Whetting Current - Page 1
Forum section: electronics / general

author: hardsoft
author_rank: Newbie
timestamp: December 14, 2013, 09:28:56 pm
content: I have a basic understanding of whetting current for switches and relays, the minimum amount of current recommended to keep the switch "healthy" by burning off oxidation. Some switches are designed for large currents and may require amps of whetting current and supposedly will not work reliably feeding very low currents such as the input of a micro. Gold plated switches, on the other hand, typically only need mAs.The details, however, are still a bit of a mystery to me. Many times the whetting current is given at a particular voltage, say 10mA at 5V. But... a closed switch is typically very low resistance, and the voltage drop across the switch will only be mV. So why would the rail voltage matter?Also, does anyone have any sense for the time required for the wetting current? If I want to use a capacitor to surge current through the switch, for example,, how long is typically required to accomplish the required "cleaning"?Any insight appreciated.

author: PedroDaGr8
author_rank: Super Contributor
timestamp: December 14, 2013, 10:44:38 pm
content: Because you want a certain amount of heat produced to break down the oxides. This means you want a minimum wattage which means you need to worry about both the current and the voltage. Also said oxides can raise the resistance of the contacts so that the voltage do is more than a few millivolts.Sent from my EVO using Tapatalk

author: robrenz
author_rank: Super Contributor
timestamp: December 15, 2013, 12:00:34 am
content: This does not answer your question but is related. Many micro Ohm meters have a "dry contact" test setting that limits the drive voltage to 50 or 20 mV so that the oxide films on the contacts being tested are not punctured. This is to ensure the resistance reading is relevant for small signal applications.

author: TMM
author_rank: Frequent Contributor
timestamp: December 15, 2013, 06:52:35 am
content: Quote from: PedroDaGr8 on December 14, 2013, 10:44:38 pmThis i have been wondering about myself. If the load doesn't produce an inrush current the contacts will never see anywhere near the 'wetting voltage'.Say we are using this relay:http://www.farnell.com/datasheets/60753.pdfMinimum switching load is specified as 1W/10V/10mAWe switch a 12Ohm load @ 12V, the initial contact resistance would have to be 60ohms to achieve a 10V drop. This is unrealistically high. The actual initial resistance of this relay is about 0.5-1ohm.I have this actual relay and i can report that using it in the above manner is easily enough to wet the contacts and achieve <0.002ohm contact resistance.

author: Unknown
author_rank: Unknown
timestamp: December 15, 2013, 07:39:08 pm
content: If there is a very thin oxide layer on the contacts, then the contacts may arc (puncturing the oxide layer) if the voltage across the oxide layer is sufficiently high. I would expect this to be a very fast process, but have never quantified it.

author: AG6QR
author_rank: Frequent Contributor
timestamp: December 17, 2013, 09:54:39 pm
content: Quote from: TMM on December 15, 2013, 06:52:35 amWhen the contacts are open, with any reasonable load, they'll have the full supply voltage across them. For that to be true, the load resistance must be small compared to the resistance of the open contacts, but that's usually the case.When the contacts close, the closure may be fast, but it's never instantaneous.The contacts have the full supply voltage across them up until the time when the first few electrons start to initially bridge the gap between the contacts. From the time the first electrons bridge the gap to the time the gap is fully closed is when the wetting takes place.You don't need to worry about the exact details of what happens as the contacts close. According to that data sheet, if you switch a 10mA load at 10V, things are going to work out OK. That doesn't necessarily mean that the relay will have 10V across the contacts at the exact same instant as it's carrying 10mA.