Spring reverb driver theoretical issue

ltournell

TDPRI Member
Joined
Jun 28, 2003
Posts
30
Location
Lille France
Hello,

I am making a DIY spring reverb and I refer to existing written material on the subject beginning with ACCUTRONICS specifications here :

I also read Rod Elliott's pages about spring reverbs here :

What I am doing in particular is sizing the input current headroom above the nominal saturation current.

Given the following impedances chart from the ACCUTRONICS site,

impchart.jpg

impchart.jpg

impchart.jpg

the first issue concerning a contradiction in their specs is easily solved : the second paragraph on the ACCUTRONICS page quotes a nominal current of 2,5 A-T (amperes per turn) whereas the diagram below ( and the text) above give 3.5 A-T. On the impedance chart we can read on the E range at the far right 1100 turns x 3.1 = 3300 mA-T end on the F range 1700 turns x 2.0 = 3400 mA-T. Obviously the truth is 3.5A-T !

In spite of the obviousness of this result the Rod Elliott's page still quotes this 2.5 A-T limit in the note 1 below his own chart here .
Extract_Rod_Elliot_1.png


This would be of no point if the rest was consistent but it is not.
The ACCUTRONICS spec gives a 3.5 A-T saturation point with a headroom above it of 10dB and all this is consistent even with the writer's experiments who misinterprets the figures.

The writer gives a peak current of around three times the nominal current and says "the values shown are based on an allowance for 6 dB headroom" but 6dB means twice the current . It is 10 dB (=3.16) that he means, exactly the figure given by ACCUTRONICS. Then he says "Accutronics claims a 10dB margin for headroom, which (at least in theory) means that up to 250A/T is 'acceptable', but this is way too high IMO. " !! No , it means that up to 3.16 x 3.5 A-T = 11 A-T is acceptable , not 250 (the cores would be huge)!
Now ACCUTRONICS either reasons in terms of current or power and both are acceptable if we consider a 10 dB headroom. Doubling the current means doubling the voltage etc.... So here we have I x 3 x V x 3 = P x 9 and 10 log (9) = 9.5 =~ 10 dB.

"I will use up to double the 'rated' RMS current, perhaps a little more if it still sounds clean" : in my opinion this is illusory because the dynamics of guitar PU exceed by far a factor of 2 or 3 or ...more !

This underlines the utility of a DWELL control in the driver, limiting the current not too high above the comp uted headroom here.
 

Jon Snell

Tele-Afflicted
Joined
Aug 31, 2015
Posts
1,093
Location
Jurassic Coast, Dorset. Great Britian.
Don't over think this. You will spend more time chasing your tail.
The 'data' you show has no real meaning.
For a reverb tank to run properly you need the drive the transmitter without over driving it and the recovery circuit must filter a lot of the signal out to give just the effect.
It is not a science.
Select your tank, decide whether it will be transformer fed; Fender style with an ECC81 or cheap Fender style with a 4558 or TL072 chip. In other words, decide on the driver impedance.
Recovery can be either solid state or valve, it make little difference as the frequency range is limited by design.
Have fun playing around with it and enjoy the challenges.
There is no point in reinventing the wheel.
 

ltournell

TDPRI Member
Joined
Jun 28, 2003
Posts
30
Location
Lille France
@Jon Snell Thank you for answer. Be sure that I am aware of all that and that I agree. I was just trying to find information to juice up the reverb unit I made in a DIY SE EL34 fine small amp. I wrote that because I was a bit irritated by the seemingly scientific approach of this guy who copyrights his "precious" article without even giving an email to join him and help correct knowing that this guy does not even understand what a dB means !

I made the reverb unit with an ECC82 driver and transformerless with a high impedance tank and it works beyond everything I expected !
 




Top