Fresh Fizz

The "this" link looks interesting. Thanks Col, I'm gonna take a look at it. Very theoretical. I'm just a simple guy. I've thought of a way measuring the efficiency of the driver by using a sine wave generator connected to my diy-sustainer amp into the driver (installed on guitar) and measure the output of the guitar pickup in the bridge position. Tune in to a specific tone on a selected string and use different voltage levels. It's a lot of work but it can be done. Yes the input of the CA3080 as well as the LM13700 are easy to overdrive. I can keep my CA3080s clean only by attenuating the signal. The LM13700 uses biasing diodes which for me makes things only more complicated. But I don't know if this takes place in our driver coil. There is no DC current going through the driver coil (blocked by output cap). But when switched on there is half the supply voltage on the + side of that cap. When switched off that voltage goes to 0. Maybe that is causing the popping noise. Could a voltage divider solve this problem? (2 resistors of equal value, one from + voltage to + side of output cap, one from + side of output cap to ground) 2 x 4,7 kOhm would "cost" you about 1 mA. The extra 2,35 kOhm load is a piece of cake for Ruby! Like you reasoned diodes wouldn't work. I'm installing the driver and amp in my guitar so tomorrow I will do some testing. It's a bit ridiculous to act the wise guy when I myself don't even have an operational sustainer! If everything works (more or less) I'll let you know. Wish me good luck!

I only mentioned it as a way of improving efficiency. It could be that when in action the driver coil measures much more than 8 ohm. The F/R amp performs best, gives most power, when driving an 8 ohm load. If the load is higher, than lower current, lower power. Like a 50W lightbulb. When the lightbulb is on the restistance is about 1000 ohm, but cold the resistance is much lower. When a high current is abruptly switched off a huge induction voltage is being created. That's why the coil of relays are bridged (?) with a diode in opposite direction of the voltage supply to suppress that induction voltage. Fizz

Ok, I understand. It's flux density that matters. So what I actually want is some kind of device I can stick my driver onto that will measure the driver's flux density. Even if it is only to get a rough idea. If I have a driver and then modify it has it become a more efficient driver or not? Now it is like modifying, installing into the guitar, listening en guessing if it's a real improvement. Let me tell you first that I have been working a lot on my very own Ross compressor modifications. It works with a CA3080, a predecessor of the LM13700 you're using. Actually it's a limiter. I understand that you want the notes to go into sustain quicker? I think that you have to distinguish two things, how the limiter works and eq. I agree with you that harmonics only wil dominant for a very short time, but even so what can happen? When hitting the strings hard the compressor starts limiting the signal. If the attack time of the compressor is short it means that signals at the very beginning of the tone (let's call it attack) will be used to limit. Compared to the attack the fundamental is relatively weak, assume it is -20 dB then if the attack is used as a reference for limiting the fundamental will still be 20 dB below the attack level. It can be that this attack is only very shortlived but what if the release time of the compressor is long? Say the attack only lasts 20 mseconds and the release time is 200 mseconds. Right after the attack you will experience a softer signal (sagging). So it is important to have a short release time. Rapid reaction is more important than whether pumping or not. Eq (taming the highs) can make the attack softer what makes that the fundamental is relatively louder after limiting. The data you are asking depends on the guitar, type of pickups, position of pickup. Maybe it's better to soundcheck and listen if the compressor sags. Cheers The Fizz

I suppose it's the amount of posts that sets the forum regulars apart from the newbies. A tone around the 12th fret has a belly right above the driver. At the 20th fret the driver is close to a node, but on the other hand the string is much closer to the driver. And there is a symmetry, driver and pickup are as close to a node. You could argue that in such case the driver gets the ideal type of sound to keep the string moving. Lower on the neck, like the 3rd fret, means more room to swing, less stiffness. But then there is some kind of split personality thing. Both fundamental as 2nd harmonic (or 1st harmonic, depends on how you count) are in phase. So in feedback both tones may occur. It will depend on sound characteristics of the guitar and eq (output elco of f/r as low-cut) which tone will feed back. I think it's very difficult to reason why some notes feed back and others don't. I still am not buying the stuff about Helmholtz motion. I am convinced that this motion happens, but I don't see what it has to do with the fundamental. When you pick a string you get a lot of harmonics and even sounds that have no mathematical relationship with the fundamental. The fundamental is masked by these harmonics. These harmonics and noise could cause the diy-sustainer to set in slowly, but don't blame the fundamental for it! Col, what I understand you are saying is that the Helmholtz motion is causing the fundamental to do strange things. But how on earth would that be possible, the nodes of the fundamental are fixed points on the guitar (nut/bridge or fretted position/bridge), they even make traveling kinks bounce back. I think that you find the solution for your problem in pre-agc eq, taming the highs.

