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While determing the AC load line, the author takes Rc to be parallel to Rl. How is it possible since the two resistors have different nodes?

enter image description here

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    \$\begingroup\$ Please inline images, or reproduce the circuit schematic using the on-site schematic tool. \$\endgroup\$
    – Hearth
    Commented Aug 24, 2019 at 11:42
  • \$\begingroup\$ Where is the tool? \$\endgroup\$
    – user29463
    Commented Aug 24, 2019 at 11:43
  • \$\begingroup\$ When editing your question, there should be an icon at the top with a little circuit diagram and a pencil. Click that. \$\endgroup\$
    – Hearth
    Commented Aug 24, 2019 at 11:44
  • \$\begingroup\$ I am using a phone so can't see \$\endgroup\$
    – user29463
    Commented Aug 24, 2019 at 11:45
  • \$\begingroup\$ Try read here electronics.stackexchange.com/questions/298560/… \$\endgroup\$
    – G36
    Commented Aug 24, 2019 at 15:10

2 Answers 2

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During the small signal analysis, you need to replace the DC power source by ground because a DC component has no AC component.

The next thing you do is replace the capacitor by short circuit. Doing so, your circuit looks like :

schematic

simulate this circuit – Schematic created using CircuitLab

Now you can see both of the resistor's nodes are connected to BJT's collector and the ground. So, they are in parallel.

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  • \$\begingroup\$ How can we determine the current through El in this case? Is it Ie or Ic? \$\endgroup\$
    – user29463
    Commented Aug 25, 2019 at 6:34
  • \$\begingroup\$ When two resistors share a common node, they are in parallel if seen from this common node and they are in series if seen from one of the "outer" nodes. \$\endgroup\$
    – LvW
    Commented Aug 25, 2019 at 8:51
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When performing small-signal analysis, DC sources have no AC component so their variation in time is zero i.e. they represent a short circuit, towards ground in this case.

Therefore, \$R_C\$ and \$R_L\$ share the same potentials: the BJT collector and ground.

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  • \$\begingroup\$ I don't understand. Which DC source are you talking about? \$\endgroup\$
    – user29463
    Commented Aug 24, 2019 at 12:12
  • \$\begingroup\$ @user29463: \$V_{cc}\$ in your case. \$\endgroup\$
    – edmz
    Commented Aug 24, 2019 at 12:13

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