Tube Active HPF/LPF

Started by Snodgrass, February 04, 2025, 08:55:13 PM

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Snodgrass

Greetings,
This is my first post here so I thought I would share something I recently successfully built.  I posted it on another forum (to very little interest) but I cannot help but think it might be interesting or useful to some one and that maybe this place is more likely to have that some one.



This is an active, low voltage tube bass HPF/LPF I put together after a fair bit of research.  It is 4 tubes (the last three are paralleled), the first being a half-mu stage (which really gives too much gain but alas) then a fixed two pole Sallen-Key HPF at around 35 Hz, then an adjustable two pole Sallen-Key HPF which allows for a cutoff frequency of between 30 adn 180ish Hz, then a fixed two pole Sallen-Key LPF at around 2.5k.  This is tailored to my particular bass rig and my design goals.  The Q's of the first two filters I tried to make such that they were roughly in Butterworth territory when put together. (I think)

I realize this could have all been much more easily been done with op amps and whatever but I like a challenge and I happened to have the tubes and most of the parts already.  It is all powered by a (somewhat expensive) 24V power supply (Meanwell GST120A24-P1M). The power supply cost as much as everything else put together.

This circuit is for me both very quiet and extremely effective.  If you are wondering why you would want to filter lows on bass there is a lot of reading to do over on the talk bass forum which I could summarize poorly if you like.

merlinb

Nice to see something out of the ordinary, looks a bit like the various active crossovers that appear on Tubecad.com. BTW if you're interested in active tube EQ, I stumbled upon this the other day:
https://www.preservationsound.com/2011/05/a-few-interesting-diy-audio-projects-c-1955/

Snodgrass

The first stage is indeed ripped straight off from Mr Broskie, or at least he is the only person I have ever seen utilize that 'half-mu' stage that I know of. It really probably isn't the most appropriate circuit as it has too much gain but it works. I was shooting for roughly unity gain and I lose I think about 2 dB through the three filters.  U1 gives me I think ~16 dB which makes it overkill and as it stands I keep the volume set on about 2 to get what I think is unityish
The rest is my attempt to hollow-state-ify existing opamp HPF circuits (the fdeck hpf specifically). Reading over the abridged version of the original Sallen and Key paper was quite useful here, and extremely humbling.
I am certain my feeble attempt here, which nonetheless works, could be improved and simplified massively by some one who knew what they were doing.

That active EQ is awesome but man the unlike double pot really is a stick in the mud for me. I've modified pots before and swapped wafers and stuff but aftewards I look at the thing and really wish I hadn't.

mozz

Nice but not a common tube (automotive), could probably sub a 6DJ8 or 12au7 in there.
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Snodgrass

I chose it for the low voltage performance specifically, yes. It is also designed primarily as an RF tube I understand so isn't ideal in a few ways. Still not hard to find, works for audio and affordable —and the circuit works.

amptramp

Taking C1 to the top of R3 rather than the bottom would change the circuit from a µ-amp to an SRPP (shunt regulated push pull) circuit that might have interesting properties.  With SRPP amplifiers, R3 should be the reciprocal of the upper tube transconductance to maintain equal swing in the positive and negative direction, so the 5200 µmho transconductance of the 6GM8 would result in R3 being 192 ohms.  Your 180 ohm R3 resistor is pretty close and would not need to be changed.

In an SRPP amplifier, the current through the upper and lower tubes is equal until signal comes in, so although it is not a power amplifier, it needs a load since the difference in current in the upper and lower stage has to go somewhere or it will clip.  Where you have capacitive coupling to the next stage, you would get bass clipping, so you would need a coupling capacitor at the input to avoid overloading the stage with bass.

Try this out, it will operate completely differently from a µ-amp but a real µ-amp would have a FET or bipolar transistor or a pentode as the upper stage since it is designed to be a current source for highest gain.

Snodgrass

Quote from: amptramp on February 05, 2025, 08:48:14 AMWhere you have capacitive coupling to the next stage, you would get bass clipping, so you would need a coupling capacitor at the input to avoid overloading the stage with bass.

I am confused by this sentence.  U2 is capacitively coupled to the previous stage.  Are you saying I would need a coupling cap at U1 grid circuit to shave off some bass, to avoid said bass overload?

Your proposed SRPP conversion is interesting but probably not a good fit for my use case where I was trying to avoid distortion and only included a voltage gain stage at all to try to compensate for losses through the filters. Perhaps it would be more useful if this was for guitar where distortion was not an issue or desirable. It is also entirely possible I am misinterpreting your suggestion which, if this is the case, I apologize.

