Closed-enclosure part discovery - Page 1
Forum section: electronics / rf-microwave

author: babysitter
author_rank: Frequent Contributor
timestamp: February 09, 2018, 08:46:12 pm
content: Today I brought home a recent ebay find, a homebrew T-match tuner (C-L-C.) Some loving care was applied (cleaners, contact lube), but some things need to be changed to be more babysitter-y (one really long single wire and switches to lock out the Cs.)I couldn't press myself into measuring min and max capacitance of the air variable capacitors or the rollercoaster inductance, but this will be done later.Anyway, that led me to a nice question, triggering my curiosity but incompatible with my math allergy:Imagine having a VNA (say a FA-VA 3) and nearly infinite part inventory to build any load, how can i figure out the min and max values of the innards without opening the case considering the known topology?(Keep in mind the switches in parallel to the Cs are not here yet)Vy 73Hendrik

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 09, 2018, 11:39:01 pm
content: A VNA does that alone. The network can be reconstructed from reflection/transmission properties and assuming the network is linear and passive (which should be a fine assumption here).Curiously, this works even if the network is not made of RLC lumped components; in that case, you get an approximation (e.g., a transmission line as a huge LC ladder).Real components aren't ideal RLC lumps either, so you can learn quite a lot about the components in the tuner for the same reason (e.g., not just the capacitor's capacitance, but ESR, ESL and higher order aspects as well).You can replace the VNA with a signal source, reflectance bridge, and various loads; and a lot more fiddling.In either case, solving the network requires simultaneous equations in N variables for N elements, and also something about least squares regression for overdetermined, noisy systems (which is to say, the VNA grabs a huge swath of data, way more points than strictly required to solve the problem, and few of which actually match up exactly with any particular solution -- that is, they're somewhat noisy). So, you're kind of SOL on the math bit, to go any farther.Tim

author: yl3akb
author_rank: Regular Contributor
timestamp: February 10, 2018, 10:29:16 am
content: Quote from: babysitter on February 09, 2018, 08:46:12 pmIf we assume that tuner consists of ideal lumped elements the problem seems quite simple. If we assume this cascade:[series C1]+[shunt L1]+[series C2]and we leave output open, we get:[series C1]+[shunt L1].At input this looks like simple series LC circuit with impedance:jZ = -jXc + jXl --> Z = -Xc + Xl = -1/(w*C1) + w*LIf You measure Z (it will be only reactive) at two different frequencies, this linear system could be used to calculate C and L (w = 2*pi*f):Z1 = -(1/w1)*(1/C1) + w1*L1Z2 = -(1/w2)*(1/C1) + w2*L1When L1 is known, C2 could be easily calculated by knowing Z at output of the tuner.You can use this calculator for linear eq. solving:http://wims.unice.fr/wims/wims.cgi

author: babysitter
author_rank: Frequent Contributor
timestamp: February 10, 2018, 10:43:12 am
content: Splitting to two L-C combinations by leaving open is a good idea, should have been obvious. Considering the 3rd part just had me seeing a bunch more equations.Vy 73Hendrik