Hello all, quick question. I'm experimenting with simple RC circuits in hopes to eventually make a simple timer. I have connected the following 5vdc -> 1MΩ -> +electrolytic cap(1uf) -electrolytic cap -> ground(5vdc return).

Just a simple RC circuit. My issue is the display on the scope, it appears that the capacitor only charges to approximately 2.5 volts. I know it will never charge to a full 5v but after 5 time constants(5 seconds in this case) it should approach 4.7 volts. Is the "loading" effect of the oscilloscope causing this issue? If so is there a way to correct it? My DMM reads the correct voltage.

Not sure if this is the best way to test the scope, but I connected the 5v directly and it reads perfectly

Thanks for the help ahead of time.

 

Your scope probably has a 1Meg input impedance. with a 1:1 probe, it will divide the input signal by 2, as you are seeing.

 

Thank you for the quick reply. That makes perfect sense, considering when I put the scope in series with my supply the entire voltage is dropped across the 1M input impedance correct?

Any link to the math behind that? For example what if on another scope the input impedance was 2Meg or a different type of probe etc etc.

Also is there anyway to offset this? I suppose I can multiply by 2 as I take readings but it's easy to forget.

 

Forgive me, just read the "operating conditions" section of the manual. "...the supplied 10x probe offers a high input impedance that minimizes circuit loading."

Tried this it reads perfectly, thanks for not telling me to RTFM right away

 

No worries, glad you have it figured out. As for the math, its a simple (now that you see it!) voltage divider with 2 equal resistances - the resistor in the circuit in series with the resistance of your scope - measured at the midpoint. So half the voltage is dropped across each R. Check out the DC circuits textbook here at AAC for more info.
Have fun.