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In Reply to: RE: UTC output transformers LS-63 posted by dave slagle on April 15, 2015 at 10:34:58
" See the attached pictures and explain to me how it is possible that a plate choke could possibly sound good at low.."
That was where I was going.
A gaped transformer (or plate choke) can't have near enough inductance (practically speaking) to allow for good performance at low frequencies.
My conclusion is to use SET amplifiers only for mid and high frequencies.
As for the high frequencies, can the shunt capacitance be held low, by the winder, if the transformer is not meant for LF duties?
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Follow Ups:
Lets look at the case of 100hz with a reasonable (50hy) amount of inductance. Granted the 5K load is a bit harsh for a 211 but it does have the narrowest ellipse and the 1Meg load (ie a plate choke) has the widest which makes me wonder if the width of the ellipse actually matters?
It is interesting to note that the Red does have the lowest current swing at 40ma, the Blue roughly double that at 80ma and the green swings 120ma of current.
I'll run some transients and see what the FFT says...
dave
That's interesting.
Isn't the width of the ellipse the indicator of the ratio, reflected//reactance, with the power into the reactance not going to the intended load, not only wasted but causing harm?
Can you do the same for a 2a3 at 15Hy and 2500 ohms and 31Hz for the LS-63 wired for 2500 ohms?
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Assume use of 2A3 tube which develops 3 watts of output power per published ideal design parameters.
pri impedance 2500 ohms
AC signal volts (full power) will be 86.6 vrms
AC signal current .0346 Arms
now let's calculate and then add in vectorially the currents wrought by the inductive reactance of the primary;using the 15 henries that you spec'd @ 31 hz and full power (86.6 vrms) we would have;
Iex .0296 Arms
Itot .0458 Arms (vector sum)
notice that our total current draw is 132% greater than the signal current.
Now let's keep all the same parameters but reduce the freq to 26.5 hz.
this is the -3db large signal point. It is where the inductive reactance equals the reflected impedance.Heres the resulting current draws;
Iex .0346 arms
I total .049 (142% greater than the signal current)
also note that the Iex equals the Iac signal current. This is because we are delivering half of our power into the shunt arm (the inductive reactance) hence the currents btwn the two "circuits" are the same value.
but now the 2A3 must work (from the current standpoint) 142% harder to drive the complex reactive loadline that the anode of the tube is looking into.
And, again, real world losses will be even greater than shown above since we are not accounting for any copper losses (remember we are now shoving 142% more current through the copper circuit) or core losses.
Not an entirely pretty picture.
MSL
Builder of MagneQuest & Peerless transformers since 1989
Edits: 04/16/15
Requested Sim... I'm not totally offended by this since in order to get a full scale signal at 30hz, the levels at higher frequencies will surely be much more of an issue.
Ultimately If I had the LS-63's I'd give it a shot but I don't think I would go on an ebay hunt to try it out.
Dave wrote:
"Ultimately If I had the LS-63's I'd give it a shot but I don't think I would go on an ebay hunt to try it out."
this trans is just begging to be parafed. It would excel at least in terms of it's magnetic curcuit behaviour.You'd have the full complement of the (as advertised) 275H not the diminished 15 or 20 henries after applying the dc (unbal) plate current
as per your graph.From the vantage point of flux density... with the AC flux and dc flux added together I would guess that this unit is probably getting close to the "redline"...
now remove the dc flux component and we have 86.6 volts across the primary... we will have tons and tons of magnetic headroom... the core will be loafing....
I did a quickie sim (posted earlier) using VoltSec's parafeed caculator;
Rp 800 ohms
Rload 2500 ohms
pri DCR 125 ohms
pri L 250H
Cparafeed 5uf
L plate choke 40H
R plate choke 325 ohmsNo output ringing/peaking
-3db large signal @ 5.8 hz
compare that to the -3db point of 26.5 hz using Tre's series fed example. The parafeed will go two plus octaves lower (for -3db point) than the series fed example.
