# Measure Voltage (Divide) on a current source

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. 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?
• The ratio is something like V=IR ... But I am not entirely sure because this is just an Electrical site Dec 17 '15 at 11:11
• 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. Dec 17 '15 at 11:44
• 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. Dec 17 '15 at 12:19