An onboard buffer for guitars
http://www.geofex.com/FX_images/Onboard_Preamp.pdf (http://www.geofex.com/FX_images/Onboard_Preamp.pdf)
So this has gain up in the hundreds of MHz.. I guess that's just a caveat of using a really low biasing current, with modern discrete devices the gain will definitely be way up there. Why not bandwidth limit this to prevent potential RF interference issues? Did you play around with bootstrapping some capacitance across the 2N5087 C-E junction in your Sziklai pair or something like this? I guess I would need to run some sims before I guess too much but I tend to bandwidth limit things like this at 40kHz-100kHz (depending on slew rate needs, which should be insignificant here even for super hard string slaps) to prevent any potential RF issues without killing needed gain in the audio spectrum.
Here you call it a buffer. The document describes it as an onboard preamp. I'm confused. Does the circuit have gain or not? And if it does, how would I adjust how much gain it has...especially if I don't want very much?
@Mark:
It has unity voltage gain, big current gain. So if you plugged into a high-z input it would not be much louder, but if you plugged into a low-z input it would be.
Quote from: Cliff Schecht on May 13, 2013, 02:28:51 PM
So this has gain up in the hundreds of MHz.. I guess that's just a caveat of using a really low biasing current, with modern discrete devices the gain will definitely be way up there. Why not bandwidth limit this to prevent potential RF interference issues? Did you play around with bootstrapping some capacitance across the 2N5087 C-E junction in your Sziklai pair or something like this? I guess I would need to run some sims before I guess too much but I tend to bandwidth limit things like this at 40kHz-100kHz (depending on slew rate needs, which should be insignificant here even for super hard string slaps) to prevent any potential RF issues without killing needed gain in the audio spectrum.
I commonly do that too. I have spent some time with a simulator trying to cleanly kill the response between 20kHz and 100kHz, which is what I usually try for if there are not specific requirements. Did I mention that this circuit is quirky? :icon_biggrin:
I messed with limiter caps on B-E, C-B and C-E on both transistors. In many of these cases, the bandwidth "reduction" resulted in a nasty peak in the AM band. I also messed with the usual base stoppers and other tricks. The most promising of these was a C-E cap on the PNP. Even that can be touchy with some values of load capacitance.
Frankly, the cleanest response - no funny RF peaks and such - was just to let it run out as far as the junction parasitics would let it. I don't trust that, of course. :icon_lol:
So, I don't yet fully understand where the sensitivity is coming from, and I noted as much. I feel pretty sure it's not going to sing with variable cable loading with the series resistor on the output. Without that, it was prone to singing with some values of cable capacitance, like many followers do.
No, it wasn't because I was ignorant of the issues. :icon_biggrin:
Quote from: Mark Hammer on May 13, 2013, 04:22:48 PM
Here you call it a buffer. The document describes it as an onboard preamp. I'm confused. Does the circuit have gain or not? And if it does, how would I adjust how much gain it has...especially if I don't want very much?
Sorry. I'll go clear that up. It's a buffer. Gain is about 0.999, thanks to the high gain of the complementary feedback pair and current source loading of the pair's "emitter".
It could be made to have gain, but that's not what it started as. This started life as the solution to a question on another forum about how to get a particular tone from a pickup that only existed with the volume control full up, but was muddied by turning the volume down. I speculated that the volume control's varying loading was interacting with the pickup to steal some treble. Hence a buffer, to stop the loading, and the "magic load" to hold the pickup at the right place in the multidimensional tone curves.
Quote from: earthtonesaudio on May 13, 2013, 05:15:46 PM
@Mark:
It has unity voltage gain, big current gain. So if you plugged into a high-z input it would not be much louder, but if you plugged into a low-z input it would be.
so could a low/high imp switch be simply added with appropriate r/c choices? pot?
I believe what he meant was that an amplifier would respond differently to the buffer on the high impedance versus low impedance inputs.
ah, yes - started to consider that possibility exactly 33 milliseconds after clicking post... ho hum...
Through the magic of revisionist history, it's now a ... buffer!
R.G. have you tried a series input resistor to help? This should have the added benefit of reducing the HF end of the bandwidth.
Followers can look simple but they can be tricky
So it looks like the Sziklai pair has a active current circuit in the "emitter" leg that gets some extra drive from the "collector' from the 100 ohm collector node to drive the bottom right 5088 base.
I am guessing you did that to not use a constant current circuit that would draw more current, where this one has the lower constant current modulated by the signal off the "collector"
Quote from: Gus on May 13, 2013, 07:15:33 PM
R.G. have you tried a series input resistor to help? This should have the added benefit of reducing the HF end of the bandwidth.
Followers can look simple but they can be tricky
I did, expecting it to be simple. It moved the RF peak a bit. I also tried a stopper in the PNP's base. No help.
