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Confusion between cap charge and discharge

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Discussion in 'Electronics Homework Help' started by Gebus, Nov 6, 2015.

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  1. Nov 6, 2015 #1
    Gebus

    Gebus

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    Nov 6, 2015
    Hi guys, first post so be gentle!

    I am struggling to answer the Art of Electronics (3rd Ed.) exercise 2.2. I have found the answer given in the text is wrong and should be 76us (found an errata page on the web). But that's not the point...

    What we have is a cap charged at 4.4V, suddenly having it's positive side switched to 0V. This means it suddenly has a -4.4V charge as seen on its other pin. Now it looks like a 10n charging through a 10k from 5V (tau=RC=100us) and the circuit output will switch when it reaches 0.6V (it's tied to an npn base). Simplified circuit below:
    WP_20151106_16_57_26_Pro.jpg

    My immediate thought was "easy, I'll just use the capacitor charge equation". It turns out this is wrong. I did manage to find the right answer, but it was through guesswork and as I don't understand it it is pointless.

    To get the right answer I had to use the capacitor DIScharge equation V=V0.ln(e^-t/RC). To confuse things even more I used an initial voltage of 9.4V and a final voltage of 4.4V (the initial and final voltage across the 10k). I feel like I am borderline understanding something but I just can't seem to make any progress!

    Thanks for any help.
    G
     
  2. Nov 7, 2015 #2
    GPG

    GPG

    452
    66
    Sep 18, 2015
    The total charging voltage is 5 + 4.4v (initial voltage across R)
    Taking the capacitors -4.4 as reference, the total change is 4.4 + .6V =5V
     
  3. Nov 7, 2015 #3
    Gebus

    Gebus

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    0
    Nov 6, 2015
    Thanks CPG, but I still don't get it. I think you intuitively understand this where I am missing something...

    Here's how I have managed to get the answer:
    Current must flow through the resistor to charge the cap, with the initial current being (5+4.4)/R = 9.4/10k = 0.94mA. The final current would be (5-0.6)/R = 4.4/10k = 0.44mA.

    The cap charging equation (for current) is
    final current = initial current * e^(-t/RC)

    Re-arrange for t
    t = -RC.ln(initial current / final current)

    t = -10n.10k.ln(0.94/0.44) = 76us which is correct.

    That's fine and all, but my initial assumption to use the voltage charge equation
    final voltage = initial voltage ( 1 - e[-t/RC] )

    charging from -4.4 to +0.6V was wrong, and I don't understand why. What am I missing?
     
  4. Nov 7, 2015 #4
    Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    First-Order-Transient.png
    Any first-order transient is going to be an exponential curve. But getting the equation for the curve takes a series of considerations, as follows:

    1. Find the time constant, τ, from the circuit.
    2. Determine whether the voltage (or current) will be rising or falling, then use the charge curve (red) or discharge curve(blue) as appropriate.
    3. Determine the total voltage (or current) change, V_T, from examining the circuit behavior.
    4. Determine the Zero-Level, i.e. initial value for charge/RED or final value for discharge/BLUE, and add that to the equation for V(t).

    Applying this procedure to your example gives:
    1. Time constant=100 usec
    2. Rising - use the charge curve (1-exp(-t/τ))
    3. Voltage changes from -4.4 to +5.0, so V_T=9.4
    4. Charge/RED initial value = -4.4

    So the equation for the capacitor voltage is V(t)= -4.4 + 9.4(1-exp(-t/τ))

    The question is to find t when V(t)=0.6 --> 0.6=(-4.4)+(9.4)(1-exp(-t/0.0001))

    1-5.0/9.4 = exp(-t/0.0001)
    ln(1-0.5319) = -t/0.0001
    0.0001(ln(0.4681)) = -t
    t = 75.91 usec
     
  5. Nov 8, 2015 #5
    GPG

    GPG

    452
    66
    Sep 18, 2015
    Similar:
    5/9.4 = .5319
    .5319 - 1 = -.468
    .468 = e^x
    ln.468 = x = -.759
    .759 * RC = 75,9 µS
     
    Last edited: Nov 8, 2015
  6. Nov 8, 2015 #6
    Gebus

    Gebus

    4
    0
    Nov 6, 2015
    Thanks,

    Can't say I fully understand it but this recipe approach does get the correct answer. Was my way of doing it using the current correct also, or did I get the correct answer by accident?
     
  7. Nov 8, 2015 #7
    Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    I would not call it a 'recipe' - it is just a recognition that all exponential curves are governed by the time constant, the total change, and the zero value. You just have to be smart enough to pick those values out of the circuit to form the equation. The capacitor current is also an exponential curve.

    Applying this procedure to the current gives:
    1. Time constant=100 usec
    2. FALLING - use the discharge curve (exp(-t/τ))
    3. Current changes from 0.94 mA to 0, so I_T=0.94
    4. Discharge/BLUE final value = 0

    So the equation for the capacitor current is I(t)= 0.94(exp(-t/τ))

    The question is to find t when I(t)=0.44 mA --> 0.44=0.94(exp(-t/0.0001))

    0.44/0.94 = exp(-t/0.0001)
    ln(0.4681) = -t/0.0001
    0.0001(ln(0.4681)) = -t
    t = 75.91 usec
     
    Last edited: Nov 8, 2015
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