And even if we're not in need of a low freq power response down to six or so hertz... the advantage is that one, two, three and so on octaves above the six hertz will benefit from lower stress and less work for the output tube... it's a win, win.
my only caveat is that push-pull transformers are designed (the better ones anyway) with the push and pull halves of the primary being mirror images of each other in terms of coil geometry...if you use a PP trans single ended you "upset" the voltage and capacitive gradients that the trans was designed for...
designing a single ended output as a dedicated SE trans... your coil design and the variables of that process can be moreso optimized for this type of operation.
But I'd go parafeed in a heartbeat vis-a-vis using this PP trans in a conventional series fed circuit.
MSL
Builder of MagneQuest & Peerless transformers since 1989
Edits: 04/16/15 04/16/15 04/16/15 04/16/15 04/16/15
It would be interesting and instructive to see the same graphs but with the inductance doubled and tripled....
and see how quickly the ellipses collapse...
L is your friend....
MSL
Builder of MagneQuest & Peerless transformers since 1989
OK...
Green 15hy
Blue 30hy
Red 60hy.
Of course L is your friend... however it doesn't come for free and I find the baggage that comes with obtaining lots of it is often not a good tradeoff.
why are there multiple traces\loops for each color/example?
MSL
Builder of MagneQuest & Peerless transformers since 1989
They were done using a LT spice simulation by Stephie Bench and I have it set to run 20 cycles for it to stabilize. I'm sure if you configured it to run 20 cycles then sample one it would be much "cleaner"
dave
nt
Builder of MagneQuest & Peerless transformers since 1989
But we can accomplish the same by limiting the LF we ask the transformer to handle.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Hi Tre:Even with use of an input filter to limit low freq response... say you use a filter for 50 hertz... you may have dug yourself out of the worse of the misbehaviour but...
none-the-less at frequencies above this... say at 100 hertz or 200 hertz... the larger inductances are going to have narrower ellipsis (i.e., a less reactive load line) than the smaller inductances...
so there is still benefit to having more as opposed to less L.
Go back and read VoltSec's article... doing this a bit from memory.... but one of the things VoltSec showed was that even two decades above the lowest frequency shown (31.25 hz) there were still discernible differences in the loadlines at 3125 hertz due to the lower primary inductances (27H IIRC).
MSL
Builder of MagneQuest & Peerless transformers since 1989
Edits: 04/16/15
I think you missed VS's point.He was running 31.25Hz and 3125Hz at the same time and then filtering out the 31.25Hz at the output to show the IMD at 3125Hz.
If you filter out the 31.25Hz before the OPT then the IMD never happens.
"How does the low frequency load line affect the high frequencies?
As the load line varies, the tube characteristics (gain) will vary. This variation causes intermodulation distortion (IMD). To keep this example reasonable, the grid drive was set to be +/ - 25V at 31.25 Hz and +/ - 25 V at 3125 Hz. This results in a +/ - 50 V waveform at the grid of the 2A3 which is the same as what we had at just 31.25 Hz.
To see the effects of the variation in tube gain I put a buffered filter on the load to strip out the 31.25 Hz.
With 27 H, the 3125 Hz waveform is modulated by the 31.25 Hz signal. The 3125 Hz signal varies from 246.6 V peak to 225.9 V peak or 20.7 V peak to peak modulation"
BTW I use a simple 6db per octave high pass filter at 130Hz at the input of my SET amps. The idea is to keep the tube from ever traversing the nasty LF load line.
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 04/16/15
Very informative thread. Thanks guys.
Tre:
Here's some quick calcs for the quantities you listed;
15H
2500 ohms
31 hz
the phase angle is 40.55 degrees
the -3db large signal (power response) will be at 26.5 hertz
your actual signal losses will be greater than -3db as we've not included any copper losses nor any core losses.
MSL
MSL
Builder of MagneQuest & Peerless transformers since 1989
There seems to be some confusion on what the inductance does. A pure inductance does not consume power, it merely draws it at a different point in time and then releases it back into the load (viewed by the opening of the ellipse). This becomes an issue when the current required by the inductance / load combo puts the source into current clipping / current cutoff. Like any other load line it also becomes problematic when it enters the nonlinear area of the tube curves but in the case of a fixed inductance and decreased load, the total current swing goes down keeping you further away from the non-linear areas.
