Re: Design of SE amplifier output stages.
- From: Andre Jute <fiultra@xxxxxxxxx>
- Date: Sun, 30 Dec 2007 09:47:10 -0800 (PST)
Patrick Turner <info@xxxxxxxxxxxxxxxxxx> wrote:
The design of the SE output stage and its OPT is a difficult and
the first time around for anyone with a good brain,
let alone for those who struggle because they are befuddled by the mass
they read on the Internet,
and if they and don't even own an osciliscope.
Here are some steps to take to designs all items which may be simpler
than at my website so far....
Patrick's netsite is here:
and a fuller description of what I say, plus illustrations, is here:
0.1 This is the most important step in building an audio chain. Make
sure you understand that the amplifier contributes perhaps 5% of the
quality of your sound and the speakers 90%, leaving 5% for the sources
if you use a turntable and preamp, less than 1% if you use CD without
02 Choose the speaker you want to marry for the rest of your life. It
is responsible for 95-99% per cent of the quality of your sound, so
0.3 Decide how much power your speaker needs. You need to drive it to
somewhere between 90dB (very loud) and 99dB (you will be deaf in a few
weeks if you use this much power -- a hydraulic roadbreaking drill
puts out 96dB at 36 feet). Ignore those who tell you the minimum is
103dB; they're non-kulturny cloth-ears.
0.4 Choose your tube(s) according to the power you need. The smaller
pentodes like the EL84 and EL34 are very sweet and almost impossible
to go wrong with when trioded in something simple like an SE amp. The
300B sounds very good too and is also foolproof unless you really work
hard at messing it up. You choose one size up or one tube more than
the minimum requirement; you will sacrifice the extra power for
freedom from harmonic distortion.
(1). Decide on the tube or parallel tubes to be used.
Eg, 4 x EL34/6CA7.
(2) Decide on the idle dissipation for these tubes.
Eg, run at 22 watts each, in triode, class A1.
You should not run tubes over 80-85% of maximum dissipation, but
neither today you should not run them at those dissipations often as
low as 50% left over from traditional practice. Somewhere between
65-80% is good, and the higher the better for control over distortion.
(3) Decide on Ea and Ia for each, then translate to
the number valid for multiple tubes if used.
Alternatively you can do this visually rather than algebraically;
drawing loadlines on the transfer curves makes what happens inside the
tube so much easier to understand:
It helps if you have this pic on your screen:
3.1 Draw the mazimum dissipation curve you have decided upon in 2
above on the Eb-Ia-Eg curves.
3.2 Notice that the load for maximum power is likely to be so
distorted as to *require* NFB to work right. Instead decide
arbitrarily on a load of *higher* than twice the plate resistance of
the tube (or the parallel equivalent). RL of about 4x Rp is a good
first approximation; at this point you will be sacrificing a
noticeable part of the possible power for freedom from distortion. A
bit lower, say RL = 3x Rp gives you more power but also more
distortion. Much higher, as in the 5K6 I load on 700 Rp 300B (that's
8x Rp) and you're into ultrafidelista territory where you pay half or
more of the realistically achievable power for an ultra-silent ZNFB
amp. See below about free lunches... (Using multiples of Rp is an
approximation but it is plenty close enough; unless you wind your own
transformers, you will anyway have to round the number arrived at when
it is "transformed" from the theoretical RL to the load the
loudspeaker reflects onto the plate via the winding ratio.)
3.3 The tangential junction of this loadline with the maximum
dissipation curve you have drawn (the lines can touch without crossing
at only one place) gives you plate voltage, quiescent current and
negative grid bias.
3.4 Check that you will have enough power. If not, you need to go to a
bigger tube, or more tubes, or draw a steeper loadline and suffer more
Eg, EL34 have pdA = 22 watts each, so 4 have Pda = 88 watts.
The Ea will be +450V, so Ia total for 4 = 88 / 450 = 196mA.
(4) Decide on the RL for maximum PO.
See comments at 3 above.
For all class A1 triodes negatively biased,
RL = ( Ea / Ia ) - 2 Ra.
( Note, if pentode of beam tetrodes are used, RL = 0.9 x Ea / Ia. )
Eg for EL34, from data anode curve sheets ses what is Ra where Ea = 450V
and Ia = 49mA, ( for 1 x EL34 ).
Say it is 1,300 ohms for 1 x EL34, so for 4 it is 1,300/4 = 325 ohms.
RL for 4 x EL34 triodes = ( 450 / 0.196 ) - 2 x 325 = 2,295 - 650 =
This is an answer that depends on maximising power delivery. It trades
off silence for efficiency. For a silent amp you must adjust the
answer upwards to a higher primary load on the plate (RL >>>> 2*Rp),
or use NFB.
(5) Find Va max into RL at anode.