QuoteSo it looks like the Sziklai pair has a active current circuit in the "emitter" leg that gets some extra drive from the "collector' from the 100 ohm collector node to drive the bottom right 5088 base.
I am guessing you did that to not use a constant current circuit that would draw more current, where this one has the lower constant current modulated by the signal off the "collector"
Yes. The signal from the collector of the CFP corrects the CCS in certain instances and prevents gain droop.
Ok, the fact that this is a buffer and not an amplifier with voltage gain (didn't notice that initially!) makes me much less worried about the bandwidth being so high.
Sort of.
The collector is most sensitive to capacitance (with regards to stability in a C-E amplifier), the source of a C-C amplifier is most sensitive to stray inductance (of which there is PLENTY in guitar cables). I am not sure about the sensitivity of the Sziklai pair to stray inductance but my gut tells me that this will hurt stability as well. How about the addition of a single bandwidth-limited stage with a known specific loading that keeps the Sziklai stage happy (i.e. optimal loading) while driving the cable/volume pot with a C-C/C-D follower?
Why can't I figure this out?
(http://i.imgur.com/Rppgo6I.gif)
Seems to me the top PNP is off, doing nothing.
If it did work: yes that connection does ring at a dozen MHz. This causes little problem in small-signal work. It can get real nasty in a class B power stage; yet many are built and few have serious problem. Yes, compensation is confounding, anti-intuitive, and often fruitless.
> pretty sure it's not going to sing with variable cable loading with the series resistor on the output.
I don't see series resistor on the output? (A few hundred ohms would be ample.)
Quote from: PRR on May 14, 2013, 02:47:19 AM
Why can't I figure this out?
(http://i.imgur.com/Rppgo6I.gif)
Seems to me the top PNP is off, doing nothing.
If it did work: yes that connection does ring at a dozen MHz. This causes little problem in small-signal work. It can get real nasty in a class B power stage; yet many are built and few have serious problem. Yes, compensation is confounding, anti-intuitive, and often fruitless.
> pretty sure it's not going to sing with variable cable loading with the series resistor on the output.
I don't see series resistor on the output? (A few hundred ohms would be ample.)
input signal is going to base of 2N5088, which means voltage/current will be bigger for base of 2N5087. Right? How much more negative is the collector of 2N5087 (because the PNP is upside down) compared to the base? That's all that matters for turning it on, right?
Quote from: PRR on May 14, 2013, 02:47:19 AM
Why can't I figure this out?
Seems to me the top PNP is off, doing nothing.
Oh. Doh.
Because I can't type. I inserted a decimal point where there should be none. That should be 33K. Schemo updated and on line.
QuoteIf it did work: yes that connection does ring at a dozen MHz. This causes little problem in small-signal work. It can get real nasty in a class B power stage; yet many are built and few have serious problem. Yes, compensation is confounding, anti-intuitive, and often fruitless.
I had a forgotten component in my sim, and left off part of the modelling of a pickup when testing input impedance.
I neglected to put in the self-capacitance of the pickup, and the inductance was angry enough to resonate with the input capacitance at the 5088. Inserting the self capacitance cured that. There's no obvious peaking on a Bode plot, and only a cycle or so of ringing with 10k square waves in.
QuoteI don't see series resistor on the output? (A few hundred ohms would be ample.)
Corrected. 47R seems to fix any possible cable interaction. Once again, I didn't clean up my sim schemo.
It's much better now. Thanks for pointing those out, Paul.
hey RG. once again you make a great post that is a total game changer. I was trying to design a low power preamp for a customer (not for production) using a low power 3v CMOS opamp. never got it off the ground cause it looked like battery sag would put me right out of my minimum power needs. this looks great. can you tell me a bit more about what watch batteries you selected?
on a side note, there are some white papers for designing 1.5v CMOS opamps but I'm not paying money to get access to those websites. I also don't have a IC factory in my back pocket.
I tried opamps first, but this is a situation where the right compromise opamp is not easily available.
The fundamental problem with onboard electronics is powering them. Sticking in a 9V battery is possible, but often requires carving a cavity in the body for it. I wanted a different solution.
I chose two 3V lithium coil cells in series for 6V, primarily to be able to swing at least 4V for the output of a humbucker. Opamps that swing rail to rail would do the job. But they also need to have low noise, low operating current, wide frequency response, and low distortion. Those requirements seem to be contradictory in this particular application.
The biggest issue was low current versus low noise. The low noise opamps were not low current. I set a goal of 100uA operating current, and found no opamps with low noise there. I could get to low noise at about 1ma per amplifier, but that gave ten times worse battery life.
An emitter follower could do the job, if the input impedance got high enough and the output swing was wide enough. I started with a darlington, didn't like the voltage available, then went for the CFP with current source loading.
Again, the central issue is power. Coin cells will give around 200-300 hours of on-time at 100uA if I calculated/guessed right. That's not good enough not to use a power switch on it. But much lower than 100uA means going to Class AB, with crossover distortion, or Class A with doubtful ability to drive cables well.
