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Old 29th July 2004, 03:34 AM   #1
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Default load line - truth or fiction

Ok my first dumb question. I am not even asking for specific answers here but could someone take a look at this site


My question is, is the following an acurate description of the slope of the load line. A simple yes or no will do I will research the rest. I just feel this may be incorrect

This is the 250V/49.5mA operating point and corresponding load line for a 5K output impedance. The Large dot is the operating point. The straight line that intersects the operating point is the load line. The slope of the load line is determined by the impedance of the load that the tube in working into - in this case the load will be a 5K output transformer. The slope of the load line is calculated by dividing the load impedance (5k ohms) into a chosen voltage - say 100V - to get the change in current - 20mA. (100 / 5000 = 0.02 = 20mA) This determines the slope of the load line. Slope is defined as the change in current divided by the change in voltage (rise-over-run as they say in trigonometry class). So with a slope of 5k ohms, for every 100V the load line drops 20mA.

j Jeff
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Old 29th July 2004, 07:24 AM   #2
Yvesm is offline Yvesm  France
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Yes !
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Old 29th July 2004, 11:59 AM   #3
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Yes, in case of a pure resistive load.

If the load is reactive in some way (eg. low or high freqs. )
the line becomes an ellypse.

Also, the dynamic shift of the working point has to be taken
into account.


P.S. I only read the quoted text. I have not read at the web site.
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Old 29th July 2004, 12:54 PM   #4
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Default slope

ThankS all I think i was getting a little confused because the way it was presented .The actual slope though to the best of my knowledge is 1/rload which in most cases is presented as rload

Thanks JEFF
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Old 29th July 2004, 03:08 PM   #5
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Depends on what dimensions you're taking. If going the strict algebra approach of y = mx + b with x horizontal (i.e. x in volts, y in miliamperes), m will be in terms of mA/V rather than V/mA, i.e. conductance (mhos) instead of ohms. It will also be negative. A strict mathematical method isn't the best way to go, just use what looks right, punch a few numbers (following algebra and logic as needed! ) and see what turns out.

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