Putting it another way, I believe the load provided by the inductance to the source is fixed and by decreasing the reflected load you actually allow the source more available current to better deal with the reactive nature of the inductance.
dave
the primary inductance (self inductance) of the output transformer is in parallel with the reflected primary impedance which is the product of the turns ratio squared times the load resistance on the secondary winding.the current consumed by the primary inductance does not contribute to power reaching the secondary. It is a shunt arm.
that's why your power response (large signal response) will be down -3db when the inductive impedance of the primary equals the reflected impedance. One half of the power is diverted from the secondary load.
in the output stage the only generator we have available is the anode of the tube. It must be capable of providing the vector sum of the necessary AC signal current as well as the current consumed by the shunt arm i.e. the inductive reactance.
the exciting current can be calculated from;Iex = Ep divided by (two times pi times F times L)
MSL
Builder of MagneQuest & Peerless transformers since 1989
Edits: 04/16/15 04/16/15 04/16/15 04/16/15
OK... here is the transient behavior for the loadlines above...
as expected, the lightest load shows the most gain and interestingly enough, it also has the lowest distortion. Now for the FFT...
I'm not sure why we have a color shift between the 5K and 16K in the FFT but here are the data points..
Direct Newton iteration for .op point succeeded.
Fourier components of V(out_16k)
DC component:0.000594292
Harmonic Normalized
Number Component
1 1.000e+00
2 1.100e-02
3 1.283e-03
4 1.987e-04
5 3.711e-05
6 6.282e-06
7 2.107e-06
8 3.145e-07
9 6.095e-07
Total Harmonic Distortion: 1.107250%
Fourier components of V(out_5k)
DC component:0.000168459
Harmonic Normalized
Number Component
1 1.000e+00
2 4.851e-02
3 1.028e-02
4 3.092e-03
5 1.113e-03
6 4.449e-04
7 1.920e-04
8 8.767e-05
9 4.107e-05
Total Harmonic Distortion: 4.970236%
Fourier components of V(out_1meg)
DC component:0.0011142
Harmonic Normalized
Number Component
1 1.000e+00
2 3.483e-03
3 2.584e-04
4 2.581e-05
5 4.752e-06
6 5.223e-07
7 7.718e-07
8 6.762e-07
9 5.302e-07
Total Harmonic Distortion: 0.349257%
Date: Wed Apr 15 23:30:12 2015
Total elapsed time: 13.688 seconds.
to keep things in perspective here is the FFT lowering the input on the 16K and 1 meg versions to normalize the fundamental.
finally, dropping the frequency down to 25hz nets the following distortions...
Circuit: * C:\Documents and Settings\Administrator\Desktop\LTspice\distortion 211.asc
Direct Newton iteration for .op point succeeded.
Fourier components of V(out_16k)
DC component:0.00186278
Harmonic Normalized
Number Component
1 1.000e+00
2 3.012e-02
3 4.780e-03
4 1.021e-03
5 2.522e-04
6 6.747e-05
7 1.962e-05
8 5.763e-06
9 1.720e-06
Total Harmonic Distortion: 3.051666%
Fourier components of V(out_5k)
DC component:0.00130548
Harmonic Normalized
Number Component
1 1.000e+00
2 7.082e-02
3 1.922e-02
4 7.692e-03
5 3.789e-03
6 2.130e-03
7 1.315e-03
8 8.702e-04
9 6.071e-04
Total Harmonic Distortion: 7.392847%
Fourier components of V(out_1meg)
DC component:0.00186514
Harmonic Normalized
Number Component
1 1.000e+00
2 2.851e-02
3 4.465e-03
4 9.272e-04
5 2.188e-04
6 5.791e-05
7 1.497e-05
8 4.934e-06
9 1.482e-06
Total Harmonic Distortion: 2.887076%
Date: Wed Apr 15 23:59:57 2015
Total elapsed time: 3.969 seconds.
tnom = 27
temp = 27
method = modified trap
totiter = 109323
traniter = 109319
tranpoints = 25692
accept = 19265
rejected = 6427
trancuriters = 0
matrix size = 33
fillins = 6
solver = Normal
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