The best buffer... or at least my favorite!

[shredgd]

Just to let you hear what I'm talking about.

In this video I  found on YouTube, you can hear the difference between a 10k vs 100k to ground in the output buffer I was referring to (those output resistors, together with the 100 e 470 ohm, are the only difference between the TS9 and TS808, right?).
I know, even the same circuit can sound different because of components tolerances, but this result is recurrent in the hundreds of TS808 vs TS9 comparisons you can find: the 808 (10k to ground) sounds more "natural" and you get a little more bass and less harshness in the tone. This is exactly the same difference I could hear comparing resistors in my buffers.

https://youtu.be/yDgQUHNvBKA

[bool]

What I like are BC550B transistors with beta cca 400-450; 6k8 emitter resistor; 470k base bias resistor; base biased from cca 0,6Vb (6v at 9v battery); and cca 22k (or 25k pot) pulldown (the one that you want to be 10k).

Works good, sounds good. I've been happy with that combo since early 90's.

[shredgd]

I think this should be the final update of my full immersion into buffers.

Although very happy with the tone I could get, my buffer did suffer from an audible compression of dynamics. Thanks to the analysis posted by Groovenut, I hypothesized this might be due to the reduced voltages at the base (and emitter, of course) I got by using a transistor with a gain of about 200 and not a higher gain one. My Ve was 3.1v, which should still be enough for a voltage swing of less than 1.0v generated by a guitar pickup: however, this proved to be wrong in real life, because raising Ve to about 4.4v (by lowering the upper resistor in the voltage divider), got me all the dynamics back!
What I learned this time is that the more the transistor is biased exactly at half the power supply voltage, the more it works good.
I measured just a 0.33 mA increase in power consumption, which is negligible in my opinion.

Another little improvement I did, after other obsessive and endless comparisons between buffered and unbuffered tone (straight to amp), was to raise the input bias resistor to 680k, which gave me a little more highs (without overdoing the guitar-cable-amp tone) and even bass.

This is the schematic of my definitive, highly optimized bipolar buffer, which I suggest you to check out. Enjoy!



PS: the parallel 1.0uF output film cap might probably be smaller (0.1uF?), and still keep its effect of compensating for the treble loss caused by the big electrolytic cap

PPS: I still didn't experiment with "bootstrapping" a la Peter Cornish, because by the demo videos I found, it still seems to me it gives too much high impedance and the tone gets brighter than my reference sound (guitar-5 meters cable-amp)

[antonis]

<What I learned this time is that the more the transistor is biased exactly at half the power supply voltage, the more it works good.>
I should totally agree in case of "good" refers on pure unistorted signal.. 

If you have the kindness to enlighten us (me, at least) about the way you managed to get +4.5V from +9V and a 10k/4k7 voltage divider we should be more than grateful .. 

Bootstrapping is a convenient way of raising input impedance which, in your case should be easily obtained with a, 4k7 say, resistor in place of 680k one and a, 10μF say again, capacitor  - but you don't practically need it..
* edit: On a second thought, you MAY need boostrapping, because your base bias equivalent resistor isn't so low compared to R_E x h_FE

Also, with 680k bias resistor value you may lower input capacitor value down to 4n7..

[shredgd]

Hi antonis,

I came out with the 4.7k in the voltage divider just by experimenting with a trimmer in parallel with the existing (and previously in place) 10k resistor: I had to use 7.4k to get +4.4V at the emitter of my idle transistor, which in parallel to 10k gives about 4.2k (4.7k was the closest value I had). Oh, now I see your doubt: of course the voltage is higher at the 10k/4.7k junction, I will correct the picture as soon as I can (I worked on a previously made schematic and didn't remember of that "+4.5V"...).

Regarding your last sentence, if you look at the simulations run by groove nut you can see how the bass frequencies are (gently) rolled off with a tiny but existing involvement of frequencies which are still in the audible range, and how the lower the gain of the transistor, the higher the cut off frequency. So I will keep the 0.1 uF in place, because I don't want to risk to lose any bass.

