From:
Subject: reverb2
Date: Fri, 8 Apr 2011 10:30:04 -0700
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reverb2
Using Op Amps to Interface to Spring-type Reverb =
Units
Copyright 1999
Roy Mallory
roym@world.std.com=20
Revised November, 2001
Although digital reverb units have achieved widespread use, the =
spring reverb=20
is still popular in guitar amplifiers. Many such amplifiers use tubes =
for their=20
sonic and overdrive characteristics, but for interfacing to a spring =
reverb, the=20
monolithic op amp offers superior capabilities. This tech note, which =
discusses=20
how to optimally interface a spring reverb to an op amp, is divided into =
three=20
sections, one describing the drive circuit, another discussing the =
optimal drive=20
level, and the final describing the signal-recovery circuit. In all =
sections, an=20
Accutronics model 8DB2C1B or 8BB2C1B reverb unit is used as an =
illustrative=20
example. The circuit design concepts presented here are equally =
applicable to=20
other reverb units.
The Drive Amplifier
In a spring reverb, a solenoid (coil of wire) is used to generate a =
magnetic=20
field that imparts vibrations to the unit's springs. As Ampere's Law =
shows, the=20
strength of the field is proportional to the current through the coil, =
not the=20
voltage across it. Yet, many reverb drive circuits produce a voltage =
rather than=20
a current output. Why is this a problem? Well, the solenoid that =
produces the=20
magnetic field is also an inductor, and the impedance of an inductor =
rises in=20
direct proportion to the frequency applied to it. Therefore, given a =
constant=20
voltage input, the current produced in, and the magnetic field produced =
by, the=20
coil goes down in direct proportion to frequency. Mathematically, we can =
express=20
this with a couple of simple equations:
where: zL =3D the coil's impedance
s =3D the complex frequency variable, j2p =20
f
L =3D the coil's inductance
RL =3D the coil's resistance
Using Ohm's Law, we see that the current in the coil is:
So, using a voltage-output amplifier causes the reverb to produce =
less and=20
less output as frequency goes up, which is not at all what one needs. It =
is the=20
higher frequencies that produce a good reverb sound, and if the drive =
amp=20
doesn't provide much output at these frequencies, the signal-recovery =
amp will=20
need more gain and thus produce more noise.
Fortunately, a standard op amp allows an ideal solution to this =
dilemma. It=20
is very simple and inexpensive to implement a current-output =
(technically=20
speaking, a transconductance) amplifier with an op amp. The fact that =
the=20
8DBC1B's drive coil is not grounded makes the task especially easy. =
Consider the=20
following circuit:
Any input voltage applied to the op amp's noninverting input appears =
at its=20
inverting input as well. Thus, the current at the inverting node, and =
therefore=20
through the reverb coil, is simply Vin/R1. So, we see that the circuit =
produces=20
a current output in direct proportion to its voltage input. Two problems =
must be=20
solved before the circuit is truly useful, however. First, we should =
roll off=20
the stage's output at low frequencies, because they contribute nothing =
to the=20
reverb sound; and second, as shown, this stage is unstable or marginally =
stable=20
at best. To understand the cause of the instability, consider that =
through most=20
of its frequency range, the op amp's open-loop transfer function =
produces a=20
90o phase shift. The network formed by the inductance of the =
reverb=20
unit and R1 also produces a 90o phase shift, causing a total =
of=20
180o of phase shift around the feedback loop. The =
modifications shown=20
in the circuit below solve both problems:
Capacitor C1 reduces the gain at low frequencies. Empirically, a 3 dB =
point=20
for the R1, C1 network between 150 and 300 Hz seem about right. The test =
circuit=20
uses a 62 ohm resistor for R1 and a 10 m F =
capacitor=20
for C1. This combination produces a 3 dB point at about 260 Hz.
The network comprising C2 and R2 compensates the stage to make it =
stable. To=20
see how, consider that at high frequencies, the inductive reactance of =
the=20
reverb coil is high, and the capacitive reactance of C2 is low. =
Therefore, at=20
high frequencies, R2 effectively becomes the feedback element, which =
restores=20
phase margin to the circuit. To understand how to select R2 and C2, =
let's solve=20
for the impedance, zf , of the feedback network formed by =
these two=20
components and the reverb coil:
Looking at the denominator, we see it has the expected second-order =
form. To=20
select C2, pick a frequency at which you would like the stage to break =
out of=20
its rising feedback impedance to a constant one, using the standard =
formula:=20
A frequency between 5 kHz and 10 kHz is appropriate for a guitar =
amplifier.=20
The 8DB2C1B tank has an inductance of about 44 mH; the test circuit uses =
a 0.01=20
m H capacitor to produce a break frequency of =
about 7.6=20
kHz.
