Build your own FREE website at Tripod.com
Share: Facebook | Twitter | Digg | reddit

Using Op Amps to Interface to Spring-type Reverb Units

Copyright 1999

Roy Mallory

[email-masked]

Revised November, 2001

Although digital reverb units have achieved widespread use, the spring reverb
is still popular in guitar amplifiers. Many such amplifiers use tubes for their
sonic and overdrive characteristics, but for interfacing to a spring reverb, the
monolithic op amp offers superior capabilities. This tech note, which discusses
how to optimally interface a spring reverb to an op amp, is divided into three
sections, one describing the drive circuit, another discussing the optimal drive
level, and the final describing the signal-recovery circuit. In all sections, an
Accutronics model 8DB2C1B or 8BB2C1B reverb unit is used as an illustrative
example. The circuit design concepts presented here are equally applicable to
other reverb units.

The Drive Amplifier

In a spring reverb, a solenoid (coil of wire) is used to generate a magnetic
field that imparts vibrations to the unit's springs. As Ampere's Law shows, the
strength of the field is proportional to the current through the coil, not the
voltage across it. Yet, many reverb drive circuits produce a voltage rather than
a current output. Why is this a problem? Well, the solenoid that produces the
magnetic field is also an inductor, and the impedance of an inductor rises in
direct proportion to the frequency applied to it. Therefore, given a constant
voltage input, the current produced in, and the magnetic field produced by, the
coil goes down in direct proportion to frequency. Mathematically, we can express
this with a couple of simple equations:

where: zL = the coil's impedance

s = the complex frequency variable, j2p
f

L = the coil's inductance

RL = 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
less output as frequency goes up, which is not at all what one needs. It is the
higher frequencies that produce a good reverb sound, and if the drive amp
doesn't provide much output at these frequencies, the signal-recovery amp will
need more gain and thus produce more noise.

Fortunately, a standard op amp allows an ideal solution to this dilemma. It
is very simple and inexpensive to implement a current-output (technically
speaking, a transconductance) amplifier with an op amp. The fact that the
8DBC1B's drive coil is not grounded makes the task especially easy. Consider the
following circuit:

Any input voltage applied to the op amp's noninverting input appears at its
inverting input as well. Thus, the current at the inverting node, and therefore
through the reverb coil, is simply Vin/R1. So, we see that the circuit produces
a current output in direct proportion to its voltage input. Two problems must be
solved before the circuit is truly useful, however. First, we should roll off
the stage's output at low frequencies, because they contribute nothing to the
reverb sound; and second, as shown, this stage is unstable or marginally stable
at best. To understand the cause of the instability, consider that through most
of its frequency range, the op amp's open-loop transfer function produces a
90o phase shift. The network formed by the inductance of the reverb
unit and R1 also produces a 90o phase shift, causing a total of
180o of phase shift around the feedback loop. The modifications shown
in the circuit below solve both problems:

Capacitor C1 reduces the gain at low frequencies. Empirically, a 3 dB point
for the R1, C1 network between 150 and 300 Hz seem about right. The test circuit
uses a 62 ohm resistor for R1 and a 10 m F capacitor
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
see how, consider that at high frequencies, the inductive reactance of the
reverb coil is high, and the capacitive reactance of C2 is low. Therefore, at
high frequencies, R2 effectively becomes the feedback element, which restores
phase margin to the circuit. To understand how to select R2 and C2, let's solve
for the impedance, zf, of the feedback network formed by these two
components and the reverb coil:

Looking at the denominator, we see it has the expected second-order form. To
select C2, pick a frequency at which you would like the stage to break out of
its rising feedback impedance to a constant one, using the standard formula:

A frequency between 5 kHz and 10 kHz is appropriate for a guitar amplifier.
The 8DB2C1B tank has an inductance of about 44 mH; the test circuit uses a 0.01
m H capacitor to produce a break frequency of about 7.6
kHz.

Component R2 is selected to produce an appropriate Q according to the formula

A 2 KW resistor should produce a Q of about 1, but
core losses will lower the Q to that of a maximally-flat response. With these
components, the magnitude of the feedback impedance is:

