Tuning the Gain
In an op-amp circuit, the resistors in the feedback loop (R3 and
R4 in this case) set the gain: the amount by which the output voltage
is multiplied compared to the input voltage. The formula for gain in
this amp is

To change the amp’s gain, you can change whichever of the
two resistors is most convenient. You should pick values somewhere
in the 1 KΩ to 100 KΩ range. Lower is better, for a couple
of reasons. First, high values will increase your circuit’s
suspectibility to stray RF and electromagnetic noise. Second,
higher resistor values are inherently noiser. (You can explore
this phenomenon with my op-amp noise
calculator.) If you go much below 1 KΩ, the current wasted
through the feedback loop will become significant.
The default gain of 11 is fine to start with. It won’t let you
use the full sweep of the volume knob unless you have very inefficient
headphones or a very weak source, but it does provide good, stable
operation of the op-amp. As you lower the gain, you make it more and
more likely that the op-amp will become unstable. Indeed, some op-amps
won’t work reliably at low gains. The OPA13x series chips are
“unity gain stable,” meaning that you can take them down
to a gain of 1, if your circuit is built soundly. (You get a gain of
1 by connnecting the output straight to the inverting input.)
I have used a gain as low as 2 with inefficient Sennheiser 500
series and AKG K x01 headphones, when the source is strong
enough. Similarly, a gain of 2 or 3 with efficient headphones should
work fine even with weak sources. A gain of 5 or 6 is practical for
almost all combinations of sources and headphones.
For your first amp, build it with a gain of 11 at first. Then
once it’s working correctly, try out different feedback
resistor values, checking carefully for audible distortions with each
change. Once you get to a gain level you like and it’s stable,
box it up.
Quieting Noises (R5)
Normally the R5 position in the CMoy amplifier is shorted.
(0 Ω) This gives the amplifier the best control over the
headphone drivers, and results in the best sound in most circumstances.
Sometimes, however, putting some resistance in the R5 position is
necessary to quiet some sort of noise you hear in the headphones.
Before I go into more details, please read all the information
on noise in the companion article Basic Troubleshooting for Headphone Amplifiers. The
solutions in that article, if they work, are superior. Adding R5
should only be done as a last resort.
Having tried everything else to get rid of noise, try putting
10 Ω to 100 Ω in the R5 position. Higher values will be
more effective, but because lower values have fewer bad side effects,
your incentive is to use the lowest value you can get away with.
Adding a DC Power Jack
Wall Power Only
The simplest solution is to wire up a standard DC power jack to the
power supply. I like the barrel type connectors, since they’re
the safest for this application. I’m a particular fan of the
5.5mm/2.5mm size, because it’s the sturdiest readily available
type, and my favorite line of commercial audio-grade power supplies
(Elpac’s WM series) uses this connector style. My second choice
would be the 5.5mm/2.1mm style, which is similar but with a smaller
center pin.
Batteries Plus Wall Power: Easy Method
The main problem you have to solve if you want to run your amp
from both batteries and a wall supply is how to prevent the wall
supply from trying to charge the batteries. Regardless of whether
you use alkalines or rechargeables, simply connecting a wall supply
across them will kill them.
The easiest way to solve this problem is to use a “closed
circuit” DC power jack. These add a switch and a second circuit
to a regular power jack. This second circuit is normally closed; thus
the name. When you stick the DC power plug into the jack, the second
circuit opens. If you connect the batteries through the second circuit,
they get cut out of the circuit when you plug the wall supply in.
I show how to wire a closed-circuit DC power jack in the companion
article, How to Wire Panel Components.
Unfortunately, the switch inside some closed-circuit jacks is rather
delicate. The DC power plug can snag on the switch while you’re
inserting it and bend it. If you’re lucky, you’ll just
bend the switch contact a little bit, so that you can bend it back
with a small screwdriver. If you’re unlucky or ham-handed,
you’ll bend the switch contact completely out of shape and
be forced to replace the jack. To some extent, you can solve this
problem by paying particular attention to the DC power plugs you use:
the better kind for this application have chamfered leading edges,
so they’re less likely to snag than the blunt kind. The high
road is to avoid closed circuit jacks if you don’t absolutely
have to use them.
