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{{done|Resolved}} This issue has been resolved, and I have therefore removed the tag, if not already done. No further action is necessary.—[[User:Cyberbot II|<span style="color:green;font-family:Neuropol">cyberbot II]] [[User talk:Cyberbot II|<sup style="color:green;font-family:arnprior">Notify]]<sub style="margin-left:-6.1ex;color:green;font-family:arnprior">Online</sub> 06:09, 12 December 2013 (UTC)
{{done|Resolved}} This issue has been resolved, and I have therefore removed the tag, if not already done. No further action is necessary.—[[User:Cyberbot II|<span style="color:green;font-family:Neuropol">cyberbot II]] [[User talk:Cyberbot II|<sup style="color:green;font-family:arnprior">Notify]]<sub style="margin-left:-6.1ex;color:green;font-family:arnprior">Online</sub> 06:09, 12 December 2013 (UTC)

== Is the load line always a "load line"? ==

"Load line" is a much wider concept than the shown here. In most cases, it is really a "load line"... but it can be also a "source line"... and then "load" is a misleading word... And what is surprising, it can be both a load and source line... And what is even more surprising, it can be a nonlinear curve... and then "line" is another misleading word...

[[Image:Load line diode.png|250px|right]]
Look for example at the simple diode circuit on the top of the article. What is the load here - the resistor or the diode? And what is the input? To answer this question, we have somehow to separate the circuit into two parts ("source" and "load") and decide what to vary as an input.

First, we can combine the voltage source V<sub>DD</sub> and the resistor R in a composed "source part", isolate the diode as a single "load part" and vary the voltage V<sub>DD</sub> as an input. The straight line will move horizontally (remaining parallel to itself) and will picture the diode exponential IV curve. So in this case, the straight line is a "source line" and the diode curve - a "load line" (actually, nonlinear).
Second, we can combine the voltage source V<sub>DD</sub> and the diode in a composed "source part", isolate the resistor R as a single "load part" and vary, as above, the voltage V<sub>DD</sub> as an input. Now, the diode exponential IV curve will move horizontally (remaining parallel to itself again) and will picture the straight resistor line. In this case, the diode curve is a "source line" (actually, nonlinear) and the straight line - a "load line".

Third, we can isolate the diode as a single "input part" (it would be too strange to call it "source part"), combine the voltage source V<sub>DD</sub> and the resistor R in a composed "output part" (also, it would be too strange to call it "load part") and somehow vary the diode IV curve (e.g., by heating the diode:) The diode IV curve will move (really, in a small region) horizontally and will picture the straight resistor line in this region. This is the usual case where a nonlinear element (a diode here) moves (horizontally here) its (constant-voltage here) IV curve as a "source line" thus picturing the straight "load line" of the composed load network (V<sub>DD</sub> + R).

Of course, to heat the diode was too fancy:)... but to control a transistor is absolutely real arrangement (the second picture in the article). Here, a nonlinear element (a transistor) moves (vertically) its (constant-current) IV curve as a "source line" thus picturing the straight "load line" of the composed load network (V<sub>CC</sub> + R<sub>C</sub>).

More sophisticated version of this transistor arrangement is when the collector resistor is replaced by another transistor (with fixed base voltage) acting as a constant-current "dynamic" or "active" load. Here, the characteristic curve of the input transistor acts as a "source line" while the characteristic curve of the load transistor acts as a "load line".

And the most sophisticated version (widely used in op-amp input differential stages) is when both the transistors are controlled by equal but opposite signals. Now, the transistor characteristic curves act both as "source" and "load" lines that simultaneously move in opposite directions...

I hope I conviced you the load line is not simply a "load line"... --[[User:Circuit dreamer|Circuit dreamer]] ([[User talk:Circuit dreamer|talk]], [[Special:Contributions/Circuit dreamer|contribs]], [[Special:EmailUser/Circuit dreamer|email]]) 16:05, 27 April 2014 (UTC)

Revision as of 16:05, 27 April 2014

this is a small circuit but we can learn more things about linear & non-liner resistance. what is load line?load+line as electric ko load and splay to line — Preceding unsigned comment added by 14.194.49.65 (talk) 04:36, 10 October 2012 (UTC)[reply]

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 Resolved This issue has been resolved, and I have therefore removed the tag, if not already done. No further action is necessary.—cyberbot II NotifyOnline 06:09, 12 December 2013 (UTC)[reply]

Is the load line always a "load line"?

"Load line" is a much wider concept than the shown here. In most cases, it is really a "load line"... but it can be also a "source line"... and then "load" is a misleading word... And what is surprising, it can be both a load and source line... And what is even more surprising, it can be a nonlinear curve... and then "line" is another misleading word...

Look for example at the simple diode circuit on the top of the article. What is the load here - the resistor or the diode? And what is the input? To answer this question, we have somehow to separate the circuit into two parts ("source" and "load") and decide what to vary as an input.

First, we can combine the voltage source VDD and the resistor R in a composed "source part", isolate the diode as a single "load part" and vary the voltage VDD as an input. The straight line will move horizontally (remaining parallel to itself) and will picture the diode exponential IV curve. So in this case, the straight line is a "source line" and the diode curve - a "load line" (actually, nonlinear).

Second, we can combine the voltage source VDD and the diode in a composed "source part", isolate the resistor R as a single "load part" and vary, as above, the voltage VDD as an input. Now, the diode exponential IV curve will move horizontally (remaining parallel to itself again) and will picture the straight resistor line. In this case, the diode curve is a "source line" (actually, nonlinear) and the straight line - a "load line".

Third, we can isolate the diode as a single "input part" (it would be too strange to call it "source part"), combine the voltage source VDD and the resistor R in a composed "output part" (also, it would be too strange to call it "load part") and somehow vary the diode IV curve (e.g., by heating the diode:) The diode IV curve will move (really, in a small region) horizontally and will picture the straight resistor line in this region. This is the usual case where a nonlinear element (a diode here) moves (horizontally here) its (constant-voltage here) IV curve as a "source line" thus picturing the straight "load line" of the composed load network (VDD + R).

Of course, to heat the diode was too fancy:)... but to control a transistor is absolutely real arrangement (the second picture in the article). Here, a nonlinear element (a transistor) moves (vertically) its (constant-current) IV curve as a "source line" thus picturing the straight "load line" of the composed load network (VCC + RC).

More sophisticated version of this transistor arrangement is when the collector resistor is replaced by another transistor (with fixed base voltage) acting as a constant-current "dynamic" or "active" load. Here, the characteristic curve of the input transistor acts as a "source line" while the characteristic curve of the load transistor acts as a "load line".

And the most sophisticated version (widely used in op-amp input differential stages) is when both the transistors are controlled by equal but opposite signals. Now, the transistor characteristic curves act both as "source" and "load" lines that simultaneously move in opposite directions...

I hope I conviced you the load line is not simply a "load line"... --Circuit dreamer (talk, contribs, email) 16:05, 27 April 2014 (UTC)[reply]