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Old 14th September 2005, 02:49 PM   #1
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Default Help - tube driver for spring reverb...

Would someone please help me understand a tube driver circuit for an Accutronics spring reverb?

peace,
memphissound <><
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Old 14th September 2005, 06:01 PM   #2
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Traditionally, it is just a simple driver stage into a single ended outout stage with recovery/mixer stage after it. I built one for a vox clone that was 2 cascaded 12ax7 gain stages into a 6V6 single ended stage to supply the power and current to the tank, with a hybrid 6922/2sk170 recovery stage...kinda like a modernized 6g15 standalone unit. you don't need extremely high output to drive the tank...only 3w or so. Fender used a parallelled 12at7 as an output tube for some tanks. Check out the bassmans for that one. Those accutronics tanks have an 8ohm input.
I have even seen some bargain bin circuits that are simpler than that.
You can think of the reverb driver circuit as being a fender champ running inside a larger amp circuit. I even used an 8ohm speaker to test mine out. I have also used audio opamp chips in current drive mode to feed a reverb tank too... no transfeomer there, BUT you have to use a tank with a higher input impedance... like 1.5K or so to be safe.
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Old 14th September 2005, 07:26 PM   #3
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Default further explaination...

I know that it's essentially a single ended amplifier with a tranny. It has been recommended by Accutronics to use a 5kohm:8ohm tranny, but how do I size it? How do I determine the tube driver characteristics? The specs for it, again from Accutronics, say that the core needs to operate near saturation, ~3.5A-T peak. But then they also say it needs a driver current of 28mA.

I have the RCA "Red Book" and the "TAB" by Aspen Pittman. I understand the basics of the amp design. But not quite how to calculate/understand the drive parameters for the reverb.

I could just copy some of the circuits from the Pittman book, but I really want to UNDERSTAND it.

peace,
memphissound <><
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Old 14th September 2005, 08:47 PM   #4
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I see, you want to get down and dirty.
I picture the transducers like one little speaker and one little microphone...which are basically the same thing, just backward. One jiggles the springs when current is passed through it-- like a speaker, while the other picks up the vibration-- like a microphone, and converts it back injto the signal chain, where it is re-amplified. So design the driver like a single ended amp that drives a speaker that has the same specs as the input transducer of the tank.


Here is a little article on theory, but from a solid state point of view, and an article on single ended amp design for triodes (could be used for the 12at7 driver type) and a fun little calculator script.
http://members.tripod.com/~roymal/reverb.htm
http://www.valveheart.com/theory/triodeSE.html
http://www.sengpielaudio.com/calculator-ohm.htm

Hope that is helpful
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Old 15th September 2005, 12:15 AM   #5
Enzo is offline Enzo  United States
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And really, don't obsess on the details of the circuit. A single ended small power tube - 6V6 or EL84 for example - drives a small output transformer. The output of that goes to the reverb transducer rather than a speaker. ANy small output transformer capable of a couple watts will work fine.

But the easiest thing to do is simply snag a Fender reverb transformer. Why reinvent the wheel? They are readily available from many places.

For example, look up stock number P-TF22921 at www.tubesandmore.com

In the Accutronics line, the part number also describes the pan. The second character in the number is input impedance. For the tu8be drive you want the low impedance input signified by the letter A. SO a pan like the 4AB2C1B or 9AB... or 8AB... would be appropriate.

The ones with higher impedance inputs have letters like E or F. So a 4EB2C1B would be a pan similar to the first but with high imopedance drive needs. These would be used with an op amp drive for example.
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Old 15th September 2005, 02:23 AM   #6
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Yeah, if you can design a single ended circuit, you can design a reverb circuit. You can get fancy and put pre and post volumes and a little tone circuit in there too if you want. You can even loewr the output a bit more than normal so that you don't overdrive the transducer..unless you like that sound of course.
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Old 15th September 2005, 05:52 PM   #7
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Default thanks

Thanks for the help. Between you guys and Scott at Accutronics/Sound Enhancements I think I'm getting it. See if this is right...

Accut suggests using the reverb pan 4AB3C1B, meaning:
4=pan length/style; A=8ohm input Z; B=2250ohm output Z; 3=decay time; C=connector type; 1=no locking device; B=mounting orientation.

