An onboard buffer for guitars

[Labaris]

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:



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.

[R.G.]

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.

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?

[Labaris]

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 

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.

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)

I'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.

[PRR]

> 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?


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> 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.

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> 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.

[R.G.]

Pay to play?

One way or another, we always do. 

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? 

[PRR]

> 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?

[Gus]

11 posts in runs at higher current with an emitter resistor
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.

[R.G.]

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?     Back when men were men and they could make batteries from coins...   

[artifus]

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? 

[gritz]

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? 

A kinetic wiggle stick - I like it!

[Jdansti]

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.

[R.G.]

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.

Double 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.

[Jdansti]

2-Cell CR2032 Holders

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.

Double 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.

[earthtonesaudio]

I was thinking something like this:

[Jdansti]

^ 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. 

[Labaris]

^ 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. 

xD

[wavley]

^ 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. 

Honestly I would rather it do this

If we're really being honest, I was looking for the sample from 720

[Ben N]

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.

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).

[R.G.]

Mmmm.

Coin cells would fit the whammy cavity. The controls cover on a LP would also accomodate them.

OK. That's two guitar models...   

Seriously folks - the electronic design isn't all that difficult. It's dealing with the pesky humans that is tough.

[Ben N]

It 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?