AndrewCE Posted January 4, 2009 Report Posted January 4, 2009 An Op Amp Voltage Buffer, a.k.a. Voltage Follower, is when you connect the output of an op amp directly to the inverting input. The voltage at the noninverting input is copied directly to the output. But there is infinitely high input impedance and infinitely low output impedance. Google it for more info. So here's the Question: Say you've got a voltage source, "A", before the Voltage Follower. In parallel with all of that, you've got another voltage source, "B". How does B affect the operation of the voltage buffer? Would the 2 voltages just be added? Would the total output just be equal to B or A? Would B invade the inverting input and get saturated? I ask because I am considering using op amps in some active circuitry in my guitar. Quote
psw Posted January 5, 2009 Report Posted January 5, 2009 Use of opamps for active guitar circuitry either as a buffer or with a few extra components and a preamp (the same thing but with gain) is very common although many prefer the sound of the discrete transistor circuits such as the tillman or fetzer. Opamps are the lego blocks of the electronics world...but your guitar signal is going through a lot of circuitry inside those little black chips. As for your second question...I don't really follow...it seems that you are suggesting having one pickup A buffered with another signal B non buffered going to the guitar output...probably not a good idea...as you google quote will have told you...there are massive impedance differences. OR...I might be completely misunderstanding the application you are proposing here. If you are designing things, I'd suggest doing a bit more homework than google or use an existing circuit as a starter...there are some good tutorials around also on this subject. Be aware also, that the control pots and such generally need to be changed to much lower values...again, because you now have a completely different impedance change. However, many have simply added a buffer at the output and run the guitar through that...generally improved highs as you eliminate cable capacitance effects... pete Quote
AndrewCE Posted January 5, 2009 Author Report Posted January 5, 2009 Use of opamps for active guitar circuitry either as a buffer or with a few extra components and a preamp (the same thing but with gain) is very common although many prefer the sound of the discrete transistor circuits such as the tillman or fetzer. Opamps are the lego blocks of the electronics world...but your guitar signal is going through a lot of circuitry inside those little black chips. As for your second question...I don't really follow...it seems that you are suggesting having one pickup A buffered with another signal B non buffered going to the guitar output...probably not a good idea...as you google quote will have told you...there are massive impedance differences. OR...I might be completely misunderstanding the application you are proposing here. If you are designing things, I'd suggest doing a bit more homework than google or use an existing circuit as a starter...there are some good tutorials around also on this subject. Be aware also, that the control pots and such generally need to be changed to much lower values...again, because you now have a completely different impedance change. However, many have simply added a buffer at the output and run the guitar through that...generally improved highs as you eliminate cable capacitance effects... pete What does all the cicuitry inside an opamp do to the sound that makes it any worse than transistors? I know that the impedance difference will cause one signal to be much more prominent, but what I am asking is: Does B affect the operation of the op amp? I was thinking that maybe since B is tied to the output, and therefore the inverting input, B would become obliterated by itself and there would be no trace of B in the final signal. Does that make sense? And also, does the impedance of B change how B affects the op amp operation? Quote
psw Posted January 5, 2009 Report Posted January 5, 2009 It's hard to give an answer or opinion without knowing what your real world proposal or thinking is. If for instance as I assumed earlier (but not confirmed) A and B are pickups and one is being buffered...you will have on the buffered pickup the equivalent of a low impedance active pickup given... there is infinitely high input impedance and infinitely low output impedance. so...it will accept any input and give out an exceptionally low output... Now mixing a passive pickup with an active pickup is notoriously troublesome and the control pots are likely to need scaling back to work properly for the active, but not for the passive...if you buffer after the controls...you are likely to induce noise and scratching from the high pots. A common strategy with opamp designs is to bridge the input to ground with a 1M or so resistor, this gives the input a 1M input resistance similar to the impedance of a guitar amp. This is just one of the many little design subtlties with working with circuits and guitar impedances. Again...not sure I am understanding the application... SO...opamps are often regarded as cold...at the least "overdone". If you find a diagram from a data sheet say of the inside of an opamp...TL071 and family being very common...you