Next, we must look at the noise performance of this circuit. Several noise
sources exist. First, let's consider the effect of op amp voltage noise
(en). This noise source, which can range anywhere from 1 to 20 nV/rt.
Hz for commonly available op amps, acts just like an input voltage. So, the
current noise produced is simply en/R1. If we take a typical op amp
noise of 5 nV/rt. Hz, this source causes output current noise of 0.08 nA/rt.Hz
in the test circuit. The Johnson noise of R1 also contributes. The output
current noise it produces is simply its Johnson noise divided by its resistance.
For the test circuit, this noise is 0.016 nA/rt. Hz, which is 5 times lower than
the noise caused by en. Therefore, we can ignore it. The Johnson
noise of the reverb coil's resistance causes voltage noise at the output of the
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
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
from this source, so io = inRout/R1. If the previous stage
had an output impedance of 20 kW , which is not
uncommon for a tube stage, and if one used a bipolor-input op amp whose input
current density were 1 pA/rt. Hz, which is typical of many such op amps, the
output current noise would be about 0.3 nA/rt. Hz in the test circuit; this
would be the dominant noise source. The lesson is clear: drive the stage from a
low impedance or use a JFET-input op amp. Such amps typically have current noise
densities of about 0.01 pA/rt. Hz. If either approach is taken, the noise from
this source will be so low that it can be ignored. Op amp current noise at the
inverting input simply goes into the reverb coil. This noise is also much lower
than that caused by en, so it too can be ignored. Thus we see that in
a properly designed stage, the op amp voltage noise is the only significant
noise source.

What signal-to-noise ratios can be expected from such a stage? Let's say we
choose an op amp whose voltage noise is 5 nV/rt. Hz, because such amps are cheap
and easy to find. Let's assume a bandwidth of 10 kHz, which is conservative,
because most guitar speakers don't produce much output beyond 5 or 6 kHz. Almost
any op amp can put out 10 to 20 mA peak; let's be conservative again and assume
that our stage can put out 7 mA rms. A quick calculation (peak rms output
divided by the product of the noise density times the square root of bandwidth)
shows a maximum signal-to-noise ratio of 119 dB!

Follow these simple design guidelines:

For good stability,

Choose R2 and C2 appropriately.

Properly bypass the supply voltages to the op amp.

Connect the ground side of R1 so that its current does not cause
unintended feedback paths.

For low noise:

Use the lowest input-noise op amp you can afford.

Use a JFET-input op amp, or drive the stage from a low-impedance source.

Make sure that the stage is driven sufficiently hard to ensure maximum
signal.

Designing For the Optimal Drive Level

Designing your drive amplifier so it can provide sufficient current to the
reverb coil is a deceptively simple task. First, one must determine the optimal
drive current that your particular model of reverb coil requires. Unfortunately,
Accutronics' specifications can be a bit misleading. First, they publish a
"nominal" drive level recommendation for their models. I'm not sure just what
this number means, but I've concluded that it is best ignored. The number they
provide that is meaningful is the maximum amp-turns that their drive coils can
withstand before going into saturation. They spec this value as 25 amp-turns
r.m.s.. Their use of an r.m.s. value here is also confusing, but let's assume
they had sinusoidal excitation in mind, and therefore the peak current value
would be 25*1.414 or about 35 amp-turns. I was able to verify by experiment that
their drive coils (actually, of course, the cores of the coils) do go into
saturation at about this level. So a design goal of a maximum drive level of
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
current at this level, let's proceed simply and say that our guitar amp should
be able to operate at full output up to 10 kHz. The 8DB2C1B's input coil has 688
turns and an impedance of 310 ohms at 1 kHz. So, at 10 kHz the required drive
voltage is

A required drive level of 135 volts peak is not encouraging, but do
we really need that much? Well, the highest note in a typical guitar is the D
that is two octaves and a whole step above middle C. This note's frequency is
1175 Hz. In addition, a string that is fretted that high doesn't put out much
power in its harmonics. To add to the good news, a guitar puts out maximum
volume when a chord and not just a single string is played. By definition, all
other notes in any chord containing the high D must be lower in frequency. So,
for an undistorted guitar, the reverb doesn't need any more headroom than that
required to reproduce a 1 kHz signal at full blast. For the subject reverb unit,
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
use a high-voltage op amp if a higher-voltage supply is available or switch to a
lower impedance reverb tank. An 8BB2C1B might be an obvious choice if the latter
tactic is used. Its drive coil has 538 turns and an impedance of 190 ohms at 1
kHz. To drive it fully at 1 kHz requires

Note however, that the tanks require 44 mA and 56 mA respectively to drive to
peak level. Therefore a high-current op amp of some sort is required. I've found
that an OPA551 works very nicely and can also take supplies up to ± 30 volts for those who want extra headroom or who are
amplifying signals with more high-frequency content.

If you would like to use the 8BB2C1B tank, the drive-circuit component values
can be changed to the following:

R1 = 56 ohms

R2 = 1.6k

C1 = 20 uF

C2 = .033 uF

The attentive reader will note that the requirements discussed above were for
an undistorted guitar. If a heavily distorted guitar signal is fed into the
reverb tank, the amount of drive voltage required will increase substantially,
because the clipped guitar signal contains much more power at higher
frequencies. Interestingly enough, not being able to reproduce the hard edges of
a clipped guitar signal doesn't sound objectionable to my ears. It just means
that the reverb part of the amplifier's output will sound a bit softer and less
distorted than the main signal. I kind of like that, but another listener may
not.

