|
Post by JohnH on Jul 7, 2007 16:57:07 GMT -5
I found a new toy yesterday. Previously, I’ve found it very useful to be able to run simulations of circuit designs using pSpice, of which I had a free version. Unfortunately, I can not get that to work on my new Vista computer. So I came across 5Spice, and its much better. The current free version has been updated for Vista. It will run any kind of guitar circuit, and by downloading component data from manufacturers, will deal with active circuits too. The main limitation seems to be the size of the models. It’s downloaded here: www.5spice.com/Its easy to set up a model, once you find the toolbar on the right with the components, and get the knack of <Ctrl> click, click, right click to connect up a wire. Here’ s the input screen: I’ve set up a test circuit with a Vintage humbucker, 500k tone and volume controls, a 20’ cable and 1M amp input impedance. There’s treble bleed cap and resistor (C2 and R4) on the volume control. Lets try it.... Tone ControlFirst, with full volume, sweeping the tone control, R5: You can see how the top-end peak is gradually reduced, to be replaced with a lower resonance at just over 500hZ, at the very end of the pot travel. Also, these steps of resistance change are equal linear increments of 50k each. Most of the tone change is happening in the last 50k. It shows why a log tone pot is often considered prefferable to linear, to get more sensitivity in that low range. Treble bleed effectsSweeping the volume pot, with tone at full 500k. First, without the treble bleed parts: The cable is changing the shape of the tone contour as volume reduces. Now with the treble bleed: The shape of the tone is more consistent, particularly in the higher ranges of volume. The 220k is also causing the volume to drop less quickly as the pot is turned. A simpler treble bleed just has a cap and no parallel resistor. Here’s one with a 500pF treble bleed cap. At lower volume, too much treble is bled in relation to bass, hence the benefit of the extra 220k resistor. No-load Tone potThis is with no treble bleed, and the tone pot taken out, assuming it disconnects at 10. The initial peak in treble is about 2-3db higher at full volume than with a normal tone pot. It also seems to allow a further peaking at low volume, due to the pup winding capacitance. Useful program I think! And great value too. If anyone wants to play with that input file, its here: people.smartchat.net.au/~l_jhewitt/circuits/bleedtest1.SchCheers John
|
|
|
Post by sumgai on Jul 7, 2007 20:36:11 GMT -5
John, (Did you win the bet? The one where you were betting I would chime in? ) I see that 5Spice has followed in pSpice's footsteps, so it's time to re-evaluate my long-held position (going on 50 years worth). The question is, why is 0dB set to some arbitrary number instead of the peak value? In days of old, it was recognized that a passive circuit, while conducting an AC signal, cannot produce a gain. Indeed, it can only resonate, given all the proper components for doing so, or else it is impeding the signal to some degree. (Frequency dependence is a given.) Such a circuit will deliver the maximum power/voltage/whatever at one point on a scale, what we usually call the peak. Since we're already agreed that there's no gain (something for nothing ain't gonna happen), then it follows that the peak is, by definition, 0dB - the maximum output value, providing there was no loss in the signal path. (0dB is supposed to mean "no gain, no loss".) Nowadays, it would appear that either a non-engineer has written the Spice screens, or else the whole of the IEEE has changed the definition of 0dB, and forgot the teensy little detail of telling me about it. I don't know which scenario applies, and I can't get a straight answer from the IEEE website, so I'm asking you - do you know which way the wind is blowing here? HYCH sumgai p.s. Unk, if a tank circuit can resonate and produce more power than what came in to it, such a thing would be the Holy Grail of Perpetual motion and Free Energy. And we'd all be living high off the hog, with no energy costs involved at all. That would be the epitome of 'sumpin for nuttin', don't ya think? For anyone else, Unk and I discussed this last year, hence the note to him. But I'd like to remind you all that even when a feedback occurs (essentially a resonance), it is not truly a runaway condition, it will either reach a maximum peak (probably one that we humans can't withstand), or it will destroy any components that can't withstand the stress. Given strong enough componentry, and no humans about, the feedback (resonance) will eventually reach a maximum level. It's in the Rule Book.
