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Old 29th June 2007, 06:27 PM   #1
oshifis is offline oshifis  Hungary
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Default Determining optimal OT turns ratio

Hi tube DIY-ers,

I built a push-pull amp with 807s (triode connected) some time ago. Althogh it sounds great, I am thinking if it could be done more optimal. I have a theoretical question:

Is there any way to determine the optimal turns ratio of the output trasformer? I consider optimal which gives the lowest distortion at a given output power in open-loop (NFB disconnected). Or which gives the maximum output power at a fixed (say 3%) distortion, at fixed anode voltage and cathode current.

There must be some method to determine the best turns ratio from the anode voltage/anode current vs. g1 voltage curves.

Here are the design values of my amplifier:

Vb = 425 V anode voltage
Ia = 30 mA cathode current
T = 25 : 1 output transformer turns ratio
Vg1 = -45 V grid bias
Pout = 32 W output power
RL = 8 ohms load

Any suggestions welcome.
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Old 29th June 2007, 07:13 PM   #2
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Go to Steve Bench's website: http://members.aol.com/sbench101/. Study the sections on loadlines.
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Old 29th June 2007, 08:50 PM   #3
Yvesm is offline Yvesm  France
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Hi

Here is my suggestion.
http://www.dissident-audio.com/PP_6L6/DA112.gif
I use it with 2 x 6L6G (not GC) at 400V anode, screen regulated at 270V.
It will be difficult to reach 32 Watts in triode mode, expect 15, at most.
http://www.dissident-audio.com/PP_6L6/Page.html
The 807 IS a 6L6 with a top cap.

Yves
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Old 1st July 2007, 11:58 AM   #4
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hi you appear to be asking a couple of things...

optimal secondary and

optimal primary impedance

that will give the turns ratio, but its actually quite involved,

as you hopefully know, a speaker isn't generally constant but varies anything from say 3 ohms to perhaps 64 ohms and looks like a roller coaster impedance.

so you try to average that out, and make that the secondary.

the thing is, that's ok at one frequency, say 8 ohms, but if your speaker is presenting 4 ohms, or 16 ohms, your load on the primary will be changed, as the turns ratio is fixed.

also, its not a pure resistive load line, but a reactive one.

And your anode impedance is changing, too.

the interaction between the output valve, its transformer, and speaker is a fascinating relationship.

I have an equation for the precise no. of turns for the primary for both set tx's and pp ones, if that's any help.

but you need to know your primary Z first, I haven't quite simplified that, its a little too variable, as you say depending on power, vs distortion

traditionally, set amplifiers were limited by good design guidance that said 5% 2nd harmonic and 1% 3rd harmonic dist.

generally, hi primary Z gives low distotion and low power, low Z gives hi dist and power, RELATIVE to the plate resistance of the tube, and its the designers discretion which to go for, power or linearity or a mix of the 2.

just had a look at your spec. up there, you are running the tube to cool, I would up the current to 50mamps, with self bias and the kathode resistor about 720 ohms.

but not far out.

There are at least 3 resistances in teh output stage.

ac tube impedance

'operating resistance' which is idle volts/idle current, I think that's the same as dc resistance

and load impedance, you can derive relations between them of a sort

actually, I diden't read your post properly, its a pp amp...so ignore the figures for it

but for pp, primary tunrs should be 33.7x sq root of primary Z
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Old 1st July 2007, 12:42 PM   #5
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assuming core area in cm squared = 4x root power out

that relation will hold for m6 silicon steel scrapless IE core.

and no I didn't pick it out of the hat, I have spent days deriving transformer relations for optimal parameters.

some of the info you are after is in the old books, its well understood, but I am only just scratching the surface of that atm, I don't know if I can derive any simpler relations based upon it, but the transformer relations are coming out well.

http://216.239.59.104/search?q=cache...lnk&cd=4&gl=uk
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Old 1st July 2007, 04:52 PM   #6
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http://www.turneraudio.com.au/loadma...p-triodes.html

more good info here
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Old 1st July 2007, 05:18 PM   #7
oshifis is offline oshifis  Hungary
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Thanks for the explanation and useful info (especially the last site has some good practical guidelines). However, I think it is easier to determine the optimal turns ratio experimentally. I have an OT with 20-40-60-80% taps, originally intended for UL. I will play with the various turns ratios, using a dummy load at the secondary. Results will come...
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Old 1st July 2007, 05:37 PM   #8
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The approximate class AB1 minimum PP load for triodes = 5 x Ra,

Where Ra is measured for the working point.

RL for PP class A triodes = ( 2Ea / Ia ) - 4Ra.

where ea and ia are the dc values of the operating point, and ra is the tubes ac resistance, measured at the operating point.

from turner audio

if your dummy load is fixed, then the turns ratio is fixed and you will be looking for the optimal primary load, Z.

for your 807 at about 1k5 ra, that gives 7500 ohms for class AB1, and 22k for class A at your figures.

anywere between those, assuming 8 ohm load, turns ratio between 937 and 2590, sorry that's impedance ratio, turns ratio is between 30 and 51

with my formula for primary turns, that makes about 2919 primary turns, and around 4894 for 22k

make core area 25 cm squared, meaning 5cm x 5cm, or around 2.5 inch each way.
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Old 2nd July 2007, 01:09 PM   #9
oshifis is offline oshifis  Hungary
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Thanks for the suggestion. Based on all the above recommendations, I will set the following values:

Vp = 425 V
Vg2 = 300 V
Vg1 = -40 V (adjustable)
Ip = 60 mA
N = 30 : 1
Npri = 3000
Nsec = 100
A = 5 cm2

I will use two primary windings with the same number of turns. One will drive the anodes, with the center tap tied to 425 V. The other will drive the screen grids, with the center tap tied to 300 V. This way I get true triode mode, whilst keeping the g2 voltage within the safe area.
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Old 2nd July 2007, 01:45 PM   #10
Yvesm is offline Yvesm  France
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Quote:
Originally posted by oshifis
Thanks for the suggestion. Based on all the above recommendations, I will set the following values:

Vp = 425 V
Vg2 = 300 V
Vg1 = -40 V (adjustable)
Ip = 60 mA
N = 30 : 1
Npri = 3000
Nsec = 100
A = 5 cm2

I will use two primary windings with the same number of turns. One will drive the anodes, with the center tap tied to 425 V. The other will drive the screen grids, with the center tap tied to 300 V. This way I get true triode mode, whilst keeping the g2 voltage within the safe area.
Mmmmh !

I fear it won't be possible to roll all that wire on a such small core.
Unless you make a typo, 5cm2 is by far too small for this power.
As shown above, I suggest some 13cm2.
You will need less turns (2200) to acheive both reasonable induction (1.3 Tesla at 25Hz, full power) and a 300H primary inductance.

If you use the same turns number for screen and plate but with a lower screen voltage they will go negative !

Yves.
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