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Earlier today i changed the value of the shunt resistor of my LTC4150 coulomb counter from 0.05 ohm to 0.5 ohm to get more precise readings.In the datasheet it was specified that voltage drop of the shunt resistor should not be above 50mV for proper operation, with that in mind, i connected an appropriate resistor as my load to limit the current wihtin this restriction enter image description here.

Since my Vin is 7.4v i picked 10kΩ to be on the safe side and to test my expected, low current resolution.

Oddly enough the board stopped producing interrupts. It used to signal low everytime some amount of charge(Depending on shunt resistor and voltage to frequency gain)have passed and it still remained broken even after i reverted the resistor change i made earlier .After a series of unsuccessful repair attempts (i even changed the IC with a brand new one to no avail.), i decided to check the resistance value of each resistor on the board with the help of the product schematic.

enter image description here

When i measured the resistances with a multimeter i saw that both of the 75k SMD resistors were at 39k (85C), while both of the 47k resistors were 0,01 ohms(473). 3.3k resistors were doing okey but all of the resistors except 47ks failed the continuity test .Do you think the reason i got inaccurate readings is that i didnt dismantle the components from the board ? Or what does failing continuity test while still having correct resistance mean for a resistor ? Have a nice day people :)

Edit : I removed the resistors from the circuit and it looks like they are not the cause of my problem.Interrupt pin of my IC is an open-drain output , its is supposed to provide me with a high signal whenever (which i will see as low since it's open drain) a certain amaount of charge has passed.Even after ı changed the IC still does not work. I am not asking for a direct solution but rather an approach to what to look for next in my troubleshooting scenario

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    \$\begingroup\$ Measuring resistors with them still in the circuit is almost guaranteed to give you wrong answers. \$\endgroup\$
    – JRE
    Jun 7, 2016 at 15:29
  • \$\begingroup\$ any thoughts about the continuity test ? \$\endgroup\$
    – Dogus Ural
    Jun 7, 2016 at 15:30
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    \$\begingroup\$ A continuity test is pretty much just the same as using an ohmmeter - it just beeps instead of showing a (low) number. They will be wrong in the presence of ICs and capacitors. \$\endgroup\$
    – JRE
    Jun 7, 2016 at 15:36
  • \$\begingroup\$ It is esd gloves by the way not dirty socks , +1 for making me giggle tho :) \$\endgroup\$
    – Dogus Ural
    Jun 7, 2016 at 15:41
  • \$\begingroup\$ The schematic shows jumpers in parallel with R4 and R8 (the two 47K resistors.) If the jumpers are closed (shorted) then the continuity test would beep. \$\endgroup\$
    – JRE
    Jun 7, 2016 at 15:57

3 Answers 3

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While measuring components on the board keep in mind that there are different paths the current can flow. Take for example this schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

Trying to measure R1 may not give a value of 100 on a pcb, because the volt meter current has the potential to flow into other paths and this lowers the measured resistance from the volt meter. In this example putting DMM leads across R1 could result in current flowing through R1 AND through Rsensor to ground and through Radc (the resistance of an analog digital converter ), through R2 and back to the DMM leads across R1. Most IC's these days also have protection diodes that turn on when you try and make measurements with the circuit off. If you really need to verify a passives value, then unsold er it from the board and measure it. A continuity tests only looks for small amounts of current so the same rules apply, and you still have lots of potential paths in any given circuit because real world parts are not ideal and don't have infinite input impedance AND function differently with no power applied AND have protection diodes.

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You appear to be using the demo board

If you have chosen 10k as the output load and not changed any other circuit element (particularly the current sense resistor, R4), then we get the following:

\$I_{load} = \frac {7.4V} {10k\Omega} = 740\mu A\$

\$V_{sense} = 740\mu A \cdot 0.01\Omega = 7.4 \mu V\$

For a nominal \$ G_{VF}\$ of \$32.55 \frac {Hz} {V}\$ then using the equation on page 8 of the datasheet, \$f = 32.55 \cdot 7.4\mu V = 240.87 \mu Hz\$ , which yields a pulse every 4,151 seconds.

The interrupt hasn't stopped, you will simply have to wait just over 69 minutes to see one.

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  • \$\begingroup\$ No i am using FRDM K64F , and i changed the sense resistor to 0.5 ohm from 0.05 ohm it stopped working after i made that change. Even though i reverted the resistor change, the breakout board still remained out of action . \$\endgroup\$
    – Dogus Ural
    Jun 8, 2016 at 9:04
  • \$\begingroup\$ And from the ONE INT = 1 / (Gvf x Rsense) equation on page 8 , i should expect an interrupt everytime 0.0625 coulombs have passed, rather than a fixed time. \$\endgroup\$
    – Dogus Ural
    Jun 8, 2016 at 9:10
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Since my solder heat was around 360 celcius and the fact that i am a newbie smd solderer i might have burnt the conductive roads. Or it was static electricity even though we had anti-static mat on the table. Either way it looks less painful to use my spare LTC4150 rather than fix the current one. Don't forget to check absolute maximum rating for soldering heat next time ! :)

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