|
|
|
Post by JohnH on Jul 7, 2007 21:00:58 GMT -5
...the whole of the IEEE has changed the definition of 0dB, and forgot the teensy little detail of telling me about it. sumgai Oops! - slight oversite! letter's in the post! Im not an EE, but I think its all cool from a physics standpoint. The 0db is relative to the input source voltage. There is a resonant peak, and it is higher than the source voltage and that is what is happening when those graphs go higher than 0db. It is true, even in this all passive circuit. Theres no laws of physics violated however, because the conservation laws apply to power, volts x amps ie energy/time, rather just voltage. At the frequency of those peaks with higher volatge, the current that could be drawn from a passive pickup would be correspondingly less, maintaning conservation of power. Its in some ways analogous to a transformer, which without further energy input exept from the source, can increase a voltage. But available output current will be less than that supplied to the input, with further reduction due to losses in the system. ps - on those 5Spice graphs, im thinking the db's are calc'ed in the power form of 20Log(Vo/Vi) rather than 10Log(Vo/Vi) - not to further confuse matters. But using a high impedance amp to just sense that voltage, drawing negligle current from the system, will indeed give us higher power output downstraem at the speakers. the extra power is being generated by the amp, which is not part of the conservation balance. John
|
|
|
Post by ChrisK on Jul 7, 2007 23:05:46 GMT -5
John, One can step various values thru the use of this command: .step param X LIST 1 100 1k 2k 4k 8k 16k 31k 62k 125k 250k 500k and replace component values with: {500001-X} and {1+X} to do a tone pot with any arbitrary stepping interval. Of course, multiple variables/statements can be used as well. While your model is fairly complete, a 1 meg Ohm input resistance is somewhat optimistic for many amps. So when are you gonna come up with a real, harmonically correct (as a function of distance along a string) pickup model. Don't forget the magnet response as well as the effective DC offset in the core due to the magnetization.) 0dB isn't set to some arbitrary number, it's set to the output level of the generator [Vs1] as the reference. The level peaking above 0dB occurs since this is a series resonant reactive circuit. This peaking above the drive level is normal and expected. The pickup is modeled as an AC generator driving a series RLC circuit, with the output taken across the capacitive component. The inductive reactance is an energy storage device (not unlike the energy (not charge) storage component in a switching regulator). Nope, nada, no way, no agreement, 'cuz in a series resonant voltage gain IS, and in a parallel resonant circuit current gain IS (duality strikes again). Not power, but voltage or current (not and). In essence, in a series resonant circuit, there will be voltage ( not energy) gain. This is why the gain (well, Q) must be evaluated in order to properly size the voltage rating of said components, which will exceed the input voltage rating of the generator, sometimes by many times. In the parallel resonant circuit, one need mind the current increase due to the gain (Q). After all, a Tesla coil isn't just a step-up transformer, it's a series resonant step-up circuit. en.wikipedia.org/wiki/Tesla_coilen.wikipedia.org/wiki/Electrical_resonanceen.wikipedia.org/wiki/Tuned_circuitAnd, in this link: "A series resonant circuit provides voltage magnification." "A parallel resonant circuit provides current magnification." en.wikipedia.org/wiki/LC_circuit
|
|
|
Post by JohnH on Jul 7, 2007 23:21:49 GMT -5
John, One can step various values thru the use of this command: .step param X LIST 1 100 1k 2k 4k 8k 16k 31k 62k 125k 250k 500k and replace component values with: {500001-X} and {1+X} to do a tone pot with any arbitrary stepping interval. Of course, multiple variables/statements can be used as well. That would be useful. I have only tried this program at the graphical level. I dont know where to put such a command. Quite a few amps have a 1M resistor to ground at the front end, but I suppose this ignores current going into the first tube. What do you think would be a better generic value for tests like this? Of course I hadnt forgotten that effective dc offset. I just misplaced it somewhere. I'm sure it will turn up eventually. John
|
|
|
Post by sumgai on Jul 8, 2007 2:32:44 GMT -5
Chris, My pre-engineering textbook pretty much says what you just did, and I now see the reason why.... it was your labeling separately the generator itself. I agree that one can model just the pickup, and make measurements. Then one can plug those numbers into some kind of circuit model, using them as an arbitrary source, and the circuit will respond accordingly. However, when both are modeled together, then we're more or less forced to watch the two parts act as a whole, they are reacting to each other. I contend that this distorts, or possbily ruins, any former readings from the pickup-only model. And that's the crux of the whole thing...... if you're gonna have a circuit that resonates, either it's self-excited, or it's not, and if it's not, then you need to source it with a steady state "black box" that doesn't interact wholeheartedly. That would be the separate pickup model, taken as a set of numbers only, and not as part of the circuit under test. It truly does seem that the older I get, the more behind I get. Perhaps this is why I retired early, and let the whippersnappers blouse each other's lips. ;D sumgai
|
|
|
Post by ChrisK on Jul 8, 2007 23:50:48 GMT -5
Perpetual motion isn't. All resonant circuits that are a'resonating are externally excited. And we do, and it is, and it does. This is why the pickup is modeled as a generator and a series resistance and an inductor ans a parallel (to said model) capacitor. This must be in order to fully analyze the resulting circuit and its inter-relationships. Thevenin/Norton IS. JohnH, I place these statements on the schematic. There is a pull-down menu somewhere that allows placing Pspice directives.
|
|
|
Post by sumgai on Jul 9, 2007 2:53:41 GMT -5
Chris, Well, now I'm no longer happy at all. If the model adds a capacitor to the inductor, then Vs1 is now tunable, and in turn, the output signal strength is dependent on the chosen component values. We are now looking at a device that can produce 0dB (maximum signal strength, presumably no loss) at only one point on the frequency spectrum (at the point of resonance between the capacitor and the inductor). So feeding this 0dB frequency into the Spice engine, we get a response, plottable as a single point on a graph. But one point doesn't tell us how the circuit performs across the frequency spectrum, so we feed in more frequencies. Question: what are the dB values for these other frequencies? IOW, what is the strength of the non-resonant frequencies coming from Vs1? Answer: it sure ain't gonna be 0dB. And if that's the case, then what values were plugged into the Spice engine for each frequency? I'd hazard to guess that they are as close as possible to the real world, but to show them on screen would be excruciatingly difficult. Better to just display the maximum value, and work from there, eh? And it devolves (Q.E.D.) that the circuit shows a peak at some frequency because it has "tuned" itself to resonate with Vs1 at the frequency that is most likely to be it's true 0dB point. Now I'm happy. ;D sumgai
|
|