will see that it is chock full of diodes, caps and resistors and quite a few transistors. Now...a transistor will give you the same effect, high input, low output...do you really want the guitar signal going through all these extra components? Well...maybe you do...it will give a very cold solid state kind of sound perhaps (different opamps have different characteristics) however, a design like the tillman or fetzer uses a single transistor and a few components to keep it happy, to provide the same kind of effect but a warmer "tube like" distortion characteristic. This may be even more desirable. Really horses for courses I guess...and again depends on the application and what kind of effect you are trying to achieve. Personally, I don't mind opamps for some things and once you start wanting to add variable gain (by altering the feedback ratio) or tone controls or adding numerous stages (a TL072 is a dual opamp in the same size pack for instance and is often easier to work with than the single 071 for the same price even if you only use half of it!)...they also have their own bias circuitry inside that gets rid of pesky biasing trimpots and stuff...they do have a tendency to draw a bit more power, but then it is very little anyway so it hardly matters that much...with more complex circuitry that might have active tone controls (cut and boost) or overdrive (added diodes to produce clipping as in the tube screamer and such) or other processing...there is a lot to be said for opamp circuits. I am not following however the application that your question is asking...generally, you would buffer all of the pickups or signals A & B I suppose, either collectively or individually, or both (as in a mixer circuit) due to potential issues with mismatched impedances, control problems and loading issues. There can also be issues with "popping" when active circuits are switched in and out of a signal...so say you have two pickups and one is active, selecting the non active pickup may well generate a mini power surge that will be loud through a wall of marshalls!!! pete Quote
AndrewCE Posted January 5, 2009 Author Report Posted January 5, 2009 It's hard to give an answer or opinion without knowing what your real world proposal or thinking is. If for instance as I assumed earlier (but not confirmed) A and B are pickups and one is being buffered...you will have on the buffered pickup the equivalent of a low impedance active pickup given... there is infinitely high input impedance and infinitely low output impedance. so...it will accept any input and give out an exceptionally low output... Now mixing a passive pickup with an active pickup is notoriously troublesome and the control pots are likely to need scaling back to work properly for the active, but not for the passive...if you buffer after the controls...you are likely to induce noise and scratching from the high pots. A common strategy with opamp designs is to bridge the input to ground with a 1M or so resistor, this gives the input a 1M input resistance similar to the impedance of a guitar amp. This is just one of the many little design subtlties with working with circuits and guitar impedances. Again...not sure I am understanding the application... SO...opamps are often regarded as cold...at the least "overdone". If you find a diagram from a data sheet say of the inside of an opamp...TL071 and family being very common...you will see that it is chock full of diodes, caps and resistors and quite a few transistors. Now...a transistor will give you the same effect, high input, low output...do you really want the guitar signal going through all these extra components? Well...maybe you do...it will give a very cold solid state kind of sound perhaps (different opamps have different characteristics) however, a design like the tillman or fetzer uses a single transistor and a few components to keep it happy, to provide the same kind of effect but a warmer "tube like" distortion characteristic. This may be even more desirable. Really horses for courses I guess...and again depends on the application and what kind of effect you are trying to achieve. Personally, I don't mind opamps for some things and once you start wanting to add variable gain (by altering the feedback ratio) or tone controls or adding numerous stages (a TL072 is a dual opamp in the same size pack for instance and is often easier to work with than the single 071 for the same price even if you only use half of it!)...they also have their own bias circuitry inside that gets rid of pesky biasing trimpots and stuff...they do have a tendency to draw a bit more power, but then it is very little anyway so it hardly matters that much...with more complex circuitry that might have active tone controls (cut and boost) or overdrive (added diodes to produce clipping as in the tube screamer and such) or other processing...there is a lot to be said for opamp circuits. I am not following however the application that your question is asking...generally, you would buffer all of the pickups or signals A & B I suppose, either collectively or individually, or both (as in a mixer circuit) due to potential issues with mismatched impedances, control problems and loading issues. There can also be issues with "popping" when active circuits are switched in and out of a signal...so say you have