Batteries Plus Wall Power: Clever Method
It’s possible to have our cake and eat it, too. We can use a
plain, switchless DC power jack and still have both battery and wall
power. The easiest way to do this is to use a diode OR bridge:

Diode OR Bridge
The name comes from the fact that it allows only one power source
or the other to run the circuit, never both at the same time.
This prevents the wall supply from trying to charge the battery when
both are connected.
The diodes can be any generic type, such as the common 1N4001.
The bar on the diode symbol in the schematic represents the
diode’s cathode, and the corresponding terminal on the physical
diode is marked with a stripe. So to build this circuit, you simply
connect the two cathodes together with the V+ line to the amplifier,
connect the positive side of each power source to one of the other
diode legs (the anodes), and the negative side of each power supply
to V- in the CMoy amp.
There are a few restrictions with this method. First, because a
typical diode has a forward voltage drop of around 0.7 V (the
“diode drop”), this can prevent you from getting the full
use out of your batteries. (Whether this is the case depends on matters
discussed in the companion article, Op-Amp Working Voltage Considerations.) The
second restriction is that the wall power supply must be at least
one diode drop higher than the voltage of a full battery, or else it
will not reliably cut out the battery. Beware that fresh alkaline
batteries have a higher terminal voltage than they are rated for;
a fresh “9 V” battery actually measures more like
9.6 V, for instance. Rechargeables similarly have higher
terminal voltages than their nominal voltage when fresh out of the
charger.
The way the diode OR bridge works is actually quite simple. A
diode only conducts current when its anode (positive) voltage is at
least one diode drop above its cathode (negative) voltage. Since
the cathodes are tied together, connecting the wall supply pulls
the common cathode up so it’s less than a diode drop below the
battery voltage, so only the wall side diode conducts. (In practice,
you usually overdesign it so the common cathode is actually higher
than the battery voltage.) When the wall supply is removed, the
common cathode point falls back down to one diode drop below the
battery voltage, so the circuit still gets power.
Let’s work a concrete example. We have 2×9 V
for our battery supply, which we’ll conservatively estimate at
19.2 V when fresh. We’ll use standard 1N4001 silicon
diodes, which have a diode drop of 0.7 V at the low currents
you’ll see with a CMoy amp. And finally, we’ll use a
24 V wall power supply. So, with only the batteries connected,
the common cathode point is 0.7 V less than 19.2 V,
or 18.5 V. When we connect the wall supply, the wall side
anode goes to 24 V, which raises the common cathode to
23.3 V. The battery can never run the circuit while the wall
supply is connected, because the battery side anode is always at
least 4.8 V lower than it has to be for the battery side diode
to conduct.
Picking a DC Wall Power Supply
If you do the above tweak, you’ll need a DC power supply of
some sort. (As opposed to an AC power supply. Powering a CMoy with
AC will fry it right quick!)
The most important feature to look for in the power supply is
isolation. This means that there is no direct connection between
the AC side of the power supply and the DC side, which is necessary
for the CMoy’s virtual ground circuit to work properly in many
common situations. You can test whether a supply is isolated by using
an ohmmeter to test for conductivity between all combinations of input
and output pins on the supply; it’s most common in non-isolated
supplies for the V- output to be tied to either earth ground or AC
neutral on the AC input side of the supply. All linear-regulated and
unregulated supplies are isolated. The other main type of supply is
switching-regulated, which are often not isolated. If you’re
not sure whether a particular power supply is isolated and can’t
test it, assume that it isn’t.
The next most important spec is whether it’s regulated
or not. You can count on an unregulated supply to put a lot of noise
on the amp’s power rails, which may get through to the output
in audible amounts. (See my article Op-Amp Power Supply Quality Considerations
for further details.) Regulation fixes this problem. The two most
common types of regulation are linear and switch-mode.
Switch-mode regulation also puts out a fair bit of noise, though
of a different type than you get from an unregulated supply. Whether
or not this is audible is a complicated issue. See the above-linked
article for more discussion on this topic.
