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PP output transformer for an SE 845
PP output transformer for an SE 845
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Old 18th June 2019, 06:21 PM   #1
kward is offline kward  United States
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Default PP output transformer for an SE 845

Hello All,

I inherited some large EI type output transformers that were used in an SE 845 amp. I've examined and measured one of the OPTs and find that it is actually a PP output transformer. Brand is Northlake Engineering Inc. Stamp marks on the side of the transformer are A139-103, 454216, 17227297.

The transformer has three leads on the primary and 4 leads on the secondary. Primary: Blue, Red, Blue/White; Secondary: Black, Yellow, Orange, Purple.

DCR on primary measures as:
Red - Blue: 68Ω DCR
Red - Blue/White: 70Ω DCR

DCR on secondary measures as:
Black - Yellow: 1Ω DCR
Black - Orange: 0.8Ω DCR
Black - Purple: 0.5Ω DCR

Impedance measures as:
54.1V AC (60 Hz) applied between Blue and Blue/White leads produces the following voltages on the secondary:
Black-Yellow: 3.34V AC
Black-Orange: 2.23V AC
Black-Purple: 1.68V AC

Also, the AC voltage measured between Red and blue primary leads (when AC input is hooked between blue and blue/white leads) is 27.05V, confirming that the red lead is tapped half way on the primary.

The secondary voltages are approximately a factor of root(2) apart, so I assume:
Yellow = 16Ω
Orange = 8Ω
Purple = 4Ω

This implies a primary impedance of approximately 4.2KΩ.

These look and act like big PP output transformers, that guess is supported by the red primary lead being mid position on the primary, and the fact that it's red in color, where the end leads are blue and blue/white.

The lamination stack is huge on these: 2" width, 4.5" height, 3.75" depth. They look like they could support 80 to 100 Watts of PP power. Size-wise they are at least as big, perhaps slightly bigger, than Dynaco Mk III (A431) output iron, if that helps for the visual on how big they are.

I don't know if this will work, but if I load the 4 ohm tap (purple) with an 8 ohm speaker, it'll reflect approximately 8.4KΩ to the primary. Load line analysis of the 845 tube with 8.4K primary says with 800V B+ I can get 25 watts output, if I drive them +40V into A2.

The real questions right now are:
1. Can I really use these PP OPTs in an SE amp? (certainly the previous owner thought so)
2. Can I load the "4 ohm" tap with an 8 ohm load? What negative effects will that have on bandwidth and power?

Picture (as best I could get) of the actual unit is attached.
Attached Images
File Type: jpg 845 SE OPT.jpg (67.0 KB, 299 views)

Last edited by kward; 18th June 2019 at 06:23 PM.
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Old 19th June 2019, 12:31 AM   #2
smoking-amp is online now smoking-amp  United States
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Was this 845 amplifier a completed and working SE amplifier? Is there a big inductor on the chassis connected to this OT by a large cap? (connected to one end of the primary, and B+ connected to the other end of the primary) In which case it is probably operating as a Parafeed arrangement, using possibly the entire P-P primary as the SE primary. (so no DC in the OT)
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Old 19th June 2019, 01:46 AM   #3
kward is offline kward  United States
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The amps haven't been powered in probably 15 years. But they are supposedly fully in tact and working, although I haven't dared power them up yet since they came tubeless and I wasn't willing just yet to shell out the bucks for the 845 tubes to try them. They are BIG BIG mono blocks (probably weigh 75 to 100 lbs each) with three channels per block (meaning three 845's per block and three output transformers per block, all fed off of one rather monster power transformer). Apparently they were used in a vertical bi-amped scenario, where two of the three channels per block are paralleled at the output transformer secondaries to power the woofers and the third channel feeds the tweeter. This is per side. The other side is identical for the other monoblock.

But anyway, The power supply for the three output stages is diode bridge rectified, the pos terminal of the bridge goes into a choke for a choke input power supply, and as near as I can tell, the negative lead of the bridge goes into a series connected triplet of capacitors, lower neg lead eventually going to ground. It almost looks like some sort of doubler configuration.

It is definitely using the entire PP primary as the SE primary. I can't quite tell yet if there would be Quiescent current flowing at tube idle. The fact that there is a fairly large capacitor between negative lead of the bridge and ground suggests this may be true. This part of the power supply feeds only the output stage. There is a separate diode rectified power supply feeding the frontend tubes from a different high voltage secondary winding.

