ((as interpreted or guessed per frequency response from manufacturer -3db frequency.)) / ( cut off -3db frequency in Hz * 6.283)
You can on the other hand easily calculate the point of -3db frequency from lundahl datasheet as example:
LL1663/50ma 35H, 100ma/17H
see as the bias increases the transformer loses inductance because of magnetization by the bias?
In class AB this magnetization becomes variable, enabling transformer L to rise and this counter-interacts the tubes Ra rise!
Lets do simple math from a Ra of our own calculation, lets say we trace out load lines etc and the tube Ra is 1100. In PP it will be:
2200ohm/6.28*35H = 10hz -3 db point.
If you have a //PP with same tube same transformer bias doubles, Ra halves, inductance is 17H due to higher magnetization: 1100ohm/6.28*17 = 10.3hz -3 db point.
You have to trace your Ra and multiply by 2. This is the tube in PP in class A.
When the tube goes into Class AB this value will be higher.
However for the transformer for which we wish to solve L, we have to assume/guess the manufacturer intended voltage/tube/UL/ = Ra.
-->not our calculation from our design!!!
L of transformer is then = Ra of tube (Z) ((as interpreted or guessed per frequency response from manufacturer -3db frequency.)) / ( cut off -3db frequency in Hz * 6.283)
You can on the other hand easily calculate the point of -3db frequency from lundahl datasheet as example:
LL1663/50ma 35H, 100ma/17H
see as the bias increases the transformer loses inductance because of magnetization by the bias?
In class AB this magnetization becomes variable, enabling transformer L to rise and this counter-interacts the tubes Ra rise!
Lets do simple math from a Ra of our own calculation, lets say we trace out load lines etc and the tube Ra is 1100. In PP it will be:
2200ohm/6.28*35H = 10hz -3 db point.
If you have a //PP with same tube same transformer bias doubles, Ra halves, inductance is 17H due to higher magnetization: 1100ohm/6.28*17 = 10.3hz -3 db point.
The other confusing thing is that
Hammond recommends a primary impedance of 3400 for parallel PP (4 tubes), and 6600 for PP 2 tubes (non-parallel), hence my confusion about the CT (center tap) designation. Is there a logical explanation that you guys can provide, or should I give them a call to try to clear things up?
Hammond recommends a primary impedance of 3400 for parallel PP (4 tubes), and 6600 for PP 2 tubes (non-parallel), hence my confusion about the CT (center tap) designation. Is there a logical explanation that you guys can provide, or should I give them a call to try to clear things up?
All push-pull tube OPTs are CT, of course. I do not understand your question.
*IF* a happy point for two tubes is 6,600r (not clear), then at the same supply voltage a 3,300r load would be fine for four tubes. Simple impedances. If two lamps turns out to be 66 Ohms, then four lamps would be 33 Ohms. 3300 and 3400 are the same for all practical purposes.
Have you actually looked at suggested conditions for EL34? They range ALL over the place. These are known-good operating points left to us by the Dead Men.
http://www.mif.pg.gda.pl/homepages/frank/sheets/010/e/EL34.pdf
*IF* a happy point for two tubes is 6,600r (not clear), then at the same supply voltage a 3,300r load would be fine for four tubes. Simple impedances. If two lamps turns out to be 66 Ohms, then four lamps would be 33 Ohms. 3300 and 3400 are the same for all practical purposes.
Have you actually looked at suggested conditions for EL34? They range ALL over the place. These are known-good operating points left to us by the Dead Men.
http://www.mif.pg.gda.pl/homepages/frank/sheets/010/e/EL34.pdf
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IF you were designing the amplifier from scratch, on a clean sheet of paper, then you would draw a couple load lines and choose one based on your preferences.
That design procedure would show you that there is not a single "best" value, for the very sobering reason that none is "perfect", since tubes are quite lossy and current limited
, so you will find a *range* of possible values.
Which one to choose?
Depends on your preference and worst case flip a coin ... within that range that is.
Personally I prefer the higher impedance values within a range since tube is less stressed, but that´s me.
Now if you don´t want to go through it, fine, you can follow other designer´s advice ... but choose one and follow it in full, don´t pick A configuration with B supply and C OT suggestions because it will only work (as it should) by chance.
That design procedure would show you that there is not a single "best" value, for the very sobering reason that none is "perfect", since tubes are quite lossy and current limited
, so you will find a *range* of possible values.Which one to choose?
Depends on your preference and worst case flip a coin ... within that range that is.
Personally I prefer the higher impedance values within a range since tube is less stressed, but that´s me.
Now if you don´t want to go through it, fine, you can follow other designer´s advice ... but choose one and follow it in full, don´t pick A configuration with B supply and C OT suggestions because it will only work (as it should) by chance.
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