two pickups and one is active, selecting the non active pickup may well generate a mini power surge that will be loud through a wall of marshalls!!! pete OK yeah let's say A and B are pickups. And lets say they both have voltage followers on their hot leads. And they are in parallel. Would the output of this circuit be simply A+B? Or would the two voltage followers affect each others' performance? oh, then consider the same circuit using transistors instead of opamps. Would the output be A+B? Hmmm I'm not trying to use them for any distortion, so "tubelike breakup" isnt important, but other than that, how would an opamp make a signal sound "cold"? .....Limited bandwidth? and by the way, I'm not just asking all this just for one specific application, it's mainly about a deeper understanding about how opamps work. I KNOW how opamps work, but to know and to understand are entirely different Quote
psw Posted January 5, 2009 Report Posted January 5, 2009 OK...I personally only have a teach yourself understanding... OK yeah let's say A and B are pickups. And lets say they both have voltage followers on their hot leads. And they are in parallel. Would the output of this circuit be simply A+B? Or would the two voltage followers affect each others' performance? Well...a voltage follower is just that...there is no gain...so...ummm...A & B...if that makes sense.... just like combining both pickups in parallel does not increase volume...series would increase volume and dramatically affect frequency response...so...in series it would be A PLUS B as in more output... so...where are we...since a buffer or voltage follower adds no gain, then there is no additional gain. You could have separate buffers...though this is a little redundant, one buffer would have much the same effect. A common way to do this is to add it just before the output jack on the whole guitar. When they speak about op amps they speak in terms of a theoretical ideal....they have infinite gain! The feedback loop...wiring the output to the input negates this gain and so faithfully follows the voltage...but the benefits are in the impedance matching. This comes up in the sustainer thread as the sustainer circuits require something like this to avoid loading effects...i seem to have trouble getting this explained...so maybe it's me! ... Anyway...there is a theoretical ideal...and there is real life! oh, then consider the same circuit using transistors instead of opamps. Would the output be A+B? transistors or a chip that contains transistors...it is much the same thing really...but the distortion characteristics are different...as this is not an ideal world, distortion tends to be the outcome. Hmmm I'm not trying to use them for any distortion, so "tubelike breakup" isnt important, but other than that, how would an opamp make a signal sound "cold"? .....Limited bandwidth? It is a little tricky to explain this...the opamps are a universal device used for everything from controling motors to making clocks and stuff...they are not necessarily an audio things. Some are optimized more for that purpose, or power consumption...some hi-fi things are rediculously expensive... Once you get into opamps...you have lost control of what is in them and a lot of stuff may not be what you are after...a transistor is a simple voltage follower and smaller, cheaper and often has a more pleasant distortion characteristic...it is too do with the harmonics...distortion to some degree is inevitable...ugly distortion is avoidable. Perhaps an extended bandwidth makes it cold...or the amount of processing...part of it may simply be that for the first time you are hearing the pickups as they truely are..without the cable inductance and control loading and such. Conventional pickups...if they are done well...tend to have all these things factored into them...so a buffered guitar while not being any louder...and sound more Hi-Fi. EMG's are an example. When guitar players talk about "tone" they are not generally talking about hi-fi kind of thing...they are talking about a particular mix of harmonic content and pickups being a big inductor with lots of resonant and other factors involved. So...evening out these things may not be the best "tone"...hence a lot of people hear the true sound of active guitars as "sterile". On the other hand...people often like them for exactly those qualities, especially with a lot of effect or distortion. Of course...active pickups like EMG's also have lower impedances, higher frequency responses and less magnetic strength and, most importantly, are designed for active circuitry. So...i don't know if I am helping at all...but in the real world, things will distort and besides the transitors inside the opamp, what more do you need to run the signal through than a simple transistor circuit. On the other hand, they are like lego to design with and when using them as a building block for more complex things...they can be excellent. Remember, some will be really down on opamps...but then love a tubescreamer...a pure opamp diode design...so nothing is clear cut. Not sure though that there is any massive benefit or what you are seeking. I'd like to see something like this that includes things like an active cut and boost tone control as well as a buffer...or even some gain potential...someone even suggested a noise gate circuit to cut out the strat noise which was really interesting and another simple opamp application that would be kind of cool!