I had an idea about how to put batteries in. Imagine a coin slot in the pickguard of a guitar.
Quote from: R.G. on May 13, 2013, 05:19:17 PM
... This started life as the solution to a question on another forum about how to get a particular tone from a pickup that only existed with the volume control full up, but was muddied by turning the volume down. I speculated that the volume control's varying loading was interacting with the pickup to steal some treble. Hence a buffer, to stop the loading, and the "magic load" to hold the pickup at the right place in the multidimensional tone curves.
That's seem to me like a great concept to work on. Thanks for sharing.
What criteria do you use for picking the 1M resistor at the output? I always choose 10k and a 1u capacitor before that, because I assume the input of the next circuit will be 10k or higher, hence the high-pass filter would be formed mainly by the 10k and 1u. Anyway, I only see it as a filter, maybe I should see it like something else too.
I use a fairly similar (cfp based) buffer since quite some time (since early 00's) and it works better than expected. I have it at a higher current and at 9v, but it's not that far off (sans the "white" thing on a ccs).
Quote from: Labaris on May 14, 2013, 11:22:45 AM
What criteria do you use for picking the 1M resistor at the output? I always choose 10k and a 1u capacitor before that, because I assume the input of the next circuit will be 10k or higher, hence the high-pass filter would be formed mainly by the 10k and 1u. Anyway, I only see it as a filter, maybe I should see it like something else too.
It's there just to ensure that the output is firmly pulled to 0Vdc. The actual value is immaterial. I tried the circuit (in a simulator, mind!) with loads from 4.7K to 1M, with generally good results. If the load gets too low, the CCS doesn't have enough current flow to pull it down fast enough on negative going signals. But the typical application would be into another pedal with modern designs at 1M or more, and even vintage designs at 100K or more. None of those are a problem.
The only problematic pedal I can think of is the fuzz face. Expecting a raw pickup as it does, a buffered guitar of any kind will cause harder clipping distortion. But reinserting a pickup impedance in the form of a resistor, an inductor, and a cap will allow one to reset the impedance and get the softer clipping back.
The real choice on the output is the 1uF cap. That was chosen to be able to drive full audio bandwidth (16Hz in this case) into a 10K load resistor. With a 1M amp or pedal load after the guitar, that could be cut back to 10nF for the same 16Hz. On the input, the DC-blocking cap only needs to be 4.7nF to get full (and way over) guitar and bass bandwidth because of the high input impedance.
Notice that there are no caps over 1uF. The actual target of this design is SMD, where normal electrolytics are too big. A quick and dirty SMD with 0805 parts and SOT-23 transistors will fit entirely on a USA quarter, that being about 24mm diameter, using two electro caps. I'm actually looking at making this in 0402 SMD parts with smaller SMD transistors. I expect that will be the size of a thumbnail.
Quote from: R.G. on May 14, 2013, 02:45:41 PM
Quote from: Labaris on May 14, 2013, 11:22:45 AM
What criteria do you use for picking the 1M resistor at the output? I always choose 10k and a 1u capacitor before that, because I assume the input of the next circuit will be 10k or higher, hence the high-pass filter would be formed mainly by the 10k and 1u. Anyway, I only see it as a filter, maybe I should see it like something else too.
It's there just to ensure that the output is firmly pulled to 0Vdc. The actual value is immaterial. I tried the circuit (in a simulator, mind!) with loads from 4.7K to 1M, with generally good results. If the load gets too low, the CCS doesn't have enough current flow to pull it down fast enough on negative going signals. But the typical application would be into another pedal with modern designs at 1M or more, and even vintage designs at 100K or more. None of those are a problem.
The only problematic pedal I can think of is the fuzz face. Expecting a raw pickup as it does, a buffered guitar of any kind will cause harder clipping distortion. But reinserting a pickup impedance in the form of a resistor, an inductor, and a cap will allow one to reset the impedance and get the softer clipping back.
The real choice on the output is the 1uF cap. That was chosen to be able to drive full audio bandwidth (16Hz in this case) into a 10K load resistor. With a 1M amp or pedal load after the guitar, that could be cut back to 10nF for the same 16Hz. On the input, the DC-blocking cap only needs to be 4.7nF to get full (and way over) guitar and bass bandwidth because of the high input impedance.
Notice that there are no caps over 1uF. The actual target of this design is SMD, where normal electrolytics are too big. A quick and dirty SMD with 0805 parts and SOT-23 transistors will fit entirely on a USA quarter, that being about 24mm diameter, using two electro caps. I'm actually looking at making this in 0402 SMD parts with smaller SMD transistors. I expect that will be the size of a thumbnail.
