From: mrk@gvgspd.gvg.tek.com (Michael R. Kesti)
Newsgroups: rec.music.makers
Subject: Re: ==> +30db GAIN Circuit for Your Guitar: $25 <==
Date: 29 Dec 92 03:11:39 GMT
Organization: Grass Valley Group, Grass Valley, CA
NNTP-Posting-Host: gvgspd.gvg.tek.com

In article <5931@maserati.qsp.UUCP>
                              danb@zx.qsp.UUCP (Daniel Benbenisty) writes:
>In article <Bznp8B.Iu6@dcs.glasgow.ac.uk>
                              dam@dcs.glasgow.ac.uk (David Morning) writes:

>>                                          30dB gain is far too much for a
>>puny 9 volt battery.

This statement, by itself, is potentially incorrect.  The gain of a circuit
is not limited by the supply voltage alone.  If, for example, one starts with
a 1 microVolt signal, an amplifier powered by a single 9 volt supply would
certainly be capable of 30 dB of gain without clipping.  I think that David
intended to say that, given a typical pickup's output level, and using a
single 9 volt supply, one cannot achieve 30 dB of gain without clipping.

>>                     I've got a 26dB first stage gain in my home made amp
>>running off 30 volts(!) and that clips occasionally with a Telcaster. 

>1. the spec "30dB gain" has no implicit power requirements.

This is correct, although it is not clear to me whether Daniel intended to
say that the spec implies no amplifier output power requirement or that it
implies no power supply requirement, but it doesn't really matter.

>                                                             It just means 
>that you're multiplying the signal by 10^3 (1000) times.

Not quite.  Power gain, in bels, is given as:

                                 Power(out)
         gain (in bels) = log(10)----------                        (1)
                                 Power(in)

where "log(10)" means "log base 10".  Bels are typically too large for
everyday use, so, in the same fashion that we prefer to use microfarads
rather than farads for the unit of capacitance, we prefer to use decibels,
and equation (1) becomes:

                                    Power(out)
     gain (in decibels) = 10 log(10)----------                     (2)
                                    Power(in)

In this discussion, however, we are not interested in power gain.  Rather,
we are interested in voltage gain.  Noting that the power dissipated in a
resistance is:

       power (in watts) = (Volts*Volts)/R                          (3)

we can substitute for power in equation (2):

                                    (Volts(out)*Volts(out))/R
     gain (in decibels) = 10 log(10)-------------------------      (4)
                                    (Volts(in)*Volts(in))/R

If we assume that the input and output resistances are equal, they divide
out, and, because volts times volts equals volts squared, equation (4)
becomes:

                                    Volts(out)          
     gain (in decibels) = 20 log(10)----------                     (5)
                                    Volts(in)         

Therefore, a voltage gain of 30 decibels becomes:

                                    Volts(out)          
                     30 = 20 log(10)----------                     (6)
                                    Volts(in)         

or:
             Volts(out)          
             ---------- = 31.6                                     (7)
             Volts(in)         


So, a voltage gain of 30 dB is the same as multiplying the input voltage
by 31.6.

>                                                          Now, the signal 
>from any pickup is pretty damn weak (anybody have any numbers?).

Sure.  A typical pickup outputs a nominal level of on the order of 100
millivolts RMS.  Some output more and some output less, and nominal level
is usually far less than maximum level, but 100 millivolts RMS is a good
typical level with which to work.  Note that I specified RMS voltage.  This
is because we are talking about an AC voltage.  If we assume a sinusoidal
waveform (Humor me, please!), the peak voltage of the positive half cycle
will be:

                   0.1
                  ----- = 0.14 volt                                (8)
                  0.707

With this input level, and a circuit gain of 31.6, the output level will be:

       0.14 volt * 31.6 = 4.4 volts peak                           (9)

Any AC voltage amplifier powered by a unipolar supply, such as our 9 volt
battery, will need to be biased in such a way that the peak output swings
are limited to slightly less than (depending on the design) half the supply
potential, or 4.5 volts in our case.  We can conclude then, that a 30 dB
circuit gain will produce an unclipped output with a 100 millivolt input,
but that it would leave no headroom for anything hotter, nor for, shall we
say, exuberant playing.

>                                                                  Maybe I'm 
>wrong, but I wouldn't imagine that 1000 times this power would be beyond a 
>9 V battery.
>
>2. The aforementioned home-made circuit is running off of 30V, but how 
>many amps does it draw?  Neglecting a battery's internal resistance, you can 
>still draw the same (or more) amps from a 9V source as from a 30V source.

The above are irrelevant because, again, we are dealing with voltage gain
here, not power gain, so the power supply's current capacity is not an issue,
except perhaps when considering other factors, such as battery life.

It might be interesting to do an analysis similar to the above for David's
circuit, but working backwards to determine the maiximum input level before
clipping.  He said he uses a 30 volt supply, but doesn't specify whether it
is bipolar.  I'll assume unipolar, which means his output is limited to
15 volts peak, or 15 times 0.707 = 10.6 volts RMS.  He specified a gain of
26 dB, which is a voltage gain of 19.95.  10.6 / 19.95 = 530 millivolts.
I believe David when he says that he can make it clip occasionally with
his Telcaster. 

>>I'm pretty certain this product WILL change your tone  :-)
>
>Hmmm, well, anyone on the net have a frenquency spectrum analyser? Is the 
>inventor of this device willing to have it subjected to scientific scrutiny?

As long as we limit our discussion to clipping, there's no need for a spectrum
analyzer.  If it clips, it clips, and that is certainly a change in tone.

Another question, though, is why would one desire 30 dB of gain at the guitar?
The typical reason for an "active" guitar is to isolate the instrument from the
effects of the cable capacitance, so that changing the guitar's volume control
doesn't result in a high frequency roll-off.  This can be accomplished by a
unity gain follower, which is much more likely to pass an unclipped signal.
What's more, excessive gain at the guitar is likely to cause interface problems
for the next device in the signal patch, be it an amplifier, stomp box, or
whatever, as their nominal input level, and circuit gain, were designed for
typical signal levels, so that even if the guitar preamp doesn't clip, one has
a better than average chance that one of these components will.
 
I'd recommend not using this circuit if one's goal is to increase gain without
changing one's tone.

-- 
============================================================================
Michael Kesti  Grass Valley Group, Inc. | "And like, one and one don't make
    mrk@gvgspd.GVG.TEK.COM              |  two, one and one make one."
    !tektronix!gvgpsa!gvgspd!mrk        |         - The Who, Bargain
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