Single ended EL34 amp design

[bepone]

From transfer curve, it is possible to make other curves via interpolation, in 20% error which allowes for tube difference, and what will be corrected later in the first powering on, but for the start it is ok

On the left side we take transfer curve, and on the right side we need to put blank pentode graph with the same current scaling, here i did simple in paint.
then for dedicated Vgs and anode voltages, we can mark spots onthe right graph where the curve will pass,

All the curves looks the same for dedicated tube, only what makes them different is Vg2.. so Vg2 is the main voltage in the pentode valve, and it is "pulling" set of anode curves up or down..it is enough to have 1 set of Vg2 curves to extract others..

Example Vg2 and Va are 250V.. on the right graph on Va=250V we can have the currents where curves will pass, from transfer curve (left)

el34 transfer.png

[bepone]

small error in result is normal, because transfer and plate curves are made with averaging from many tubes , not from 1 tube

[bepone]

In upper El34 graph is little bit more difficult, because on transfer curve left, we dont have Vg1=0 values.

When we have, it is easier to reconstruct, because we have max current curve, when valve is most open, Vg1=0

Here is example of PCF802 small signal pentode (allready existing for Vg2=150V, but just to show how it is):

PCF802.png

[Stephen1966]

First, a big thank you... I know it takes some time to prepare these charts and explain. It is very much appreciated.

Time is my enemy now, but I hope to return to this again this evening.

[bepone]

no problem.. with this technique you can make any curve for any valve without curve tracer.. ofcourse for pentodes/tetrodes...

you just need to move forward from triodes...when you think that you have active element (pentode) and you can change curves of it via altering Vg2 voltage, this is another world

big advantage for tubes vs transistors.. almost anything is possible with pentodes, distortion, compression, clipping, one side, both sides, high gain, low gain, ... i'm using them a lot

[Stephen1966]

Thinking about the voltages I am leaning towards a higher Va plate voltage for the sake of efficiency. Again, I'm not looking to get more than about 10W out of the amp,but rethinking the voltages I've also had to recalculate the output impedance of the tube and the kind of output transformer that would be suitable.

Plate voltages of 390V along with a cathode voltage of 26V ask for a raw B+ of 416V. Dividing this by the DC power scaling factor of 1.3 we arrive at a transformer that can provide somewhere in the region of 320-0-320 (the CT will be useful for the designing the bridge rectifier.

One transformer that fits this scenario is the Tube Town FenderR Style 5E3 High Power Transformer: https://www.tube-town.net/ttstore/tt-po ... ower.html

With 390V on the plates I now recalculate the output impedance of the tube according to:

Zout = Va^2/Pa

When Pa is the maximum power output of the EL34 = 25W

Zout = 390^2/25 = 6084 Ohms

Most single ended 15/20W OTs offer primary impedance of 3/3.5k or with universal taps 2.5/5/10k.

I want the primary impedance (Zpri) to be slightly higher than the output impedance of the tube (Zout) to offer maximum power with a minimum of THD so to get this, I could take an output transformer rated at 3.5k, like the 300B: https://primarywindings.com/product/300b_20w_se/ and wire it so the reflected impedance is 7k.

There are formulas for finding the impedance ratio of transformers, like this: http://www.maxmcarter.com/classecalcs/tratiocalc.html This gives us results in terms of the turns ratio for primary and secondary coils but the formula holds good for understanding the reflected impedance of, for example, what happens to the primary impedance when tap an 8 ohm speaker off a 4/8/16 ohms secondary.

The turns ratio is the square root of the impedance ratio,

i.e., Npri/Nsec = √(Zpri/Zsec)

or Zpri/Zsec = (Npri/Nsec)^2

to be continued...

