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25th March 2009, 10:41 PM  #1 
diyAudio Member
Join Date: Jan 2009

Dumb Question Brain is numb today!
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)??? 
26th March 2009, 03:26 AM  #2 
diyAudio Member

Plot it with a single plate characteristic using half the current and twice the load resistance.

26th March 2009, 03:41 AM  #3 
diyAudio Member
Join Date: Jan 2007
Location: Cypress Texas

Twice the plate load,,, in parallel,,,, wouldn't it be half the plate load?
Or did I miss something? Nick 
26th March 2009, 03:54 AM  #4 
diyAudio Member
Join Date: Nov 2007

doesn't twice the load resistance = half the plate load?

26th March 2009, 04:01 AM  #5 
diyAudio Member
Join Date: Jan 2007
Location: Cypress Texas

If so I've learned something, which would be a good thing LOL.
Nick 
26th March 2009, 04:19 AM  #6 
diyAudio Member
Join Date: Nov 2007

Well, I've always thought of a large load as a small resistance. I mean its like gear ratios, high gears are lower numerically. Are we talking about a large number or a heavy load for the amplifier stage?

26th March 2009, 05:02 AM  #7 
diyAudio Member
Join Date: Oct 2004
Location: Maui, Hawai'i, USA

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 
26th March 2009, 01:44 PM  #8 
diyAudio Member
Join Date: Jan 2009

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? 
27th March 2009, 04:57 AM  #9 
diyAudio Member
Join Date: Jan 2007
Location: Cypress Texas

Ok ,, I could be wrong but, let go back to simple ohms law. How do you double the load resistance and double the current draw. These 2 are inverse of each other,,,, aren't they???
Nick 
28th March 2009, 06:25 PM  #10  
diyAudio Member
Join Date: Nov 2007

Quote:


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