0
\$\begingroup\$

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?
\$\endgroup\$
3
  • 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
2
\$\begingroup\$

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.

Added

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

\$\endgroup\$
3
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.