Linear-regulated power supplies are best for high-performance
audio. Unfortunately, there aren’t that many commercial linears
suitable for audio DIY on the market. The most readily available ones
are a few models in the WM series (not all of them!) from Elpac Power
Systems, carried by Digi-Key, Mouser, Newark and Allied Electronics. By
far, the most popular is the WM080 (24 V/330 mA), Newark
part number 16F126.
It’s possible to DIY your own linear power supply, with a
bit of work. I offer a few different power supply
boards which are designed with headphone amp circuits in mind.
If you want to pick a different power supply from the ones
I recommend above, it should supply at least 12 VDC,
and no more than 30 V. (See the companion article
Op-Amp Working Voltage Considerations for further information on that
topic.) The common voltages are 30, 24, 18, 15, and 12, in that
order of preference. Check the power supply voltage specs in your
op-amp’s datasheet to be sure it can accept the voltage you
choose! Historically, most op-amps work with supplies up to 30 V,
but increasingly, modern chips have lower voltage tolerances. Check
the datasheet to be sure! You’re looking for the recommended
maximum voltage, not the absolute maximum.
The amount of current your supply can put out really doesn’t
matter for this application — even the crummiest little wall
wart can supply 100 mA, whereas your CMoy amp shouldn’t
ever draw more than about 20 mA.
Using Different Caps
There are two main numbers that describe a capacitor: its voltage
rating and a measure of how much energy it can hold, given in fractions
of a farad. (One farad is a very large amount of capacitance.)
Voltage Rating
The voltage rating simply tells you how many volts the cap will
tolerate. You should use a cap with a rating that’s greater than
the largest voltage you expect it to see. However, higher voltage
ratings mean physically larger cases and higher cost; there are no
secondary performance benefits from using caps with a higher voltage
tolerance than necessary. Consider the power caps: if we’re
using a 2×9 V battery supply (18 V), a 16 V
capacitor in the power supply would be insufficient, a 25 V
cap would be fine, and a 35 V cap wasteful.
You might think you could actually get away with a
10 V cap for our example, because the virtual ground
divider is supposed to divide the power supply voltage by 2,
putting only 9 V across each power supply cap. The first
problem with this logic is that the divider in the CMoy will be imbalanced in actual use. If you to use
10 V power supply caps with our 18 V example power supply,
it’s very likely that one of the two caps will see more than
10 V and be damaged. The second problem stems from the fact that
you should never optimize something like voltage tolerance assuming
that the circuit is working properly. Let’s say you’re
using 10 V caps with a 15 V power supply instead. In
this case, it’s quite unlikely that you will get more than 10V
across one side of the supply in normal operation. But, what happens
if something goes wrong? There are several ways the virtual ground
divider could fail, putting all 15 V across one side of the
power supply. Bye-bye power cap.
Bottom line, it’s prudent to make your power supply
caps’ voltage tolerance at least equal to your rail-to-rail
power supply voltage in this amp.
Capacitance Rating
The caps’ values will change the way the amp performs.
For the power caps, 220 µF is adequate, but bigger ones will
provide a bigger current reserve, which can be useful in handling high
instantaneous loads, like big drum hits. I’ve tried 470 µF
caps in my CMoy amps before, and it does improve the bass handling
significantly and they’re not all that much bigger than 220s.
As for the input caps (C2), a larger cap will improve bass
handling, but it’s by a different mechanism than with larger
power caps. See the companion article, Input Capacitors for Headphone Amps
for a full discussion. Bottom line, the default 0.1 µF is a bit
on the low side. Try 0.22 µF, 0.47 µF, or even 1.0 µF
instead.
With film caps, the smallest voltage tolerance you’ll likely
be able to find is 50 V. Although even 50 V is way too
high, don’t worry about it: just get the lowest-voltage caps you
can find. If you go with polypropylene caps, you probably can’t
find caps with less than 200 V tolerance. If that’s the
lowest voltage tolerance you can find, go ahead and use them. Just
don’t think that getting the 800 V versions will do
anything more than waste board space and money.
