Measuring capacitance using oscilloscope. - Page 1
Forum section: electronics / beginners

author: shteii01
author_rank: Frequent Contributor
timestamp: January 29, 2017, 10:38:06 pm
content: Someone earlier linked this youtube video:I thought it was simple enough that even I could do it. Plus I got a bunch of capacitors from ebay and I wanted to check some of the values.Oscilloscope: HP 1740A (100 MHz analog)Square signal: this scope has calibration square wave signal of 1 volt at 1.4 kHzCircuit: 1 kOhm resistor in series with unknown capacitor (22 pF capacitor according to the markings)Procedure:I connected one leg of 1 kOhm resistor to the calibration stud on the 'scope. The other leg of the resistor was inserted into the breadboard. One leg of capacitor was inserted into breadboard to form connection with the resistor. The other leg was connected to the ground stud on the 'scope. I verified that the calibration feature of the scope is working and working correctly by attaching HP 10006D (10:1) probe to the calibration stud and attaching probe ground alligator clip to the ground stud. I saw nice square wave at correct frequency. I then attached the probe to the junction of the resistor and capacitor. I observed the same square wave.What did I do wrong?P.S.: My time constant is going to be 1ee3*22ee-12=22ee-9, 22 nanoseconds. Too fast for this 'scope?

author: tggzzz
author_rank: Super Contributor
timestamp: January 29, 2017, 11:16:08 pm
content: Look at your probe specification, and then draw the compete schematic. Don't forget wires are 0.8nH/mm, and who knows what the breadboard's characteristics are.What was the time based of the scope, relative to the 22ns time constant? What is the risetime of the Cal out?

author: Paul Rose
author_rank: Regular Contributor
timestamp: January 29, 2017, 11:23:44 pm
content: With a 1.4khz cal output, you are going to have something like a period of 714 microseconds. To fit a whole period on the screen your horizontal timebase will be something like 100 microseconds per division.With a time constant of 22 nanoseconds, you will not be able to see the difference.A higher frequency square wave would be better.You could raise the resistor to 1 meg, but then your probe will start loading it down.

author: f5r5e5d
author_rank: Frequent Contributor
timestamp: January 30, 2017, 12:17:18 am
content: the picture shows split time base display, you need that or just change the main time base to see just the rising edgeand the comments about parasitics are on target - read some classics on probing fast signals

author: David Hess
author_rank: Super Contributor
timestamp: January 30, 2017, 02:47:23 pm
content: The time constant is not too fast for a 100 MHz oscilloscope which has a transition time of 3.5 ns but the calibrator output is probably not fast enough and the x10 probe capacitance of about 15 pF in parallel with the 22 pF capacitor will add considerable error although this could be calibrated out.

author: SingedFingers
author_rank: Frequent Contributor
timestamp: January 30, 2017, 03:32:14 pm
content: Job for a Bob Pease active probe and a 74AC pulse generator!

author: David Hess
author_rank: Super Contributor
timestamp: January 30, 2017, 03:53:50 pm
content: Quote from: SingedFingers on January 30, 2017, 03:32:14 pmThat would do it.The probe's input capacitance combined with any test fixture capacitance if used can be measured separately and then the formula for parallel capacitance used to find the value of the test capacitor. Then just a fast pulse generator is needed but practically any logic family is fast enough.While measuring the RC time constant is better for manual measurements, if I have automatic measurements available then I usually use a sine wave source and adjust the frequency until the phase lag through the RC circuit is 45 degrees. Then the capacitive reactance is equal to the resistance. Besides most DSOs, some universal frequency counters can measure phase.