reverb driver transformer primary impedance question.

[pula58]

Let's say I take a standard fender blackface/silverface reverb driver circuit. Both plates of a 12AT7 drive through the little reverb driver transformer whose primary Z (with an 8 ohm load at secondary) is 20k - 25k.

Now, If I add a center tap to that transformer and drive each end of the primary with the (now separated) plates of the 12AT7 (to drive it push/pull) should the pimary impedance of the tranny be the same, or, should it be doubled to have the same signal transfer from the 12AT7 tube to the reverb tank?? Quadrupled?


Thanks!

P.

[John_P_WI]

I built a p-p reverb circuit 15+ years ago using a Hammond transformer (125C ?) with the multi-taps and think I configured it to around 25k p-p. Sorry I don't remember the specifics but is was based on KOC's push pull reverb circuit in one of his TUT's. Worked as expected.

A quick search on Powerscaling.com came up with this answer from Daviddeakin:

The optimum load for max output power is equal to the plate resistance, which is about 10k for a 12AT7. So you would initially look for a 20kplate-to-plate OT for push-pull, or 5k for single ended.
However, triodes are very forgiving of loading, and many designs use much higher load impedances than this (and sometimes lower too).

Have fun.

[martin manning]

I have to disagree with the above from Daviddeakin on one point: If you wanted each side of a p-p pair to work at 10k, then the a-a load would be 40k. If you look at the Fender single-ended set-up with a nominal 22k load for two triodes in parallel, that is like 44k to each. To run that same load in p-p would require an a-a load of 176k.

[pula58]

I have to disagree with the above from Daviddeakin on one point: If you wanted each side of a p-p pair to work at 10k, then the a-a load would be 40k. If you look at the Fender single-ended set-up with a nominal 22k load for two triodes in parallel, that is like 44k to each. To run that same load in p-p would require an a-a load of 176k.

your reasoning makes sense to me. if the Fender circuit has two plates in parallel driving a 25k/20K reverb driver tranny, then, each plate separately would be twice the impedance, your 44k. But, if the two primary taps of the pp reverb tranny had z=44k, the load each plate sees whould be 1/4 that, 11k. That's where you get that the total p-p Z should be 4 * 44k = 176k. Right?

Anyone else want to reverberate on this?

[John_P_WI]

Guys,

I just talked to a coworker who is a fantastic analog EE. He recently built a combo unit - his own design - and used a 12ax7 to 8 ohm pp transformer. As he recalled the primary was 22k5 to 8 secondary and has full drive potential to the reverb tank through the entire bandwidth. See here:

http://www.ebay.com/itm/OT5PP-Tube-Outp ... 0548924482

After full analysis and scoping the unit out he explained to me that the tank had a reflected impedance of 8 ohm @ 1khz, but 0.8 r @ 100 hz and consequently 80 ohm @ 10,000 hz. The 22k5 ohm primary p-p transformer worked well. I played the amp and can attest to it.

As a side bar, one could use the 12a_7 to 8 ohm p-p circuits such as the "Firefly" as precedence to the required impedance.

As Daviddeakin referred to which I referenced, triodes in this application are very forgiving.

John

[pula58]

anyone else want to chime-in on this?

[Firestorm]

I can't say anything terribly intelligent here, but I'm interested in doing the same thing without having to spring for KOC's custom tranny. It seems to me, tho, that you wouldn't want to design for full bandwidth: the input is high-pass and the Fender circuits historically use higher and higher high-pass filters to keep "stage rumble" out of the reverb mixer. So I think that implies erring on the high-Z side.

[John_P_WI]

I can't say anything terribly intelligent here, but I'm interested in doing the same thing without having to spring for KOC's custom tranny.

Firestorm, no doubt the custom tranny is first rate, although pricey. If you reference some of the TUT designs, there is a 33k resistor between the p-p plates to limit drive. This is not necessary, especially if using a Hammond 125C.

[Dingleberry]

This thread solved my problem what to do with an old OT from late 40's tube radio with one EL41 power tube in it.
I gutted the radio some time ago and used the PT for a 6V6 SE amp.
I measured the OT and found out that it had fairly high primary Z (2k5:1ohm) so it has been sitting on the shelf for a year and a half now.
Reading this thread I remembered that OT and realized: 2k5:1ohm times eight...
That makes it 20k:8ohm!

It will probably be a good candidate for parallel 12AT7 driven reverb.
It's bigger in size comparing those tiny reverb transformers found in Fender amps, but when you build from scratch and from recycled iron (as I normally do ) bigger size is just a another challenge among others.

Let's re-verb!

-T

[martin manning]

DB did you measure dc resistance or AC voltages to determine the impedance ratio?

[Dingleberry]

DB did you measure dc resistance or AC voltages to determine the impedance ratio?

Is it even possible to to determine the impedance ratio just measuring dc-resistance? I used AC voltages aka. "signal generator and fluke" -method.

-T

[martin manning]

Just checking... You quoted secondary at 1 ohm, which is about what the DCR would be.

[Dingleberry]

Just checking... You quoted secondary at 1 ohm, which is about what the DCR would be.

Ah, that confused you. I just usually write down ratios like that. Something:1ohm.
-T

[pula58]

I guess this comes down to basics, and here is how I am thinking about it. The considerations here are the same as when calculating what type of turns ratio you need in any output transformer. Something which I am not 100% confident at in terms of how to make correct calculations to come-up with a new design, not re-cyling old ones (not that there is anything wrong with that).And, by the way, thanks for all the replies everyone!

In the Fender ab763 type of reverb circuit, the cathodes of the reverb driver tube sit at about 8V (with the 2.2K cathode resistor, 450V plate V). So, if I look at the grid=-8V curve, the inverse of the slope of the plate current vs. plate voltage curve is about 58k ohms. But in the Fender circuit the two plate are in-parallel, so, the effective plate resistance is then half that, or, 29K. So, the Fender 25K:8 ohm transformer makes sense (to me), it's in the right ballpark.

Then, to get the same amount of power transfer with a push-pull 12AT7 driver, it seems to me that the primary-to-center tap Z should be the value of the plate slope-resistance of one triode: 58K. Then, the total primary-primary Z should be 4 x 58k = 232k. So, the center tapped reverb driver p-p impedance should be 232K: 8 ohms to get the equivalent drive as the Fender circuit.

Am I cracking up here, or is this, strictly speaking, the correct way to think about it?

[martin manning]

Yes. I mean yes that is the correct way of thinking about it ;^)