Zero Ohm link 1/4 watt for power to circuit. - Page 1
Forum section: electronics / projects

author: JanJansen
author_rank: Frequent Contributor
timestamp: June 09, 2017, 04:13:25 pm
content: Hi, i use these zero ohm links 1/4 watt for 12v and - .I have a row cards made they take together 44mAh, the power to the rails for the cards goes via these 1/4 zero ohm links.Now i went searching :http://www.electronics-tutorials.ws/resistor/res_7.htmlAccording to Resistor Power Rating Example No1 i dont have enoug with 1 link, i need 3 instead.Now i read Wikipedia :https://en.wikipedia.org/wiki/Zero-ohm_linkThey say : For example, a surface-mounted 0805 size resistor of 0.003 ohms, rated at 1/2 watt can, in theory, safely pass up to 12.9 amperes of current.That is a lot more.If these 1/4 watt links are not good enough, can someone suggest me what to use instead ?i have these :http://nl.mouser.com/ProductDetail/Yageo/MFR-25FTE52-0R/?qs=sGAEpiMZZMu61qfTUdNhG0IXHLFuiNndEiango%2ft3K4%3dthey also have these with a bit higher watt rating ( 400 mW ):http://nl.mouser.com/ProductDetail/Vishay/SFR25000Z0000ZA100/?qs=oyz6OjtwqKSFE2PwBzTDHg%3d%3dThanks in advance

author: rcbuck
author_rank: Frequent Contributor
timestamp: June 09, 2017, 05:35:31 pm
content: Check these 1/4 ones out at Mouser:71-FRJ50-0-E3 - 25A @ 25 degrees C, 10 mohm603-ZOR-25-R-520R - 10A @ 25 degrees C, 20 mohmWhy are you concerned? You are only using 44 mAh per card. Even if you have 100 cards in a row, that is only 4.4 amps.

author: thm_w
author_rank: Super Contributor
timestamp: June 09, 2017, 09:56:13 pm
content: Quote from: JanJansen on June 09, 2017, 04:13:25 pmIt looks like you did the calculation incorrectly.QuoteThere is not 12Vacrossyour resistor, there is 12V on one side and very slightly less than 12V on the other. So the actual voltage across is on the order of microvolts (0.0001V).

author: Richard Crowley
author_rank: Super Contributor
timestamp: June 09, 2017, 10:21:13 pm
content: Assuming you are talking about thru-hole (vs. SMD):Remember that "1/4 watt" only refers to the SIZE of the body.It has NOTHING to do with the actual electrical rating of what amounts to a solid piece of wire.

author: Brumby
author_rank: Unknown
timestamp: June 10, 2017, 05:38:32 am
content: Quote from: Richard Crowley on June 09, 2017, 10:21:13 pmStrictly speaking - it actually does refer to the power dissipation capability of the device.However, considering that it IS pretty much the equivalent of a solid piece of wire, you'll be having other issues with the magnitude of the current required to get to that power level ... like frying PCB tracks.

author: JanJansen
author_rank: Frequent Contributor
timestamp: June 10, 2017, 02:44:48 pm
content: Thanks, so the voltage difference only counts for selecting parts ?, nice to know.So 1 is enough for the 44 milliampere, i was worried about nothing, next time i let em fry before asking.btw : i have these ones to be exact :http://nl.mouser.com/ProductDetail/TE-Connectivity/1623934-1/?qs=%2fha2pyFadugkd5x9VElL9rBQfoUflu8yomTrfzEP61BzKubyve0uGw%3d%3d

author: JanJansen
author_rank: Frequent Contributor
timestamp: June 10, 2017, 02:54:58 pm
content: 71-FRJ50-0-E3 - 25A @ 25 degrees C, 10 mohm603-ZOR-25-R-520R - 10A @ 25 degrees C, 20 mohmDoes it not matters for the ( 0v ) ground to have 20mohm ?Does mine also have 20mohm ?http://nl.mouser.com/ProductDetail/TE-Connectivity/1623934-1/?qs=%2fha2pyFadugkd5x9VElL9rBQfoUflu8yomTrfzEP61BzKubyve0uGw%3d%3d

author: Richard Crowley
author_rank: Super Contributor
timestamp: June 10, 2017, 04:00:04 pm
content: Quote from: Brumby on June 10, 2017, 05:38:32 amNo, it isn't "the equivalent", it is ACTUALLY a solid piece of wire. It just has a "body" molded around it for convenience of handling (automated pick-n-place, etc.)

author: kayvee
author_rank: Regular Contributor
timestamp: June 10, 2017, 04:04:12 pm
content: I also think it's quaint how they are also sometimes specified at 0 ohm +- 5%

author: ebastler
author_rank: Super Contributor
timestamp: June 10, 2017, 07:03:49 pm
content: Quote from: JanJansen on June 10, 2017, 02:44:48 pmYes, that is essentially correct. The power rating tells you how much power the resistor can dissipate (i.e. turn into heat inside the resistor), not how much power it can transmit to a load downstream. Because your resistor value is very small in this case, the dissipated power will be very small: P = R * I².