Signal-Recovery Amp

The most basic decision for this stage is whether to use an inverting or
noninverting topology. The noninverting topology shown below is preferred
because it allows the values of the gain-setting resistors to be optimized for
lowest noise while allowing the input resistance of the stage to be set
independently.

The schematic above shows all the necessary circuit elements. Resistors R1
and R2 set the gain of the stage. The requisite gain will depend on the rest of
the system's gain, but gains between 50 and 100 are common. The test circuit
uses a gain of 101, with R1 set at 68 ohms and R2 set at 6.8 kohms. The value of
C1 is chosen to roll off the low-frequency gain of the circuit. The test circuit
uses a 10 m F capacitor, producing a 3 dB point of
about 230 Hz. There is considerable leeway in the selection of the 3 dB
point-let your ears be the guide. C2, which is 1500 pF in the test circuit, and
R2 form a low-pass filter at 15.6 kHz; with such high gain, it is a good idea to
keep possible out-of-band noise from subsequent stages.

The inductance of the reverb unit's pick-up coil will resonate with the
capacitance of the coil's winding plus the capacitance of the shielded cable
used to connect the coil to the circuit. In the case of the 8DB2C1B unit, a
short cable will produce a resonance above 20 kHz, and R3 is not needed. For
other units, or if a long cable is used, the resonance can move into the audible
range. R3 can be used to damp this resonance. Its value can be calculated using
the equation presented earlier for determining Q. Alternatively, it may be
easier to determine its value empirically by checking for an impedance peak.

Some designs call for placing a capacitor across the reverb unit's output to
purposely place the resonance at some audible frequency. Although this technique
does produce more output at resonance, it does so at the expense of raising the
input impedance at that frequency. In my experience, any increase in output is
more than offset by an increase in noise. In addition, the response emphasis is
not pleasant to this author's ears. For lowest noise, it is best to damp any
in-band resonance with R3.

The most important consideration for the design of this stage is low noise.
As previously stated, the stage needs quite a bit of gain. As we did for the
drive circuit, let's investigate all the noise sources. First, the Johnson noise
of the reverb unit's output coil is connected directly to the stage's input and
will be amplified along with the signal. The 8DB2C1B has an output resistance of
200 ohms. The Johnson noise of this resistance is 1.8 nV/rt. Hz. In the test
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
must be kept smaller than the resistance of the reverb coil if it is not to add
appreciable noise. Because noise sources add in an r.s.s. (root sum squared)
way, if R1 is less than one-third the value of the reverb coil's resistance, the
noise it will add will increase the total noise by less than 1 dB. The 68 ohm
resistor used in the test circuit meets this criterion. Note, however, that some
resistors have noise considerably in excess of Johnson noise. Carbon composition
units are notorious in this regard and therefore shouldn't be the first choice
for R1. The Johnson noise of R2, because it is not amplified, is negligible.

Because impedance at the noninverting node of the op amp is low, op amp
current noise at this input causes negligible output noise. Current noise at the
inverting input is multiplied by the value of R2. Still, this noise source will
also be negligible, even for a bipolar-input op amp. Finally, the op amp's
voltage noise is amplified along with its signal. Unless one uses a very quiet
amp, this noise will be higher than the Johnson noise of the reverb coil, and so
will be the dominant noise source of the stage. If an op amp with an
input-referenced voltage noise 5 nV/rt. Hz is used, the total output noise of
the stage is about 540 nV/rt. Hz. Over a 10 kHz bandwidth, total noise is about
60 mV r.m.s., quite a respectable figure.

In many designs, the reverb unit is attached so it does not electrically
contact the amplifier's chassis. As such, no ground loops will be produced. If,
however, the reverb unit electrically contacts the chassis, the ground
connection at the op amp input may have to be eliminated to avoid hum. If so
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
resistance.

Choose C1 to roll off the low-frequency response around 200 to 300 Hz.

If necessary, install R3 to damp in-band resonances.

Use the lowest voltage-noise op amp you can afford.

Break the ground loop if the reverb tank is connected to a grounded part
of the chassis.

Conclusions

The test circuits have been auditioned by several guitarists, all of whom
have pronounced the quality of the reverb to be excellent. I have seen many
statements that claim that op amps should not be used for interfacing to spring
reverbs because they are too noisy or are unsuitable in some other way. I can
only assume that such statements are based on misconceptions or poor circuit
design. I believe that this tech note shows conclusively that op amps provide a
nearly ideal mate to a spring reverb. In the combination of performance, size,
and cost, no other circuit element can do such a superb job.

Return to
home page