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Dumb Question Brain is numb today!

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I know I will get a bunch of posts calling me a flake but I need help.

Using a triode curve for a SINGLE section of a dual Triode. I am trying to plot some load lines for the two sections in Parallel. Voltage stays the same, current doubles, but what do I plot for the Load ? Twice the OPT or Half or Same (assuming I double the plate current on the graph)???
 
It seems that there's some confusion as to what you mean in postulating your paralleled pair of tubes.  If you are looking at a set of plate curves, and have settled on an operating point of x volts and y milliamps, and have figgered the gm, rp and µ for this condition, with the two devices in the envelope operating in parallel (assuming they are identical):

rp will halve

gm will double

µ, and thus circuit gain will thus remain the same

optimum load, though (which is a function of rp) will halve

current drive into the load (if it is a difficult one like an 845, this is a real improvement) will double.

Did I explain that okay?

Poinz
AudioTropic
 
I think I understand,

Using a single plate curve,

Plot with twice the load resistance and then I can double the currents for what will actually flow at any given grid voltage?

Example,(made up numbers)

if Idle current is 30mA @ Plate Volt 200V with -30 Grid
Plot it out on the curve using an 8K load.

Gives Imax of 50ma and Imin of 10ma
So in parallel into the ACTUAL Impedance of 4K I have

Imax 100mA Io 60mA and Imin 20mA...correct
Voltages stay the same so Pout is 2x the single with twice the load
Distortion should be the same?
 
Plot with twice the load resistance and then I can double the currents for what will actually flow at any given grid voltage?

This statement is a bit confusing. You can plot load lines and find a good one for a single section, then put a plate load of half the resistance that you found in your load line (twice the load). Your load line will be correct for either half. Current for the pair will obviously be double what your line for one half indicates. (since there are two)
 
nhuwar said:
Ya I will admit I'm getting confused myself.

It really isn't that hard.

Loadline

Here's a loadline for a 6J5. This is like one half of a 6SN7, but let's suppose we don't have any 6J5s. You could parallel the sections of a 6SN7 and get the same operation. For two sections in parallel, Ipq would double, as would g(m), and r(p) would be cut in half. Since you'd be pulling twice the current, you would need to cut the cathode bias resistor in half, and you'd need to use a 30K resistor between the positive rail and the paralleled plates. The gain should stay substantially the same, and if both sections were perfectly balanced, the THD would likewise remain the same.

If you wanted, you could improve the balance by degenerating individually the cathodes with separate cathode resistors.
 
Miles Prower said:


It really isn't that hard.

Loadline

Here's a loadline for a 6J5. This is like one half of a 6SN7, but let's suppose we don't have any 6J5s. You could parallel the sections of a 6SN7 and get the same operation. For two sections in parallel, Ipq would double, as would g(m), and r(p) would be cut in half. Since you'd be pulling twice the current, you would need to cut the cathode bias resistor in half, and you'd need to use a 30K resistor between the positive rail and the paralleled plates. The gain should stay substantially the same, and if both sections were perfectly balanced, the THD would likewise remain the same.

If you wanted, you could improve the balance by degenerating individually the cathodes with separate cathode resistors.

No Miles it's trying to read the instruction on making a load line then implementing it "per instruction from Morgan Jone's book" I find tricky. I just haven't spent enough perfecting it.

Nick
 
nhuwar said:
No Miles it's trying to read the instruction on making a load line then implementing it "per instruction from Morgan Jone's book" I find tricky. I just haven't spent enough perfecting it.

I don't know about that Morgan Jones book. May be he's making it harder than it needs to be? You might want to take a look at the RCA Receving Tubes Manual and/or Steve's Site (scroll down a bit -- under the "Technical Reports" column you'll find his five part series on loadlines).
 
What I was getting at was this.

I can plot a loadline on a Triode Curve for a given Load and any operating point I want with no problem.

I can calculate Bias resistor etc NO PROBLEM.

It was when I wanted to PARALLEL 2 sections that I ran into trouble.
The issue was this.

I wanted to use 6AS7/6080's that I have from my junkbox.

Wanted SE and have only some junkbox OPT's either 4K to 8 or 2K to 4.

I have numerous PT's so I wanted to play with different combinations of Voltage/current/load and check on Pout and Distortion.

Here is my method of using plate curves.

I get a fairly clear copy of the curve I want from the net. I "cut and paste" it into a bitmap sized around 1024x768 or 800x600. I open it in MS Paint and use colored lines to plot the loadlines.

My issue is when I wanted to "parallel" the two sections. I can double the currents on the curve or plot different loads. For the sake of accuracy I settled on plotting a 4K load on the single section curve. Since I was ACTUALLY going to use a 2K OPT impedance. This way once I found the Operating point I wanted to use (keeping the fact that the currents will double in mind) I could simply double the current values for Imax, Idle and Imin for parallel sections into the 2K load for the Pout and Distortion calculations. (even thought you could simply double the single section Power and distortion should be the same.)
 
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coldcathode said:
My issue is when I wanted to "parallel" the two sections. I can double the currents on the curve or plot different loads. For the sake of accuracy I settled on plotting a 4K load on the single section curve. Since I was ACTUALLY going to use a 2K OPT impedance. This way once I found the Operating point I wanted to use (keeping the fact that the currents will double in mind) I could simply double the current values for Imax, Idle and Imin for parallel sections into the 2K load for the Pout and Distortion calculations. (even thought you could simply double the single section Power and distortion should be the same.)

Imagine you had designed the two sections independently. Perfect valves would swing identical voltages, so you could connect all corresponding points together. You would then have a single cathode bias resistor of half the value and your anode load would be half the value. Job done.
 
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