preamp triode replication

[makepeace]

Got Merlin's preamp book (eventually!). Have been going skimming through it and trying to get to grips with the theory, but haven't had much time - so don't have a comprehensive understanding as of yet.

I have a question though: Say you have an example triode preamp stage (with all relevant voltages and resistor values given) and you want to replicate this stage as closely as possible (WRT gain, operating point etc) with a valve of similar characteristics, but not exactly the same as the original valve. How would one go about this? For example 12AX7 -> 6SL7. I know the 6SL7 has slightly lower gain and is more comparable with the 5751, but if I'm not mistaken you should be able to get similar output characteristics to a 12AX7.

I would assume that you would try to work with the same grid curve - as this will largely influence your overdrive characteristics. Then, I would think that you would look at a load line with a similar gradient, this is where my problem comes in: Would you position it on the plate characteristics plot using a similar plate current value, or a similar plate voltage value? Or have I got everything completely wrong?

Pre-thanks for any insight anyone can give!

[Steve Conner]

I think the 6SL7 is close enough to the 12AX7 that you could just try it with the same component values. If the DC bias point is a little off, you could tweak the cathode resistor until you get the desired plate voltage at idle. That is the most important thing IMO.

If you use the same plate resistor, then similar plate voltage necessarily leads to similar plate current. The plate resistor is what determines the load line.

[makepeace]

Thanks for your reply!

I would like to know how to work it out though - the 6SL7 was maybe a bad example, its too similar. Could you perhaps roughly describe a step-by-step as to how one would do it?

edit: eg. the input stage of the 1959 superlead - bright channel. Vgk ~ -1V, Vsupply = 250V, Rp = 100k

This leads to quiescent Vp of ~ 140V and Ip of 1.1mA.

If you take those same original conditions and apply them to the 6SL7, you get:

Vp ~ 120V and Ip of ~ 1.25mA.

This may be close to that of the 12AX7 stage, but its not the same. How to get it the same, preserving output characteristics.

[soundguruman]

Well you can reduce the value of the plate resistor for less gain.
Plate Bypass Capacitor Calculator
6SL7 has gain of about 70?, 12AX7 has gain of about 100? or there about.

Oh I see, I'm no longer a member, I'm old timer...very funny.

[makepeace]

Nope sorry soundguruman, think you've missed the point .

In short: I have a 12AX7 stage - with a particular quiescent setup. I now want to replicate this quiescent setup as closely as possible (with maths), using a different type of valve ie 6SL7. Obviously it's never going to sound the same, but the aim here is to get the theoretical understanding involved with substituting one tube for another and adapting the stage to produce a theoretically indistinguishable result, or get verifiably as close as possible.

oh, and congrats on being an old timer, i suppose

[soundguruman]

Nope sorry soundguruman, think you've missed the point .

oh, and congrats on being an old timer, i suppose

ha ha ha very funny.

[philbo]

This might seem obvious, but keep in mind the 6V filament of the 6sL7 won't last long if you run it at the 12V level used by the 12AX7.

Another point to bear in mind: the quiescent current is just one point on an angled line (the fabled 'load line'). Even if you get the quiescent current to match, it doesn't necessarily mean you would have the same angle on the load line. It just means the two lines intersect where input voltage is at zero.

If it helps to visualize it, the vertical axis is plate current and the horizontal one is plate voltage. The line intersects the tubes characteristic curve to determine the operating limits for the tube (cutoff and saturation points). The angle of the line is proportional to the gain of the stage, and is determined by the bias, characteristic gain of the tube (the 'mu'), plate load, and negative feedback (if present) caused by an unbypassed cathode resistor.

Here's a little light reading for you about it:
http://valvewizard1.webs.com/Common_Gain_Stage.pdf

[makepeace]

Thanks for pointing out the obvious philbo, I always feel someone needs to do it. If you had read the first line of my post, you would have seen that I have just got merlins book. Guess I should have implied harder that I've read that chapter (the first one in the book) a couple of times.

