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Capacitor Lab - Charging/Discharing Paths

  1. Feb 17, 2007 #1
    I performed a lab this past week that went over charging/discharing time constants, and I was asked the following about the circuit below:

    http://synthdriven.com/images/deletable/EEN204-Lab3.jpg [Broken]

    Calculate the charing time constant, (Tau_c), and the discharging time constant, (Tau_d). I was asked to them compare these results with those that I measured in the lab, for three instances. (C=0.1uF)

    Instance #1)

    I measured:
    Tau_c = 52.76us (<-us=microseconds)
    Tau_d = 16 us

    Instance #2)

    I measured:
    Tau_c = 35.18us
    Tau_d = 35.18us

    Instance #3)

    I measured:
    Tau_c = 24us
    Tau_d = 48us

    As far as the both the time constants went, I came up with the following formulas:

    And these are my calculations:
    Instance 1:
    Tau_c = 52us
    Tau_d = 47us

    Instance 2:
    Tau_c = 105us
    Tau_d = 47us

    Instance 3:
    Tau_c = 52us
    Tau_d = 100us

    These results from the calculations are very different from what I measured in the lab. So something must be incorrect. We measured the time constants by setting up the circuit as shown (I'm 99% certain that the circuit we set up was correct), and measuring the voltage via an oscilloscope hooked up to a computer... That way, we were able to freeze the curve and perform precise calculations with the cursor on the oscilloscope. We were asked to reproduce the curves by hand to turn in with our reports. And by looking at those graphs now, the values the computer gave us for the time constants look to be correct... By the behavior of the curve on the graphs anyway.

    So I'm writing because I figure that the error is in the formula I put together about the charging/discharging paths. I must be calculating the wrong thing because I probably set the equation up incorrectly.

    Could someone tell me if the equations I derived for Tau_c and Tau_d are correct according to the figure?

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 17, 2007 #2


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    Your charging time constant looks wrong.

    As the voltage across the capacitor increases, an increasing amount of current will flow through R2 instead of charging the capacitor, so the charging rate will be slower than if R2 was not in the circuit. Therefore, the correct formula will have R2 in it as well as (R1+RS).

    The discharging time constant is OK, because the diode blocks any current through Rs and R1
  4. Feb 18, 2007 #3
    My thought concurs with AlephZero and I believe the expression for the charging time constant is ((R1+Rs)||R2)C.
  5. Feb 18, 2007 #4


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    Maybe I should have just gone with "different" in my previous post, not "slower" :rolleyes:

    You are right. If you removed the diode, the discharge time constant would obviously be C(R2||(R1+Rs)). The charging time constant must be the same as that, because the diode doesn't affect the charging time constant.
    Last edited: Feb 18, 2007
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