[Stephen1966]

The formula above is not strictly necessary for our purposes because we already know the impedance ratio from the published datasheets but it can be useful when we have an unidentified transformer and we want to work out its primary and secondary taps, or when we have an unusual transformer with an odd number of turns for each tap. For example, we could take an unidentified transformer and apply a voltage across one of the coils - if the voltage reading on the opposite coil is lower, it is stepping down and you are applying the voltage to the primary coil. Vice versa if you apply the voltage to the opposite coil first, the voltage reading will be higher, you are applying voltage to the secondary side and the voltage is stepping up. BEWARE! If you apply 10V to the secondary side and the turns ratio is 20:1 you could be in line for a visit to Accident and Emergency if you touch the primary coil's wires! Don't count count on the inverse current not to kill you! If you know your voltage in (Vin) and voltage out (Vout) you can also find its inverse the current ratio Iout/Iin and you have:

Vin / Vout = Iout / Iin

Giving you the turns ratio:

N:1 = Vin / Vout = Iout / Iin

If you then square the turns ratio, you get the impedance ratio:

Z:1 = (Vin / Vout)^2

Using the impedance ratio as a working datapoint is a good way to confirm the obvious.

Problem
What is the reflected impedance (Z) when an 4 or 8 or 16 Ohm speaker load (L) is connected to the 4, 8 and 16 Ohm secondary taps (Zsec) of an output transformer with a primary impedance of 3500k (Zpri)?

First we find the impedance ratio for each of the secondary taps, using Zpri/Zsec when Zpri = 3500k

i) Z = 3500/4 = 875; (4 Ohm tap)
ii) Z = 3500/8 = 437.5; (8 Ohm tap)
iii) Z = 3500/16 = 218.75; (16 Ohm tap)

It's important to note that it's not as if the secondary coil expects to see anything; it is simply what it has been designed to apply when the manufacturer optimized the windings for production and operation. There will be losses, in efficiency, in conductivity, you name it when we step outside those parmaters, but we can apply any load on the secondaries and they will reflect back onto the primary winding in a - more or less - consistent manner. Cue the impedance ratio!

When L= 4 Ohm, and Z = reflected impedance, we get:

Z = 4 x 875 = 3500 Ohm
Z = 4 x 437.5 = 1750 Ohm
Z = 4 x 218.75 = 875 Ohm

When L = 8 Ohm, and Z = reflected impedance, we get:

Z = 8 x 875 = 7000 Ohm
Z = 8 x 437.5 = 3500 Ohm
Z = 8 x 218.75 = 1750 Ohm

When L = 16 Ohm, and Z = reflected impedance, we get:

Z = 16 x 875 = 14000
Z = 16 x 437.5 = 7000
Z = 16 x 218.75 = 3500

To get us to the reflected impedance of 7k for our tube we can see there are a couple of possibilities here. In order of options available and usefulness though, with least first, we can see that a 4 Ohm speaker doesn't work for us. The maximum reflected impedance of the 4 Ohm load on the 4 Ohm tap is just 3500k and it gets smaller as we plug it into higher impedance taps. Already though, you can see the pattern. For each step up you double, the reflected impedance is halved. We might leave it there; it isn't important that we devise an extended formulaic proof with a known transformer such as this. But let's take it further anyway...

The two useful values of an 8 Ohm load are when we plug it into the 4 Ohm tap where Z = 7k and the 8 Ohm tap where Z = 3k5. Likewise, we could plug a 16 Ohm load into the 8 Ohm tap giving us the 7k reflected impedance or into the 16 Ohm tap giving us the rated 3k5 primary impedance.

Here's an article I googled lazily: https://musicstrive.com/8-vs-16-ohm-speakers/ Perhaps the leading piece of advice they should have headlined better is

Your amp should never be set to a higher ohm resistance than your speakers – this can quite literally blow your speakers!

Not just the speakers though because as we will see, this would probably fry your tubes as well. It would ruin your day!