Thanks for the answer. So my criteria is not that bad. I always calculate "f0" at the input and output just to ensure full bandwidth response.Another question (maybe off-topic): I'm designing an opamp buffer right now and I chose to do the biasing through a 10M resistor, like this:
(http://i148.photobucket.com/albums/s21/Labaris/buffer_bias.jpg) (http://s148.photobucket.com/user/Labaris/media/buffer_bias.jpg.html)
It is better to do it this way or put VB directly on the opamp +input instead? With the second choice you get 3x10M resistors in parallel, with the first you only have 2, am I right?
So the choice of a right capacitor value at the input depends on this too.
Quote from: Labaris on May 14, 2013, 04:55:51 PM
So my criteria is not that bad. I always calculate "f0" at the input and output just to ensure full bandwidth response.
That works fine at the input, where you get to choose the loading the input capacitor sees by choosing the design of the input stage. It's not so good at the output where almost anything can be plugged in. The lowest load a 1uF on an output can drive to 20Hz is R = 1/(2*pi*F*C) = 1/(2*pi*20*1E-6) = 7962 ohms. Looked at another way, a 1uF/10K load has a half power point of 15.9Hz. A 100nF and a 100K is the same 15.9Hz, and so is a 10nF and a 1M.
I used this on the input stage to get 4.7nF and around 1M for a roughly 30Hz input bass cutoff. But on the output, you have to guess what will be plugged in.
Quote
Another question (maybe off-topic): I'm designing an opamp buffer right now and I chose to do the biasing through a 10M resistor, like this:
It is better to do it this way or put VB directly on the opamp +input instead? With the second choice you get 3x10M resistors in parallel, with the first you only have 2, am I right? So the choice of a right capacitor value at the input depends on this too.
Your input source sees a load of the first 10M in parallel with whatever the rest of the circuit does. But for choosing the input capacitor, the first 10M to ground may be ignored, as it plays no part in the frequency rolloff.
If you choose to bias with a 10M to some decoupled bias voltage, then the input resistance seen by the input capacitor is 10M in parallel with the input resistance of the opamp; this may or may not be "nearly infinite". The 5532, for instance, has an input resistance of only about 100K. JFET input opamps may have input resistances of hundreds of megohms. The details of the opamp matter here. However, this is a quiet way to bias.
Using simply a 10M to V+ and another to ground is usually not as good from the standpoint of noise from the resistors.
I'm curious - why do you need 10Ms and a 100nF, which gets to an input bass cutoff of 2Hz?
Quote from: R.G. on May 14, 2013, 05:35:28 PM
Quote from: Labaris on May 14, 2013, 04:55:51 PM
So my criteria is not that bad. I always calculate "f0" at the input and output just to ensure full bandwidth response.
That works fine at the input, where you get to choose the loading the input capacitor sees by choosing the design of the input stage. It's not so good at the output where almost anything can be plugged in. The lowest load a 1uF on an output can drive to 20Hz is R = 1/(2*pi*F*C) = 1/(2*pi*20*1E-6) = 7962 ohms. Looked at another way, a 1uF/10K load has a half power point of 15.9Hz. A 100nF and a 100K is the same 15.9Hz, and so is a 10nF and a 1M.
I used this on the input stage to get 4.7nF and around 1M for a roughly 30Hz input bass cutoff. But on the output, you have to guess what will be plugged in.
I guess that finding an input impedance of less than 8k is not likely to happen in "modern-sounding" designs. So that's my bet :icon_cool:
QuoteQuote
Another question (maybe off-topic): I'm designing an opamp buffer right now and I chose to do the biasing through a 10M resistor, like this:
It is better to do it this way or put VB directly on the opamp +input instead? With the second choice you get 3x10M resistors in parallel, with the first you only have 2, am I right? So the choice of a right capacitor value at the input depends on this too.
Your input source sees a load of the first 10M in parallel with whatever the rest of the circuit does. But for choosing the input capacitor, the first 10M to ground may be ignored, as it plays no part in the frequency rolloff.
That's new for me. I always thought that the first resistor had something to do with the filter too.
Quote
If you choose to bias with a 10M to some decoupled bias voltage, then the input resistance seen by the input capacitor is 10M in parallel with the input resistance of the opamp; this may or may not be "nearly infinite". The 5532, for instance, has an input resistance of only about 100K. JFET input opamps may have input resistances of hundreds of megohms. The details of the opamp matter here. However, this is a quiet way to bias.
Using simply a 10M to V+ and another to ground is usually not as good from the standpoint of noise from the resistors.
Ok, that sounds good. I'm using TL072 just beacuse it has JFET input (I'd like to find something more hi-fi)
QuoteI'm curious - why do you need 10Ms and a 100nF, which gets to an input bass cutoff of 2Hz?
Why 2Hz? I got 0.2Hz using the RC filter formula. Maybe I'm wrong.
You're right, I just need 820pF for 19Hz using 10M for bias.
> Because I can't type.
