DIY questions

You don't need to know how the electrical parts inside opamps work in order to make a useful circuit.

When you read basic opamp material you can treat it as a 'black box'.
Resistors are there to define gain and operating points (ground levels)

Capacitors in C'Moy circuits can be there for power supply decoupling (across power supply lines) or as coupling caps or they are used to limit the frequency response.
It depends on where they are.

As a C'Moy is THE most basic opamp circuit (together with a signal follower and comparator function) any opamp text book will be able to educate you on the application of them (not about what each component inside does)
http://www.electronics-tutorials.ws/opamp/opamp_1.html

 

A good place to start would be look at simple 1 transistor circuits and learn how (why) theses work.
Wikipedia is one easy source: https://en.wikipedia.org/wiki/Common_emitter

An opamp is just many transistors, resistors and sometimes diodes arranged in different stages and topologies. By looking at opamp's circuit it can be often seen with experienced eye why or what these different sections are there for. It takes some practice and maybe writing/calculating through to get a good sense how a simple one transistor circuit works. I wouldn't try that for opamp though, there are software packages for this.

 

I love the Idea. I love DIY. I also love electronics!

CMOY is pretty easy to go through because there aren't many parts, so lets go through the schematic posted on his site.

I'll do this in 2 sections, the Power section and the Amplifier section.

Power Section Basics

When we get to the amplifier, we're going to need power. The op-amp uses both positive and negative voltages in amplifying the signal, so we need to "make them".

2 9V batteries are used to generate 18V.

RLED and D1 are just there to let you know that the power is on. Nominally, the LED when its conducting will have (for round numbers) 2V dropped across it (this value is due to the material properties of the semiconductor that the LED is made from. Depending on the color it may range from 1.7ish to almost 4). Below 2V, the LED doesn't really conduct at all (very high resistance). Above 2V, it behaves almost like a short circuit (it has very low resistance). RLED is put in place to prevent huge amounts of current flowing through the LED. If D1 has 2V dropped across it, the resistor has 16V. From ohms law, we see that there is 1.6mA running through the resistor and LED.

R1+ and R1- are there for a somewhat subtle reason - creating a "0V" reference voltage, sometimes called "ground". There is some subtlety here that even some electrical engineers do not properly grasp (though they should!). Voltage is an electrical potential difference between two points. You MUST have 2 points at different potentials in order to have "voltage". In our circuit, for example, there is 9V between the + and - terminals of each battery (nominally). There is 18V between the + of the top and - of the bottom battery. You see that how we observe "voltage" depends on where we measure.

This is important because by creating a voltage across the voice coil in our headphones, we make the transducer move. When there is no signal, we want the voltage difference to be 0V. BUT! In order to have 0V across something, we need to attach both ends to something! It is common to attach one end of the voice coil to "ground" (the little pointy 3 line arrow symbol). The R1+ and R1- resistors exist to "create" a "ground". Ground is simply a node in our circuit we are designating as and designing around as that 0V reference point.

In reality, R1+ and R1- are a voltage divider. Current flows through R1+ and R1- and because they are of equal value, the voltage dropped across each is equal. If we start with 18V, each has 9V dropped across it. We have decided to call the node between the two "ground" and so we will refer to its voltage as "0V". If we measure from "ground" to the - terminal of the lower battery, we will get -9V. Likewise the + terminal of the top battery we will get +9V. The decision of which point to call ground is technically arbitrary, but it matters a lot when we get to the amplifier section.

Using a voltage divider to set a voltage reference is fraught with possible problems, but I'll follow up on those later in an "advanced" post.

The capacitors C1+ and C1- are there to help stabilize the voltage across each resistor. Capacitors store energy in an electric field between two conductors. Because they store energy, you cannot change the voltage across them instantaneously. You can think of their inability to change voltage instantly like a big bucket of water with an inlet at the top and a drain at the bottom. Initially, it takes a while for the bucket to fill with water, but once it's full, it takes a while for it to drain. The Thus, when power is applied, the capacitors "fill up". They are attached between GND and their respective power rail (+9 and -9), and they keep the voltage across themselves stable. Thus our power delivery system (+9, -9, and GND) is stabilized. If all of the sudden, +9V requires a bunch of current, the capacitor will give up some of its stored charge to help, which keeps the voltage steady. When this need for current subsides, it fills back up.

