50 ohm termination again - Page 1
Forum section: electronics / beginners

author: Phaedo
author_rank: Contributor
timestamp: January 17, 2020, 11:28:37 am
content: Hi ,I read this thread :https://www.eevblog.com/forum/beginners/50-ohm-terminator/and watched Alan Wolke's video there as well.I am a tad confused ( as usual )...Suppose we have a square wave (fast rising edge , i guess ) generated by a digital IC running at 5V.This square wave is connected by a 50Ω coax into the BNC of the scope.Does it mean that the IC has to be capable of sourcing 100mA (5V/50Ω),if it is getting terminated into a 50-ohm terminator (to get the 5V amplitude) ?TIA

author: tggzzz
author_rank: Super Contributor
timestamp: January 17, 2020, 11:53:58 am
content: Firstly, except for femtoamp and photon counting applications, all electronics is analogue. A "digital" circuit is merely an analogue circuit that interprets analogue voltages (or currents) as digital signals - provided the analogue voltages are within the specified limits.Don't rely on comments thrown together on bulletin boards; do search out decent literature e.g. application notes. One starting point ishttp://www.ti.com/lit/pdf/snla043Initially the transmitter "sees" a load of 50ohm+50ohm in series. If the scope input is 1Mohm//20pF, then after a "long" time the transmitter will see a load of 50ohms+1Mohm in series.

author: RoGeorge
author_rank: Super Contributor
timestamp: January 17, 2020, 11:56:27 am
content: Quote from: Phaedo on January 17, 2020, 11:28:37 amYes.

author: langwadt
author_rank: Super Contributor
timestamp: January 17, 2020, 12:04:38 pm
content: Quote from: tggzzz on January 17, 2020, 11:53:58 amwhere do you get the extra 50 Ohm from?

author: tggzzz
author_rank: Super Contributor
timestamp: January 17, 2020, 12:09:55 pm
content: Quote from: langwadt on January 17, 2020, 12:04:38 pmGood question.I speedread the OP's question and inferred a 50ohmsourcetermination. Doh.OTOH, it is an excellent example of "Don't rely on comments thrown together on bulletin boards"

author: Caliaxy
author_rank: Frequent Contributor
timestamp: January 17, 2020, 02:12:33 pm
content: Quote from: RoGeorge on January 17, 2020, 11:56:27 am... and that’s way series termination (at the source end) is preferred for sending logical signals through coax cables.

author: David Hess
author_rank: Super Contributor
timestamp: January 20, 2020, 02:24:58 am
content: Quote from: Phaedo on January 17, 2020, 11:28:37 amThat is right. A parallel terminated transmission line is actually a pretty difficult load for a typical logic output.Quote from: tggzzz on January 17, 2020, 11:53:58 amThat is the series termination case. The driver initially sees the 50 ohm series termination in series with the transmission line impedance which generates a 1/2 amplitude signal in the transmission line and in this case requires 50 milliamps. The 1/2 amplitude doubles at the end of the transmission line because of the reflection from an open load and returns to be absorbed into the series termination after which no current is required.In the parallel termination case, the driver initially sees the 50 ohm transmission line impedance and drives it to full amplitude requiring 100 milliamps for 5 volts. No reflection occurs from the load because of the parallel termination after which the driver sees the 50 ohm parallel termination at DC and low frequencies. 100 milliamps is required in this case at all times. In practical circuits, sometimes the parallel termination resistor is placed in series with a capacitor (AC termination) to reduce current to zero at DC and low frequencies.Parallel termination is often used with a thevenin equivalent termination to reduce the current requirement.Quote from: Caliaxy on January 17, 2020, 02:12:33 pmSeries termination is preferred when there is no parallel termination at the other end or low power is desired. For digital logic, double termination is problematical because it cuts the signal amplitude in half so either series or parallel termination is preferred alone unless other provisions are made at the receiving end to handle a half-amplitude signal like a comparator or line receiver.

author: Phaedo
author_rank: Contributor
timestamp: January 21, 2020, 06:56:04 pm
content: Thanks a lot Gents.Just one question more :Why is the impedance of the co-ax not included in the current calculations ?Quote from: tggzzzFor a moment I thought one of the 50Ω,was actually the impedance of the co-ax,being considered.Sorry about the noobie questions!

author: Vovk_Z
author_rank: Super Contributor
timestamp: January 21, 2020, 09:19:23 pm
content: "50 Ohm" for coax cable or conector means it's characteristic (wave) impedance but not real wire resistance (it's very small).But when we are talking about terminator (or load) then 50 Ohm is real electrical resistance or impedance.

author: Benta
author_rank: Super Contributor
timestamp: January 21, 2020, 09:33:01 pm
content: The completely basic model for a 50-ohm transmission line or 50-ohm coax is:A transmitter with 50-ohm output impedance -> 50-ohm coax (or transmission line) -> receiver with 50-ohm input impedance.The transmitter could be an amplifier with low output impedance and a 50-ohm resistor in series with the output.The receiver could be a high-impedance input with a 50-ohm resistor to ground.Then you'll have impedance matching and signal integrity.

author: radiolistener
author_rank: Super Contributor
timestamp: January 22, 2020, 12:36:50 am
content: Quote from: Phaedo on January 21, 2020, 06:56:04 pmBecause this is the input impedance of the coax line before reflected wave will be returned back. It has some impedance, because coax line can be represented as a lot of LC circuits in series. Distributed inductance of coax line limits current change speed, this is why coax line has some impedance. But this is not heat loss resistance. When reflected wave will be returned back it will affect the source, so the input impedance of the coax line will be changed and will depends on coax line length and coax line load.In other words electric wave has a limited speed of distribution in the cable, the wave needs some time to travel from cable input side to cable output side. At this time period, the input of the cable will looks like simple resistor equals to cable impedance. For example if coax cable is 75 Ohm, it will works like 75 Ohm resistor until wave is traveling from input side to output side.But the things may be changed when the wave reach the output side of the cable. At this point the wave may be reflected back if the load on the cable is not equals to coax cable impedance.When reflected wave will return back to the input side of the cable it will start to affect the source. It equals to change of the cable input impedance. This change will depends on the frequency, cable length and the load impedance on the output of coax cable.So, when you enable generator, first you will see 75 Ohm on the input of the coax cable, but after some short period of time (which is needed for wave propagation through cable back and forth) the impedance will be changed.If the load impedance is equals to the coax impedance, it will consume all wave energy and reflection will not occurs. So the cable will always work with 75 Ohm on the input side.This 75 Ohm is not a heat loss resistance, this is a reactive resistance which consumes energy needed to charge the coax line with RF energy. This energy is stored in the cable, the same as it happens with LC circuit at resonant frequency.

author: Phaedo
author_rank: Contributor
timestamp: January 22, 2020, 07:00:23 pm
content: Thanks a lot folks !! Most helpful and erudite set of replies.