~ So...nothing wrong with op amps per se, but if actually designing things...it is not always as simple as you may be led to believe. Generally opamps require a plus and minus power supply...to achieve this you normally use a voltage divider circuit and some power filtering caps...often you will want a small input cap and some output cap...maybe even a circuit protection power diode...it starts to add up, still it is a simple device. What opamps are good for is where you want gain...it is a simple matter of ratios of resistors in the feedback loop to get what ever gain you like...but now the buffer is a preamp! As things are moving more miniture, most new devices are only in surface mount devices that make it hard for the DIYer and experimenter. There are whole ranges of specialized devices that include equalization, noise reduction and such things on the chip...you will find them in use in your ipod and phones...but unfortunately, not practical for the experimenter. pete ps...maybe I am just confusing things even more! Quote
bluesy Posted January 5, 2009 Report Posted January 5, 2009 An Op Amp Voltage Buffer, a.k.a. Voltage Follower, is when you connect the output of an op amp directly to the inverting input. The voltage at the noninverting input is copied directly to the output. But there is infinitely high input impedance and infinitely low output impedance. Google it for more info. So here's the Question: Say you've got a voltage source, "A", before the Voltage Follower. In parallel with all of that, you've got another voltage source, "B". How does B affect the operation of the voltage buffer? Would the 2 voltages just be added? Would the total output just be equal to B or A? Would B invade the inverting input and get saturated? I ask because I am considering using op amps in some active circuitry in my guitar. You really can't connect ideal voltage sources together. There needs to be some resistance, otherwise interaction due to loading will occur. This is how a basic mixing circuit works. Each voltage source is "mixed" together by feeding it to the common point via a relatively large resistor. The result is then usually further buffered by a gain stage (another opamp if you like), and it is A + B although, as these are complex waveforms of various frequencies, sometimes peaks coincide, sometimes they don't and the result is not twice A or B alone, unless your voltage sources are identical (which they are not if they are coming from different pickups for example) Quote
AndrewCE Posted January 5, 2009 Author Report Posted January 5, 2009 Well...a voltage follower is just that...there is no gain...so...ummm...A & B...if that makes sense.... but there is theoretically infinite current gain, that's how the voltage follower has low output impedance ...right? and why would there be distortion, if a pickup's signal is only a few millivolts and I supply the opamp with +-9v? Quote
AndrewCE Posted January 5, 2009 Author Report Posted January 5, 2009 You really can't connect ideal voltage sources together. There needs to be some resistance, otherwise interaction due to loading will occur. hmm i'm not sure i'm understanding your explaination here. What is "interaction due to loading" referring to? What is loading what? You have 2 low impedance sources in parallel, why would the 2 voltages not just average each other? Quote
bluesy Posted January 6, 2009 Report Posted January 6, 2009 You really can't connect ideal voltage sources together. There needs to be some resistance, otherwise interaction due to loading will occur. hmm i'm not sure i'm understanding your explaination here. What is "interaction due to loading" referring to? What is loading what? You have 2 low impedance sources in parallel, why would the 2 voltages not just average each other? Let's try it this way. An "ideal" voltage source has zero impedance. If you connect two zero impedance voltage sources that are different in voltage, the voltage difference divided by the impedance (zero) means it would try to deliver infinite current. In the real world, this is not a good thing. They will NOT "average each other". They MIGHT try to destroy each other :-) except that built in protection in opamps and the fact that they are less than ideal voltage sources, might save them. They will both try to control the voltage to the value they want, and as that is different - they cannot both succeed. How's that for interaction? Quote
