# RC Circut, strange behavior

Discussion in 'General Electronics Chat' started by jjtjp, Apr 16, 2014.

1. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
So in my lab last week we were supposed to make an rc circuit and plot the current and voltage over time. As we all know, this is an exponential decay function and current should follow
$I(t) = I_{0}e^{-t/\tau}$.
In this lab, a time constant was chosen to be so large so that the students could observe the change over time. In our case, we had a time constant
$\tau=RC=20k\Omega \times 2200\mu F= 44s$.
Yet when plotting the circuit, the first plot followed the formula while the subsequent points were linear. I've attached a graph of our data points as well as the circuit used. I am really just curious what physical phenomenon could explain why our data was as is. We tried the experiment several times with different capacitors and all followed the same trend. Thanks!

Sorry for all the edits. My internet connection was causing troubles.

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• ###### RCCapture.JPG
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Last edited: Apr 16, 2014
2. ### JoseJ28 New Member

Nov 10, 2013
2
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First you should verify that the RC value is not 44s but 0.44s

3. ### JoseJ28 New Member

Nov 10, 2013
2
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I did a quick graph and its working using RC= 0.44s

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4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,976
617
Ok. I am confused now. You calculation uses 22 uF. Your schematic shows 2200 uF. One of them is right. Which one?

5. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
Sorry that should be 2200μF as in the schematic. I cannot edit my original post for some reason. The Time constant is, in fact 44 seconds.

Last edited: Apr 16, 2014
6. ### ifixit Distinguished Member

Nov 20, 2008
647
114
You have a 35 Volt source, so what is the 2200uF cap voltage limit? Large electrolytic caps have some leakage current as well. This leakage current may reduce as you leave the voltage on the cap, but the voltage source must always be less than the cap voltage rating.

Ifixit

7. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
The capacitor was rated at 35V. The lab called for a 50V rated capacitor but there were none in stock. Could that discrepancy cause such drastic issues?

8. ### #12 Expert

Nov 30, 2010
17,899
9,319
Stable at more than 1 ma? That's within the realm of an old capacitor.
A new cap is guaranteed to .01 CV after 2 minutes of stabilizing.
.01 x 2200E-6 x 35V = .00077A
That's 770 ua for a good aluminum electrolytic AFTER you stabilize it.

• ###### e-he-28896.pdf
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9. ### ifixit Distinguished Member

Nov 20, 2008
647
114
Your graph shows a fairly flat current going somewhere at 1.4 to 1.2 mA over approximately 5 seconds. This is obviously the capacitor drawing the current since there is nothing else in the circuit. If you leave the cap on the voltage for a day or so, it should form up to some minimum low value of leakage current. Leakage current is equivalent to having a resistor in parallel with the cap.

The supply is 35 V and the cap rating is 35 V so this is okay, but the cap may not have been charged to its max in many years and will have to "get used to" that value again. This behaviour is typical of electrolytic caps.

You can investigate online and look up the spec of the cap you have. How old is the cap?

Ifixit

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10. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
This capacitor was found in an obscure place in a back room by the lab manager. So it could possibly be very old. We tried with two from the same bin, but the results were very similar, although not identical. That graph is actually the average of the two trials. Thanks all for the input, although I'm not sure I understand what AnalogKid was saying about a voltage drop. This is a graph a charging capacitor. It was initially shorted and time began when the short was removed.

11. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
One more quick question, if you will entertain it...
I was always under the impression that the leakage current through a capacitor was highest at higher stored charges, but from what you are saying, it is apparent at low charge and improves (lessens) as potential difference increases. Is this a correct understanding of capacitors? Also, do you believe that if I had fully charged the capacitor once or twice with voltage directly applied to its terminals before conducting my experiment that I would have had more typical results?

12. ### ifixit Distinguished Member

Nov 20, 2008
647
114
If you "form" the cap first before using it in an application then the leakage will be minimum so long as nothing else is wrong with it. To form it, you apply full working voltage to it and let it sit like that for awhile until the leakage current is no longer reducing. At that point, it is as low as it will get.

All caps have leakage current. For some caps leakage is so low it can be safely ignored, but in your case it can not. Leakage goes up as the voltage goes up and leakage values are typically higher for higher values of capacitance. The type of dielectric also makes a big difference, as does temperature.

However, if your resistor was 35Ω then you could ignore the leakage of 1 measly mA and call the charge complete since the current is not going to get much better. The resultant charge curve would look much like the text book, but the TC would only be 77 mS.

There are low leakage tantalum caps but I don't think the values go as high as 2200uF. You could put ten of 220uF in parallel though.

Having fun learning new things?

Ifixit

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13. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
Yes I'm having a blast learning new things. Thank you!! I only wish my instructor could have provided me with some of the great answers I've received here. Thanks again!

14. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,074
338
It could be that the second time constant is just too low to measure accurately. Try increasing the initial voltage. The curve becomes asymptotic to the horizontal axis pretty quickly.