Path: utzoo!utgpu!news-server.csri.toronto.edu!rutgers!dimacs.rutgers.edu!seismo!uunet!isavax.isa.com!cliffb From: cliffb@isavax.isa.com (cliff bedore*) Newsgroups: comp.protocols.tcp-ip Subject: Re: What is the voltage spec for thinnet? Message-ID: <1991Feb8.111625.23635@isavax.isa.com> Date: 8 Feb 91 11:16:25 GMT References: <1991Jan30.155606.21529@dartvax.dartmouth.edu> <1991Feb5.131333.2047@isavax.isa.com> <1991Feb6.033424.21632@usenet.ins.cwru.edu> Reply-To: cliffb@isavax.isa.com (cliff bedore*) Distribution: usa Organization: ISA Inc. Arlington, VA Lines: 87 In article <1991Feb6.033424.21632@usenet.ins.cwru.edu> jb@falstaff.mae.cwru.edu (Jim Berilla) writes: >In article <1991Feb5.131333.2047@isavax.isa.com> cliffb@isavax.isa.com (cliff bedore*) writes: >>In article <1991Jan30.155606.21529@dartvax.dartmouth.edu> wbc@moose.dartmouth.edu (Wayne B. Cripps) writes: >>> >>> >>>What should the voltage be on thinnet? - I get readings of >>>-1.8 to -2.0 volts and .2 to .3 volts - is this in the >>>range? is 1.2 volts ok? >>> >>> Wayne >> >>At the risk of stepping into something soft and gooey, I thought I'd put on >>my engineers hat for a while and comment on this. >> >>First. It will be very traffic dependent (assuming you're using a voltmeter >>that does some averaging). > >True. Don't use a voltmeter, use an oscilloscope. You'll see many interesting >things. > >>Having stated that and not knowing the details of an ethernet board, but >>knowing something about transmission lines, we can get a wag of ranges for >>the voltages. > >Not true. The voltages on the ethernet are clearly defined. I didn't say they weren't clearly defined. I said we could get a range of what was reasonable. We engineers call this a worst case analysis. > >>The lines are 50 ohms and are terminated in 1/4 or 1/2 watt resistors. (Mine >>is cause I did it myself and haven't had problems). >> >>Power (watts) = voltage ^2 / resistance or >> >>voltage = sqrt( power * resistance) >> >>voltage = sqrt ( 1/4) * 50 ) or 3.5 volts. for 1/4 watt power >> >>voltage = sqrt ( 1/2 * 50 ) or 5 volts for 1/2 watt power > >What *are* you talking about? (And take off that silly hat.) >In the case of ethernet, the voltage depends on the amount of current >pulled out of the tap. It's independant of the power rating of the >terminating resistors. Remember that the tap appears as a 25 ohm load, >i.e. it's connected to two 50 ohm transmission lines. > What I am talking about is how much voltage you can put through a 1/4 watt 50 ohm resistor without exceeding its power limits. Not circuit impedance! The above analysis is true. If you put 2 50 ohm 1/4 watt resistors in parallel,you get a 25 ohm 1/2 watt resistor. This reduces the resistance but keeps the power dissipation the same so the maximum voltage will be the same. (Ohms Law; He was a hero to us engineers) The 2nd calculation was assuming the specs call for 1/2 watt terminators. The idea of this was to get an idea of what would be the max voltage you should see on an ethernet line using a voltmeter. Given that the above calculations do that, I'll keep my hat on; thank you very much. The point is sometimes circuits don't pull current, they get pushed voltage. If the voltage-current combination (power) exceeds the rating of the resistor, it will burn up. If you think this isn't important get a 50 ohm resistor (47) from Radio Shack and plug it into your 110 outlet (but wear eye protection and gloves). Or run 15-20 volts down your thinnet and be prepared to look for a new job >For the AM7996 transceiver (common in a lot of Sun's), the voltages are >specified as follows: High level is between 0 and -.1 volts, low level >is between -1.625 and -2.2 volts. As stated above, this voltage is >generated by a current sink (to -9V) by the chip. An ideal current sink >has infinite impedance, and doesn't load the transmission line. If two >or more stations transmit at the same time, the voltage on the line goes >below -2.2 volts. The chip detects collisions on this basis. Had you been as quick to answer the original question as you were to critcize my answer, we could have saved a lot of bandwidth on the net. I waited 4 days to see if someone would post an answer. Since no one did, I gave the above estimate of what would be reasonable. . . > Jim Berilla / jb@falstaff.cwru.edu / 216-368-6776 >"My opinions are my own, except on Wednesday mornings at 9 AM, > when my opinions are those of my boss." In spite of my above comments, I do appreciate knowing what the true voltages are for ethernet. I've filed them away for the next time the question comes up. Cliff