I need to use a constant current source 20mA, the current flows through multiple diodes and resistors which are in series. The resistor value can vary from 0 to 1.6k and also a possibility of Open.

Basic Circuit Diagram

As the measuring device can read only upto 10V(Not multimeter, it is a 10V Analog Input Card), and at present 20mA*1.6k=32V+(diode drop). Hence I need to reduce this voltage before passing to the measuring device.

  1. Is there a simple way to do it?(Current Source is a must)
  2. To protect Open Condition of the resistor is a Zener Diode Sufficient?
  • 2
    \$\begingroup\$ The ratio is something like V=IR ... But I am not entirely sure because this is just an Electrical site \$\endgroup\$
    – Josh Jobin
    Dec 17 '15 at 11:11
  • \$\begingroup\$ Ok I get it you are a genius. I have a problem since there are diodes in series to the resistor, so I take the ratio of the Voltage and Current, I manipulate the Output. The constant current source is a must for me. \$\endgroup\$ Dec 17 '15 at 11:44
  • \$\begingroup\$ Are you trying to measure the resistance of R? Why would you want to put a resistor in parallel with R? Use a digital multi-meter to measure the voltage if you are worried that your voltmeter's resistance is going to affect the measurement. \$\endgroup\$
    – Vinod
    Dec 17 '15 at 12:19

To measure the voltage across the resistor under test, you just measure it, like by connecting a voltmeter across it.

With 20 mA thru the resistor under test and 10 V max, the highest resistance you can measure is 500 Ω. Even a crappy voltmeter will have so much higher resistance that it won't distort the measurement. For example, let's say you're measuring a 500 Ω resistor. According to your spec, it has 20 mA going thru it, so 10 V across it. A 1 MΩ (that's really crappy) voltmeter will draw 10 µA, which will distort the measurement by (10 µA)/(20 mA) = 0.05%. Is your voltmeter good to 0.05%? I didn't think so.


You have now shown a schematic. The voltmeter is in the wrong place. Put it directly across the resistor under test.

  • \$\begingroup\$ Hello, I've edited my question. When I read it even I could not Understand it myself. Hope this helps. \$\endgroup\$ Dec 17 '15 at 14:38
  • \$\begingroup\$ Sorry, it is not a voltmeter (but a 10V analog Input Card). And I have to read the voltage from the source side as the resistor would be 160m away from the source. \$\endgroup\$ Dec 17 '15 at 14:56
  • \$\begingroup\$ OMG!!! A simple voltage divider was sufficient to do the job. I was worried about the overvoltage which it would create during open condition, \$\endgroup\$ Dec 17 '15 at 15:24

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