[shredgd]

On a second thought, you MAY need boostrapping, because your base bias equivalent resistor isn't so low compared to R_E x h_FE

Thank you for your replies, antonis. Can you explain this observation you did?

[antonis]

On a second thought, you MAY need boostrapping, because your base bias equivalent resistor isn't so low compared to R_E x h_FE

Can you explain this observation you did?

Let us see..

You should have a very "stiff" (compared to what it "feeds") voltage divider comprised of 2 x 10k resistors in the absence of 680k resistor..

+4.5V point is considered as a voltage source in series with it's internal resistance which is simply voltage divider's equivalent resistance - Thevenin theorem.. 
(in cases of substantially lower equivalent resistance, compared to h_FE x R_E, it's simply ingored..)   

Of course, a 5k (10k//10k) input impedance isn't what you will, so you've placed a 680k to effectively raise input impedance to (680k +5k) // h_FE x (R_E +r_e)..
(I hope you follow me - neither my educational nor my English skils are adequate..  )

The above resulted in drastically raising the analogy between voltage source and load resistances..
(it should be convenient in case of biasing a FET gate but for a BJT base bias current DOES matter..)

Your circuit may be quite stable ONLY under specific circumstances (temperature, V_BE, BJT beta, resistors tolerance, etc.)

That's the reason for significantly lowering 680k resistor, keeping it's impedance high by bootstrapping it.. 

[shredgd]

I have read a couple of pages about this topic and the problem you raised is true when you bias your transistor directly with two resistors (one linked to +9V and one to ground): in that case, the equivalent resistance of the two resistors is suggested to be less then 1/10 the value of hFE x Re, to be stable enough.
But voltage dividers are really there to avoid this limitation. If you think about it, the standard value of 510k isn't that low compared to hFE x Re either, but it's been used nonetheless, in countless circuits and for decades! This is thanks to the low value of the resistors in the voltage divider (10k), which make it more difficult for a load to move the bias voltage. Correct me if I'm wrong!

[antonis]

<I have read a couple of pages about this topic and the problem you raised is true when you bias your transistor directly with two resistors (one linked to +9V and one to ground): in that case, the equivalent resistance of the two resistors is suggested to be less then 1/10 the value of hFE x Re, to be stable enough.>

You're right but in the above case there isn't anthing interposed between voltage divider and the base of the transistor..
Voltage divider output "sees" nothing in front of its load - you may actually consider it as a voltage source in series with its internal resistance (parallel combination of divider resistors).

In your case, a quite big resistor is placed in bias current path (as seen from base) so you need higher enough voltage divider out for setting Emitter at same voltage as before..(high voltage drop on bias resistor).

For any load variation, bias is upset propotionaly to bias resistor/load value..


<If you think about it, the standard value of 510k isn't that low compared to hFE x Re either>

I've thought it Giulio and I've resulted that 510k is about the HALF of h_FEx R_E (2N3904 current gain at 500μΑ collector curent is slighty over 100 in best case scenario - worst case of h_FE of 40 leads to voltage source's resistace bigger than load's one..) 

What I'm trying to tell you is, despite of your Vref trimming (to effectively bias Emitter on Vcc/2) your bias circuit is highly sensistive to even slightest parameters upset/deviation..!!

Just imagine two different voltage sources feeding two identical loads where first source internal resistance is 1/10 of load resistance and second source internal resistance is equal to (or greater than) load resistance.. 
Which of the above is more stable against load variations..?? 

P.S.
I don't say you've wrongly biased your circuit - if it works, it works..!! 

I'm just trying to tell you that you could also obtain equally high (or even higher) input impedance together with the benefits of a stiff voltage divider bias by significantly lowering bias resistor value and simultaneously bootstrapping it.. 