Component R2 is selected to produce an appropriate Q according to the =
formula=20
A 2 KW resistor should produce a Q of =
about 1, but=20
core losses will lower the Q to that of a maximally-flat response. With =
these=20
components, the magnitude of the feedback impedance is:
Next, we must look at the noise performance of this circuit. Several =
noise=20
sources exist. First, let's consider the effect of op amp voltage noise=20
(en ). This noise source, which can range anywhere from 1 to =
20 nV/rt.=20
Hz for commonly available op amps, acts just like an input voltage. So, =
the=20
current noise produced is simply en /R1. If we take a typical =
op amp=20
noise of 5 nV/rt. Hz, this source causes output current noise of 0.08 =
nA/rt.Hz=20
in the test circuit. The Johnson noise of R1 also contributes. The =
output=20
current noise it produces is simply its Johnson noise divided by its =
resistance.=20
For the test circuit, this noise is 0.016 nA/rt. Hz, which is 5 times =
lower than=20
the noise caused by en . Therefore, we can ignore it. The =
Johnson=20
noise of the reverb coil's resistance causes voltage noise at the output =
of the=20
op amp, but no current noise, so it too can be ignored.
The current noise at the noninverting input of the op amp produces an =
input=20
voltage that is simply the product of the current noise times the output =
impedance of the previous stage. This noise divided by R1 is the output =
noise=20
from this source, so io =3D in Rout/R1. If the =
previous stage=20
had an output impedance of 20 kW , which is =
not=20
uncommon for a tube stage, and if one used a bipolor-input op amp whose =
input=20
current density were 1 pA/rt. Hz, which is typical of many such op amps, =
the=20
output current noise would be about 0.3 nA/rt. Hz in the test circuit; =
this=20
would be the dominant noise source. The lesson is clear: drive the stage =
from a=20
low impedance or use a JFET-input op amp. Such amps typically have =
current noise=20
densities of about 0.01 pA/rt. Hz. If either approach is taken, the =
noise from=20
this source will be so low that it can be ignored. Op amp current noise =
at the=20
inverting input simply goes into the reverb coil. This noise is also =
much lower=20
than that caused by en , so it too can be ignored. Thus we see =
that in=20
a properly designed stage, the op amp voltage noise is the only =
significant=20
noise source.
What signal-to-noise ratios can be expected from such a stage? Let's =
say we=20
choose an op amp whose voltage noise is 5 nV/rt. Hz, because such amps =
are cheap=20
and easy to find. Let's assume a bandwidth of 10 kHz, which is =
conservative,=20
because most guitar speakers don't produce much output beyond 5 or 6 =
kHz. Almost=20
any op amp can put out 10 to 20 mA peak; let's be conservative again and =
assume=20
that our stage can put out 7 mA rms. A quick calculation (peak rms =
output=20
divided by the product of the noise density times the square root of =
bandwidth)=20
shows a maximum signal-to-noise ratio of 119 dB!
Follow these simple design guidelines:
For good stability,
Choose R2 and C2 appropriately.=20
Properly bypass the supply voltages to the op amp.=20
Connect the ground side of R1 so that its current does not cause=20
unintended feedback paths.
For low noise:
Use the lowest input-noise op amp you can afford.=20
Use a JFET-input op amp, or drive the stage from a low-impedance =
source.=20
Make sure that the stage is driven sufficiently hard to ensure =
maximum=20
signal.