To determine the drive current at 1KHz you would take the
3.5AmpTurns (needed for near saturation) and divided by the number of coil turns. The 8ohm "A" input coil number of turns is 124 so 3.5/124 equals ~28mA drive current.

So the current output on the secondary of the output x-former needs to be 28mA. Right?

Accut suggest using a 5k:8 ohm out put tranny (8watt rating).
Considering this x-former:
Z ratio 5k:8 or 625:1; resulting N ratio 25:1
and I2/I1=N1/N2 so 28mA/I1=25/1; I1=1.12mA
Therefore, I will need a Plate current output of 1.12mA.

Right? If I need more schooling on this let me know please. Thanks again.

peace,
memphissound <><
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Old 15th September 2005, 07:36 PM   #8
cerrem is offline cerrem  United States
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OK...Lets look at this in a structured method...

AT 1kHz you have 8 ohm load looking into the tank....
They say drive it "close" to sat. @ 28mA AC signal current...

OK...SO that is a 225mV AC drive signal....

You have a OPT that is 5K to 8 ohms..... That is a turns ratio of 25:1 ......

Therfore, you want to see 5.625 V across the primary of the OPT....at 1K.. I leave it to you to check this at other frequencies..

Due to the impedance rise of the input of the tank....they say to scale the voltage with respect to the frequency...which I find to be troubling, since they are assuming a LINEAR rise that doubles itself???
Whatever???... I would have to measure the input Z to see for myself what the heck is really going on....

Fender used a doubled 12AT7.... AT that operating point the plate resistance for each 12AT7 is roughly 10K...so figure a 5K source resistance for the paralleled 12AT7 valves..... driving a 5K plate load... This makes a voltage gain of .5 of whatver mu...figure 57..so figure Gain of 28... I guess someone thought of MAX power transfer and not concerned with the distortions at hand....Not the best loading for Hi-Fi.....but works for the reverb....

When you have a 1kHz 200mV RMS signal at the grids of the paralleled 12AT7 ...this would make the 5.625V across the plates of the transformer, thus 225mV at the input of the tank.... This is simple first order analysis that should be in the ballpark....

Hope this helps...

Chris
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Old 16th September 2005, 05:51 AM   #9
PRR is offline PRR  United States
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> current output on the secondary of the output x-former needs to be 28mA. Right?

That is what they say.

> Z ratio 5k:8 or 625:1; resulting N ratio 25:1 and I2/I1=N1/N2 so 28mA/I1=25/1; I1=1.12mA Therefore, I will need a Plate current output of 1.12mA.

The math is right.

HowEVER: most real-world reverb drivers make more power than that. I have seen chip opamps with discrete push-pull boosters, suggesting much more than 28mA. Tubes with ~5K:8 iron tend to be 5mA-30mA, not the couple-mA we calculated. Yet tanks don't burn up. I suspect we DO like to push the tanks past the 3.5AT rating. If not, all tank drivers would be one 12AU7 into 10K:8, not small power tubes (or doubled-up 12A_7) as they usually are.

> they say to scale the voltage with respect to the frequency...which I find to be troubling, since they are assuming a LINEAR rise that doubles itself??? Whatever???...

The input impedance is not a resistance, it is an inductance with resistive losses. IIRC it does rise from the bottom of the audio band to around 1KHz, then levels-off due to core losses. (It must also level-off in the bass, but it seems to slope to below the audio band.)

Because of this semi-inductive impedance and the saturation, you get different sound if you drive it low-Z or hi-Z. One high-tech plan used with chips is to generate a current-source drive. The same is possible with pentodes, but we often see triodes which would act as medium-low impedance drive. In chips we sometimes see a resistor for a similar medium-Z drive. The other extreme is a zero-Z drive. Each different drive condition needs a different EQ for "flat" response. I think if you use low-Z drive, you need to cut some bass out in the recovery amp (typically just an indersize coupling cap). Of course from the "flat" response (which can never be really flat), we normally deviate all over the place to dial-in the "sound" we want. So in the end, it is simplest to just build it, laid out so modifications are easy, and play.
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