[shredgd]

I see your point! Maybe bootstrapping just doesn't convince me because I cannot calculate input impedance that way, I only know it will be "much higher"... but the starting point of my almost endless research about buffers, beside the fact that I can't stand the tone of FET, MOSFET and IC based buffers (yes, they absolutely have different tones each), is that I can't stand too much high input impedance in buffers (it gives way too much highs compared to my reference guitar-cable-amp tone, with unpleasant results if you use overdrive/distortion after them)... so now I feel the need to be able to anticipate the exact input impedance of my buffer, and be able to fine-tune it to my reference tone!

PS: for the few ones who read all this thread, the above schematic is absolutely verified and working! It sounds and feels (under your fingers) as I always wanted a buffer to do!

[PRR]

> biased exactly at half the power supply voltage

You have 10K emitter DC resistor, and 10K bleeder(why?), and whatever it actually drives.

With 10K load and 10K pull-down, and a high current pull-up (any BJT), the ideal best-swing bias would be 2/3rd of supply. Set emitter at 6V. The BJT will easily pull-up 10K to 8.9V. But pull-down with 10K pulling 10K is only half of 6V, or 3V. So 2/3rd is about optimum.

However I suspect the 10K bleeder is not needed. Make it 100K. Assume another 100K of actual load. Now we have 10K pulling 50K. Will swing about 5/6 of whatever the bias is. So setting emitter near 4.5V gives 3.7V down, 4.4V up, which is close enough to both-ways-equal for any speech/music application. (E at 4.85V may be be insignificantly more perfect math.)

[shredgd]

According to this http://m.electronicdesign.com/analog/avoid-clipping-emitter-follower-ac-coupled-resistive-load interesting explanation of the effect of a load after the decoupling output cap, the effect of the load resistor isn't to be considered as a voltage sink like you did. It must be considered in terms of current to subtract from Ie, which is a function of the voltage swing of our guitar signal divided by the resistor itself (Vampl/10k in my case): even for the strongest pickup-on-steroids voltage swing of 1.0V, this would be 0.1mA with 10k, which multiplied again for Re (10k) gives 1.0V (to subtract from Ve). Not 3.0V or half-whatever Ve we have, but a "fixed" value of (Vampl/Rload) x Re.

I agree that a 100k resistor would reduce this value by ten, but as you can read at the beginning of this thread, what "started it all" was the amazing improvement in tone I got from reducing this pulldown resistor to 10k a la TS808! Don't ask me why...

[antonis]

I see your point! Maybe bootstrapping just doesn't convince me because I cannot calculate input impedance that way, I only know it will be "much higher"...

I also see your point now... 

Just for your info, impedance calculation of bootstrapped resistor isn't complicated..

Using a cap of relatively high capacitance (i.e. 10μF) and supposing a working current of 450μA in your case, you have a r_e (intrinsic emitter resistance) of about 55R (25/I_C(mA))

Your emitter follower gain is A = R_E / (R_E + r_e) => A = 0.9945
(actually, R_E should be the parallel combination of Emitter resistance and any driven load..)

Bootstrapped resistor value is calculated as: R/(1-A) => 180 times its actual value..
(i.e. 850k for a 4k7 resistor..)

[bool]

Your calculation dosent factor in the resulting cumulative load "Z" which in this case would be the combination of a Re || (Xc + Rpulldown) ... simplified; so will be frequency dependant - but for the actual passband you should factor in the Re || Rpulldown instead of just Re; because that's what will reflect back to the transistor base ...

The easiest method is to inspect the Zin with a simulator like a ltspice. You get a nice curve to look at and to tweak.

[DavidRavenMoon]

- electrolytic DC decoupling caps DO eat some of your high frequencies (presence): a film cap parallel to the 10uF output cap is not just an option, but really necessary to "take the blanket off" (without adding any frequency you could already get with the guitar-cable-amp setup, which is your reference)

You do realize that putting two caps in parallel changes the value, right? You get the sum of the two caps. So the value increases. That passes more low frequencies.