 

Amplifier Section Basics

The meat and potatoes of the CMOY is the op-amp. Op amps are magical little electronic devices that have a very specific set of characteristics. Ideally, a perfect op amp would have infinite gain and infinite input impedance. All op-amps amplify the difference between their + and - terminals. Thus, if the + terminal is more positive than the - terminal (by even a microvolt), the output would theoretically be + infinity. If the + terminal is more negative than the - terminal, the output would theoretically be - infinity. Further, no current at all flows into either the + or the - terminals, because they have INFINITE impedance (they are as an open circuit).

In reality, there are some limitations on op-amps. There's no infinite gain. Gain is VERY high, but still not infinite. Input impedance is not infinite (it is usually very high, like MΩ high). Further, there's no way to output more voltage than is available at the power supplies. Thus if + is slightly more positive than the - terminal, the output will go to the positive voltage rail (or close to it, depending on the limits of the op-amp's output stage). If + is slightly more negative than the - terminal, it will go to the negative rail (or close to it).

What good is this, to have an amplifier with so much friggin gain??? If even the tiniest input voltages will cause the output to swing wildly between the + and - rail????? Well, my friend, what you need is feedback!

Looking at the schematic, we see that the input (through some components we'll talk about in a bit) ultimately ends up at the + terminal. Let's say that the input starts to become positive with respect to the - terminal. The output will start to go positive as well. We also see that there is a connection between the output and the - terminal. Thus, when the output starts to go positive, the - terminal also starts to become positive. Thus, some of the difference between the + and - terminals starts to get canceled out! This is called "negative feedback". We use a portion of our output to counteract itself!

What will the output value ultimately become, now that we have negative feedback? Good question, lets look. If a tiny difference between + and - causes the output to change, by the inverse we can assume that if there is absolutely no difference between + and -, the output will stop changing. Thus, the output voltage will continue to rise until the voltage at the - terminal EXACTLY MATCHES the value at the + terminal. Exactly what voltage this is depends on R4 and R3.

You'll note that R4 and R3 form a voltage divider, with the node between them attached to the - terminal. In our case, with R4 being 10k and R3 being 1k, if we have an output voltage of 11V, the - terminal voltage will be +1V. Thus, if the + terminal is at +1V, the output will rise to +11V, making the - terminal voltage +1V. At this point, the output voltage will stop rising! By this, we also know that the gain of the circuit is 11. Gain is Vout / Vin. 11V / 1V = 11! Keep in mind that our maximum output voltage is only 9V. If we try to output more voltage than we have, our waveform will clip and be flattened off causing all manner of badness.

Now lets talk about the pot, R2, and C2.

The pot, or potentiometer, is a variable voltage divider. If we turn the pot so that the wiper is all the way at the input side of the resistor, the value at the wiper will be the full input voltage. If we turn it the opposite way so that the wiper is at the ground side of the resistor, the value at the wiper will be ground. If it's anywhere inbetween, it will be a scaled version. This is your volume control. If the input voltage is nominally 1V peak to peak and you turn the wiper so that the resistance above and below the wiper is equal, the output voltage will be 0.5V peak to peak.

C2 is a coupling capacitor. Remember that I said voltages across capacitors cannot change instantaneously. If you apply a voltage across them, current will flow into the capacitor (or out of, depending on polarity) until the voltage across the capacitor exactly equals the input voltage. Once the voltages are equal, current stops flowing (if there is no voltage difference, ohms law tells us there can be no current flow). This is important because it means that non-varying current ("DC current" as the Department of Redundancy Department would say) cannot flow through the capacitor. Because the voltage cannot change instantly, it appears that time-varying current will flow. Thus our audio signal, which varies with time, will just shake the capacitor voltage back and forth without changing it. Our audio signal will pass through the capacitor, but any evil DC offsets will not! R2 is in place to ensure that the + terminal side of the capacitor on average remains at 0V (e.g. imagine that there is no signal on the input any necessary current will flow through R2 until there is no voltage dropped across it.)