Capacitor Type and Quality
There are many types of capacitors, because we have so many
reasons for using capacitors. No single capacitor type works for
all situations.
For the CMoy power supply, electrolytics are the best choice.
Only they have the high capacitance values and low cost needed for
this task. Different brands of capacitors can sound different.
Many companies make good electrolytics: Panasonic, Nichicon, ELNA,
Rubycon, Sanyo...
Electrolytics do have a downside, which is that their value
tolerance is very bad, and they have high distortion. Neither problem
matters in a power supply, but it would be foolish to use these for a
CMoy’s input caps (C2). For this job, plastic film caps are best.
Good brands are Panasonic, Vishay, Wima...
Not all capacitor lines from the above listed companies are
equal. Do some searching online for advice if you want to use
different capacitors than the one I recommend in the parts lists. Here
is a typical thread. Also, there is some advice in the companion
article, Input Capacitors for Headphone Amps.
Before you go deviating from caps recommended to you, especially
if you’re swapping them into an existing layout, check their
datasheet to see how large they are. If you’re staying with the
same product line and just getting a different value, the replacement
could be larger in diameter, or height, or both. Or, you might get
lucky and find another cap in the same line that’s the same
size as the one it’s replacing. (It happens!) If you’re
switching to a different manufacturer or product line, it’s
doubly important to check the datasheet because it’s common for
a higher quality cap to be larger than a lower quality cap, all else
being equal.
Tweaking the LED Resistor
The RLED resistor is there to limit the current going through the
LED. If you don’t limit the current, the LED will burn itself
up. (Yes, they actually get a bit hot to the touch before they stop
glowing. If you have a spare LED and want to see what happens, just
touch the leads to a 9 V battery’s terminals.)
The formula governing current through the LED is:

Or, we can rearrange that to calculate the proper resistor value,
given the current we want:

These equations might look a little funky at first, but
they’re actually very simple. Let’s say we’re using
a 2 V green LED; that’s Vf, the “forward voltage
drop.” We’ll use an 18V power supply; that’s Vs,
the “supply voltage”. We’ll stick with the default
value for RLED here, 10 KΩ. We’re calculating the forward
current through the LED, I. The top part of each equation
subtracts the LED’s forward voltage drop from the supply voltage,
which gives the voltage across RLED. So in this example, 2 of the 18
supply volts is dropped across the LED, leaving 16 V across
the RLED resistor. After that, it’s just Ohm’s law:
I=V÷R. This tells us that the current through the LED will
be 1.6 mA if we use the standard RLED value of 10 KΩ. The
second equation is just another rearrangement of Ohm’s Law:
R=V÷I.
LED brightness is proportional to the current going through it.
Common LEDs can handle at least 20 mA, but you only need 1 to
2 mA for efficient red, green and amber LEDs to seem adequately
bright. So for our example green LED, 1.6 mA will be fine. Even
with a piggy blue or white LED, I never need more than 5 mA
to get adequate brightness.
In this circuit, the LED voltage and the RLED value are constant.
So if you look at the first equation, current can only vary if
the supply voltage varies. This will happen with battery power
supplies. Taking our example above a step further, what if our
“18 V” power supply is actually 2×9 V in
series? In that case, the battery voltage will drop to about 10 V
by the time the batteries are dead. Now there will only be 0.8 mA
going through the LED. From the “1 to 2 mA” advice
given above, you’d be correct in saying that 10 KΩ is a
little high for RLED in this situation. Dropping it to 7.5 KΩ
or even 4.7 KΩ would help.
Beware that you’re trading battery life for LED brightness
here. You might decide that you can tolerate a dim power LED when the
batteries are nearly dead in order to squeeze a bit more run time out
of them. I find that efficient LEDs reach minimum useful brightness
with about 0.2 mA. Let’s use the second equation, with our
minimum supply voltage of 10 V, our 2 V green LED, and our
0.2 mA minimum current target. We find that RLED should be no more
than 40 KΩ to get minimum brightness at minimum battery voltage.
We might choose 33 KΩ as a compromise, then, between getting
reasonable brightness and getting absolute maximum battery life.