Oo-key. ~~18uA makes much more sense. Good for super-Z impedance. Also weakens the input transistor enough to stagger the fT against the PNP and lessen the 10MHz overexcitement.
hmmmmmm..... fT of the NPN at 18uA seems to be around 3MHz. Which sure will take the peak off 10MHz action. The 100uA PNP may be fT~~18MHz. 6:1 stagger often does tame a peak.
And the 47 ohms works now that the emitter impedance is OTOO 15 ohms. Whatever you put on the far side of the 47 ohms, the buffer's naked output still has "gain" (0.7 out of 0.99) and isn't too strained.
I thought of some of the other ways. There's some very good low-volt opamps now. Some TOO lo-volt for a 6V system. Many sacrifice noise. The low-hiss ones may never sell enuff to be readily available, now or a few years down the road. The LM302 topology allows internal compensation, and you can demonstrate with three transistors, but for large swing it needs several current sources and the device-count soars. Class B saves power, and can be clean, but a clean implementation is heavy work and not readily replicable. Diamond buffer 'cancels' its input current for a little better input Z, and will push-pull, but you lose 1.2V right away and more with bias sources.
> Imagine a coin slot in the pickguard of a guitar.
Pay to play?
------------------------------------------------
> input signal is going to base of 2N5088, which means voltage/current will be bigger for base of 2N5087. Right? How much more negative is the collector of 2N5087 (because the PNP is upside down) compared to the base? That's all that matters for turning it on, right?
"voltage/current will be bigger"-- this isn't assured, you have to make-it-so. *In yesterday's (mis-typed) plan* the total current is 0.12mA (or 0.1mA per R.G.; depends on the parts). If this flows in a (mis-typed) 3.3K resistor it drops 0.4V. This appears across the PNP's B-E junction. Typ Vbe for 2N5087 is 0.6V at 100uA. If Vbe is 0.4V, 200mV lower, the current is about 10^3.3 times lower, or 1/2,000th of 100uA, which is 0.05uA (0.000,05mA!). Since this is significantly less than the 100uA in the rest of the circuit, I call it "off" because it can have no effect.
With today's 33K value, it does have effect. The NPN only has to pass 18uA to turn-on the PNP. PNP bootstraps the NPN and holds its current pretty constant at 18uA. Both of them together must absorb the 120uA coming up from below. Counting on thumbs we see ~~20uA in NPN and ~~100uA in PNP.
The PNP current will vary with signal and load. The NPN, not so much.
-------------------------------
> 1M resistor at the output? I always choose 10k
When you have a chunky battery (or juicebox) and an opamp which will drive 2K easy, 10K is a fine value.
R.G. design is very thrifty on battery. Only 0.1mA of flow. In our dreams the peak voltage from 6V supply is 3V; R.G. likes 4Vpp or 2V peak. 2V/0.1mA is 20K minimum load. It's a tenth as strong as a 5558 or TL072. That is plenty for all our purposes, but we shouldn't hang a heavy bleeder also. I could see 100K, or 1Meg, or whatever I find far-right in my resistor drawers. Assuming that cap is low-low leakage, 1Meg is what you put down so builders don't have to wonder.
Quote from: PRR on May 14, 2013, 10:08:52 PM
Pay to play?
One way or another, we always do. :icon_biggrin:
Convenient way to stick batteries into a guitar. A stack of two 3V cells is about 3/16" thick. It's probably easier to slip them into a slot in the pickguard than to remove a cover or otherwise hide them. Removal becomes the issue then, but hey, there ought to be some problems, yes? :icon_wink:
> stick batteries into a guitar.
What about a Lemon Battery? If you play bars, you may get small-change tips, and the bartender will give you a lemon slice and a napkin. Sandwich a penny, a nickle, and a napkin-tear juiced from the lemon. That's maybe 0.7V so you need several. But the coins will last a long time, or can be rinsed and spent if you need 54 cents before the juice eats all the way through Lincoln's face.
Maybe not on a finely lacquered guitar?
11 posts in runs at higher current with an emitter resistor
http://www.diystompboxes.com/smfforum/index.php?topic=98244.0 (http://www.diystompboxes.com/smfforum/index.php?topic=98244.0)
If you want to drive a load in a known range and want to minimize the current needed a CC in the Sziklai pair "emitter" is needed like R.G. posted.
If you want to be able to drive a load with a EF using a emitter resistor you need to up the current and offset the emitter voltage from 1/2 the supply voltage.
Quote from: PRR on May 15, 2013, 01:29:22 AM
What about a Lemon Battery? If you play bars, you may get small-change tips, and the bartender will give you a lemon slice and a napkin. Sandwich a penny, a nickle, and a napkin-tear juiced from the lemon. That's maybe 0.7V so you need several. But the coins will last a long time, or can be rinsed and spent if you need 54 cents before the juice eats all the way through Lincoln's face.