psw Posted January 6, 2009 Report Posted January 6, 2009 Again...it might be easier to understand with some real circuits and aplications even if hypothetical...I'm glad bluesy has joined in and I am not the only person confusing you... above are some extremely basic building blocks with op amps...the last one being a voltage follower...the wire from the output back to the negative input gives it a gain of +1 or basically no gain or loss. The diagram on the left bottom is a voltage divider required to power the chip from a single battery. But these are not really real life circuits but building blocks and principles...they are very close though...the chips contain very many components. The two top diagrams could be preamps...with plus or negative gain...the gain is et by the ratio of the resistors (rA and rB) in both circuit types. But in reality you may wish to add a couple more components to make a decent simple working circuit. SO...loading...hmmm...this is very difficult to explain. The little poweramps used to drive a typical sustainer for instance will have an imput impedace of about 10K but if you try and run the guitar pickup signal into this...it will make a sound...but it will lack power and highs due to loading effects...add a buffer in front of it and there is no resistance or very low "impedance" meaning that it is quite happy to share the signal between these two sources. The same kind of thing happens if you try and plug a guitar into the line in of a computer or stereo...it is expecting a low impedance buffered or preamped signal...guitars will make a sound, but not a very good one. It is very similar in effect and sound to turning down the volume of the guitar to about two...loss of tone and volume...often this is compensated for with a treble bleed cap to allow the highs to continue to go through un-impeded... I know...really bad explanations...but it is something you need to study I guess to both explain well and to understand in a deep sense...it can be very anti intuitive. Still...that shouldn't put you off doing something if you have a particular application in mind. Find a simple opamp buffer circuit, it would only cost a couple of buckes to make...and run your guitar into it like an effects box even (don't even need to build it in) and see what the effect is. It should be noted that every effects box effectively has buffers or preamp stages in them and you get the benefits of this as soon as you plug one in, usually even if it is off...hence many effect builders want true bypass to completely remove these effects when the thing is off! Quote
AndrewCE Posted January 6, 2009 Author Report Posted January 6, 2009 Let's try it this way. An "ideal" voltage source has zero impedance. If you connect two zero impedance voltage sources that are different in voltage, the voltage difference divided by the impedance (zero) means it would try to deliver infinite current. In the real world, this is not a good thing. They will NOT "average each other". They MIGHT try to destroy each other :-) except that built in protection in opamps and the fact that they are less than ideal voltage sources, might save them. They will both try to control the voltage to the value they want, and as that is different - they cannot both succeed. How's that for interaction? hmm, thats exactly what i was afraid of. sorry but i still don't understand your "If you connect two zero impedance voltage sources that are different in voltage, the voltage difference divided by the impedance (zero) means it would try to deliver infinite current." Do you mean voltage follower A would try to deliver infinite current into the output of the voltage follower B? now what if you put a resistor after each voltage follower, and THEN brought them together. Then the impedances would not be zero. Would that produce "A+B"? Quote
AndrewCE Posted January 6, 2009 Author Report Posted January 6, 2009 psw, do you mean that the signal produced by part A (pickup A in series with voltage follower A) will try to run through part B? I don't think a signal can enter the output of a opamp. Unless, as I was thinking earlier, signal A entered the output of voltage follower B via opamp B's inverting input. Would this happen? And could it be stopped by adding resistors after each voltage follower? Quote
bluesy Posted January 6, 2009 Report Posted January 6, 2009 sorry but i still don't understand your "If you connect two zero impedance voltage sources that are different in voltage, the voltage difference divided by the impedance (zero) means it would try to deliver infinite current." Do you mean voltage follower A would try to deliver infinite current into the output of the voltage follower B? now what if you put a resistor after each voltage follower, and THEN brought them together. Then the impedances would not be zero. Would that produce "A+B"? For goodness sake yes - that's exactly what I said and provided the diagram for above. You are describing a mixer !! Quote
bluesy Posted January 6, 2009 Report Posted January 6, 2009 I don't think a signal can enter the output of a opamp. Of course it can. Any voltage will try to produce a current through any non-infinite impedance it is connected to. If the "signal" voltage is different from the output voltage of the opamp, current will flow. Quote
AndrewCE Posted January 6, 2009 Author Report Posted January 6, 2009 For goodness sake yes - that's exactly what I said and provided the diagram for above. You are describing a mixer !! But your diagram had resistors before the op amp. I'm saying if a resistor was after each parallel voltage follower. Surely they wouldn't act the same way. Quote
bluesy Posted January 6, 2009 Report Posted January 6, 2009 For goodness sake yes - that's exactly what I said and provided the diagram for above. You are describing a mixer !! But your diagram had resistors before the op amp. I'm saying if a resistor was after each parallel voltage follower. Surely they wouldn't act the same way. V2, V3, V4 etc, are the outputs of your opamps. The diagram shows the resistors, R2, R3, R4 that mix the outputs into the following buffer (the op-amp shown in the diagram) Quote