Obviously I will not be running the valve at more than required heater voltage, even if I could. In most amps that I've seen, the 12AX7's are supplied by 6.3V (which is allowable). Come to think of it, I don't know if a popular commercial valve amp PT with 12.6V taps even exists.

Another point to bear in mind: the quiescent current is just one point on an angled line (the fabled 'load line'). Even if you get the quiescent current to match, it doesn't necessarily mean you would have the same angle on the load line. It just means the two lines intersect where input voltage is at zero.

^

Then, I would think that you would look at a load line with a similar gradient

The example that I gave in my second post was only to express the notion that there *is* a difference between the results obtained using the same component values for each the 12AX7 and the 6SL7.

If it helps to visualize it, the vertical axis is plate current and the horizontal one is plate voltage. The line intersects the tubes characteristic curve to determine the operating limits for the tube (cutoff and saturation points). The angle of the line is proportional to the gain of the stage, and is determined by the bias, characteristic gain of the tube (the 'mu'), plate load, and negative feedback (if present) caused by an unbypassed cathode resistor.

really?

Sorry, its early - I get frustrated when prefer to not read properly.

Question still not answered! Please guys.. Tis quite an easy one I would think?

[Steve Conner]

What makes you think it is even possible? If one tube could replicate the performance of another, why were so many different types of tube made?

If you take any particular tube and set it up in the datasheet recommended circuit, you'll be getting the optimum performance from it, and probably close to the best sound in a musical instrument application. But no amount of math will make them "replicate" each other. A 12AU7 will always have less gain and lower output impedance than a 12AX7 no matter what you do.

[makepeace]

Well, I've seen countless build posts where people take a design that calls for popular commercial (relatively expensive) tubes, and redesign it using less popular tubes that you can find for peanuts. I realize that you will never get it exactly the same. But hypothetically, as this will usually yield a good starting point - how would you go about doing the maths?

-Do you need the load line gradient to be constant between the two designs?
-Do you need to look at the same grid curve?
-Does the angle between the grid curve at the operating point and the load line need to be the same?
-Does the plate current need to be the same?
-Does the plate voltage need to be the same?

-Is there a 'most important' factor? If so, which is it of the above?

[Steve Conner]

I think the two most important things are:

The idle point on the loadline should be placed the same with respect to grid current clipping and plate current cutoff, as it was in the stage you're "replicating". Most tube amplifier stages have the idle point about halfway between these two extremes, as this gives maximum headroom.

The ratio between plate load resistor and tube's internal plate resistance should be preserved. (This defines the load line gradient.)

These two conditions should allow you to calculate values for the plate and cathode resistors, and provide answers to most of your questions.

There are only two degrees of freedom for the designer of a triode tube to play with. There are three fundamental parameters, but they're not independent: Mu = gm*rp.

[makepeace]

Thanks for your post Steve, on the mark of what I'm aiming for.

Now more explicitly, could one say that one needs to know the grid bias voltage, the load line gradient and the gain of the stage? And then using the relationships at hand you can work backwards to find the plate voltage/current and thus the Rp?

edit: Another thing that has me boggled, the angular relationship between the grid curve and the load line; does this have ANY effect whatsoever on anything? I noticed that if I draw load lines of similar gradients on the 12AX7 and 6SL7 plate graphs, the angles between the loadline and grid curve differ (ie 12AX7: loadline<gridcurve <90deg, 6SL7 loadline<gridcurve >90deg.

[Steve Conner]

As I said, the triode only has two degrees of freedom, so the two constraints I gave you should be enough to answer all your questions. It just becomes a geometry problem if you like load lines.

For instance, the angle between the load line and the grid curve is a function of the ratio between the plate load resistor and Rp. I'm not entirely sure how to work out the other parameters, but I'm pretty sure it can be done!

[Enzo]

Or pick a tube 6SL7 or whatever, and use the table of recommended part values for that type in the back of the RCA manual.

[makepeace]

Which manual would you recommend that I look for? Are there different editions and what not?