# How to set up signal generator in lab to obtain an output voltage?

In one of my laboratory assignments, I am to study how to work with square signals in RC circuits (low pass filter). Please keep in mind this is an introductory Electronics course. Basically, I am thinking of using a 1k$\Omega$ resistot and a 6.8 nF capacitor. My square signal has an amplitude of 5 voltage peak-to-peak. Then, I am asked to start by setting the frequency of the generator such the output voltage as measured accros the capacitor is 63% of the input voltage.

This seems fairly easy: I need the Vpp of the output signal to be $0.63*5V = 3.15 V$, so then I just move the frequency button on the generator until I get this value on the screen, right? And the time constant would be half the period since a capacitor charges to 63% in one time constant, right? Another part asks whether we can measure the peak to peak voltage of the output. To me, this is confusing since the capacitor has not enough time to completely charge and discharge, so wouldn’t the peak to peak voltage change continuously? It would just make sense to define these limits for one period.

I appreciate any help I can get. I know these are basic questions but I have to be at the lab tomorrow and I want to be prepared to work as well as possible. Thank you.

• Sure you’re not supposed to find the 70.7% output? 63% seems like an odd value. – Blair Fonville Apr 11 '18 at 23:47
• "the capacitor has not enough time to completely charge and discharge, so wouldn’t the peak to peak voltage change continuously?" - no, not continuously. after a few cycles the peak values will stabilize. But don't take my word for it, do the experiment and find out for yourself! – Bruce Abbott Apr 12 '18 at 0:16
• @Blair: See the opening paragraphs of en.wikipedia.org/wiki/RC_time_constant. 63% is correct. You may be mixing it up with the RMS value relative to peak voltage of a sinewave (which this is not). – Transistor Apr 12 '18 at 5:49
• @Transistor No, I was assuming he was to find the -3 dB point of the filter: 20*log(0.707) – Blair Fonville Apr 12 '18 at 5:57
• I think this part of the experiment is to show that at 63% output the half-wave duration will be equal to the time constant. Don't forget that s/he's working with a squarewave and not a sine so the -3 dB point will have a more complex relationship to the raw squarewave. – Transistor Apr 12 '18 at 6:29

To me, this is confusing since the capacitor has not enough time to completely charge and discharge, so wouldn’t the peak to peak voltage change continuously? It would just make sense to define these limits for one period.

Figure 1. RC charge curve. Source: InterfaceBus.

The bit of insight you might be missing is that you can start anywhere on the curve and the time taken to get from there to 63% of the way to V-max will be the time constant.

In the graph of Figure 1, for example, if we start at 63% we have 37% more to go to be fully charged. 63% of that = 37 x 0.63 = 23.3% so at the end of the second time constant the charge will have risen to 63 + 23.3 = 86.3% as shown on the graph. You can pick any starting point and the voltage will have risen 63% of the way from there to V-max after one time constant.

• Thank you. I did not realize this and I am thankful you pointed it out. Then, in my experiment, so I just have to set the period to two time constants? – Bee Apr 13 '18 at 6:13
• While my explanation of the RC curve is right I'm don't think that I'm right about your experiment. See Blair's comments to your question. I messed around with a simulation last night and got the same result as his calculations but haven't had time to follow through. Please un-accept for a few days to encourage other answers. – Transistor Apr 13 '18 at 6:17