Each triode can have 30% max efficiency, so expect PO max at clipping =
30% of Pda = 0.3 x 88 = 26.4 watts.
You can calculate this on the graph too. If you do it algebraically,
and power is at all a critical element, better to work with 25%
Va = square root of ( PO x RL ) = s.rt ( 26.4 x 1,645 ) = 208 Vrms.
There is no such thing as a free lunch. Nor is there any such thing in
hi-fi as taste-free engineering. It isn't that my approach is more
valid than Patrick's, it is merely different, and better suited to my
taste and engineering inclinations (I want the meaning of every
decision laid bare, not hidden in algebra -- Patrick is happy with the
The rest of Patrick's detailed design steps moves on to transformer
design. I am not qualified to comment. This is where I pick up the
A little inaccuracy sometimes saves tons of explanation. --H.H.Munro
Visit Jute on Amps at http://members.lycos.co.uk/fiultra/
"wonderfully well written and reasoned information
for the tube audio constructor"
John Broskie TubeCAD & GlassWare
"an unbelievably comprehensive web site
containing vital gems of wisdom"
Stuart Perry Hi-Fi News & Record Review
(6) What is the core selection for OPT?
(6a) Transformer weight in Kilograms determined by PO x 0.18 approx at
Larger than required is always perfectly acceptable!!!!!!
Eg, 26.4 x 0.18 = 4.72 Kg.
Core will be 0.8 x total transformer weight approx because wire will be
approx 20% of the weight.
(6b) Decide on transformer core material.
Eg, core weight = 0.8 x 4.7 = 3.76 Kg = 3,760 grams.
(6c) find volume of iron for core.
Density of iron approx = 8 grams per cu.cm.
The wanted iron volume = 470 cu.cm.
(6d) Let us try a core size we have found out is available.
Select tongue width GOSS E&I lams, wasteless pattern.
Eg, Select wasteless 38 tongue material.
This has plan area = 87.6sq.cm, not including windows for wire winding
Therefore height of the stack of laminations = volume wanted / plan area
= 470 / 87.6 = 5.36 cm, = 54 mm.
54mm is above a standard bobbin height allowing 50mm stack, so
proceed to use the next size up for bobbin size above the 54mm, = 62mm,
( 2.5 inches ).
(6e) Will the the selected core give an acceptable aspect ratio,
ie, stack height / tongue width of less than 2:1???
Eg, 65 / 38 = 1.71 :1, and is OK.
(6f) What is the Afe, or core centre leg cross sectional area?
Theoretical Afe = actual calculated stack height x seected tongue width.
Eg, in this case theoretical Afe = 54mm x 38mm = 2,052sq.mm.
Eg, But we have selected stack = 62mm, so actual Afe = 62 x 38 =
(6h) What other alternatives are better?
Any core material with larger tongue width will be OK.
The Afe must be at least the theoretical Afe as calculated above.
Eg, we could have 47mm stack of 44 tongue, so use bobbin 50 x 44 if
or we could have 40mm stack of 51mm tongue so use bobbin 50mm x 50mm,
although with a slight turn increase a 38 stack x 51 tongue is OK and
WILL BE BEST
because the winding window will be largest in relation to Afe, and give
plenty of room for wire layers, insulation layers, and the interleaving
(7) Find primary turns.
For all SE amps, we will want the ac flux density = dc flux density at
and to be = 0.8 Tesla, so that with full PO, Bac = Bdc = 0.8 Tesla = 1.6
and above this level of operation the OPT will saturate.
Bac in Tesla = ( 22.6,constant x Vrms across anode RL x 10,000,constant
all divided by ( Afe in ACTUAL sq.mm. x Np,primary turns x frequency in
Eg 0.8 Tesla
= ( 22.6 x 208Vrms from (5) x 10,000 ) / ( say 2,400 x Np x 17 )
0.8 = 1,152 / Np.
Np = 1,152 / 0.8 = 1,440 turns.
(8) What is the effective permeability, µe?
The iron when fully interleaved E&I may have a maximum µ = 10,000+ with
no dc present.
But when gapped to prevent dc current saturation, the actual µ becomes
because the air gap has vastly increase the
effective iron path magnetic length.
It is called µe, and it determines the primary inductance.
Under the dc conditions Bdc in Tesla
= ( 12.6,constant x µe x Idc in amps x Np) / ( 10,000,constant x iron
magnetic L without gap in mm )
Eg, we have 0.8T = (12.6 x µe x 0.196 x 1,440 ) / ( 10,000 x 230mm [for
38 tongue material] )
= 0.00155 x µe.
µe = 0.8 / 0.00155 = 517.
(9) Is this acceptable?
The µe of a gapped E&I core will vary widely depending on the gap size
dc flow. With closely butted E&I without gap material the µe with
wanted dc flow can be around 100 for GOSS laminations,
and maybe less for lower grade iron. The µe without dc
with closely butted material can be up to around 6 times higher, showing
the dc has a drastically lowering effect on µe when gap is minimal.
