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I am building a device that measures current in various ranges. For this I want to switch in different current sense resistors. To do this I see two ways: the resistors in series or the resistors in parallel.

Series:

schematic

simulate this circuit – Schematic created using CircuitLab

Parallel:

schematic

simulate this circuit

They produce slightly different values, but that doesn't really matter all that much, because I have to calibrate it for the actual values anyways.

But I am wondering if there are any inherent advantages or disadvantages of one method over the other for current sense purposes.

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    \$\begingroup\$ Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. \$\endgroup\$
    – Transistor
    Apr 28, 2019 at 21:33
  • \$\begingroup\$ Oh, didn't notice that ;-) I'll make an edit \$\endgroup\$
    – Jens
    Apr 28, 2019 at 21:33
  • \$\begingroup\$ It's just for future reference. Your circuit is very simple. \$\endgroup\$
    – Transistor
    Apr 28, 2019 at 21:36
  • \$\begingroup\$ Fairly simple I think, you just need to find the equivalent resistances and compare them each other. \$\endgroup\$
    – Unknown123
    Apr 28, 2019 at 22:16
  • \$\begingroup\$ Read up on Kelvin sensing. Yes, from THE Lord Kelvin. \$\endgroup\$ Apr 28, 2019 at 23:47

1 Answer 1

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The shunt resistors are big enough to neglect trace resistances, switch contact resistances, etc. Therefore, there is no real need to use Kelvin sensing.

The shunt resistors are small enough to neglect isolation resistances, etc.

So, there is no significant difference between placing the switches in parallel or in series.

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  • \$\begingroup\$ Sounds good! Does anything change given that the cmos switches i want to use have about 100 ohm on resistance? \$\endgroup\$
    – Jens
    Apr 29, 2019 at 7:37
  • \$\begingroup\$ @Jens The \$R_{DS(on)}\$ of analogue switches depends on supply voltage, signal voltage (or: input voltage) and temperature. 100\$ \Omega\$ is only 10% of 1k\$ \Omega\$, so 10% deviation on \$R_{DS(on)}\$ gives a 1% measurement error. Since they are also cheaply available, I would have picked an analogue switch with a lower ON resistance. \$\endgroup\$
    – Huisman
    Apr 29, 2019 at 9:23

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