They look so slim, they must have been starved to death! Don't Pete. We still need you. Maybe you could even reanimate some of the rats for further testing. I was wondering if it's the best choice to use coils with a dc resistance of 8 ohm. In operation the driver works with currents of 350mA (1W output). So the coil must getting hot. How hot? If the coil gets hot the resistance rises. Think of lightbulbs or heater in electronic valves/tubes. So how much? This could be tested with a 9V battery and a resistor in series with the driver (preferably a lab rat). With a 18 ohm resistor in series with the 8 ohm driver we would get a initial current of 350mA. It's nice to see if that current drops when the driver gets hot or melts (no more current anymore). Measure the voltage over the driver R_driver = V_driver / I = V_driver / (9 - V_driver)/R If the result is a lot more than 8 ohm we could try to use thicker wire, but same amount of windings. That's why I like to talk about test procedures. At which current do we test? Bye

Pete and Col, thanks for replying Maybe it was a bit much for a first post. Maybe it was better first to say what I was aiming at. The way I see it the production of a driver is to pass a series of test procedures. So actually you should think of a ratrace, the drivers and electronics are the rats. Failing a test the rat dies. This is the way how you've been improving your drivers, Pete. You could tell everybody that 0,20 mm copperwire works best because you've experimented with various thicknesses of copperwire. My first post was to show you a manner of how to measure the electrical properties of the produced driver. This would give some indication of the quality of the driver (it's eveness of frequency response). Col, when I spoke of the ideal conductor, I ment it as electronic component, not as driver coil. The 90 degrees phase shift will only occur at very high frequencies. Indeed the power transfer is 0. That's not what we need, we need energy to move the strings. What we need is 0 degrees phase shift. But that's exactly the problem. Building a driver is finding the best compromise. Pete knows all about that. For o degrees phase shift we could stick a big resistor underneath the strings? Nice to keep your picking hand warm, but no magnetic flux at all. Col, you were quick to point out that the core materials, the whole construction, dimensions matter. That's true, that is the quantitive aspect of the driver. We should have a test procedure for measuring the strength of flux produced by the driver. Anybody? How about revolting e-strings? I don't think you can blame the driver for e-strings being so skinny. This has to be solved electrically (eq, agc, overdrive, more power). Threaten to turn it into a f# or g-string! Combining both test procedures would enable us to experiment with coil, magnets, core material and improve the driver design. And I wouldn't rule out the option to include test procedures to tame down EMI. It only looks a bit more complicated to me because it could be driver dependent but how bad the EMI is can only be experienced and evaluated after a complete installation. Col, thanks for the information of strings motion and so on. The clip reminds me of a Pantera video clip. But I don't see why the Heimholtz motion is that important to the functioning of the DIY sustainer. Let me try to convince you. A complex sound consists of multiple frequencies. According to fourier (analysis) a sound can be broken into it's separate frequencies (sine waves). Al these sine waves can have their own life span and phase shift compared to the fundamental. Even a picked guitar string can be considered as a fundamental (standing wave) and harmonics (even heimholtz motion) superimposed on top of it. The harmonics can be so dominant that because of subtraction from the fundamental there appears to be an out of phase situation. But the out of phase situation only counts for the harmonics. Because it's out of phase it won't feed back at all. The fundamental that was buried underneath this powerfull complex of harmonics is in phase and will feed back. Maybe your design experiences some complication because of use of agc. Right after the attack your design limits the signal. It uses the louder higher harmonics as reference. As a result the fundamental is attenuated and slowly released after the higher harmonics have died out. I'm more in favour of using overdrive for sustaining. Better chop off the attack! Having said that I like the low battery sound clip. There are more ways that lead to Rome! The link shows that things aren't as bad as they look. Signal seems to have the wrong phase, but as you can see, the fundamental is completely healthy! (second picture) http://www.geocities.com/SunsetStrip/Studio/2987/fatts.html

