Ohms converter?

Status
Not open for further replies.

throbscottle

Well-Known Member
Okay, so I've worked out most of how my dmm is going to work now (with some help, thanks folks!), but I'm trying to understand ohms converter circuits from the couple I've found on the web. What I'm seeing is a constant current source connected to an op-amp's inverting i/p, nfb via a fixed resistance, a reference resistor (the dmm's input ladder) connected from it's o/p to non-inverting i/p, and the dut goes between the non-inverting i/p and ground. The adc is connected across the dut (or to a further op-amp connected to the same point)

It looks like a modified Howland current pump, and I understand that a constant current will produce a voltage proportional to the resistance, but I don't really understand what is going on in this circuit. Can anyone give an explanation? Also, this is now quite an old method - are things done differently in modern meters? I'm looking to produce something very stable, in the long run.
 
Here's a good example with explanation from a high end DMM.
 
Last edited:
Oh, okay. I just couldn't work out what the op-amp was doing, before. Makes more sense now. Nice reference to Keithley too - I was wondering how to do four wire measurements.

Thanks to both of you.
 
best of all, it's written by Bob Pease
The Czar of Bandgaps was a really smart guy.

At least when you read a Pease app note, you know the circuits were actually built up and tested and the data is real..... not just garbage spit out by a simulation. Too bad he was the last of that dying breed.
 
Last edited:
Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…