One thing that occurred to me when I was designing this thing is that is seems to be somewhat difficult to get very small amounts of voltage gain (say 2 to 5 dB) in tube circuits.  I can get big jumps (common cathode) or slight losses (cathode follower) easily.  Perhaps I could have made U1 a common cathode stage with local negative feedback but I was worried about having large series resistances and the associated noise. Or maybe I should have studied Merlin's preamp book more closely.


merlinb

#7
Quote from: Snodgrass on February 05, 2025, 07:47:45 AMThat active EQ is awesome but man the unlike double pot really is a stick in the mud for me. I've modified pots before and swapped wafers and stuff but aftewards I look at the thing and really wish I hadn't.
It's not too difficult to mod the circuit to use normal stereo pots. Here's a version I drew up for the bass section, based on discussion from the original article.



Quoteit seems to be somewhat difficult to get very small amounts of voltage gain in tube circuits.
You're not wrong! Maybe I should write about this somewhere. A trick you can do with an ordinary gain stage is to take some of the load from the anode and stick it in the cathode instead (you will need an input cap to block DC). When half the total is in the anode, and half in the cathode, the gain will be 1. So you can get any gain from max down to 1 (or even less!). Linearity improves as you reduce the gain, but the output swing is reduced and PSRR gets worse, which is the price you pay.



Snodgrass

Quote from: merlinb on February 05, 2025, 11:24:36 AMA trick you can do with an ordinary gain stage is to take some of the load from the anode and stick it in the cathode instead.
Very clever! Does this do anything to lower the output impedance?  I imagine not as your bias point stays roughly the same? I was worried in my design about I/O impedances and how they might effect the individual filters' performances. With opamps you don't have to worry about this so much obviously but moving over to glass I felt I had to be much more careful. The half-mu (overkill though it is) seemed like the best compromise between low output impedance and moderate gain that I could dig up. I am not educated enough to design from first principles so I have to sift through existing examples I can dig up and cobble them together.

Quote from: merlinb on February 05, 2025, 11:24:36 AMIt's not too difficult to mod the circuit to use normal stereo pots.
Not too difficult. A comedian as well?
Am I correct in thinking that the operating principle here is that we are selecting for a center frequency which is shifted 180 degrees through the three RC stages like in a Fender phase-shift oscillator tremolo circuit? Apologies if this is elementary.

merlinb

#9
Quote from: Snodgrass on February 05, 2025, 11:46:33 AMDoes this do anything to lower the output impedance? 
Output Z initially gets worse as you put more resistance in the cathode, but it starts falling again if you keep going, since output Z cannot be larger than Ra itself.

BTW, if you convert your input stage to an SRPP (no changes needed, just move the output from the bottom of R3 to the top) you get lower output Z for free!

QuoteNot too difficult. A comedian as well?
A circuit simulator can make an expert of anyone  ;)

QuoteAm I correct in thinking that the operating principle here is that we are selecting for a center frequency which is shifted 180 degrees through the three RC stages like in a Fender phase-shift oscillator tremolo circuit?
Exactly! See the original article on page 20:
https://www.worldradiohistory.com/Archive-All-Audio/Archive-Audio/50s/Audio-1954-Nov.pdf

Snodgrass

Quote from: merlinb on February 05, 2025, 11:53:58 AMif you convert your input stage to an SRPP (no changes needed, just move the output from the bottom of R3 to the top) you get lower output Z for free!
Don't I then run into distortion issues like amptramp seems to have suggested above? I thought the half-mu was already pretty low output impedance as it is defined by the two triodes in parallel. I had read the SRPP was not suitable for driving light loads.

merlinb

#11
Quote from: Snodgrass on February 05, 2025, 12:31:21 PMDon't I then run into distortion issues like amptramp seems to have suggested above?
No that's only when driving a heavy load like headphones. When driving a high-impedance load, the SRPP functions basically identically to the half-mu, except for having lower output Z. (Under these conditions I call it a quasi-SRPP, since it stops being truly push-pull). IMO this makes the half-mu kinda redundant, since the quasi-SRPP is "the same but better"! I think amptramp is getting different concepts mixed up.

PRR

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Snodgrass

I might give that a go then though I am not entirely sure this thing can get much better for my humble needs.  I was massively relieved when I first built it a few days ago and it actually seemed to do what it was designed to and the tone improvements I was hoping for actually materialized.

This thing was kind of a hail mary for me as I had never actually seen any one use a tube Sallen Key before or make anything like this*. Probably in the era when this thing would have been conceivably built, the sort of people who needed such a thing were the sort of people who had money for massive well made inductors, I guess.

*Not true, actually.  I did find a web site where a fellow made a tube crossover which used Sallen Keys (Sallens Key?)

PRR

#14
Quote from: Snodgrass on February 05, 2025, 03:03:36 PMI had never actually seen any one use a tube Sallen Key

They are around. Not really a fabulous idea. Get your tone from amplifiers and attenuators. Get frequency filtering from good op-amps (chips very much better).

