Thick film hybrid reverse engineering and a mathematical challenge - Page 1
Forum section: electronics / repair

author: vtp
author_rank: Regular Contributor
timestamp: August 15, 2018, 12:52:02 pm
content: I have a thick film hybrid with three resistive voltage dividers on it, all three being in parallel with each other.This of course means that the resistors can be individually measured but since the other chain resistors are in parallel and series with the resistor being measured the result is incorrect.Say one of the chains is R1 and R2. The second R3 and R4 and third is R5 and R6. Each of the resistors have been measured in-circuit and the result is the following set of equations:R1 || (R2 + (R3+R4) || (R5+R6)) = 10890R2 || (R1 + (R3+R4) || (R5+R6)) = 3830R3 || (R4 + (R1+R2) || (R5+R6)) = 10660R4 || (R3 + (R1+R2) || (R5+R6)) = 8800R5 || (R6 + (R1+R2) || (R3+R4)) = 10000R6 || (R5 + (R1+R2) || (R3+R4)) = 670Where || means parallel operator for resistances, A || B = 1 / (1/A + 1/B). It takes precedence over summation above.There is an additional constraint (R1+R2) || (R3+R4) || (R5+R6) = 9800. Meaning that each chain is about 30kohm total in average. I made an assumption that upper resistors would have been the same but they were not (checked by measuring voltage of each divider and then calculating for the rest).So, how would anyone go about solving for each resistor value?Of course each chain could be isolated by cutting a trace but I would rather not do that, it being a thick film hybrid.This is also a repair topic as the last chain breaks (or I assume so at the moment) when in equipment and letting it run for a while. Of course I could not make it break on desk with a hot air pencil.

author: bd139
author_rank: Super Contributor
timestamp: August 15, 2018, 12:55:56 pm
content: This is just a system of equations. Any decent symbolic mathematics package should be able to solve this (mathematica etc). There are probably negative solutions so I would force a constraint of positive real numbers.However it may find more than one solution or none of you’re unlucky

author: james_s
author_rank: Super Contributor
timestamp: August 16, 2018, 01:36:04 am
content: If they're voltage dividers can you just put a known potential across them then measure the voltage from the tap of each one? Then you'll have the ratios and you can probably take a pretty good guess at the individual values.

author: coromonadalix
author_rank: Super Contributor
timestamp: August 16, 2018, 02:36:56 am
content: what is the thick film part numberdo you have an schematic of it, where it goes ??

author: The Electrician
author_rank: Frequent Contributor
timestamp: August 16, 2018, 02:47:28 am
content: Here's what Mathematica gets:

author: Zucca
author_rank: Unknown
timestamp: August 16, 2018, 08:07:38 am
content: Interesting,so the only possible solution is the last row, since all the others have negative values.EDIT: Why I did not read the comment on the top?

author: bd139
author_rank: Super Contributor
timestamp: August 16, 2018, 08:19:30 am
content: @The Electrician: nice one. I was hoping someone would do this because my Mathematica license expired ages agoI imagine these are most likely in E96 series and possibly binned after that. Fun times!

author: Zucca
author_rank: Unknown
timestamp: August 16, 2018, 08:30:17 am
content: Maybe Matlab could also help here?

author: Ian.M
author_rank: Super Contributor
timestamp: August 16, 2018, 08:46:42 am
content: Assuming you can inject a voltage to let you measure each divider tap to get its ratio, then the next step is to short the common top end of all the dividers to the common bottom end, and measure the Thevenin resistance of each divider tap (to the shorted ends). Its then trivial to calculate the individual divider resistances as you don't have to consider the influence of the other dividers.

author: vtp
author_rank: Regular Contributor
timestamp: August 16, 2018, 08:47:36 am
content: Quote from: The Electrician on August 16, 2018, 02:47:28 amThank You.I was about to start typing that to Reduce which I installed yesterday but you got there first and Mathematica seem very nice with that equation editor. Anyway, very impressive.There is another set of 8 resistors that are in circuit with each other on the hybrid. However, four of those are JFETs so I'll just lift some JFET pins or remove them instead of writing eight equations to solve for the values.But this technique looks to be good in solving resistor networks measured in-circuit.The hybrid in question is a YHP multislope AD-converter part, used in many YHP products. This one is from 4284A LCR meter. The resistor chains discussed above provide HI, LO and ZR (as in zero) comparator levels for the integrator. LO and ZR are actually taken from the integrator with gain so that their actual tripping point could be set low enough.This ADC system also includes a PAL for which the read result is attached.For schematics of the hybrid application, see for example 4278A service manual.

author: doktor pyta
author_rank: Frequent Contributor
timestamp: August 16, 2018, 09:10:45 am
content: nice.where to find such edge-soldered pins ?

author: bd139
author_rank: Super Contributor
timestamp: August 16, 2018, 09:11:35 am
content: Was just about to ask that

author: vtp
author_rank: Regular Contributor
timestamp: August 16, 2018, 09:21:50 am
content: Quote from: doktor pyta on August 16, 2018, 09:10:45 amIt is called SIL leadframe, available from Mouser and Digikey and other usual sources. I do not have the package at hand to give you actual part number but I am sure you can find those with the proper search term.It actually took me a while to figure out what that was called...

author: vtp
author_rank: Regular Contributor
timestamp: August 16, 2018, 09:29:47 am
content: Can someone have a look at that PAL equations file?I am a bit concerned about this line and the feedback term there in:/f14 = /i2 + i5 * /f14I am also attaching the jedec file from which the PAL equations were derived from.