They go on to talk about 16 Ohm speakers as sounding brighter - you may agree or not with that - but as amp builders we know it isn't as if the speaker is the whole equation here, the circuit before it matters and when you plan to use an EL34 as I do, you temper your expectations with the possibilities it typically affords. The tube isn't known for its crystalline highs. For that, you might go over to a KT77 or a KT88. That article glosses over the fact that speakers are reactive loads as well... Really the three situations that we avoid are a) using any of the 4 Ohm taps with an 8 or 16 Ohm load, b) using the 16 Ohm tap with an 8 Ohm speaker load, and c) using the 4 Ohm tap with a 16 Ohm load. A quick glance over at the Celestion site for 10 inch speakers tells us that 8 Ohm speakers outnumber 16 Ohm speakers by 6:5 and 12 inch speakers by 28:23 - there is just one 12 inch 4 Ohm product. This isn't exhaustive by any means then, but I think it's fair to say 8 Ohm speakers offer more in the way of choice. It seems six of one and half-dozen of the other but perhaps in choosing either the 8 Ohm or the 16 Ohm speaker, it is about which would work in the most efficient way with the transformer.

Which would you choose? It would seem safer to outmatch the amp with a 16 Ohm load tapped into 8 Ohm and 16 Ohm secondaries, but does this also lead to better frequency response and fewer losses than an 8 Ohm speaker tapped into 4 Ohm and 8 Ohm secondaries?

[Stephen1966]

Before I go any further there is a datasheet that gives a mutual characteristics graph that goes as far as Vg1 = 0V

Brimar EL34.pdf

I'm totally into the Old School method of drawing load lines now though so here is my clean copy of the Mullard charts:

EL34 loadlines.pdf

[Colossal]

Nice work, Stephen. I've been enjoying this thread.

[bepone]

Plate voltages of 390V along with a cathode voltage of 26V ask for a raw B+ of 416V.

i recommend use of well known and explored operational point from datasheet.. so 300-350V on anodes.. also Z load is then lower, 2-3k, means primary of the transformer is standard and you can find that easily..

using higher loads (5-7k or more) means for the same performance of OT you need to have 1.41x turns more, means bigger and more expensive transformer! and ofcourse if you go for cheap version, you will have other problems.. they need to use thin wire and for result you will have much more compression and losses in transformer)..

lower load is always better, easier to make good transformer (primary inductance , lower number of turns)..

[Stephen1966]

Nice work, Stephen. I've been enjoying this thread.

Thank you. I'm learning a lot... If it can help others as well - Mission Accomplished.

[Stephen1966]

Plate voltages of 390V along with a cathode voltage of 26V ask for a raw B+ of 416V.

i recommend use of well known and explored operational point from datasheet.. so 300-350V on anodes.. also Z load is then lower, 2-3k, means primary of the transformer is standard and you can find that easily..

using higher loads (5-7k or more) means for the same performance of OT you need to have 1.41x turns more, means bigger and more expensive transformer! and ofcourse if you go for cheap version, you will have other problems.. they need to use thin wire and for result you will have much more compression and losses in transformer)..

lower load is always better, easier to make good transformer (primary inductance , lower number of turns)..

Lower load is always better

Thanks, I wasn't sure about that. Makes perfect sense though when I consider the winding has both parallel and series resistance and parasitic capacitance and leakage inductance. All factors that stand in the way of an "unfiltered" AC signal.

In terms of the trade-off though, it isn't really about increasing the gain or really, about increasing the amount of current at the operating point. It may end up being down to a choice between the readily available components. I can run preamp tubes at around 200V quite effectively but can I get a PT that will provide the lower voltage without looking at some kind of dropper for the power tube? I like the PT I linked above, because it has enough Amps on tap for the heater supply - which after power factor losses in a DC regulated supply may require twice as many Amps on tap than for a regular AC supply. I might end up spending more on a good OT and I'm up for that, but if it also means I don't need to buy an extra filament transformer, I'm possibly saving weight, and money.

A reflected impedance of 7k is possible with that OT and it looks better quality compared to the Hammond for instance. I think for safety's sake it is better to forget something like alternative jacks for 8 Ohm and 16 Ohm for instance. Running an 8 Ohm speaker into the 4 Ohm tap and taping the others off seems to be what you are suggesting. The lesser of two evils

[Stephen1966]

i recommend use of well known and explored operational point from datasheet.. so 300-350V on anodes.. also Z load is then lower, 2-3k, means primary of the transformer is standard and you can find that easily..