I don't know if this works well with today's coins. I did build a coin battery once in my childhood from a stack of alternating pennies and dimes. But this was when pennies were copper, not copper plated zinc, and dimes were silver alloy, not culpronickel/copper/culpronickel sandwiches. I suspect that if culpronickel (what nickels are made from) and copper were electronegatively-different enough to make batteries that the layers of dimes, quarters, and half-dollars would corrode badly.
Dang! That's another "back in the good old days" story, isn't it? :icon_biggrin: Back when men were men and they could make batteries from coins... :icon_wink:
while we're at it, why not hook up a hacked wind up radio mechanism into the whammy bar to recharge the battery a little during every townsend-esque wind mill propelled crescendo? :P
Quote from: artifus on May 15, 2013, 11:25:10 AM
while we're at it, why not hook up a hacked wind up radio mechanism into the whammy bar to recharge the battery a little during every townsend-esque wind mill propelled crescendo? :P
A kinetic wiggle stick - I like it!
CR2032 batteries are 225mAh, and at 0.1mA current draw, that would be about 2,250 hours, or about 94 days of straight operation (if my arithmetic is correct). The batteries would last about 2 years playing 4 hours every day. Not a bad interval to have to remove the pick guard.
Double CR2032 holders are available.
Quote from: Jdansti on May 15, 2013, 08:19:41 PM
CR2032 batteries are 225mAh, and at 0.1mA current draw, that would be about 2,250 hours, or about 94 days of straight operation (if my arithmetic is correct). The batteries would last about 2 years playing 4 hours every day. Not a bad interval to have to remove the pick guard.
What is it with me and decimal points these days. I came to about 200+/- hours on that calculation.
Oh. Wait. Now that you mention that, I think did the calc on an opamp with 1ma per amplifier, then didn't update it. Doh!! That's one of the motivators for going to 100uA. What a ditz I am.
I think you're right - 2250 hours.
QuoteDouble CR2032 holders are available.
Got a link?
I found some that would enable the coin-slot opening. In a strat, this could be used on each side of the ditch routed from controls down to the jack, I think. But an internal two-cell is probably OK at that many hours.
2-Cell CR2032 Holders
Quote from: R.G. on May 15, 2013, 08:45:25 PM
Quote from: Jdansti on May 15, 2013, 08:19:41 PM
CR2032 batteries are 225mAh, and at 0.1mA current draw, that would be about 2,250 hours, or about 94 days of straight operation (if my arithmetic is correct). The batteries would last about 2 years playing 4 hours every day. Not a bad interval to have to remove the pick guard.
What is it with me and decimal points these days. I came to about 200+/- hours on that calculation.
Oh. Wait. Now that you mention that, I think did the calc on an opamp with 1ma per amplifier, then didn't update it. Doh!! That's one of the motivators for going to 100uA. What a ditz I am.
I think you're right - 2250 hours.
QuoteDouble CR2032 holders are available.
Got a link?
I found some that would enable the coin-slot opening. In a strat, this could be used on each side of the ditch routed from controls down to the jack, I think. But an internal two-cell is probably OK at that many hours.
2-Cell CR2032 Holders:
http://www.digikey.com/product-detail/en/BH800S/BH800S-ND/221549
http://www.digikey.com/product-search/en?x=30&y=16&lang=en&site=us&KeyWords=Keystone+1026
http://www.mouser.com/ProductDetail/Keystone-Electronics/1026/?qs=%2fha2pyFadujp6VwXBQ0HmMawfeGvkaMjNKQRVlsmEjRSUkf0FcD7xw%3d%3d
Ignore the "1 battery" on the Mouser page. The data sheet shows that Keystone part #1026 holds two CR2032 batteries.
I was thinking something like this:
(http://cdn.arstechnica.net/wp-content/uploads/2012/06/insert-quarter.jpeg)
^ The only down side is when you're in the middle of a tune and the guitar sound switches off and you hear, "Game Over" from your amp. ;D
Quote from: Jdansti on May 16, 2013, 11:06:16 AM
^ The only down side is when you're in the middle of a tune and the guitar sound switches off and you hear, "Game Over" from your amp. ;D
xD
Quote from: Jdansti on May 16, 2013, 11:06:16 AM
^ The only down side is when you're in the middle of a tune and the guitar sound switches off and you hear, "Game Over" from your amp. ;D
Honestly I would rather it do this
If we're really being honest, I was looking for the sample from 720
Quote from: R.G. on May 15, 2013, 08:45:25 PMIn a strat, this could be used on each side of the ditch routed from controls down to the jack, I think. But an internal two-cell is probably OK at that many hours.
In a Strat, there's always the whammy rout in back (4 screws are better than 11, if you even have the back cover on).
Mmmm.
Coin cells would fit the whammy cavity. The controls cover on a LP would also accomodate them.
OK. That's two guitar models... :icon_lol:
Seriously folks - the electronic design isn't all that difficult. It's dealing with the pesky humans that is tough.
