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19th October 2004, 11:20 PM  #1 
diyAudio Member
Join Date: Sep 2004
Location: Bungawalbyn, NSW

Output transformer application help please.
Hi,
I've salvaged some output transformers from an old 100v line amp with solid state drivers. I've made new bobbins and have rewound the primary and secondary with enamelled wire. I was looking at some transformer calculation pages on the web but didn't like the discrepencies between sites so I took them as a guide and guessed from there on. To be safe I,ve wound 1200 turns of 0.2mm wire for the primary, and to make it versatile for experimentation, I put in ultra linear taps, and further ultra linear taps of the first taps! So the primary has 6002501000100250600 turns (overall DCR=105 ohms) and should be versatile for SET amps using these as required. The secondary has 80 turns of 1mm wire tapped 010204080 turns. Can anyone advise where would be a good place to start for 2500 ohm impedance for a 2A3 SET amp? Any info would be appreciated. 
19th October 2004, 11:35 PM  #2 
diyAudio Member

I'm not sure about winding ratio etc but I think you may have a more fundamental problem!!
I'm guessing that the transformers are not airgapped due to their previous application. This would make them unsuitable for an SET application as the dc current would cause them to saturate. SE transformer design is a bit of an art. You might want to consider using something ready made like a hammond 125ESE which is inexpensive and works well. 
19th October 2004, 11:42 PM  #3 
diyAudio Member
Join Date: Sep 2004
Location: Bungawalbyn, NSW

I totally rebuilt the (previously lapped core) laminations including a sheet of waxed paper for the air gap.
This meant re alligning the E laminations in line as opposed to opposing as they were before. Thanks for the warning anyway. 
20th October 2004, 12:32 AM  #4 
diyAudio Member
Join Date: Jun 2002
Location: Denver, CO

impedance is sq of turns ratio
I don't think you are going to get exactly a 2500 ohm impedance but if I did the calcs right you might get something like either 1800 or 4500 into 8 ohm speaker.
Z= turns ratio^2; so using the full 1200 turns / 80 turns = 15 = turns ratio. Then 15^2= 225 and loaded with an 8 ohm speaker (8*225) you get 1800 Z primary. Tying other taps, I think that I got something like 4500 Z primary for 8 secondary. You can play with the numbers and perhaps different secondary loadsyou'll get something out of the speakerwho knows what? Congratulations for trying this! Hope it works out for youlet us know. This is diy! Rick 
20th October 2004, 12:48 AM  #5 
diyAudio Member
Join Date: Sep 2004
Location: Bungawalbyn, NSW

Thanks for the support!
I will draw up a table and work out the impedance ratios for various arrangements of taps. I calculated about 2.1k using one arrangement and 3k using another but I'm more concerned about letting too much current through the 2a3. Anyway, I'm sure the actual performance will be a long way from what I calculate. Theory and practice are seldom the same... Trial and error is my only realistic option here! 
20th October 2004, 02:47 AM  #6 
diyAudio Member
Join Date: Mar 2004
Location: AB

This is what posted on other thread for 2A3 OPT (Primary: 2.5K; Secondary: 4, 8 & 16 ohms). Hope this will help you...
ok here is one design for output traffo you specified: 1. core is ei 1 1/4 , available from deeco, you will need about 3kgs for set of two outputs.. 2. a standard plastic bobbin for 1 1/4 sqaure core stack is used, this is available from deeco... 3. primary winding will have 5000 turns awg 32 heavy formex split in two sections of 2500 turns each, primary will have estimated dc resistance of about 230 ohms... 4. secondary will have 400 turns awg 22 enamel wire for 16 ohms, 200 turns for 4ohms and 280 turns for 8ohms, if you like, you can also split this into two coils, so that primary and secondary will interleave... the secondary winding will have estimated dc resistance of about 2.2 ohms... 5. adequate insulations must be used between primary and secondary coils.... 6. once the completed traffo is wired up and finished, you can install the ei cores, all the e's go together and there must be a gap between the e and the i to prevent dc saturation, you can use fish paper as a gapping material. 7. completed traffo with brackets installed can be dipped in polyurethane insulating varnish, you can use the air drying type if you do not have oven for this job. 8. if you want to find out the primary inductance, you can use a capacitor of 1ufd in series with the primary coil, and feed a 50volt 60hz voltage taking measurements for voltage, you can determine inductance, bearing in mind that in a series ckt current flowing thru the cap is the same as that on the coils.... 
20th October 2004, 02:07 PM  #7  
diyAudio Member
Join Date: Jul 2004
Location: Ardeche

Re: Output transformer application help please.
Quote:
Quote:
But there are many other factors to consider, specially trade off between core saturation and sufficient primary inductance for good bass response, along with winding interleaving for clean highs Gimme mechanical data of your iron and I'll propose you an "argumented" solution. Yves. 

