[SI-LIST] Re: Transmission lines reflections again
- From: "Loyer, Jeff" <jeff.loyer@xxxxxxxxx>
- To: <scott@xxxxxxxxxxxxx>, <lalexman@xxxxxxxxxxx>
- Date: Wed, 19 Oct 2005 11:41:03 -0700
Here's the derivation (stolen from a class of long ago)
VL, IL
Vinc, Iinc ---> Characteristic Impedance =3D Z0 ---> ZL
<--- Vrefl, Irefl
=20
Vinc, Iinc =3D Incident Voltage and Current
Z0 =3D Characteristic impedance of transmission line
ZL =3D Impedance of load (or a subsequent transmission line)
VL, IL =3D Voltage and Current at the load (or subsequent T-line)
Vrefl, Irefl =3D Reflected Voltage and Current
Relationship between Incident, Reflected, and Load waveforms:
VL =3D Vinc + Vrefl =3D IL * ZL
IL =3D Iinc - Irefl (Irefl travels in opposite direction)
Vinc =3D Z0 * Iinc; Vrefl =3D Z0 * Irefl
Substituting equations gives:
ZL =3D (Vinc + Vrefl)/(Iinc - Irefl)
=3D (Vinc + Vrefl)/(Vinc/Z0 - Vrefl/Z0)
Solve for Vrefl:
Vrefl =3D Vinc * [(ZL - Z0)/(ZL + Z0)] or rho * Vinc, where rho is the
reflection coefficient
VL =3D (1 + rho) * Vinc
-1 <=3D rho <=3D 1
When rho =3D 1 (open), Vrefl =3D Vinc, and VL =3D Vinc * 2
When rho =3D -1 (short), Vrefl =3D -Vinc, and VL =3D 0 (which it better)
Hope this helps.
Jeff Loyer
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
On Behalf Of Scott McMorrow
Sent: Wednesday, October 19, 2005 11:03 AM
To: lalexman@xxxxxxxxxxx
Cc: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Transmission lines reflections again
This is basic circuits theory.=20
You can use Kirchhoff's laws to solve for the voltage at the end of an
open transmission line.
Scott McMorrow
Teraspeed Consulting Group LLC
121 North River Drive
Narragansett, RI 02882
(401) 284-1827 Business
(401) 284-1840 Fax
http://www.teraspeed.com
Teraspeed(r) is the registered service mark of
Teraspeed Consulting Group LLC
Leonard Alexman wrote:
>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=20
>
>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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