The trick with buffers is that if you want a brighter tone, use a higher input impedance. That prevents your pickups from being loaded. Most amps are about 1M. The capacitance from your cable effects the treble too, but you need an awfully long cable, or a poor quality one, to hear that.

If you don't want the tone brightened (which is the real tone), use a lower input impedance. I find about 150k doesn't sound like a buffer.   


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[shredgd]

You do realize that putting two caps in parallel changes the value, right? You get the sum of the two caps. So the value increases. That passes more low frequencies.

As you can read in the same quote of my post you did, I said that putting a parallel film cap restores the high frequencies (presence) lost by the electrolytic cap, I didn't talk about bass frequencies in that paragraph.

Also, nobody talked about 150k input impedance (where did you read that?).

The point about all this long talking about buffer is that if you have a real quality amp, quality guitars, a good guitar playing and a good ear, you will clearly hear that guitar-cable-buffer-cable-amp, usually doesn't sound as good as guitar-cable-amp.

This is true for the most popular buffer schematics on the web, and also for most commercial buffer pedals. This is "sometimes" due to a too high impedance of buffer circuits (over 1M), which makes the tone shrill (we all know why), but most often if you lower that input impedance (e.g. if you put a 1M resistor from signal to ground to lower the 10M input impedance of the popular MOSFET buffer) that buffer still doesn't quite right/faithful to the original tone.

Thanks to the other contributors to this thread, and thanks to experimenting, I also discovered that a lower bias with a bipolar buffer (which is the case of a standard TS buffer with a mid gain transistor, not a wrong variation I did), although still sounding great tone-wise, adds compression: this is something I've never read about in buffer-oriented threads! I invite everyone who is following this thread to check this phenomenon (you obviously have to use a very clean amp and try some wide dinamic escursions in your playing) and test how lowering the +9V tied resistor in the voltage divider restores dynamics!

So, "the trick about buffers" doesn't only rely upon setting the right impedance...

[DavidRavenMoon]

You do realize that putting two caps in parallel changes the value, right? You get the sum of the two caps. So the value increases. That passes more low frequencies.

As you can read in the same quote of my post you did, I said that putting a parallel film cap restores the high frequencies (presence) lost by the electrolytic cap, I didn't talk about bass frequencies in that paragraph.

You don't seem to understand how caps work. As I said putting two caps in parallel increases the value (the sum of the two caps). With a coupling cap that will increase the low frequency response. Electrolytic caps don't lose high frequencies. But I prefer tantalum over aluminum cans.

Also, nobody talked about 150k input impedance (where did you read that?).

Why would I need to read that? Lol. I'm giving you a tip. You don't seem to understand how impedance works in regards to passive guitars.





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[shredgd]

Electrolytic caps don't lose high frequencies. But I prefer tantalum over aluminum cans.

Type "equivalent series resistance" on google 

You don't seem to understand how impedance works in regards to passive guitars.

If you are right about saying the above buffer circuit has 150k input impedance, then nobody in this thread/forum does, only you do. You lucky boy! 

[anotherjim]

I was wondering if the AC load on the emitter was being considered. Effectively close to 5k, and an unlucky gp transistor  has, say hfe 70, and your input impedance is 50% lower than you might accord from the Re value. I don't know what pickup/onboard pot combination you are working with, but we know already it might darken the tone or not depending.

The 150k was as I read it, mooted by David as a minimum, but that does in my experience depend on the guitar. My Telecasters neck pickup is quite happy into 100k. Me personally, I don't like the idea of relying on loading to control tone, but it goes on all over the place in guitar land.

[shredgd]

As this is being a highly "fine-tuned" buffer, gain here is known, and it is a measured hFE of about 200. I probably should have specified it on the schematic. However, don't forget we're talking about a couple small variations (one being a bigger bias resistor, which increases input impedance) from the good old tubescreamer buffer... it can't be so bad, history has proven it!