Designing For the Optimal Drive Level
Designing your drive amplifier so it can provide sufficient current =
to the=20
reverb coil is a deceptively simple task. First, one must determine the =
optimal=20
drive current that your particular model of reverb coil requires. =
Unfortunately,=20
Accutronics' specifications can be a bit misleading. First, they publish =
a=20
"nominal" drive level recommendation for their models. I'm not sure just =
what=20
this number means, but I've concluded that it is best ignored. The =
number they=20
provide that is meaningful is the maximum amp-turns that their drive =
coils can=20
withstand before going into saturation. They spec this value as 25 =
amp-turns=20
r.m.s.. Their use of an r.m.s. value here is also confusing, but let's =
assume=20
they had sinusoidal excitation in mind, and therefore the peak current =
value=20
would be 25*1.414 or about 35 amp-turns. I was able to verify by =
experiment that=20
their drive coils (actually, of course, the cores of the coils) do go =
into=20
saturation at about this level. So a design goal of a maximum drive =
level of=20
about 30 amp-turns to allow for a safety factor seems about right.
To determine how much voltage we need to supply the reverb coil to =
produce=20
current at this level, let's proceed simply and say that our guitar amp =
should=20
be able to operate at full output up to 10 kHz. The 8DB2C1B's input coil =
has 688=20
turns and an impedance of 310 ohms at 1 kHz. So, at 10 kHz the required =
drive=20
voltage is
A required drive level of 135 volts peak is not =
encouraging, but do=20
we really need that much? Well, the highest note in a typical guitar is =
the D=20
that is two octaves and a whole step above middle C. This note's =
frequency is=20
1175 Hz. In addition, a string that is fretted that high doesn't put out =
much=20
power in its harmonics. To add to the good news, a guitar puts out =
maximum=20
volume when a chord and not just a single string is played. By =
definition, all=20
other notes in any chord containing the high D must be lower in =
frequency. So,=20
for an undistorted guitar, the reverb doesn't need any more headroom =
than that=20
required to reproduce a 1 kHz signal at full blast. For the subject =
reverb unit,=20
we are only talking about a 13.5-volt peak signal.
13 volts is cutting things close for an ordinary op amp, so one might =
want to=20
use a high-voltage op amp if a higher-voltage supply is available or =
switch to a=20
lower impedance reverb tank. An 8BB2C1B might be an obvious choice if =
the latter=20
tactic is used. Its drive coil has 538 turns and an impedance of 190 =
ohms at 1=20
kHz. To drive it fully at 1 kHz requires
Note however, that the tanks require 44 mA and 56 mA respectively to =
drive to=20
peak level. Therefore a high-current op amp of some sort is required. =
I've found=20
that an OPA551 works very nicely and can also take supplies up to =B1 30 volts for those who want extra headroom or =
who are=20
amplifying signals with more high-frequency content.
If you would like to use the 8BB2C1B tank, the drive-circuit =
component values=20
can be changed to the following:
R1 =3D 56 ohms
R2 =3D 1.6k
C1 =3D 20 uF
C2 =3D .033 uF
The attentive reader will note that the requirements discussed above =
were for=20
an undistorted guitar. If a heavily distorted guitar signal is fed into =
the=20
reverb tank, the amount of drive voltage required will increase =
substantially,=20
because the clipped guitar signal contains much more power at higher=20
frequencies. Interestingly enough, not being able to reproduce the hard =
edges of=20
a clipped guitar signal doesn't sound objectionable to my ears. It just =
means=20
that the reverb part of the amplifier's output will sound a bit softer =
and less=20
distorted than the main signal. I kind of like that, but another =
listener may=20
not.
Signal-Recovery Amp
The most basic decision for this stage is whether to use an inverting =
or=20
noninverting topology. The noninverting topology shown below is =
preferred=20
because it allows the values of the gain-setting resistors to be =
optimized for=20
lowest noise while allowing the input resistance of the stage to be set=20
independently.
The schematic above shows all the necessary circuit elements. =
Resistors R1=20
and R2 set the gain of the stage. The requisite gain will depend on the =
rest of=20
the system's gain, but gains between 50 and 100 are common. The test =
circuit=20
uses a gain of 101, with R1 set at 68 ohms and R2 set at 6.8 kohms. The =
value of=20
C1 is chosen to roll off the low-frequency gain of the circuit. The test =
circuit=20
uses a 10 m F capacitor, producing a 3 dB =
point of=20
about 230 Hz. There is considerable leeway in the selection of the 3 dB=20
point-let your ears be the guide. C2, which is 1500 pF in the test =
circuit, and=20
R2 form a low-pass filter at 15.6 kHz; with such high gain, it is a good =
idea to=20
keep possible out-of-band noise from subsequent stages.
