[SI-LIST] Transmission lines reflections again
- From: "Leonard Alexman" <lalexman@xxxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Wed, 19 Oct 2005 10:53:58 -0700
Hi ,
I am still trying to figure out how in a simple open transmission line the
voltage gets doubled at the end of the line. I have seen the formulas and
rope drawings but only found one article that kind of goes into what I what.
The article I read had a battery connected to a serries50 ohm resistor and a
50 ohm transmission line. The equivelant circuit of the transmission
linsmission line is a series inductor with a capacitor to the return path to
the battery
When the last capacitor in the line is charged, there is no voltage across
the last inductor and current flow through the last inductor stops. With no
current flow to maintain it, the magnetic field in the last inductor
collapses and forces current to continue to flow in the same direction into
the last capacitor. Because the direction of current has not changed, the
capacitor charges in the same direction, thereby increasing the charge in
the capacitor. Since the energy in the magnetic field equals the energy in
the capacitor, the energy transfer to the capacitor doubles the voltage
across the capacitor. The last capacitor is now charged to the battery
voltage and the current in the last inductor drops to zero.
My question is
1. Since the second to the last cap is charged to 1/2 the battery voltage
where does the current flow from the left end of the last inductor to the
bottom of the last cap in order to double the voltage on the last cap ?
Can anyone point me to an article that explains the above in detail ?
Leonard Alexman
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