Low Frequency Two-Port Network Characterization - Page 1
Forum section: electronics / projects

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 26, 2014, 10:20:34 pm
content: So I'm working on a project now that requires me to evaluate an array of passive electrical components at low frequency -- basically low/band pass filters. There are a number of ways to characterize such a system (voltage transfer functions, etc), but I would like something that takes a variable input and output impeadace into account. Enter two-port network modeling....http://en.wikipedia.org/wiki/Two-port_networkIt goes without saying I don't have access to a HP network analyzer. Nor would it be that helpful, as I'm interested in the 0.01 Hz - 100kHz region.So I understand that the basic strategy here is to evaluate how the DUT functions when its input/output is bridged or open... e.g. I=0/V=0. Once you get those two parameters (for both input and output) -- four complex numbers per frequency -- you can linearly combine them (kind of in a load-line method) to model how the DUT will function with any combination of input/output load. Awesome... awesome to the max.That's all well and good for the mathematician, but I need to *actually* perform the test. And there are two problems with performing this test "à la mathématiques" ... 1) some devices don't like their output bridged, 2) even if you're using all passive components the dynamic range between 'open' and 'bridged' measurements can be huge.... I mean how much current will be passed though a capacitor at 0.1Hz vs a short circuit? And again at 100kHz?So I'm wondering if anyone is familiar with taking these types of measurements? Any tips?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 26, 2014, 11:07:50 pm
content: If it's a linear, impedance matched system, it won't at all mind ports being open or shorted. The short circuit current can only be twice the nominal (matched) output current, and the open circuit voltage can only be double that of the same condition.Just Thevenin's theorem (or Norton if you prefer), extended for a circuit with two arbitrary terminals and ground.If you're using a linear test on somethingnonlinear, well...Tim

author: CaptnYellowShirt
author_rank: Frequent Contributor
timestamp: February 26, 2014, 11:17:35 pm
content: ------> 0----------------------------< <----| || |[1.0uF] | | || |------> 0----------------------------< <----This is a simplification of the problem I'm running into.What's the ratio of open/shorted input currents at ... DC? 0.1Hz? 100kHz? On one hand, at DC the input impedance looks like OPEN/SHORT, but then at 100kHz you have something that looks like SHORT/SHORT. So trying to accurately quantify the input currents in these cases becomes a dynamic range problem... You need an electrometer on one hand, and then a heavy duty current shunt on the other.The question is more about how to preform a particular measurement, e.g. using white noise in the 0.01Hz-100kHz band. Its no fun if you have to test every single frequency piece-meal style and reset the experiment's shunt resistor to a new value in every decade.My first thought here is to use a set of resistors in place of the short. That way, the the values of the input impedance stay in some ball park (lets say 4 orders of magnitude that my ADC can accurately measure). Then in post processing use a linear system of equations (load-line method) to estimate what the impedances should have been with an open/closed circuit.