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Is it possible to use a shunt monitor to measure current for low values(10 mA-100mA) and if is it possible, should I use low resistive values like in most of the power circuit applications? I know that I need low resistance because it shouldn't effect my load(I don't know the value of it but probably in the order of magnitude of M(\$ 10^6 \$)Ω) but from this pdf I learned that for my application I need a resistive value of 2kΩ(200V/100mA[estimated]). I am starting to getting confused because there are also other parameters like tolerance, temperature coefficient and type(wirewound, metal, smd etc.) and I do not know which one should I take into account.

I am sorry if I can not point the question clearly, but I am a bit lost!

Edit: I just noticed that I didn't tell about the application itself. It will be an I-V tracer for samples of SiO2 wafers etc. that can log current data(using the above configuration with an instrumentation amplifier) and voltage data with a daq.

Edit2: Also as Neutronned suggest I also add the pieces of information that can be needed: Voltage range is 0-200V and the input impadance of the sourcemeter is bigger than 10GΩ as datasheet suggests.

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  • \$\begingroup\$ Re: "(I don't know the value of it but probably in the order of magnitude of MΩ)". MΩ means megaohms (millions), i.e. \$10^6\$. Do you mean milliohms mΩ, i.e. \$10^{-3}\$? Given that you estimated your sense resistor as 2kΩ, I assume your load is the former though (MΩ). \$\endgroup\$
    – got trolled too much this week
    Commented Nov 22, 2015 at 23:28
  • \$\begingroup\$ Nope I mean mega not milli, let me edit it the post. \$\endgroup\$
    – user86045
    Commented Nov 22, 2015 at 23:33
  • \$\begingroup\$ That makes sense given the application. I'm guessing you'd be using thin layers. \$\endgroup\$
    – got trolled too much this week
    Commented Nov 22, 2015 at 23:40
  • \$\begingroup\$ There will be other materials on it as well but I do not think they will effect the overall resistance that much. \$\endgroup\$
    – user86045
    Commented Nov 22, 2015 at 23:42
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    \$\begingroup\$ Nope it is a Keithley 2400 series. The resistivity can change for different substances but yes this is what I am doing :) but isn't 1kΩ resistor will dissipate too much heat and can have a higher tolerance? \$\endgroup\$
    – user86045
    Commented Nov 23, 2015 at 0:13

2 Answers 2

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We need a couple of other pieces of information - what is the voltage range and input impedance of the method you'll be using to measure the voltage on the shunt resistor? I'll make a couple of assumptions and use those to calculate something close.

Let's that you're using an analog input on an Arduino (or other micro). These are very high impedance (megohms) and have a typical range of 0 to 5V. It sounds like the maximum current you expect to measure is 100mA, so you'd want the max current to be somewhere close to 5V. Ohm Law V = I x R or R = V / I or 5V / .1A or 50 ohms. So a 50 ohm resistor will develop 5V across it when 100mA is flowing.

If you're sticking with standard resistor values, a 47 ohm would work fine.

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  • \$\begingroup\$ Voltage range is 0-200V and the input impadance of the sourcemeter is bigger than 10GΩ as datasheet suggests. I am also using an instrumentation amplifier before feeding the arduino also. \$\endgroup\$
    – user86045
    Commented Nov 22, 2015 at 23:40
  • \$\begingroup\$ He said he was using a DAQ; those can have even 100Gohms input impedance. \$\endgroup\$
    – got trolled too much this week
    Commented Nov 22, 2015 at 23:47
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    \$\begingroup\$ i will both try a daq card and arduino so it is half true :) \$\endgroup\$
    – user86045
    Commented Nov 22, 2015 at 23:55
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Anyway, Ohm's law states V=IR. V is voltage, I is current through your sense resistor, and R is the resistance of your sense resistor. If we re-arrange it, we get:

R = V/I.

Let us say you want to have 100 mV when I = 10 mA. Plug that in and you will get:

R = 100 mV / 10 mA = 10 Ohms

If your load is 1 Meg, 10 Ohms will not affect it. Now you just need to measure the Voltage across your 10 Ohm sense resistor. If you can arrange to put the sense resistor on the low voltage side of the load, you can sense the voltage with a volt meter.

Just re-arrange Ohms law to understand the relationship between sense voltage and current:

I = V/R (where V is meter voltage and R is value of sense resistor) I = V/10 = 0.1 * V

Addendum The power dissipated by the current sense resistor can be calculated using one of three forumulas: P = I^2 * R P = I * V P = V^2 / R

If the peak current through the resistor is definitely going to be 100 mA, then the peak power will be: P = 100 mA * 100 mA * 10 P = 0.1 * 0.1 * 10 = 0.1W or 100 mW

So you can use a resistor rated for 1/8 W or more. If you change the resistor value, or change your estimate of the maximum current, you should recalculate the power dissipation of the resistor.

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  • \$\begingroup\$ I would recommend that the current sense resistor be a 1%. It is not like 1% resistors are expensive. \$\endgroup\$
    – user57037
    Commented Nov 22, 2015 at 23:52
  • \$\begingroup\$ We simply try to transfer other materials and see if it effects the resistivity of it so 100mA is rather an optimistic guess :) As I understand I should use 20W(200V*100mA) or higher values of resistor. \$\endgroup\$
    – user86045
    Commented Nov 23, 2015 at 0:01
  • \$\begingroup\$ I will update to provide advice about power rating. \$\endgroup\$
    – user57037
    Commented Nov 23, 2015 at 1:22
  • \$\begingroup\$ I think this thread is died, thanks for all the help. When I (hopefuly) solve the problem I will update the page. \$\endgroup\$
    – user86045
    Commented Nov 27, 2015 at 12:31

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