Thanks again. I'm looking into this now but there are now so many numbers floating around, it's starting to get out of hand. Time for a bit of Excel wizardry

[Stephen1966]

Taking Va = 350V as my quiescent voltage I've drawn up the loadline for Zpri = 5k. 5k was arrived at by first calculating the Zout of the tube:

Zout = Va^2 / Pout

When Pout = 25W

Zout = (350)^2 / 25 = 4900 Ohms

For a lower THD I have chosen the operating point based on an OT with a Zpri of 5k

EL34 - Va 350 - Zpri 5000.jpg

(Q) the operating point

The blue line sets the gradient and I simply transposed this up to the red line with (Q) operating at 90% of maximum dissipation = 70mA x 0.9 = 63mA. I was able to confirm this below to within +/-4V and +/- 4mA

(B) the theoretical maximum negative Va swing

When Q, Ia = 63mA, (B) is equal to 2 x Q or

B, Ia = 2 x 63 = 126mA with Va at 38V marked on the chart.

Deducting B, Va from Q, Va gives the theoretical maximum negative voltage swing: -312Vpk

(D) the theoretical maximum Va swing

When Q, Va = 350V and B, Va = 312V, the sum of these gives us point (D) on the chart: the theoretical maximum positive Va swing as the tube approaches 0mA, cutoff.

350 + 312 = 662Vpk

If we weren't able to easily transpose the load line from the initial blue gradient, this would give us our peak Va. As it is, the transposition of the load line in a graphics app. makes the calculation moot but it at least confirms we are on the right path at (more or less) the right maximum Va swing.

(A) Ia, when Va = 0V

Again, we've already found this by transposing the gradient but if you are working with pencil and paper, to find this we only need to draw a line through points BQD. If they don't line up, we need to figure out what went wrong and start again!

(C) the peak Vg1 swing

Assuming Vg1's swings around the operating point are linear and equal, point (B) is almost directly on the -12V curve and (Q) sits easily on the -26V curve which I've interpolated here in magenta between the -24V and -28V curves

To find point (C) we sum the Vg1 of (B) and Vg1 of (Q):

C = -12 -26 = -38V

I have interpolated another line corresponding (roughly) to Vg1 = -38V and plotted point (C) at the intersection with the load line.

We now have all the points ABQCD and this part of constructing the load line is complete.

[Stephen1966]

I used the EL34 with Vg2 = 360V chart because I wanted to be able to calculate the power output and the second harmonic with a reasonable degree of accuracy; the other anode characteristics chart with Vg2 = 250V can be used for working out the screen voltage parameters. To help confirm my findings, I also used the 'Universal loadline calculator' Martin linked earlier: https://www.vtadiy.com/loadline-calcula ... alculator/. This was also good for playing around with things like Va, different operating points, Zpri, screen voltages and +/-V headroom. This also provides information on the output power and the THD. All useful, when it came to checking my calculations. Namely in...

Power output, W(rms)

First, the theoretical maximum load value of Vpk between points BQ when Ia = Q:

Vpk = 312V
Ia = 63mA

Rpk = 312/0.063 = 4952.4 Ohms

Which I am rounding up to 5k and corresponds nearly exactly with the Zpri of the output transformer - our load.

Next the peak to peak difference between points BC when B = 38V and C = 600V

600 - 38 = 562 Vpp

V(rms) = Vpp / (2 x √2) = 562 / 2.83 = 198.59 V(rms)

Squaring that and dividing by Rpk gives us the power output in W(rms)

198.59^2 / 5000 = 7.89 W(rms)

This is a little less than the value given by the calculator for the same load, operating point and Va (10.93W) but these are theoretical values and the only real way to be sure of the actual power output is to run the circuit physically using a known resistance value as a dummy load with a measured Vrms at different frequencies and watching for clipping on the scope. Higher frequencies with lower voltages and higher output with lower voltages and vice versa. Around 8-11W here, agrees with Merlin's original figures and as working average it is good enough.