Quote from: R.G. on May 13, 2013, 05:19:17 PMIt could be made to have gain...
Do tell. A bit of switchable boost would be a nice option. Or would any gain necessarily drive us back into the arms of the 9v battery?
Horses for courses.
You can put in a modest amount of gain by inserting some resistors in the complementary feedback pair. However this begins to dilute some of the advantages of the CFP. You can correct for that, probably eating some more battery in the process. And you're still stuck with the voltage ceiling of 6V. You could add a third - or fourth! - coin cell for more volts.
At some point the tinkering gets complex enough to make an opamp a more palatable solution. There are opamps that would do something useful with gain at about 175-200uA per stage, and give you a lot of gain.
However, unless you really need switchable boost inside the guitar, you can put as much gain outside the guitar as you can stand. The buffer has eliminated the problems of control and cable loading, so whether the gain stage is inside the guitar or not depends on other issues, like whether you can stand to have the gain external to the guitar, or must have it coming out of the raw guitar.
A lot of design work comes down to objectives. Change an objective or feature and that ripples back into the design as changes. Some changes, not even huge ones, can ripple all the way back to the start of the design.
Two more (possibly obvious) questions:
In terms of how to incorporate this buffer, I'm guessing that standard tone controls could be part of one's magic loading, with additional R+C in parallel to taste, to duplicate the loading of the volume pot and cable; then a single volume control, such as on a Tele or Strat would go post-buffer. In that case, should the volume pot should be reduced in value to 25k-100k?
And, in an LP style guitar, do you need to have two buffers?
R.G. why did you use "magic loading"
There is no magic loading there is the interaction of the guitar/bass R, L and Cs with the cable and the input of the amp or effect after the guitar.
There is science to the resulting EQ and with software like LT spice etc one can model what they want to happen.
Did you use "magic loading" to start another internet term that will be repeated again and again.
Quote from: Gus on May 19, 2013, 01:08:26 PM
Did you use "magic loading" to start another internet term that will be repeated again and again.
I hope so! Then I can tell my grandchildren that I was there when Sorceror Keen conjured the magic loading! :icon_cool:
Seriously though, maybe RG is paraphrasing Clarke ("any sufficiently advanced technology is indistinguishable from magic") and sparing the less scientific readers from the intricacies of of pickup behavior.
Hi
I'd hope that people listen to both Gus AND Brian. Do we need to subscribe to EITHER science OR magic?
I love science, but I also love simplicity and myths.
Magic load? Cool name, I like it personally, but it's smoke and mirrors for sure!
cheers
Magic is just science that's too darn difficult to figue out yet. Or something. :icon_lol:
Quote from: Ben N on May 19, 2013, 11:37:33 AM
In terms of how to incorporate this buffer, I'm guessing that standard tone controls could be part of one's magic loading, with additional R+C in parallel to taste, to duplicate the loading of the volume pot and cable;
All this started with a complaint/question from a guitar owner that the tone from his guitar suffered with changes from the volume control, and was worst at middle volume control settings. A little thought said that the pickups were loaded most in this position by the volume/tone controls themselves. I opined that a buffer ought to eliminate this, and designed a buffer to meet this need. I believe this would work.
The term "magic loading" cam about when I realized that the loading of the volume and tone control at certain positions was part of the tone the player wanted to preserve, and that it was possible that each player might want a different setting to get just the tone they wanted to preserve. Hence, there was a magic setting that pleased their ears. Magic is real - but it only exists inside the human mind!
Quotethen a single volume control, such as on a Tele or Strat would go post-buffer.
That's one solution.
QuoteIn that case, should the volume pot should be reduced in value to 25k-100k?
Depends on the pesky humans. Guitar players have this predilection for demanding that their guitars should be defaultable back to the basic magnetic pickup with no electronics. Valid or not, they do. So the right setup for cleanly driving a cable to the amp is a lower volume control. 25K would be good. But then the guitar isn't easily convertable back by flipping a switch.
QuoteAnd, in an LP style guitar, do you need to have two buffers?
Quote from: Gus on May 19, 2013, 01:08:26 PM
R.G. why did you use "magic loading"
There is no magic loading there is the interaction of the guitar/bass R, L and Cs with the cable and the input of the amp or effect after the guitar.
There is science to the resulting EQ and with software like LT spice etc one can model what they want to happen.
Did you use "magic loading" to start another internet term that will be repeated again and again.
See above. It was a semi-satirical reference to each guitar player thinking that THEIR pet setting is the magic one.
Quote from: B Tremblay on May 19, 2013, 02:15:32 PM
I hope so! Then I can tell my grandchildren that I was there when Sorceror Keen conjured the magic loading! :icon_cool:
Seriously though, maybe RG is paraphrasing Clarke ("any sufficiently advanced technology is indistinguishable from magic") and sparing the less scientific readers from the intricacies of of pickup behavior.
