EEVBlog #72 - question RE uCurrent 10 milli-ohm connection - Page 1
Forum section: electronics / general

author: cube1us
author_rank: Contributor
timestamp: August 10, 2013, 02:57:30 pm
content: I just started watching old blog video episodes (well, actually I watched your microchip vs. atmel one a while back, and was really impressed), and came up with something I don't quite follow. In your final uCurrent design, the connection of the 10 milli-ohm resistor still seems to go thru both switch terminals. Now at 19:55 in the video your circuit diagram shows one tap going over to the amplifier. That gets one switch contact out of the way - but if it did that, that resistor would always be in the circuit - not good at all - wouldn't work on the other ranges. By 43:06 that connection has moved over to the switch, so at least the circuit works, but as far as I can see, extra terminals on the resistor or not, the switch contacts are still in the circuit and affect the measurement. So, how did this help deal with the contact resistance of the switch? I don't get it.

author: SArepairman
author_rank: Frequent Contributor
timestamp: August 10, 2013, 03:41:19 pm
content: which video?

author: Bloch
author_rank: Unknown
timestamp: August 10, 2013, 04:00:10 pm
content: Quote from: SArepairman on August 10, 2013, 03:41:19 pmhttp://lmgtfy.com/?q=EEVBlog+%2372Just a guess :-)

author: cube1us
author_rank: Contributor
timestamp: August 10, 2013, 04:29:53 pm
content: Yes, that's the one.

author: ejeffrey
author_rank: Super Contributor
timestamp: August 11, 2013, 04:44:54 pm
content: The switches are still in the circuit, but they don't affect the measurement. The input current flows through the first switch, but the amplifier doesn't measure the voltage across that switch. The second switch doesn't have any current flowing through it except the opamp bias current, so its contact resistance doesn't matter. All three ranges actually use this trick. The only difference on the milliamp range is that the 4-wire resistor also eliminates the effect of resistance in the PCB traces.

author: cube1us
author_rank: Contributor
timestamp: August 12, 2013, 01:44:33 am
content: I see. So, essentially, the current thru the second gang is really small, so no appreciable voltage drop as compared to the voltage being measured across the resistor. OK. I get that. The first gang though, however, still presents the contact resistance in series with the shunt resistor. It would be on the order of, say, 1 milli-ohm, per what Dave said in the video. That is, somewhere around 1/10 of the value of the shunt resistor. Since we are ostensibly measuring a few milli-amps on this range, then the voltage drop should be on the order of a few micro-volts. If the circuit under test were than sensitive to such a small "burden", then the 10 milli-ohm shunt resistor would already be big trouble. So, it doesn't affect the measurement at all, and for the intended application, it shouldn't affect the circuit under test by a lot less than the shunt resistor. OK, I think I get it. Thanks. I think what tripped me up was the implication in the video about what the benefit of the 2nd set of terminals actually was. Thanks again.

author: ejeffrey
author_rank: Super Contributor
timestamp: August 12, 2013, 04:10:35 am
content: Correct. The first switch still affects the burden voltage, but it is already so much less than a standard meter that it doesn't matter. If you really wanted to redesign the ucurrent to measure on the milliamps range with the minimum possible burden voltage you could use a separate jack the same way that many meters have amps and milliamps on different jacks. That would be less convenient to use, so it isn't a good tradeoff unless you really care about every last millivolt.