Single amp tube EQ does it work ?

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mik

Well-known member
GDIY Supporter
Joined
Jun 4, 2004
Messages
461
Location
Milano
Single amp tube EQ,
this is the combination of The VT opAmp found

http://img94.exs.cx/img94/2526/singleamptubeeq7qi.png

ok, now I know how to wound inductors,:green: I've lerned how use design softwere,:green: let's talk abuot topology;
:roll: here I need, some guru can understed if this really can work ?

may I use a low inpedance like 600 ohm or less for the feedback of the OPAmp ?

what's the advantage to keep low inpedance in a circuit?

thank you
:wink:
 
I am having real difficulty seeing your picture very well. Maybe it is my old eyes but I cant see resistors or capacitors, etc, just their designation. Can you make it clearer?
 
ok, thank you Jakob, so I have to find Reactance around 1 1,5K in the normograf, am I right ?

The immage is large I' do not have problem to see it , but i can post it in E Mail, if some on are intressed in.
Mik.
try this
http://img94.exs.cx/my.php?loc=img94&image=singleamptubeeq7qi.png

clik on immage
 
The .png image comes up as purple/blue on a black background.. nearly impossible to read, actually..

Could you make it a simple B/W .GIF or something?

Jakob E.
 
I can see the image perfectly... red, black and blue on a white background.

Peace,
Al.
 
right today "maestro" FRED told me this:

"Some suggestions...

1) You need to add a 10k resistor from the transformer to the input of the
value Grid and the EQ Net. This resistor provides the loss in the "cut"
mode, while your R10 resistor provides feedback loss in "Boost" mode.

2) You may need to load the secondary of the input transformer, but that is
something your have to try when you are testing the design.

3) Short out R3.. it doesn't need to be there, Because the output of the diff-amp is single-ended, there is no need for a
plate load on the first half of the diff-amp. You can think of this somewhat
as a cascode connection where the input is the grid of the triode #1, which
drives the cathode of triode #2. The output of the circuit is the plate of
triode #2. The difference is that both grids (triode #1 and #2) are being
used as inputs.


Other than that, I think it should work.

Regards,

Fred Forssell."


I will post a new rev. soon as possible
mik

:green:
 
I agree with what Fred told you, except I prefer to keep both halves of the diffamp loaded. This may be just my anal response to wanting symetry but I think it gives a more equal set point to each half. It will work without a plate loadded equally, and It may work fine, I have never measured it. I would suggest you try it and see.
 
it can probably work in both way.
C3 is there, in VT OpAmap found in Fred's pages. maby short it out can unstable the amp..........who knows....
enyway I will let you know.
thank you all :thumb:
 
> understed if this really can work?

You do need about 10K between the pot network and the input, as noted.

Of course it will work. The dip effect doesn't even need the tubes. The boost effect is just a dip in the feedback loop to get the inverse of a dip.

R1 R2 are not needed if you keep the grids connected to the pots and input transformer.

> may I use a low inpedance like 600 ohm or less for the feedback of the OPAmp?

Yes but all the impedances must be dropped. If you have 600Ω input and feedback resistors, and 10K pots, as you turn the knob nothing will happen until nearly the end of the rotation. You get three settings: full boost, full cut, and 90% of rotation is near-flat. The pot(s) needs to be similar value to the input and feedback resistors. Also the Ls and Cs and Q-resistors need to be scaled. And using 600Ω networks will cause a HEAVY load on 12AT7-class tubes! That's really not a happy situation.

The lower impedance would also allow using smaller coils, but larger caps. Since coils are expensive, especially since these size coils will use VERY thin wire, it would be nice to use smaller coils even with bigger caps. But using tubes, I think 10K impedance is about as low as you can go unless you switch to more and more-powerful tubes, which would cost more than the savings on smaller coils. 10K is a happy compromise.
 
thank you "Maestro" PRR :thumb:

ok, It seams to me that it was a step in the rigth direction, ( do not fourget I'm still beginners) I Know that inductor cost money but now I can wound by myself, is not an easy work, but I have obtained a good result, for the ones i've wound, the real problem is to obtain a matched pair for each 5 bands a total of 10 for 2 ch.
nex step is to plan a good PCB design and the way to avoid filter interaction, maby sreening inductors and caps with an aluminum box, a cound as old fender sreening caps box
if they fits, because the cap and inductor I'm looking at are big, caps are russian type Paper In Oil (PIO) and inductor are square 50 mm.

but one thing it miss me:

How to to calculate the Q bell, I mean, if I'm a like man, I 'll make an inductor with thiny wire with the minimun DC R. optenible for the value Im looking for, and it's ok, close enough to ideal 3db bell, I suppose :roll: but I'm looking for the opportynity to broad the band as 6 and 12 db too, if it is worth ( the E+A+R doesn't have it ) What's the formula to compute R in relation of db for oct.
does anybody know ?
:oops:
2 qustions :
350V B+ does it metter? may I use 250 V Jakob PSU ? or better: May I change some component value in J PSU to obtai 350 V ?
other one:
A over load detector topology for tube operation
does anybody Have.

you know guys....seriusly I m fear of electric shocks :shock:
 
> way to avoid filter interaction,

They are sure to interact. Screening won't fix the fact that all the tuned circuits are on the same busses.

Anyway in audio we often want some interaction.

> How to to calculate the Q bell... close enough to ideal 3db bell

Why is "-3dB bell" somehow "ideal"? That's just a handy way to express a curve.

