From ic–(at)–nake Wed Apr 26 16:01:57 CDT 1995
Article: 48995 of rec.music.makers.guitar
Path: geraldo.cc.utexas.edu!cs.utexas.edu!howland.reston.ans.net!gatech!bloom-beacon.mit.edu!boulder!ice
From: ic–(at)–nake (John Mastrangelo)
Newsgroups: rec.music.makers.guitar
Subject: Re: 10pf cap in reverb mix circuit
Date: 26 Apr 1995 15:09:41 GMT
Organization: University of Colorado at Boulder
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In a previous article George Kaschner asks how he can reduce
the amount of reverb in his mix and asks whether changing
the 3.3 M resistor would be appropriate.
In the interest of time, I’m going to make this brief. If
it unclear, let me know and I will explain further.
There are at least 10 ways to achieve this, but this approach
is the simplest to implement, is the most effective, easist
to analyze, couldn’t be much cheaper (2 resistors) and most
importantly will not affect any other function of the amp.
Here we go:
In most (if not all) Fender amps w/ reverb there is a multiple
stage voltage divider which controls the dry gain of the reverb
channel and the amount of reverb in the mix. This VD
consists of the 3.3M/10pF pair and the 470k/220k pair
connected to the center tap of the reverb mix control and
the grid of the reverb recovery triode.
Lets look at the reverb level first. The ratio
220k/(220k + 470k)
controls the max amount of reverb available. Changing the
ratio of these two resistors will control the max amount of
reverb in the mix.
George mentioned wanting to mod his reverb circuit
such that on “10” it equalled the stock control on “3”.
Since the reverb mix control is linear, “3” corresponds
to 30% of “10” which indicates that his resistor ratio
above (220k/690k) = .32 should be reduced to .32*.3 = .095.
Second is the gain of the reverb channel. Here the 3.3M/10pF
pair and the *parallel* combination of the 220k/470k pair
form a voltage divider which controls the gain of the
dry reverb channel signal. The parallel combination of
the 220k/470k pair is approx 150k ( the pot resistance
is neg.) Ignoring the effect of the 10pF cap yields:
150k/(3.3M + 150k) = .043 = 1/23.
So the reverb dry signal is being attenuated by a factor
of 23. That is why the reverb channel has only slightly
more gain than the Normal.
Bottom line:
* Decrease the ratio of the reverb signal VD from
.32 to .095
* Maintain the parallel resistance of 150k from
these two resistors.
Bottom bottom line:
* Replace the 470k resistor connected to the
center tap of the reverb control with 1.5M.
* Replace the 220k resistor with 180k.
Then:
Reverb divider ratio = 180k/(180 + 1500)k = .11
Parallel combination = 1/( 1/180k + 1/1500k) = 160k
Which is plently close for what we are doing here.
There are a lot of other effects that are connected to
this discussion and I’d like to address them all, but
I don’t have the time or the desire to write a book.
However, I’m happy to answer any questions that may arise
from this article.
John Mastrangelo
Osprey Amplification
