Newsgroups: sci.electronics Path: utzoo!henry From: henry@utzoo.uucp (Henry Spencer) Subject: Re: Transmission Lines - What Are They?? Message-ID: <1989Nov27.230049.590@utzoo.uucp> Organization: U of Toronto Zoology References: <1989Nov12.013850.7756@utzoo.uucp> Date: Mon, 27 Nov 89 23:00:49 GMT A couple of people have asked me to elaborate a bit on my explanation of reflections, so, on the chance that it will be of general interest... Briefly summarizing my previous posting: If you apply a voltage with a risetime of 1ns to a pair of wires 10ns long with a resistor across the other ends, how much current flows 2ns after the voltage is applied? It's determined by the characteristic impedance of the wires, which is a composite of their resistance (usually small) and their capacitance and inductance. When the signal reaches the other end, if the impedance there doesn't match the characteristic impedance of the wires, reflections result. More specifically, say we have a 6V signal, a 1kohm resistor, and wires with a c.i. of 500ohms. The c.i. means we initially get 12mA of current flowing. So when the pulse reaches the resistor, it arrives with 6V and 12mA. This does not match the resistor's impedance, however. The resistor will take 6mA at 6V, so we have an excess 6mA arriving. Electrons are conserved; they have to go *somewhere*. There are two places they can go: into the resistor, or back up the wire. This is a linear circuit, so it's valid to analyze current back into the wire as if it were independent of current coming out of the wire. The wire and the resistor are a classic current divider, so (500/(1000+500)) = 1/3 of the current will go into the resistor and the rest back up the wire. The voltage will have to rise a little to do this, specifically (2mA*1kohm) = (4mA*500ohm) = 2V. So a new signal, 4mA at 2V, starts propagating back up the wire, and the voltage at the resistor is 6+2 = 8V. The voltage source will have some non-zero input impedance, probably not the same as the wire's c.i. When the new signal hits it, the same thing happens again: the current splits, with some of it going back into the source (i.e. the source's net output current being reduced) and some of it bouncing back into the wire. If the source's impedance is less than that of the wire, some of the voltages and currents involved will be negative; this does not change the overall picture. Each time the signal hits an end of the wire, some of it carries on and some of it bounces back, with the bounced signal steadily getting smaller. The net effect, observing at any specific point on the wire, is that the voltage and current bounce up and down somewhat and gradually settle toward the final value. In theory the reflections continue forever, but in practice they eventually disappear into the background noise. -- That's not a joke, that's | Henry Spencer at U of Toronto Zoology NASA. -Nick Szabo | uunet!attcan!utzoo!henry henry@zoo.toronto.edu Brought to you by Super Global Mega Corp .com