But you won't be deterred? The foundation for this field is the WWII MIT Rad Lab books, specifically book 18 chap 10 pp 384-408. There is a set here: https://www.febo.com/pages/docs/RadLab/ --- Note narrow-band filter book 18 page 406, and (different) stagger-tune pg 408. This is before Sallen and Key 1955. It took a long time for filter theory to ripen.

Do not forget Don Lancaster's Active Filter Cookbook, "by far the best-selling active filter book of all time."

It got published because Silicon op-amps were down to a buck a piece but the filters don't care.
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amptramp

Quote from: merlinb on February 05, 2025, 01:03:39 PM
Quote from: Snodgrass on February 05, 2025, 12:31:21 PMDon't I then run into distortion issues like amptramp seems to have suggested above?
No that's only when driving a heavy load like headphones. When driving a high-impedance load, the SRPP functions basically identically to the half-mu, except for having lower output Z. (Under these conditions I call it a quasi-SRPP, since it stops being truly push-pull). IMO this makes the half-mu kinda redundant, since the quasi-SRPP is "the same but better"! I think amptramp is getting different concepts mixed up.

The µ-amp may not require any changes in parts from an SRPP, but the behaviour is different.  The µ-amp is basically a triode amp with a high-impedance plate resistor except in this case, the upper triode stage has a lower resistance than the typical transistor or FET stage, which act like current sources.  The upper triode stage has a resistance more like Rp which is in the few thousand to tens of thousand ohms.  With the triode top section, it doesn't run into any worse dynamic range and clipping issues than an ordinary common-cathode stage.

With SRPP, you have an output from the upper stage cathode and this can be considered as an amplifier with a  pair of current sources, one being the bottom triode and the other being the top triode, connected together.  Since the currents will be different when it is amplifying a positive input or a negative one and these points are in series, you have to have a load resistance as a third leg of the circuit at this junction that can absorb the excess current in one direction or the other.  SRPP is not a power amp stage, but it requires a resistance that can absorb the current difference without hitting the rails.

For example, if you have a stage that runs from 24 volts and has a current difference of 5 mA, you need 2400 ohms to go from +12 volts at 5 mA to +24 volts or the same 2400 ohms to go from +12 volts to 0 volts.  Bandshaping like capacitive coupling should still permit current flow if the input of the first stage has highpass coupling.  Otherwise, the output should have 4800 ohms to +24 V and 4800 ohms to ground.  Then the bandshaping would have no effect since it is in parallel with resistances that are adequate to avoid clipping under all conditions.

merlinb

#16
Quote from: amptramp on February 06, 2025, 02:51:39 PMSince the currents will be different when it is amplifying a positive input or a negative one and these points are in series, you have to have a load resistance as a third leg of the circuit at this junction that can absorb the excess current in one direction or the other. 
And what will happen if there is no load attached at all? Will there be no output signal, no amplification, or something else? Think about it... ;)

amptramp

Quote from: merlinb on February 06, 2025, 03:26:02 PM
Quote from: amptramp on February 06, 2025, 02:51:39 PMSince the currents will be different when it is amplifying a positive input or a negative one and these points are in series, you have to have a load resistance as a third leg of the circuit at this junction that can absorb the excess current in one direction or the other. 
And what will happen if there is no load attached at all? Will there be no output signal, no amplification, or something else? Think about it... ;)

The signa; will slam up against the rails, both positive and negative.  If you have, say, 3 mA going positive and 4 mA going negative, the output will go down to the lower rail or some point near it if you don't have a load that can absorb the missing 1 mA.  If you have, say, 4 mA going up and 3 mA going down, it will hit the positive rail or as close as the devices allow.  I could see a fuzz where the control was a pot connected from the output to a signal ground.  At low resistance, the stage would be linear but as the resistance got higher, it would introduce more distortion.  I believe there was a stompbox that had two SRPP stages in series followed by an output buffer stage but I have forgotten the name of it.  The pot could be AC coupled to the output of the stage and this would allow distortion to start at bass frequencies and extend up to treble frequencies.

merlinb

#18

Here's a thought experiment. You said about the circuit on the left:
QuoteThe upper triode stage has a resistance more like Rp which is in the few thousand to tens of thousand ohms.  With the triode top section, it doesn't run into any worse dynamic range and clipping issues than an ordinary common-cathode stage.
You said about the SRPP on the right:
QuoteThe signal will slam up against the rails, both positive and negative.
So with no load attached (infinite impedance), how does the circuit 'know' which one it is? How does it know whether to behave like an ordinary common-cathode stage, or whether to slam up against the rails?


PRR

Quote from: merlinb on February 07, 2025, 08:59:00 AMHow does it know

That's a question that Broskie has asked and answered at some length. Google "srpp site:tubecad.com" or similar terms.
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