If it were not for C2 and R2, and we attached our amp to a piece of crappy gear that has a 1V offset between its signal pin and its ground, our amplifier would amplify it by 11, driving our amp to its + rail! We would never get any sound out of it and our headphone speaker would fully extend! Bad bad bad!

I think this more or less covers it. As far as specific values go, the ratio of R4 to R3 sets the gain. The value of R2 and C2 don't matter that much except that C2 must be large enough with respect to R2 (bigger as R2 gets smaller) so that very low frequencies aren't attenuated more than high frequencies (read about RC circuits to find out more). The value of R2 and the POT are related such that R2 should be quite a bit larger than the POT. If it were close to it in value, you would get additional signal attenuation because it would act like a parallel resistor to the bottom half of the pot resistance. The pot value in parallel with R2 is also the input impedance of the amplifier. You want your input impedance to be high with respect to the output impedance of your source so that most of the signal shows up at your amp! Those pesky voltage dividers!

 

Power Supply Advanced

A "brief" word about R1+ and R1- as a means of setting a voltage reference.

Using a voltage divider to set a reference can be a bad idea if you are not careful about what you are doing. Imagine your load (our amplifier) as if it were a resistor- rather two resistors, one from +9 to GND and one from -9 to GND. You would intuitively see that your load forms a parallel resistance to R1+ and to R1-. For good luck, lets say your amp consumes 0.1mA on the + and 0.1mA on the - with no signal. 9V / 0.1A = 90kΩ. 90kΩ || 4.7kΩ = 4.46kΩ. A voltage divider between two 4.46kΩ resistors still produced a 9V drop across each, thus the GND is still evenly split between the two rails.

Now imagine your amplifier hypothetically consumes more power on the +9V rail than the -9V rail. Thus the resistor between +9 and GND is a little bit lower in value than the resistor from -9V to GND. 0.2mA on the + side and 0.1mA on the - side. Now we have somewhere around 45kΩ || 4.7kΩ on the top and 90kΩ || 4.7kΩ on the bottom. Now we have a voltage divider between 4.25k on the top and 4.46k on the bottom*. Instead of 9V on the top and 9V on the bottom, now we only have 8.8V on the +9V rail and 9.2V on the -9V rail! You see the problem we have! The value of our voltage rails are dependent on their load! How terrible!

This problem gets worse the higher the load is (1mA vs 0.1mA) and the larger the imbalance is.

NOW if the load on each rail is even or very small, the difference may not matter very much. Indeed that is the designer's tradeoff taken here. Why aren't we just calling the node between the batteries ground? It seems that the problems caused by very likely mismatched battery voltages are very likely worse than those caused by the voltage divider.

One note I'll make is that, provided C1+ and C1- are big enough, time-varying load differences caused by audio signals don't actually matter. Their demands are transient and supplied by the stored energy in the capacitors. We are only talking about load differences that are either time-invariant or have a longer period than the RC time constant formed between R1 in parallel with the load and C1.

What solutions do we have to this problem?

1. Make R1+ and R1- smaller in value, thus making load changes less significant (this has a negative effect on battery life as there will be a higher nominal current drain.
2. Regulate the + and - rails with LDOs. Here we would call the node between the batteries GND and regulate the + and - rails to 8V or something using a linear voltage regulator (the negative here is twofold, reduced voltage rails and additional part cost)
3. Use a circuit called a servo amplifier to set GND (the negative here is additional part cost)

*The actual value should take into account the reduction in voltage, but I chose not to due to laziness.

 

An easy solution is to use 2 x 9V battery connected in the middle to 'ground' BUT if one of them is (accidentally) disconnected some DC might appear on the output.
Some headphones may not be amused.
Most opamps have little to no problems with the positive and negative rails being somewhat different.

Another easy solution is to use a rail splitter like the TLE2426
Their low idle current (0.4mA) and decent stabilisation current capabilities (+/- 20mA) make them suited for battery operated devices such as this while maintaining a perfect divided voltage.