OK, maybe some of that too. :icon_lol:
Quote from: brett on May 19, 2013, 07:09:36 PM
I'd hope that people listen to both Gus AND Brian. Do we need to subscribe to EITHER science OR magic?
I love science, but I also love simplicity and myths.
Magic load? Cool name, I like it personally, but it's smoke and mirrors for sure!
Has anyone else been struck by the acronym for smoke and mirrors being S&M?
:icon_lol:
Clarification Question:
Are the un-labled pads for a second pick up and do I have them labeled correctly in the graphic? This is going in a 3 year-old 99$ Dean Evo to drive a 50ft cord with the original cheapo factory pick-ups. I was also wondering if I could use the stereo jack trick to turn off the battery by unplugging when not in use. Thanks.
(http://drunkenafficianado.files.wordpress.com/2014/03/buffer.png?w=640&h=480)
> un-labled pads for a second pick up
Power Input.
(Hint: those pads go right to a big capacitor. 1uFd will short-out ALL pickup signal.)
Hi not at my computer right now. I take everything you say as gospel PRR, but is this a jest? A pickup has intermittent signal so could be considered a capacitor functionally. I may have to instead go with orman's booster cable. A disappointment because I have parts, but it is big. I saw no capacitor listed.. any help appreciated.
> A pickup has intermittent signal so could be considered a capacitor functionally.
I don't understand that trail of thought.
Re stereo jack, connect the board's ground to the ring to use the jack as a switch.
I'll let PRR handle the mystery connections.
My mistake, I shouldn't have posted. I'll go back to silently lurking and learning. I will use the ghost powered booster cord from Orman.
(http://drunkenafficianado.files.wordpress.com/2014/03/plugschematic.gif?w=374)
(http://drunkenafficianado.files.wordpress.com/2014/03/phantomschematic.gif?w=374)
I will remove the images after 24 hours. If there was a decent upload process for posting images here rather than using a second URL, I would not have posted the URL in my blog, which no one reads because there is nothing there.
Actually this schematic is from J. Donald Tillman not Orman, sorry.
Quote from: JohnBlakeArnold on March 05, 2014, 05:52:50 PM
My mistake, I shouldn't have posted. I'll go back to silently lurking and learning.
Don't go getting all passive-aggressive "I'll just sit here and freeze in the dark." on us. :icon_biggrin:
What Paul was gently trying to tell you is that what you said about "A pickup has intermittent signal so could be considered a capacitor functionally." doesn't make sense electronically, and that there's a gap or mistaken impression in your knowledge that you can get filled in.
A pickup having an intermittent signal does not qualify it for consideration as a capacitor. Capacitors aren't the only things that have intermittent signals. And worse yet, pickups are heavily, heavily inductive; so considering them capacitive leads you way off into the weeds when you start designing around them.
Going back to your original question,
QuoteAre the un-labled pads for a second pick up and do I have them labeled correctly in the graphic?
No, those are where the power is provided for the circuits to operate.
QuoteThis is going in a 3 year-old 99$ Dean Evo to drive a 50ft cord with the original cheapo factory pick-ups.
In that case, put this buffer circuit after everything in the guitar, but before you connect to the input jack itself. It will make a good cable driver.
QuoteI was also wondering if I could use the stereo jack trick to turn off the battery by unplugging when not in use.
Yes.
You're doing fine - ask your questions, and appreciate being told the straight stuff in return.
I *agree* that the un-labeled pads may confuse some builders.
Problem is: experienced plumbers kinda "know" what pipes to expect. Here's where water comes into the house, there is where water (and yuk) runs out to sewer, and this third pipe may be gas to heat the water?
So they (we!) over-look labeling ALL the connections.... done it myself too often.
So as a journeyman you want to learn what the master sees. A simple buffer usually needs input, output, and power. You don't have to look for the water-meter to find the input, R.G. did note "from pickup". And the output is over there. But what's this other connection...?
There's a "drawing problem". We have an input and an output, and a power connection that often connects to *many* things between input and output. We may put the input and output left and right. But then the power circuit tends to be in-the-way in the middle. It gets tangled-up. We often try to abstract it into a corner, by using "+6V" arrows here-to-there. That's often confusing also, and I dread those drafting-program schematics which encourage young designers to splatter arrows over all pages of a 6-page schematic (often with mystery designations like "G43").
The Tillman is excellent although in this context the power scheme is broken-up over two units, not like most toyboxes which have it all in one unit. It's not perfectly clean and its uncleanness can be musically useful. It is moderately sensitive to FET selection. The Phantom power option does put the battery over-there where space weight and accessability are no-problem.
The R.G. plan linked here is compact battery and dead-clean. It is not at all parts sensitive (nearly any post-1975 transistors will work).
If you need multiple pickups, the traditional way is a Pickup Switch before the buffer. You could go for a no-knob mixer, or something more elaborate.