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I understand theres a lot online about transmission line theory it terms of why the reflection, but if say the load was open or really high, thinking of a standard circuit, the total series inpedance would be be very high so why would there be power going to the load to be reflected in the first place? I don't get the concept of “it doesnt see the open part yet” since by that reasoning wouldn't any open load just drain?

JRE
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George kirby
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  • The question yesterday was closed for the same reason. If you don't understand the answers on transmission lines then please ask about that rather than reformulate your question without saying anything different. This part (of my answer) is crucial: *When you send the pulse down the transmission line (t-line), initially the source of the pulse cannot know what load is at the far end so, in order to initially pass current, it uses the t-line's characteristic impedance (50 ohm, 75 ohm etc.) and produces a current proportionate with the voltage and that resistance (i.e. ohms law).* – Andy aka Jun 13 '23 at 10:20
  • Play around with this: http://helloworld922.blogspot.com/2013/04/online-transmission-line-simulation.html – DKNguyen Jun 13 '23 at 14:58
  • Your are confusing terms likely because you are thinking of mechanical analogies which are opposite of electrical. An "open load" is a LARGE resistance which is a SMALL load, because "open circuit". But in water pipes this is a CLOSED/blocked end. The opposite of "draining". The load that you are thinking would "drain" is, in fact, a low resistance load which is a LARGE load In water pipes this is an OPEN end. – DKNguyen Jun 13 '23 at 15:20
  • It is easy to see why reflections happen if the load impedance is too high. All the energy is already en route down the wire before the source can react and when it reaches the end it has nowhere to go but back. Just like how water will splash back at you when it reaches the end of a blocked pipe after propagating back to the end you are pumping water into. This is because of inductance, the equivalent of inertia for water. If the load impedance is too low you don't get a voltage spike. – DKNguyen Jun 13 '23 at 15:25
  • Play with the simulation and understand the terminology because answers get muddled when we need to correct you on your mixed up terminology while trying to answer. – DKNguyen Jun 13 '23 at 15:30
  • @Andyaka i did reply to further ask on the previous post, maybe i tagged the wrong person sorry. My question to your response is the initial pulse , I should have probably mentioned without coax, but why would current flow in the first place even without the coax? Looking at the eqivalent diagram for coax I understand due to capacitance and inducance. But even with just a purley open load on a transmitter they say its bad and all the online stuff shows its impedance as in series, and therfore i cant understand why any power js sent in the first place. Thanks – George kirby Jun 13 '23 at 21:26
  • @DKNguyen thankyou alot for the help, i do get the hang of transmission lines more now, so my question is even without the coax, why is it bad and why does power get sent still if theres no load, since i see the internal impedance mostly moddled as in series anyways. Thats – George kirby Jun 13 '23 at 21:28
  • Some of the answers to [this question](https://electronics.stackexchange.com/q/59208/11683) might provide some of the insight you're looking for. – Dave Tweed Jun 14 '23 at 00:18
  • Because damage. It's like dry firing a bow. The energy being transmitted has to go somewhere and without an arrow or antenna through which the energy can leave the system, it's going to to be dissipated in the system itself. – DKNguyen Jun 14 '23 at 03:17
  • @DKNguyen but why would there be energy transmited in the first place even without to coax – George kirby Jun 14 '23 at 09:22
  • It's a transmission line. So the energy starts being dumped into it before it knows there's nowhere for it to go. The transmission line isn't entirely contained to the antenna or the extension coax. It starts on the circuit board. Whether it causes damage probably depends on the circuit board trace design but the point is the user doesn't know the design and even if they did the typical user won't know enough to judge whether damage would be caused or not. Having a coax extension with no antenna certainly increases the risk. – DKNguyen Jun 14 '23 at 14:14
  • @DKNguyen ahhh i see, so the bit upto the port can be treated as coax in a way then, and that makes the impedance up then im assuming? Thanks for the help, and when you say it starts dumping the energy, is the amount determnied by the inital 50ohm impedance then? Thanks again – George kirby Jun 14 '23 at 15:15
  • The characteristic impedance of the trace, yes. Any impedance discontinuities along the way will cause some degree of ringing or reflection such as transition to an unmatched coax or connectors, etc. It's water hammer. – DKNguyen Jun 14 '23 at 15:21
  • @DKNguyen got you:). Its the fact the impedance is shows as in series on some diagrams rly throws me of, ive done some more looking into it and it seems its not really the case due to capacitances to ground, is this correct? – George kirby Jun 14 '23 at 16:31
  • That's the lumped sum model. In reality it is continuous and distributed. Free space capacitance is a thing. That's what lets you carry a static charge to shock things. – DKNguyen Jun 14 '23 at 16:31

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