The inductance of the reverb unit's pick-up coil will resonate with =
the=20
capacitance of the coil's winding plus the capacitance of the shielded =
cable=20
used to connect the coil to the circuit. In the case of the 8DB2C1B =
unit, a=20
short cable will produce a resonance above 20 kHz, and R3 is not needed. =
For=20
other units, or if a long cable is used, the resonance can move into the =
audible=20
range. R3 can be used to damp this resonance. Its value can be =
calculated using=20
the equation presented earlier for determining Q. Alternatively, it may =
be=20
easier to determine its value empirically by checking for an impedance =
peak.=20
Some designs call for placing a capacitor across the reverb unit's =
output to=20
purposely place the resonance at some audible frequency. Although this =
technique=20
does produce more output at resonance, it does so at the expense of =
raising the=20
input impedance at that frequency. In my experience, any increase in =
output is=20
more than offset by an increase in noise. In addition, the response =
emphasis is=20
not pleasant to this author's ears. For lowest noise, it is best to damp =
any=20
in-band resonance with R3.
The most important consideration for the design of this stage is low =
noise.=20
As previously stated, the stage needs quite a bit of gain. As we did for =
the=20
drive circuit, let's investigate all the noise sources. First, the =
Johnson noise=20
of the reverb unit's output coil is connected directly to the stage's =
input and=20
will be amplified along with the signal. The 8DB2C1B has an output =
resistance of=20
200 ohms. The Johnson noise of this resistance is 1.8 nV/rt. Hz. In the =
test=20
circuit, this resistance will cause output noise of 181 nV/rt. Hz.
The Johnson noise of R1 will be amplified by the R2/R1 ratio. =
Therefore, R1=20
must be kept smaller than the resistance of the reverb coil if it is not =
to add=20
appreciable noise. Because noise sources add in an r.s.s. (root sum =
squared)=20
way, if R1 is less than one-third the value of the reverb coil's =
resistance, the=20
noise it will add will increase the total noise by less than 1 dB. The =
68 ohm=20
resistor used in the test circuit meets this criterion. Note, however, =
that some=20
resistors have noise considerably in excess of Johnson noise. Carbon =
composition=20
units are notorious in this regard and therefore shouldn't be the first =
choice=20
for R1. The Johnson noise of R2, because it is not amplified, is =
negligible.=20
Because impedance at the noninverting node of the op amp is low, op =
amp=20
current noise at this input causes negligible output noise. Current =
noise at the=20
inverting input is multiplied by the value of R2. Still, this noise =
source will=20
also be negligible, even for a bipolar-input op amp. Finally, the op =
amp's=20
voltage noise is amplified along with its signal. Unless one uses a very =
quiet=20
amp, this noise will be higher than the Johnson noise of the reverb =
coil, and so=20
will be the dominant noise source of the stage. If an op amp with an=20
input-referenced voltage noise 5 nV/rt. Hz is used, the total output =
noise of=20
the stage is about 540 nV/rt. Hz. Over a 10 kHz bandwidth, total noise =
is about=20
60 m V r.m.s., quite a respectable figure. =
In many designs, the reverb unit is attached so it does not =
electrically=20
contact the amplifier's chassis. As such, no ground loops will be =
produced. If,=20
however, the reverb unit electrically contacts the chassis, the ground=20
connection at the op amp input may have to be eliminated to avoid hum. =
If so=20
required, break the connection marked with an "x" in the schematic.
Follow these simple design guidelines:
Choose R1 so that its value is not more than 1/3 of the reverb =
tank's=20
resistance.=20
Choose C1 to roll off the low-frequency response around 200 to 300 =
Hz.=20
If necessary, install R3 to damp in-band resonances.=20
Use the lowest voltage-noise op amp you can afford.=20
Break the ground loop if the reverb tank is connected to a =
grounded part=20
of the chassis.
Conclusions
The test circuits have been auditioned by several guitarists, all of =
whom=20
have pronounced the quality of the reverb to be excellent. I have seen =
many=20
statements that claim that op amps should not be used for interfacing to =
spring=20
reverbs because they are too noisy or are unsuitable in some other way. =
I can=20
only assume that such statements are based on misconceptions or poor =
circuit=20
design. I believe that this tech note shows conclusively that op amps =
provide a=20
nearly ideal mate to a spring reverb. In the combination of performance, =
size,=20
and cost, no other circuit element can do such a superb job.
Return to=20
home page
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