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Understanding load lines
Understanding load lines
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Old 18th December 2016, 04:02 PM   #1
merlin el mago is offline merlin el mago  Europe
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Default Understanding load lines

For example I want to understand the load line for 5687 tube, I have 209V as B+ and 10K as load anode resistor.

To know the plate current is to divide 209V : 10.000 ohms = 0.0209 A = 20,9mA.

Now I draw a line between these points to get the load line, If I choose the operating point at -3V as grid voltage the tube will draw 12mA, right? If I want each tube draws 10mA the grid voltage have to be -4.5V? The best OP is -7V as grid voltage?

How to know the best plate current for each tube?

Last edited by merlin el mago; 18th December 2016 at 04:05 PM.
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Old 18th December 2016, 04:09 PM   #2
merlin el mago is offline merlin el mago  Europe
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Attached simulator pic
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File Type: png load line 209V 25mA.png (183.9 KB, 487 views)
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Old 18th December 2016, 04:13 PM   #3
merlin el mago is offline merlin el mago  Europe
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Simulator says 14.2mA in place of 12mA?
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Old 18th December 2016, 04:17 PM   #4
merlin el mago is offline merlin el mago  Europe
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Attached pic manual load line
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File Type: jpg 5687 load line 209V -B+ -3V grid 001.jpg (766.3 KB, 479 views)

Last edited by merlin el mago; 18th December 2016 at 04:19 PM.
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Old 18th December 2016, 04:33 PM   #5
audiofan is offline audiofan  Canada
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This graph give the operating point ( static point ) but you must keep in mind that when you will have some AC signal you must keep some margin on both side of the operation point ... otherwise with AC signal the tube will go to saturation ( reach maximum current ) or cutoff ( have zero current )
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Old 18th December 2016, 04:56 PM   #6
merlin el mago is offline merlin el mago  Europe
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Yes, two AC volts more or less at grid right?
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Old 18th December 2016, 05:12 PM   #7
Palustris is offline Palustris  United States
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I am not certain I understand the question, but I can offer a little help with load lines. First you need to determine the voltage output that you expect which generally is a function of how much voltage you need to drive the next stage to saturation. Next you need to determine from the load lines the most linear region e.g., the load lines that are equally spaced. When the load lines start to bunch up is where you will get 2nd harmonic distortion.

You can see below that I have overlaid a 10k ohm load line in parallel with yours. Note the difference. You have a max output of 40V peak or 28.28 VRMS; my load line has a max output of 80V peak or 56.56 VRMS. The difference, then is that my load line can drive the next grid with twice the voltage.
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Old 18th December 2016, 05:48 PM   #8
merlin el mago is offline merlin el mago  Europe
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That's a very good advice: "most linear region e.g., the load lines that are equally spaced"
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Old 18th December 2016, 06:05 PM   #9
merlin el mago is offline merlin el mago  Europe
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Quote:
Originally Posted by Palustris View Post
I am not certain I understand the question, but I can offer a little help with load lines. First you need to determine the voltage output that you expect which generally is a function of how much voltage you need to drive the next stage to saturation. Next you need to determine from the load lines the most linear region e.g., the load lines that are equally spaced. When the load lines start to bunch up is where you will get 2nd harmonic distortion.

You can see below that I have overlaid a 10k ohm load line in parallel with yours. Note the difference. You have a max output of 40V peak or 28.28 VRMS; my load line has a max output of 80V peak or 56.56 VRMS. The difference, then is that my load line can drive the next grid with twice the voltage.
145V x 0.015A = 2.175W so is OK because the max dissipation both plates operating is 7.5W : 2 = 3.75W each triode, are ok my calculations right?

mA is only related to to make sure that I'm not violating the maximum dissipation allowed in the tube?
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Old 18th December 2016, 06:53 PM   #10
merlin el mago is offline merlin el mago  Europe
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The question is: I'm trying to know how many mA draws each tube (to make a CCS) with B+ 233V 10K anode load resistor so max plate current 23.3mA and -3V at grid, my calculations give me 14mA, I do it right?

Attached load line pic.
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File Type: jpg WP_20161218_003[1].jpg (568.7 KB, 414 views)

Last edited by merlin el mago; 18th December 2016 at 07:13 PM.
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