Timeline for How to choose nodes for Thevenin equivalent circuit
Current License: CC BY-SA 4.0
19 events
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| Aug 14, 2020 at 13:39 | history | edited | wbs2422 | CC BY-SA 4.0 |
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| Aug 11, 2020 at 17:48 | vote | accept | wbs2422 | ||
| Aug 11, 2020 at 3:54 | answer | added | jonk | timeline score: 4 | |
| Aug 11, 2020 at 3:07 | comment | added | jonk | @Wheatley I'll write up something. | |
| Aug 11, 2020 at 3:02 | comment | added | wbs2422 | @jonk Thank you very much, I think I now see what's going on | |
| Aug 11, 2020 at 2:52 | comment | added | jonk | @Wheatley Signals are "single-ended" or "differential." If single-ended, the implication is that there is a "signal wire" implicitly referenced to ground. (Technically, there could be a different implied reference point, but if so that usually has to be stated somewhere explicitly.) A differential signal provides both wires for the signal with the implication that neither of them are with reference to ground (or any specific voltage rail.) In your circuit, the source signal is single-ended. So yes, referenced to ground. But keep in mind not everything is always referenced to ground. | |
| Aug 11, 2020 at 2:08 | comment | added | wbs2422 | I just had a thought: could the reason simply be because we reference everything to ground? | |
| Aug 11, 2020 at 2:03 | comment | added | wbs2422 | Because we want to find something related to the base of the transistor, I can understand that the base node (node B) will be one of the two nodes of our Thevenin equivalent. However, what is the reason that we have to take the ground (node D) as the second nodes of the Thevenin equivalent? Why can't we take for example the node A? | |
| Aug 11, 2020 at 1:59 | comment | added | wbs2422 | @jonk thank you for your comments. Yes I'm talking about the pair of nodes. For the amplifier example, our goal is to find the base current and base voltage. Thus we decide to find the Thevenin equivalent circuit between nodes B and D so we can find the base current and base voltage much more easily. | |
| Aug 11, 2020 at 1:41 | comment | added | The Photon | You choose the nodes that are the output of your circuit. | |
| Aug 11, 2020 at 1:33 | comment | added | jonk | @Wheatley But perhaps you are asking about which pair of nodes? If so, then sometimes we are wondering what a circuit's input looks like (to something that is driving that circuit.) That choice decides the nodes for us. Similarly, sometimes we are wondering what a circuit's output looks like (to something that is being driven by that circuit.) That choice also decides the nodes for us. In your first case, if LOAD "wants" to know what is driving it, then first remove LOAD and see that the answer is 0.5 V with 50 Ohms source impedance. | |
| Aug 11, 2020 at 1:23 | comment | added | jonk | @Wheatley Are you looking for bright lines? My approach is to redraw the schematic (see here for details), removing any busing around of power and ground and replacing all that with markers indicating the voltage at the wire, instead, and orienting the schematic so that signal flows left to right and conventional current flows top to bottom of the page. Once that is done, the nodes are pretty easily identified by inspection. Anything "unknown" will be any node not labeled with a specific, known voltage. Those are the nodes to be selected. | |
| Aug 11, 2020 at 0:38 | history | edited | wbs2422 | CC BY-SA 4.0 |
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| Aug 11, 2020 at 0:35 | comment | added | wbs2422 | @jonk do not mind the voltage and components values, I haven't done the calculations. My question is not really how to get a Thevenin equivalent between two nodes, but what I would like to understand is how do we choose these nodes? | |
| Aug 11, 2020 at 0:34 | comment | added | wbs2422 | @BeB00 : I agree that the transistor cannot be included in the Thevenin equivalent, but what I do not understand is why do we consider the nodes B and D for the Thevenin equivalent, and not, for example, nodes A and B? | |
| Aug 11, 2020 at 0:13 | comment | added | jonk | Wheatley, I think I see how you got your equivalent in the first schematic. But it's probably not right. | |
| Aug 10, 2020 at 23:07 | comment | added | jonk | @BeB00 At least KCL works fine with BJTs. It's just that the equations are more difficult to solve. You can use that to solve a Thevenin equivalent around some operating point. If that's of value. Also here. | |
| Aug 10, 2020 at 22:21 | comment | added | BeB00 | well one answer would be that thevenins theorem is only valid for voltage sources, current sources and resistances, so including a transistor wouldn't really work. | |
| Aug 10, 2020 at 21:41 | history | asked | wbs2422 | CC BY-SA 4.0 |