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I have a simple RC circuit of 22 ohms and 3.3uF My measured values put the resistor at 22.10 ohms and the cap at 3.15 which gives me a time constant of 68.5 us. The signal applied is a 100Hz Square wave. Now as you can see from the oscilloscope image where the dotted cursor lines cross is about 63.2% which is 1 time constant however the difference between the solid vertical line and the dotted vertical line is not equal to 1 time constant is is off by A LOT! According to the oscilloscope the 63.2% mark is equal to 540us as opposed to my calculated 68.5us.

Another thing to note is that if I connect another oscilloscope probe to measure the input signal, instead of getting the signal the new channel will just match the RC output. In other words I do not get a square wave. If I measure the square wave alone without RC it measures it just fine. That is why I didn't include the input signal overlayed in the image because it wont be visible -- it will just show the same as the output. So what is going on here? Funny thing is I have done this before and it works out fine but certain cap/resistor values I cannot get the readings I am expecting.

Note schematic says 3.3F but its uF

schematic

simulate this circuit – Schematic created using CircuitLab

So I added a picture of the input versus output wave forms, when I complete the RC circuit my input waveform changes and matches the output? Why? I am sure this is why my measurements are wrong. If I break the circuit the waveform becomes a square wave like it should be.

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Okay, given input impedance, then my new schematic resembles this. And my probe measuring my input signal is essentially in between my circuit and the input impedance.

schematic

simulate this circuit

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  • \$\begingroup\$ The Rigol DG1022 output impedance is 50 Ohms at it's lowest. This impacts your calculation. \$\endgroup\$
    – Jack Creasey
    Sep 10, 2017 at 2:19

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In your calculations you are forgetting the output impedance of your signal generator. While usually it supposed to be 50 Ohms, reverse calculations suggest that it is about 150 Ohm. What do the specifications for DG1022 say about output impedance?

You need to start from measuring the ACTUAL impedance of your driver (DG1022). Start without any capacitor. Set 100 Hz square waveform, and record the amplitude of output WITHOUT ANY LOAD. Then connect any known load, 50 Ohms, and do the same. The difference will allow to calculate the actual output impedance of the generator.

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  • \$\begingroup\$ well i have it set to HIGH Z , but i can change it to 50 Ohms. If you look at the new picture I posted, i think this is the main issue of why my numbers are not adding up \$\endgroup\$
    – Edwin Fairchild
    Sep 10, 2017 at 2:14
  • \$\begingroup\$ @EdwinFairchild: Questions to you, to help you figure it out. 1) Now that you know that your signal source isn't ideal and has an output impedance, how would you draw that impedance on your schematic? 2) Once that impedance is represented, add your second probe to the schematic. \$\endgroup\$
    – jmr
    Sep 10, 2017 at 2:21
  • \$\begingroup\$ very interesting, ive uploaded a new schematic..so if i am seeing this correctly even if i did not have my own resistor and i measure the input signal i would still have an RC..and thus see that type of waveform? I have just removed my resistor and put a short in its place and indeed I have an RC... this is fascinating! \$\endgroup\$
    – Edwin Fairchild
    Sep 10, 2017 at 2:30
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    \$\begingroup\$ @EdwinFairchild - I suspect that setting "high impedance vs 50 ohms" doesn't affect the output impedance, just the interpretation of the voltage setting. When driving 50 ohms the voltage will be half of what it will be driving high impedance. The output impedance will be 50 ohms in both cases. \$\endgroup\$
    – Kevin White
    Sep 10, 2017 at 2:32
  • \$\begingroup\$ yeah it really does not seem to have much of an impact, but even so when I have output directly connected to the capacitor and i calculate the time constant using the output impedance and capacitor the numbers still are way off and not even within an acceptable margin. \$\endgroup\$
    – Edwin Fairchild
    Sep 10, 2017 at 2:49

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