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Ok so here's my question. I'm fairly new to electronics research, my background is chemical engineering, but I'm attempting to improve my understanding of basic circuits and data transmission for my work. For this issue; I have a problem, and a solution, and I'm trying to work backwards to figure out why that solution solves that problem.

What I'm working with is a device that's transmitting data. DC voltage is supplied to a device, the device then attenuates the DC voltage to create a modulated current which is sent back to a computer and interpreted as a digital information signal. An issue was occurring where a new version of these devices would not transmit any data all. The way it was explained to me was that the new versions were not drawing as much current as anticipated so the modulated current being sent back was not high enough to be interpreted as a logical 1, so the computer was essentially receiving all 0's.

I don't have specifics about the circuitry of the end device, nor any diagrams or anything like that, but I have the solution. To solve this problem one standard ~10,000 ohm resistor was installed, that's it. Again I don't know exactly where or how. I want to know how this could work.

My only theory that makes sense to me is that this resistor must have been installed in parallel to the circuit, reducing overall resistance and allowing enough current to flow that the computer could distinguish the incoming modulated current between logical 1's and 0's. Does this sound on track at all?? I'm sorry if this was confusing, but it's a bit of a puzzle and I'm very curious to see if my theory is correct and to learn a bit more about electronics. Thank you. Ask questions if you have any.

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  • \$\begingroup\$ Really no way to know without the specifics. Circuit diagrams, specs, etc. would be necessary to answer this question with anything other than wild guesses. \$\endgroup\$
    – John D
    Dec 15, 2016 at 0:34
  • \$\begingroup\$ Even knowing that only one resistor was added? Can a resistor act so differently based on surrounding pieces? \$\endgroup\$
    – Jacob
    Dec 15, 2016 at 0:35
  • \$\begingroup\$ There are lots of permutations and ways that adding a resistor can affect a circuit. I could make up a situation where adding a resistor would fix a problem like you're having in some theoretical system, but the chances of it being right are slim. \$\endgroup\$
    – John D
    Dec 15, 2016 at 0:43
  • \$\begingroup\$ It sounds like your peripheral device is something like a "4-20 mA" device, but used for a digital signal (or it's actually sending an analog signal but you're using a comparator to interpret it as digital). You can google to get lots of information about how 4-20 mA devices send and receive data. \$\endgroup\$
    – The Photon
    Dec 15, 2016 at 1:26
  • \$\begingroup\$ Sounds like a 4-20mA loop? \$\endgroup\$
    – user253751
    Oct 31, 2017 at 4:14

1 Answer 1

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It's a question of switching threshold.

For simplicity, let's suppose that the supply voltage is 10 V, so that the added 10 kΩ resistor is drawing 1 mA.

The device receiving the data (Rx) is monitoring the current being drawn by the transmitting device (Tx) from the 10 V supply. If this current is above a certain value, say 5 mA, then the Rx tells the computer that the value is "1", and if the current is below that value, the Rx tells the computer that the value is "0".

So, we'll postulate that the older Tx devices drew either 4.5 mA to send a "0" or 5.5 mA to send a "1".

The newer Tx devices, for some reason, are only drawing 3.5 mA for "0" and 4.5 for "1". Since both of these values are less than 5 mA, the Rx device always reports this as "0". But by adding the 1 mA drawn by the resistor, the total current seen by Rx is now 4.5 mA and 5.5 mA, respectively, allowing it to decode the data as intended.

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  • \$\begingroup\$ Thank you very much. All of that makes sense, the only thing that still confuses me then is how the addition of that extra resistance increases current draw, and again I'm sorry if this is very basic. It would have to be added in parallel to have the overall effect of decreasing total circuit resistance yes? Or am I missing some essential piece of knowledge \$\endgroup\$
    – Jacob
    Dec 15, 2016 at 0:54
  • \$\begingroup\$ Yes, the resistor connected in parallel with the Tx device causes an additional 1 mA of current to be drawn from the Rx device. One of the basic rules of electronics is that devices connected in parallel have the same voltage across them, and the currents through them add up. (Conversely, devices connected in series have the same current through them, and the voltages across them add up. The more-general form of these rules is known as Kirchhoff's laws of voltage and current.) \$\endgroup\$
    – Dave Tweed
    Dec 15, 2016 at 0:57
  • \$\begingroup\$ Or the resistor is added to the threshold detection circuit to shift the threshold. Lots of possibilities, but this one is certainly one of them. \$\endgroup\$
    – John D
    Dec 15, 2016 at 1:02
  • \$\begingroup\$ It was definitely added to the end device circuit so that makes sense to me. Thank you for the quick responses \$\endgroup\$
    – Jacob
    Dec 15, 2016 at 1:03

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