So if we say we have µe = 600 for GOSS without dc then there is a chance
that if we gap the core
appropriately the wanted dc flow will produce a µe that gives us enough
The phenomena we see with a core with the wanted dc flow is that with
the µe is low, then it rises as gapping material is placed across the
and the inductance rises to a peak at some value of gap. Once the gap
is increased further the inductance begins to fall.
When the ideal gap for maximal inductance is set up the removal of the
should cause less than 12% rise in inductance.
Therefore the above calculated µe should be below 600.
(10) Assuming the above calculated µe can be realised, calculate Lp.
Lp in Henrys =
( 1.26,constant x Np x Np x Afe x µe ) / ( 1,000,constant x ML of iron
in mm )
and where Np is in thousands of turns.
Eg, Lp = 1.26 x 1.44 x 1.44 x 2,400 x 571 / ( 1,000 x 230 )
(11) What is the reactance of the primary inductance at 17 Hz?
ZL = L x F x 6.28,constant.
Eg, ZL = 15.6 x 17 x 6.28 = 1,665 ohms.
(12) Is this acceptable?
For good SE amp operation we want maximal bandwidth
and the more P turns you have, the lower is the HF cut off point.
But we cannot just increase Np without worrying about the HF cut off
caused by leakage inductance and shunt capacitances due to the
The rules on leakage and capacitances are covered in detail at my
The above calculations have brought us to a minimal Np figure for the
Fitting the turns on and interleaving patterns have to be calculated
from the umpteen
selection/logic steps at my website.
But to be sure, at least we want the ZL = RL at 17 Hz or lower, and
frequency of saturation
also at 17Hz where the anode ac signal voltage is 3dB below the 1 kHz
Eg, We calculated the RL above for thre 4 x EL34 triodes = 1,645 ohms.
ZL from above = 1,665 ohms, which is so close to RL that we can say
the LOAD on the tubes is partially reactive and = 0.707 x RL = 0.7 x
1,645 = 1,163 ohms.
It is impossible to maintian the full mid F clipping level signal across
a load of
0.707 times the mid F load so signs of clipping overloading will occur
at 17 Hz
if one attempts to test with a sine wave of constant level of input and
no means to reduce its
amplitude as F is lowered.
(13) Effects of saturation.
To see the effects of saturation more plainly on an oscilloscope,
we should first find out the mid F ( 1 kHz )level of output voltage at
with RL connected, and record the voltage.
( Using voltmeters only does not reveal what is happening anywhere
clearly enough ).
Then remove RL, and then wind down the frequency applied, and make sure
the couplings in the amp or signal gene
produce no attenuation of signal above 10Hz.
At 17 Hz the level should reduce to be the same as the loaded mid F
You should see some sudden appearance of vertical spikes and some higher
on rising parts of the sine waves at 17Hz or just below as F is adjusted
between 20Hz and 10Hz.
(14) Are the conditions of operation OK as described above?
It is unlikely anyone will use an SE amp for sub-woofer powering where
F range is say 17Hz to 50Hz, -3dB points.
If this is the case, expect that if full power is used, the OPT will
Full range music including sub 50Hz info needs to have enough headroom
for ALL frequencies.
So if you have 100 different F all simultaneously present then their
amplitudes add to
make a total maximum which must be accomodated, and it is is the mid F
So you can have say 14.5Vrms total in a 26.4 watt signal fed to 8 ohms.
Within this signal you'd want to have no more than 7Vrms of any signals
which will leave another 7Vrms for all other F, which tend to have less
amplitude than the bass signals.
If ever there is some large 14V signal at 17Hz, the OPT will saturate
and cause huge IMD of other F.
But with most music, it rarely may occur, and signals below 17Hz of
large amplitude are unlikely
to ever appear.
Therefore the well designed SE amp will NOT misbehave sonically with
full range music
providing there is no clipping at any time, and this is always true of
audiophile usage where they never run anywhere near clipping levels
someone stupidly tries to use one lone EL84/6BQ5 in triode to drive a
insensitive speaker to loud levels.
If you have beam or pentodes used as such, testing without a load must
be done extremely
carefully because the voltages developed at the OPT can swing wildly
unless either NFB is used, or the load is left connected.
I have concentrated on triode behaviour because that's the tube most
folks use for SE amps.
Using all of the above logic and calculation flow I retro designed the
845 amp OPT and 13EI amp and came up with figures almost identical to
I have actually made.
The 845 amps saturate their OPTs at full mid F clipping voltage without
a load at 16 Hz.
Let me know if I have left anything out.
- Design of SE amplifier output stages.
- From: Patrick Turner
- Design of SE amplifier output stages.
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