In any case, I'm not really excited to use the amps as is, since I don't think I have any application for these monoblocks in current form. For one thing they're just too big. I was more interested in salvaging the parts and making a 20 watt two channel 845 SE amp. That's when I ran into the OPT issue discussed in the opening post.

I could probably power them up at a low mains voltage, say 25V AC or something, and install a big power resistor between plate and cathode of one of the 845 sockets and see if it draws current. I also might be able to reverse trace the power supply more carefully now that I know I might be dealing with a parafeed design.

If it is a parafeed design, it makes more sense to use these PP OPTs in SE form, but still it bothers me a bit that the primary is supposedly 4.2KΩ, as that seems way low for an 845 SE amp. I was thinking 10K would be more appropriate.

Any suggestions for next steps to try to decode this?

Last edited by kward; 19th June 2019 at 01:57 AM.
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Old 19th June 2019, 02:42 AM   #4
kward is offline kward  United States
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It's not a parafeed.

Here's the schematic of one of the woofer channels, as best I understand it at this point.
Attached Images
File Type: jpg Reverse Engineered Schematic DRAFT.jpg (445.0 KB, 259 views)

Last edited by kward; 19th June 2019 at 02:47 AM.
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Old 19th June 2019, 03:21 AM   #5
cerrem is offline cerrem  United States
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Based on the photo..... The E-I laminations look to be a 2" stack of M6 29 gauge 150 size...with a 1:1 alternate interleaving...this would indicate that it can not tolerate any significant DC offset.... Therefore I would conclude it has a LC ...ie capacitor DC blocking.. unless I am looking at the PT in that picture ???
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Old 19th June 2019, 03:22 AM   #6
smoking-amp is online now smoking-amp  United States
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Hmmmm, sure looks like that would need an air gap in the OT. There are special I laminations that have a center/side cutout gap that can be used to make an air gapped E-I xfmr. with close packing (no visible gap on the outside of the OT) But one would expect those to be stacked with the I laminations all on one side. Your xfmr appears to have interleaved laminations. ?????

The only way to tell if there is a gap would be to measure the primary inductance at some reasonable AC excitation. Typical L meters use a very small test signal where lamination permeability would be low. I would put 24 VAC 60 Hz across the primary and measure the AC current. Can calc. the inductance from that. (be sure to use the primary for this test, AC on the speaker windings could generate some dangerous high voltages on the primary side) A non-gapped OT will show hundreds of Henries of L primary, while a gapped one will likely be in the tens of Henries.

Another (distant) possibility would be the use of a DC current source across one of the secondary windings to neutralize the DC primary cu;rrent in the OT. Probably some well-heatsinked power transistor used for that and a LV DC power supply for it.

Last edited by smoking-amp; 19th June 2019 at 03:35 AM.
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Old 19th June 2019, 03:35 AM   #7
kward is offline kward  United States
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Smoking amp, how do I do this? Can I put say a 1 ohm resistor in series with the primary to measure current? I assume I need some sort of resistor load on the secondary for current to flow on the primary? Or should the secondary be unloaded? Once I measure current what math do I use to get primary inductance?
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Old 19th June 2019, 03:41 AM   #8
kward is offline kward  United States
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Quote:
Originally Posted by cerrem View Post
unless I am looking at the PT in that picture ???
It's one of the output transformers.
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Old 19th June 2019, 03:44 AM   #9
smoking-amp is online now smoking-amp  United States
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A one Ohm series resistor should work fine to measure the AC current. Leave the secondaries un-connected.


XL = Vac / Iac
XL = 2 pi f L
so L = ( Vac / Iac ) / ( 2 pi 60 )

If the Vdrop on the 1 Ohm resistor is too small to read accurately (DVM), use a higher sense resistor. With a much higher sense resistor you could also read the Vac directly across the primary to eliminate the Vdrop factor due to the sense resistor.

Last edited by smoking-amp; 19th June 2019 at 03:51 AM.
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Old 19th June 2019, 04:09 AM   #10
kward is offline kward  United States
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Right. Thanks. I'll try this...tomorrow. Thank you.
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