If you want all these railsplitters to work decent I would advise to increase the capacitance of C1 and C2 from 220uF to 1000uF minimum.
Certainly if you plan to drive low impedance headphones.
The bigger they in value are the more 'stable' the 'ground' will be.
No need to go above 4700uF.

When only driving 300Ohm headphones 470uF is more than enough.

Indeed you can also use a 'follower' circuit or DC servo amplifier but more often than not these circuits don't like to see any capacitances connected to their outputs.
C1 and C2 should be on the input side of that circuit and should not be mounted on the output of that circuit.
Punishment could be devices breaking down or oscillation.

 

It applies to a voltage splitter circuit and is regardless of the power supply feeding it.

A TLE2426 rail splitter can provide a small amount of current in both negative and positive direction but can not deliver an equal amount of current in both directions.
All audio currents (which are AC by nature) thus must come from another part.
This is where the 2 capacitors come into play connected to its output.
The Pimeta power supply uses a TLE2426 (bottom page).
Found this test about TLE2426 and C'moy as well which also touches on the subject.

You can use a simple LM317 & LM337 dual power supply in which case you don't need the rail splitter (in fact you cannot use it) nor need > 1000uF of caps on the power lines.
In this case indeed just simply feed the circuit from +/- voltage.
Mind the maximum power supply and max output current ratings of the opamps you use.
The test I linked to above shows a few (most outdated) options for opamp choices.
As the opamps need to drive headphones directly these properties are of crucial importance.

Using too much solder, or a too low temperature or not allowing enough time for the solder to 'flow' can result in bulbous solder blobs.

 

A note about regulators: they aren't perfect.

Here's an article with some noise measurements and circuits just as an aside...
http://www.tnt-audio.com/clinica/regulators_noise1_e.html

So I'd say... use huge capacitors anyway... if you can. The bigger they are, the more stable things will be, and the less noisy, too. Batteries especially can use a bit of help with regards to regulation.

You can try to omit capacitors for space-saving reasons for a portable amp, but for a desktop topology, especially for the power supply, I wouldn't skimp and think the regulators are enough. When your load draws a lot of current (such as when you're driving an ortho or e-stat headphone), the regulators can be quite strained.

 

It was my impression that an electrostatic headphone draws very little current but has very high voltages. Orthos definitely draw a lot of current.

 

An LM317 and LM337 or 'similar' regulators won't be 'stressed' in 132/134 C'Moy based amplifiers.
The currents are limited by the used opamps (around +/- 50mA/channel), not the regulators.
The max output current is thus limited by the opamp's output stage, not the headphone impedance nor the regulators.

Be careful with adding larger value capacitors or very low ESR capacitors behind regulators.
NOT all regulators like to see large amounts of capacitance on it's output.

It certainly may not improve stability nor noise levels, in fact it may even be worse as large capacitances DO stress the regulator trying to 're-fill' them with a limited current.

Bigger values = 'better' does NOT apply here.

 

Despite that the op amp may output maximum a certain amount of current, it still has to draw from the power supply, and the amount it draws is obviously going to have to be greater than whatever it needs to deliver to the load/headphone. So I think it is better to "overestimate" in this case. Also doesn't hurt to have an overkill power supply in case you wanna try going Class A...

The larger value capacitors are obviously not going to go after the regulators, but before. I was under the impression that the previous suggestion meant one can skimp on capacitors before a decent enough regulator implementation, and I don't quite agree due to the regulators themselves not being perfect.

It's strictly a power supply thing. The headphone is the load for the amplifier, but the amplifier is the load for the power supply. While the headphone may draw very little current, the amplifier may need more.

Typical wall sockets in the US are +/-110V, which isn't really enough for e-stat (SR-009 needs 580V DC for bias, for instance), so the voltage has to be "stepped up". This "step-up" process typically means drawing high current from a lower voltage and convert that to a higher voltage with lower current. In that case, power delivered is the same and is conserved (hopefully, but what I've learned so far is that any kind of conversion always has some amount of loss, no matter how small).