> What's the formula to compute R in relation of db for oct.

Simplify. Just look at the Cut section: the input 10K resistor, one pot and tuned circuit, and an infinite load. You can lash this up on the bench without the opamp, just a signal generator, an AC voltmeter, and some passive parts. Use one pot, with one cap and resistor (no coil). When the pot wiper is at the far end of the pot, there is little effect (in the complete cut/boost plan, a centered wiper has no effect, or rather its effects on cut and boos cancel out).

With the pot wiper at the other end: At very low frequency, the cap has very high impedance, so the "cut" network is unity gain. When the capacitor's reactance equals 10K, the response is down 3dB. At very high frequency, the cap is a very low impedance. If the resistor were zero, you would have very large cut. If the resistor is say 2K, you have a flat shelf at (2K/(10K+2K)) or -15dB. If the resistor is 10K, the shelf is -6dB. So the "Q" resistor actually sets the maximum depth. Note however that the 10K pot adds to this, so when the pot wiper is not at the end the depth decreases.

Now try the same thing with a coil instead of the cap. Now the high frequencies are unaffected, the lows fall off to the depth set by the Q resistor.

The term "Q" is not good here. In general audio EQ we use Qs from 0.5 to maybe as much as 2, mostly around 1. A Q of 1 is a VERY low Q, and combined with the limited depth (less than 15dB) you just do not get the same things as radio tuned circuits with Q of 10, 50, even 200. Everything is sloppy and you don't get the intense interaction of a high-Q circuit.

For our low Q equalizers, you can pretty nearly estimate where the flat part of the curve starts to dip (or boost) just by looking at where the L or C equals 10K, without considering the other reactance.

When you do the full cut/boost plan with opamp, the boosts are the mirror image of the cut (assuming the opamp has much more gain than the maximum boost).

For this multi-band EQ all on one bus, you also have to figure the effect of all the other pots. Five 10K pots with centered wipers gives about a 1K impedance for the reactances to suck against, not the 10K of a single network. It also increases noise: the overall gain is still unity, but internally it is attenuating by 10 and then amplifying by 10. This would not be the first-choice plan for a 27-band EQ.

> 350V B+ does it matter? may I use 250 V Jakob PSU?

They are vacuum triodes. Feed them whatever you have. All voltages change, nearly in proportion, and it still works fine. Maximum output level drops a little faster than supply voltage, but 250V should be fine. (And 350V is really a little high for heater-cathode insulation in the upper triode in the output stage.)

> A over load detector topology for tube operation

Audio is audio. Any level detector will tell when you have "high level". You usually do not want to wait for an OVERLOAD light, you usually want some light to come on -before- overload.

For a very rude indication that this particular amp IS overloading, wire an LED across the input grids. As long as the opamp is doing its job, the voltage across the two input grids will be a fraction of a volt. But when the opamp overloads, it can't force its inputs to be equal. Increased signal will put a couple volts across the grids, which will light the LED (dimly).

> you know guys....seriously I'm fear of electric shocks

I've taken 120V shocks all my life, but I would be afraid to live where you have 230V coming out of wall outlets. Every time you plug in a lamp, you put your finger next to 320V peaks 100 times a second.

High voltage is no more dangerous than poison snakes or power saws. Keep the snake in a tight box. If you have to play inside, unplug the saw and be sure it is completely stopped before put a finger near it.

This plan will work JUST as well with an IC opamp powered from 9V batteries. It won't have quite the same sound or output level, but you can perfect your filter components fearlessly. Then build the tube opamp separately, and be sure there is zero voltage on the input and output terminals (you do need the 1Meg grid resistors for this testing, and maybe a 10K bleeder on the output cap). Once you have the two parts working individually, you can bring it all together.
 
Many thancs Paul.... :sad: I'm discoraged, because for get 20 Herz cut boost with this impedance you need a 15 H and 4,2 uf cap, and this is quite large coil...... so maby I've to put a buffer in front of it, with the filter section in the middle........I'll let you know.

Merry Cristmast and Happy new year to all of you :green: :green: :green: :green: :green: :green: :green: :green: :green: :green:
 
thanks jackob, the challeng was to do my own design..after all I'm here to learn, not to clone, My goall was to do a good design to share with the Lab people. Cloneing gear is good to learn too, and maby is not time to invent simple topology like I've thought (there's a ambitions and real knowlage missmatch :roll: ). anyway E*A*R was the starting choose
 
> maybe is not time to invent simple topology

I bet that nearly every such EQ topology has been explored. There are not so many ways to put resistances, reactances, and amplifiers together. And someobody wrote a paper covering all the EQ topologies commonly in use or normally avoided... does someone have that reference?
 
[quote author="PRR"]> maybe is not time to invent simple topology

I bet that nearly every such EQ topology has been explored. There are not so many ways to put resistances, reactances, and amplifiers [/quote]

Yea, you're right P, but I thik the minimalism some time win, the single ampifier topology was simple enough for everyone, I've thought forgetting the impedence issue, E*A*R ain't simple; 3 unknow trafo 3 tube for channel, alot of component...
the goal was to get "cheap, neat, and nice sounding machines" ( this is my company song):green:

Jakob:
:cool:
one day I will come there in your cold land, to give to you the myown made most proud machines as a gift.....but pleas wayt for the next 5 years :wink:

Dave :

I'v got Rane page intressing reading, for sure :thumb:
but I do not think he's means those

marry Crismast
mik. :green:
 
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