psw Posted January 7, 2009 Report Posted January 7, 2009 Of course it can. Any voltage will try to produce a current through any non-infinite impedance it is connected to. If the "signal" voltage is different from the output voltage of the opamp, current will flow. yes...which is why in designing real circuits you will generally find additional input and output capcitors to prevent possible DC voltages and noise from the power source... But generally, I am getting lost...these things can be simple or complex but it would really, really help to have a scheme or real life application or a circuit for such an application to explain and learn from. Even if it is only hypothetical. One most obvious question is...why would you want to do this, what is it supposed to achieve? Are you seeking low impedance output and no cable capacitance effects for instance...this is one good reason you may consider a buffer, but it need not concern these kind of "issues"...just buffer the entire output just before the output from the jack. Could it be that you have seen a simple block of an apparent IC with a few connections and no external components and are thinking...surely it isn't this simple...but if it is I could just do this? It is almost that simple, almost...but for putting a battery into a guitar you'd want some kind of decent payoff I would assume. A lot of the confusion seems to be about the application and the terms or the silliness of it...so... If you have a signal and you run it into a voltage follower...you get the same signal out. If you run that into yet another voltage follower...you again get the same signal out....the net "gain" is nothing. Saying A+B is very confusing. Now...Bluesy is right in the mixer type of circuit...this is how you would combine a number of signals into a typical mixer circuit. Now...you may see those resistors there and think...ohh, I don't want more resistance, or I don't want to have to deal with external components...I don't know...but remember that the impedance or resistance of an average pickup will be at least 9,000 ohms and the input to your guitar amp will typically be about 1M ohms...not to mention things like your control pots and cables, etc. You could use any number of buffers I guess...but it would seem pointless really. We are not getting any gain out of a voltage follower, or you are doing is changing impedances...once you plug into an effect or an amp you are doing much the same thing at the other end of the cable. All of these things can be achieved with a single transistor also for lower cost and size and avoiding the overkill of the internal chip circuitry and possibly extra power drain. There probably isn't an easy way to answer such general non application questions without actually studying electronics from the basics up. The other way to do it is to have a goal and pick up the information required...or put up a scenario and have it explained why it will or won't work. Generally my "knowledge" comes from the two former...have in a goal or idea and then reading and questioning that specific information...but a little bit of knowledge can be a dangerous thing. Water is wet...but it is dangerous to extract from this fact that the only danger in jumping into the sea is that you will get wet...it may be deep, there may be sharks, there may be storms, there may be currents...and pretty soon we will all be out of our depth. Quote
AndrewCE Posted January 7, 2009 Author Report Posted January 7, 2009 V2, V3, V4 etc, are the outputs of your opamps. The diagram shows the resistors, R2, R3, R4 that mix the outputs into the following buffer (the op-amp shown in the diagram) OHHHHHHHHH okay, yeah then that's what I was describing. so it will work, with the resistors. K thanks! as long as the signals dont travel through the resistors anyways... Quote
psw Posted January 7, 2009 Report Posted January 7, 2009 Well...the signals will travel through the buffers and then through the transistors into the final stage buffer...so, yes they kind of will go through the resistors...but since the first buffers before them eliminated the potential loading...there will not be the effect you may otherwise be concerned with...however....I think there may be simpler ways and remember, the chip needs to be powered +/- or a voltage divider set up to run then plus a few extra bib's and bobs to keep them happy! I have been thinking again about doing some more design work on guitar circuits myself...of course then I would need to brush up on these things...but a few posts recently have had me wondering again whether guitar players are ready to perhaps consider active circuits as thire bass playing cousins have done. Personally...if you are to put a battery into a guitar I think that there is a little more than a simple buffer that could be done with reasonably low power and open up a whole array of options. Another reason is that I have a project that combines piezos and mags in the works now and such a combination does necessitate exactly this kind of thing. A universal buffer/mixer circuit perhaps using a quad opamp may well be exactly the kind of thing that is required for this and many similar operations. Other ideas that people have put forward is perhaps an on board noise gate for single