For instance, if I want 660V output from 110V input, then the ratio between them is 6:1. That means I can convert 6A @ 110V into 1A @ 660V. In this case, the power delivered (which is hopefully well-conserved during the conversion process) is (6A * 110V) or (1A * 660V) = 660W.

Obviously, most e-stat headphones on the market don't need 660W of power, but that's just to illustrate how they can still be a heavy current load on the power supply. If my load doesn't demand 1A @ 660V, then I can reduce current on the other end of the conversion accordingly.

 

The 'overhead' in this case is about 5mA !

If you want a 50mA opamp to be in 'class-A' (when not loaded it IS in class A) it would have to run at around 20mA idle current leaving an Asymmetric clipping circuit.
Some prefer 'class-A' for opamps in circuits where it doesn't have to drive any substantial loads.

Not applicable to C'Moys as the 'extra neede current is very low (mA's).
Remember OP seeks advice about the C'Moy, not power amps.

The bias voltage needed for e-stats draws almost NO current at all... a few micro-amps (on startup) only !
The audio circuit could be either an audio trafo (with a huge ratio) or a tube/SS circuit which doesn't need to handle a lot of current.
It only needs to 'charge' something between 10pF and 100pF.
Of course at full power and 20kHz this is still substantial.
Fortunately in music the HF contents are realtively small.

Irrelevant if the voltage goes up or down.
The output power of any transformer = input power + small losses is slightly more drawn power than it can output.

 

The regulator circuit you are pointing to is the 'standard' application for the positive rail LM317 power supply and is anything but optimal.
The data sheets of the LM317 and LM337 will give you pointers on how to use this chip.
R1 and R2 (their ratio) determine the output voltage where R1 always has 1.25V across it (the internal reference) and R2 the total output voltage - the 1.25V.
The circuit is not optimised for lowest noise though.

THIS is what you need to build as a power supply for a C'Moy type amplifier though.
For amplifiers that need to provide peak currents above 600mA per channel, or ar heavily biased in class-A there are much better designs.

The PCB design, however, is far from optimal.
C1, C2, C8 and C9 must be as close as possible to the pins of the regulator.
Diodes CR1 and CR2 are only needed when the power supply is loaded with 100uF capacitors or higher.
C3 may be increased to 100uF
Of course I am talking about the circuit used in the link in the word 'THIS" above.

You only need to add a transformer.
The voltage rating of the transformer needs to be based on the desired output voltage.

The power supply can deliver 1.5A so 0.12A max drawn by the C'Moy is no problem.

 

It's an advice thread, not really a contest to see who's right, so I'm not gonna go back and forth, but just to say... my original intent was simply to show that regulators are really not perfect.

Perhaps in this case, when you are implementing a power supply only for the original cmoy (with no extra modifications like biasing the op amps into Class A using a current source, adding a buffer, running said buffer also in class A, etc...), then just as T.rainman said, whatever I suggested would have been overkill.

However, I'd say... building some extra overhead never quite hurt anything IF your application can afford it. The best case IMO is to build a super overkill power supply now, then think of how you can scale your cmoy so that later on, you can build a better amp section that may just be more demanding than the cmoy. Seriously, if we're talking about overkill, whatever I suggested doesn't quite stand up to the monstrosities that are in something like the beta22... or Dynahi.

As for the LM317, the datasheet will give you some pointers, but ultimately, the trade-off calculations are up to you.

Here's a thread discussing the different values of the adjust resistors vs output capacitor on DIYAudio (note: ya need to register and log in to see pictures):
http://www.diyaudio.com/forums/power-supplies/143539-another-look-lm317-lm337-regulators.html

Here's another TNT-Audio article on ripple rejection and how bigger capacitor on the output of the regulator (up to 2200uF) has an impact on ripple rejection in the circuit (note: at this point, I'm not talking about the LM317 implemented like that, I'm suggesting a different regulator circuit...):
http://www.tnt-audio.com/clinica/regulators3_ripple_e.html

And if you're worried about capacitors drawing too much current, just put a resistor in series as the TNT article suggests. I think that's quite sufficient.