20th October 2004, 02:53 PM  #8 
diyAudio Member
Join Date: Sep 2004
Location: Bungawalbyn, NSW

Ok,
The overall dimensions of the assembled core is 114mmx95mmx33mm.(4.5"x3.75"x1.3") The bobbin slots are 57mmx18mm.(2.25"x3/4") The laminations are about 0.5mm thick. (0.02") stack 63 The overall centre core area is 1254mm^2 (33x38) Thanks for the advice. 
21st October 2004, 03:50 PM  #9 
diyAudio Member
Join Date: Jul 2004
Location: Ardeche

Hi Bob
Thanks for using metrics ! At first, your core is large enough to allow designing a 10W SE OPT, well over that could be obtained with a 2A3 or even a 300B ! Next, the 0.5mm lams indicates a standard grade iron, not bad but induction will be keep below 1 Tesla (or 10000 Gauss) The permeability is also lower that M6X offers, but since a gap will be added to void saturating iron from DC plate current, it will be lowered any way. Here is a suggestion for a 2500 to 8 ohms project: Primary: 2760 turns of 0.4mm dia wire, divided in 4 sections. Secondary: 156 turns of 1mm dia wire, divided in 3 sections interleaved with the primary. A 0.2mm thick insulation between each sections. A 0.3mm magnetic gap (in fact a 0.15mm paper sheet, since it appears twice in the magnetic path). Thus the winding is organized as follow: P/4+S/3+P/4+S/3+P/4+S/3+P/4 where P/4 is 690 turns, O.4mm dia wire, S/3 is 52 turns, 1mm dia wire, + represents the O.2mm insulation between layers. Put a thin (0.05mm) paper every two primary layers; Such a design should be usable up 10 Watts and down to 20Hz before iron starts to saturates. Hi frequency limit (3dB) above 30Khz. For a 4 ohms load, use only 2 sections from the secondary, tie the remaining one in paralell with any other one to reduce losses. !! WARNING: To do that, insure each secondary section has EXACTLY the same turns number !! Argumentation: (tentative of !) To evaluate if wires will fit in the bobbin, I've reduced the available width to 55mm (instead of 57) and the deepth to 17mm (instead of 18) to allow room for the bobbin material itself. From that, it appears that about 52 turns of 1mm dia. wire will fit neatly in a single layer and thus the secondary could be 3 layers, that is 156 turns. If loaded at 8 Ohms, and if the primary is expected to be 2500 Ohms, the turn ratio is sqr(2500/8)=17.7, so that the primary will need 2760 turns. Now, assume the DC plate current is 0.06A (too hi for a 2A3, but right for a 300b). Assume the iron permeability is around 4000. The DC induction is: µ x N x I / MPL, result in Gauss. where µ is permeability, N is turns number, I is the DC current and MPL the Magnetic Path Len. Here: 4000X2760X0.06/23 = 28800 , obviously too hi. We can either reduce the turns number, but this will reduce the inductance quickly because (see below) it depends on the square of the turns number. We can reduce the permeability by inserting a magnetic gap. A 0.3mm gap lowers the permeability down to 650, the DC induction becomes 4680 Gauss. Right ! The AC induction is U x 100000000 / 4.44 / N / F / A, result in Gauss. where U is the rms voltage (158v on 2500 Ohms for 10 watts) F is the frequency of interest in Hertz and A is the central core area. thus: 158x100000000/4.44/2760/20/12.54 = 5140 Gauss, if we add with the DC, we have 9820, showing that full power at 20 Hz won't run iron into saturation. The primary inductance is: µ x N x N x A / 100000000 / MPL, result in henrys where A is the central core area. So : 600x2760x2760x12.54/100000000/23 = 27 Hy. Good. Some remarks: Total induction must be keep under the iron saturation limit. Some 1 tesla for standard grade, up to 1.4 Tesla for hi quality grain oriented silicon steel. To reduce DC induction, turns number must be reduced. To reduce AC induction, turns number must be increased. AC induction decreases quickly with frequency. To reduce both AC and DC induction, we can add a magnetic gap. Inductance increases as the square of the turns number. Inductance decreases as the gap. Since space for the winding is limited, increasing turns number means reducing wire dia, and thus increasing losses. So long for now ! Tell me if something seems wrong or obscure. Yves. 
21st October 2004, 04:19 PM  #10 
diyAudio Member
Join Date: Nov 2002
Location: sg

super cool....would these figures change much with M6 0.35mm laminates?

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