Hi diy-comrades, I've more or less read the thread and just have built my second driver. I took your advice and used 0.2 mm copper wire. It's a humbucker with 2 blades. Dimensions: 1.4 x 16 x 57 mm. Coil height: 5 mm Windings: 2 x 60 turns Resistance: 8,4 Ohms My first driver was a single coil design and terribly sensitive to magnetic feedback (pigscreamer). The new driver seems to be able to do the job. I get string feedback without squeal. I've only done some outboard testing with the new one, but I have good hope it will work when it's properly installed. Col you seem to want a more systematic, scientific approach Electrical properties of the sustainer/driver There is an easy way to measure the inductance of the driver. You need (1) a sine wave generator with a low impedance output (mine is 50 ohms) and a constant voltage output level over it's frequency range (2) an oscilloscope (3) a multimeter that measures frequency. My oscilloscope is a very old handyprobe 1, so nothing special. This setup also comes of handy when biasing a fet or setting up voltage levels needed for your opamp(s). Create a notch filter. For that purpose I connect in series: 180 ohms resistor, 680 nFarad capacitor and the driver. Connect this chain to the sine wave generator: resistor to signal, unused driver connection to earth. Now probe the signal in between resistor and capacitor. Now you need to walk through the frequencies to find the resonance frequency where the voltage drops the most. Then measure the frequency (signal of sine wave generator, not in between resistor and capacitor where because of the voltage drop the measuring can be problematic). Calculate inductance L = 1/(4* pi^2 * F_resonance^2 * C) In my case it measured 6000 Hz: L = 1/(4* pi^2 * 6000^2 * 680e-9 ) = 0,001 H = 1 mHenry The highest fundamental is about 1100 Hz. X_C = 2 * pi * f * L = 2 * pi * 1100* 0,001 = 6,9 Ohm Now we know the impedance at 1100 Hz. Real component R = 8,4 ohm, imaginary component X_L = 6,9 ohm Z_1100Hz = (8,4 ^2 + 6,9 ^2 ) ^0,5 = 10,9 ohm There are 2 reasons why at 1100 Hz the driver works less efficiently: (1) higher resistance (10,9 instead of 8,4), lower current (I = V / Z), lower magnetic flux (2) phase shift between voltage and current arctan(X_L/R)= arctan(6,9/8,4) = 39,4 degrees (0 is in phase, 180 is out of phase) Note that 90 degrees is the maximum phaseshift. At 90 degrees phase shift it's all conductance, the driver works as an ideal conductor. The driver won't dissipate energy, it stores energy and then spits it back at the lm386 (in case of the f/r. Therefore a zobel network is used to keep the impedance low at high frequencies and prevent instability. The following is my DIY-theory. Maybe it's incorrect and completely useless, but my goal is to produce some kind of benchmark for drivers so we can compare one driver to another. Maybe there are better ways to express the quality of the driver. As long as there is a way of measuring it, otherwise i don't believe it. It would be nice if I could combine (1) and (2) in one single number, a compensation factor for high frequencies. I will use the term specificaly for 1100 hz. (1) 8,4 / 10,9 = 0,77 (2) use cosinus function normaly used for calculation of power: P = I * V * cos(phase shift between I and V) In order for the sustainer to work the string should be helped. So when the string moves down the driver should pull and when the string goes up the driver should push. Just like a swing the timing when to push and when to pull is important. Phase shift, pushing too late or too soon, reduces efficiency. cos(39,4 degrees)=0,77 (!) (1) and (2) are the same. So the compensation factor should be 0,77^2 = 0,60. If there is a linear relation between current and magnetic flux then we could use this factor instead of phase shift and impedances. The magnetic flux at 1100 hz is 0,60 times the magnetic flux at let's say 50 hz. So in order to get an even response the questions are: (1) is a compensation factor of 0,60 enough? (2) can we construct drivers with a higher compensationfactor (closer to 1). Maybe some of you have constructed a much more efficient driver, I only know what I have. If this compensation says something essential about the driver's quality maybe we can draw conclusions about the rest of the circuitry. If the driver doesn't get a good benchmark but is sustaining flawlessly then the circuitry must somehow have compensated the driver's weaknesses (eq, more power, agc). And if a good driver doesn't sustain, we should suspect the circuitry and try to improve it. Sorry for being long-winded, it must be the thread that makes me do it. Also, this contribution of mine seems to be a bit out of sync with what is being discussed at the moment. But it would be nice if every s******er builder has a way way to measure and calculate the inductance of his driver. Even more so if he shares his results with the other builders.