AndrewCE

i need to ask about why my opamp headphone amp doesn't work. and I drew up a scematic in MS Paint, but i first need to figure out how to attatch an image to this post. It only accepts URL addresses can someone tell me how to attatch the image?

he's trying to cut bass, not treble

to the left of the circuit looks very interesting, but i'm lost, looking at everything east of those diodes (that could be attributed to a sloppy schematic, or the fact that I don't know what the 4 plugs are on an oscilloscope). I guess my first question is, why so many capacitors? They are all in parallel, so can't you just figure the equivalent capacitance and use one big cap?

Well if you have the caps their in the rectifier circuit, it cuts the power, to the RMS voltage, and say you have a 5Vp at the diode, you would now be producing a 3.535 Vrms DC voltage after the capacitors, so in theory, I'm not disobeying the laws of Physics, and in fact all I'm doing is increasing the amperage with all the parallel circuits. And I'm redirecting that amperage back into the main line through a few transistors or in the case that I'm finding out, an op-amp.

are you trying to put a passive circuit inside your actual guitar itself, that will turn up your amp's gain?

i dont understand what you're talking about. are you trying to boost your signal without a power source?

so, i guess what youre all saying is, don't be a boutique pedal designer? or at least, dont go to college for it. seems like whats gonna end up happening is i'm gonna get an awesome degree in who knows what, and maybe end up going do something relatively unrelated to that degree?

EE would give you the tools you need to learn this stuff, but programs that actually teach it as part of the curriculum are few and far between. Best thing to do is learn circuit design in school and apply it to guitars on your own time. yes, that's what i figured, but what major would best teach me circuit design? is it EE? i just dont wanna end up with my only career choice as an electrician...

What college major would be best for someone aspiring to design guitar electronics, amps, and stompboxes for a living? Is it electrical engineering? Or audio engineering? P.S. My college doesnt have any "audio engineering."

and note, the key is the part about dividing the product by 10V. That way, the output unit would be V, not V^2. Or else scientifically it wouldnt work.

i havent read the whole document yet, but by the looks of it, i can make this work! how did you find this chip? if you dont mind me asking is there some sort of chip database that you went to? and it's analog! awesome!

but... then you'd have a dc voltage coming out of the first pedal in your chain? couldnt you electrocute yourself with that? can 9 or 18 volts electrocute you? it seems like a good idea as long as it's safe...

Well that covers series and parallel connections, but I'm asking about a possibly different connection, probably involving a transistor or opamp, that does multiply the two together. the idea i had originally was to connect one input to a transistor collector, and the other input to the base somehow. but the method in which you connect it gets pretty tricky. can anyone figure it out?

seems like youre describing an opamp. but no, an opamp can only add/subtract signals together, then multiply them by a constant. that is different from multiplying the two together.

How can you multiply two voltages together? I'm working on a pedal that requires taking two input voltages and outputting the algebraic product of the two. This is not the same as multiplying a single voltage times a scalar value. Also, is it possible to do the same thing, but with current instead of voltage? I imagine either of them would involve using a transistor to have one input control the other, but I can't figure how exactly to do it. For clarity, the device would follow: Vo = V1 x V2. Vo is the output and V1 and V2 are inputs. It might also be helpful if someone has a method of dividing two voltages or currents. I understand that there may be a problem with units; you would end up getting volts squared or amps squared. But maybe if there could be some sort of device that follows Vo = (V1xV2)/V3............ any thoughts on how to do any of this?

So, i'm sure this was answered somewhere in the past 306 pages, but I can't find it: is it possible to just use a cheap guitar pickup as a driver?

right. that way is much simpler than going get a multipole switch like i wouldve done. Just solder the metal cover straight to the bridge or something.

in normal wiring, the metal cover is assumed to be grounded. that way, when you touch the strings(which are also grounded) to it, there is no short circuit between hot and ground. That is exactly what is happening in your pickup. you touch the two together, and all the electric signal goes straight to the earth, through the strings instead of going to your amp. one option is to make the metal cover isolated from both wires of the pickup, but then you lose the sheilding, and it gets a bit noisier. The other option is to rewire the switches so that when you phase switch the pickup, the metal cover gets connected to the other wire. i dont have any schematics for that, though, you'll have to find some yourself

Have a look/lend an ear to David Gillmore. He does those 2 1/2 note bends and he probably cross several poles during those bends and you newer hear any volume drop or anything like that wehen he gets going. I think you missed my meaning or I didn't phrase it properly. I was referring to the high and low E strings (where the bulk of the misalignment is), that one can only bend a very small amount (to the edges) before rolloing over the fret edges, meaning the edge of the fingerboard, where pitch usually jumps up some ridiculous amount, depending on what becomes the new magic fret surface when you lose the intended fret surface. I do realize the only way to get more than the 1/8 bend on these string is to bend them inward toward the center strings or center of the fretboard. For this reason, I would venture (and yes, this is totally speculative) that most players bend the high and low E strings toward the center of the fretboard, not toward the edges, except maybe for vibrato effect, if nothing else to avoid running out of fretboard. So, that being the case, it just seemed it would make more sense to place the poles for the E's where the strings most likely will be, not outside of them, on the fretboard edge, where most players do not want to go anyway (for me, out of fear of dreadful upward pitch shift upon fretboard rolloff). right, you're supposed to bend pretty much all strings toward the center of the fretboard, or else you run the risk of going off the edge. i dont think the pickups were DESIGNED for going off the edge, they probably just make them a bit wider than they have to just in case they get misaligned a bit in assembly

yes, individually, a reversed signal will not change the sound of the pickup. Unless you connect the pickup shield to one of the wires coming out of the pickup, in which case the wire with the shield on it must be the ground wire. And impedance matching is not important either, in those single-pickup spots.

I don't usually correct grammar, but it's "just as soon," not "just assume."

sounds like the symptoms of a dead battery. i mean, i'm sure youve checked something simple like that by now, but just in case you didnt, that's my opinion

are you an advertizer or something?

Grrrrr...

But the signal MUST travel through the resistors. When you say things like that, I get the feeling you don't understand the basics of electronics. The signal is applied as V2, V3, V4 etc as I said. Simple Ohms law tells you the signal must travel through the resistor to the (virtual) ground of the opamps inverting input. If it didn't, how would the opamp ever see it to amplify it? wow, I just realized I should have worded that a bit differently. I know that the signals must travel through the resistors directly after their respective voltage followers, but I was saying that the signal from Voltage Follower A could peek throught resistor B and still effect opamp B, with the same principle as if there were no resistors. I'm sure you can minimize that effect, but you'd have to choose resistor values carefully I guess.