coils or active tone or clipping circuits. You could even have some provision for switchable gain if it were done cleverly...for instance, trim pots for each signal that would allow you to not only split, but boost the signal making perhaps more of pickup wiring options. But...a buffer is only a small building block to somthing a little more and it is difficult to speak in generalities for some reason over specifics or adapting circuits. Of course, there are specialist forums for a lot of these things and perhaps I am not the best qualified...the stompbox forum is a good place to start, the DIY layout creator free software is great for getting your ideas planed and for sharing on the net...there are a few electronics forums but they may not fully get the guitar connection...even the GN2 forum has some very cluey people on there...a lot comes down to actually building and putting these things to use. You will though still come across the people who favor discreet transistors and components over opamps for simple things like buffers...so be prepared! personally, I can see both sides and there is many applications (say where you need multiple buffers or preamp/mixers) where the opamps are very appropriate pete Quote
bluesy Posted January 7, 2009 Report Posted January 7, 2009 as long as the signals dont travel through the resistors anyways... 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? Quote
AndrewCE Posted January 8, 2009 Author Report Posted January 8, 2009 as long as the signals dont travel through the resistors anyways... 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. Quote
Amplexus Posted January 26, 2009 Report Posted January 26, 2009 It's hard to give an answer or opinion without knowing what your real world proposal or thinking is. If for instance as I assumed earlier (but not confirmed) A and B are pickups and one is being buffered...you will have on the buffered pickup the equivalent of a low impedance active pickup given... there is infinitely high input impedance and infinitely low output impedance. so...it will accept any input and give out an exceptionally low output... Now mixing a passive pickup with an active pickup is notoriously troublesome and the control pots are likely to need scaling back to work properly for the active, but not for the passive...if you buffer after the controls...you are likely to induce noise and scratching from the high pots. A common strategy with opamp designs is to bridge the input to ground with a 1M or so resistor, this gives the input a 1M input resistance similar to the impedance of a guitar amp. This is just one of the many little design subtlties with working with circuits and guitar impedances. Again...not sure I am understanding the application... SO...opamps are often regarded as cold...at the least "overdone". If you find a diagram from a data sheet say of the inside of an opamp...TL071 and family being very common...you will see that it is chock full of diodes, caps and resistors and quite a few transistors. Now...a transistor will give you the same effect, high input, low output...do you really want the guitar signal going through all these extra components? Well...maybe you do...it will give a very cold solid state kind of sound perhaps (different opamps have different characteristics) however, a design like the tillman or fetzer uses a single transistor and a few components to keep it happy, to provide the same kind of effect but a warmer "tube like" distortion characteristic. This may be even more desirable. Really horses for courses I guess...and again depends on the application and what kind of effect you are trying to achieve. Personally, I don't mind opamps for some things and once you start wanting to add variable gain (by altering the feedback ratio) or tone controls or adding numerous stages (a TL072 is a dual opamp in the same size pack for instance and is often easier to work with than the single 071 for the same price even if you only use half of it!)...they also have their own bias circuitry inside that gets rid of pesky biasing trimpots and stuff...they do have a tendency to draw a bit more power, but then it is very little anyway so it hardly matters that much...with more complex circuitry that might have active tone controls (cut and boost) or overdrive (added diodes to produce clipping as in the tube screamer and such) or other processing...there is a lot to be said for opamp circuits. I am not following however the application that your question is asking...generally, you would buffer all of the pickups or signals A & B I suppose, either collectively or individually, or both (as in a mixer circuit) due to potential issues with mismatched impedances, control problems and loading issues. There can also be issues with "popping" when active circuits are switched in and out of a signal...so say you have two pickups and one is active, selecting the non active pickup may well generate a mini power surge that will be loud through a wall of marshalls!!! pete Check out the Burr Brown OPA627, it is fantastic but not cheap, I think you would have a hard time beating it with discrete transistors. if you can afford it is should be your first choice, Analog devices make some good ones as well with a bit hotter sound, Amplexus Quote
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