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Electronics => Beginners => Topic started by: RossS on September 04, 2016, 09:15:17 pm

Title: Multimeter input protection question
Post by: RossS on September 04, 2016, 09:15:17 pm
After watching Dave's multimeter input protection video (https://www.youtube.com/watch?v=zUhnGp5vh60), I'm confused about the diode bridge used to clamp voltage across the shunts until the fuse blows.

Dave says the diode bridge in the Fluke 27 uses standard 1n4007 diodes, and indeed the service manual (http://assets.fluke.com/manuals/27______smeng0100.pdf) confirms this. However, the 1n4007 diodes I've found can only handle a peek forward current of 30-45A for single digit milliseconds (example 1 (http://www.mccsemi.com/up_pdf/1N4001-1N4007(DO-41).pdf), example 2 (http://www.vishay.com/docs/88504/1n4001gp.pdf)). Surely the current will be much higher than this if you plug your meter directly into mains? Until your mains breaker blows, you'd be limited just by the resistance of the fuse. I measured the 44/100A fuse in my Fluke 87 (same fuse as the 27) at ~800 mOhms. At 120V where I'm from, that's 150 amps.

How can the diodes deal with this huge amount of current for the amount of time it takes for the fuse to blow? I must be missing something.
Title: Re: Multimeter input protection question
Post by: retiredcaps on September 04, 2016, 09:23:22 pm
Did you watch the followup video?

https://www.youtube.com/watch?v=ne_Prn0eGQE (https://www.youtube.com/watch?v=ne_Prn0eGQE)
Title: Re: Multimeter input protection question
Post by: Kleinstein on September 04, 2016, 09:34:51 pm
Watching that 2nd video, a high current pulse should not be a problem - the fuse should be always faster than the 1N4007.

However at a current just over 1 A, the fuse might actually be rather slow. So not much spare to take into account that the diodes are rather close to each other.
Title: Re: Multimeter input protection question
Post by: RossS on September 04, 2016, 09:40:31 pm
Did you watch the followup video?

I completely missed that. Thanks for the pointer!

The video nicely answers my question. As Kleinstein also mentions, the point I was missing is that the fuse is always going to blow before the diodes.