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Cpacitor discharge

Discussion in 'Electronic Design' started by Joe G \(Home\), Feb 11, 2007.

  1. Hi All,

    I am hope some can confirm statements below.

    For a capacitor

    Q = C * V = I * t

    I am trying to work out how long an electrolytic will hold up the Vdd on
    circuit.

    nominal Vdd = 3.3v and lowest Vdd = 2.4v

    If the current drawn by the circuit is constant 10mA = I

    C * V = I * t

    Thus t (time) = (C * V) / I Where V = 3.3v - 2.4v =
    0.9 v

    Hence for a 10,000uF the time the Vdd rail is held up is

    t(time) = (10,000 E-6 * 0.9) / 0.01 = 900mS


    QUESTIONS

    Q1
    Are my statements above correct?

    Q2
    Should I use the natural e discharge model V = 3.3v * (1 - e(t))
    style of equation?
    any ideas on this equation would be appreciated.


    Joe
     
  2. I think they are.
    Not if the current is constant. That form applies to a
    capacitor discharged by a resistor, so that as the voltage
    falls, so does the current.
    Another handy form of the capacitor equation that relates
    voltage and current is I=C*(dv/dt). The current is
    proportional to the time rate of change of voltage. If the
    current is not a constant (where the C * V = I * T solution
    fits) or proportional to the total capacitor voltage
    (where the exponential solution fits) this completely
    general form may allow a solution.
    Note that there are non electrolytic capacitors for low
    voltages that have capacitances very much higher than 10,000
    uF for low current supply hold up applications. e.g.:
    http://www.elna-america.com/products/dynacap/dynacap_frame.htm
     
  3. mpm

    mpm Guest

    Should I use the natural e  discharge model      V = 3.3v  * (1 - e(t))
    Yes. For a constant current charge or drain, the charge/discharge
    rates will vary with instantaneous voltage. See equations here: (Do I
    don't have to type them in....)

    http://www.rowan.edu/colleges/lasold/physicsandastronomy/LabManual/labs/Capacitor.pdf

    I actually just researched this exact same question two weeks ago...
    Keep in mind electrolytics can easily vary +/- 20% or more in value,
    and also age.
    This will very likely affect your timing calculations.

    Also, I happened upon a neat little online simulator where you could
    (for free) "virtually" build your circuit online and run some simple
    simulations. This might help you verify your assumptions / math.

    Visit: http://www.falstad.com/circuit/index.html

    -mpm
     
  4. Eeyore

    Eeyore Guest

    Not if the load is indeed constant current.

    Graham
     
  5. Eeyore

    Eeyore Guest

    Er NO !

    Graham
     
  6. mpm

    mpm Guest

    Er NO !
    Did I screw this up??
    Maybe I didn't understand his question.

    Take a cap, charge it, disconnect it and let is sit there static.
    Now, let's say it's charged to 3.3 votls.

    He wants to run a uPC (or some such) that will operate from 3.3 Volts
    down to about 2.4 Volts.
    He wants to know how long it will take for the cap to reach 2.4 Volts.

    And he's going to draw a constant 10mA current.
    (And the resistance of the uPC is not changing.)

    Therefore, the discharge function of the capacitor must be a function
    of the instantaneous voltage on the capacitor. As the capacitor
    drains, less voltage is applied to the load, so the current through
    the load is likewise reduced.
     
  7. Tom Bruhns

    Tom Bruhns Guest


    As stated, the current is constant, not dependent on voltage. In that
    case, the capacitor voltage will be a linear ramp. However, if indeed
    he's running some cmos circuit whose power supply current is
    practically all used to charge gate capacitances, the current will
    almost certainly be even more strongly dependent on power supply
    voltage than it would if the load were a resistor.

    Cheers,
    Tom
     
  8. Eeyore

    Eeyore Guest

    You said -A. " he's going to draw a constant 10mA current "
    then B. " As the capacitor drains, less voltage is applied to the load, so the
    current through
    the load is likewise reduced. "

    Either it's constant current or not.

    The exponential discharge is with a resistive load. Most ICs tend to demand a
    near constant current regardless of supply voltage btw.

    Graham
     
  9. HKJ

    HKJ Guest

  10. Thanks all who replied.

    My assumption is the load will draw a constant 10mA.

    My load is a microcontrolller circuit.... and I have made a big assumption -
    constant 10mA.

    I will have to either build or model the microcontroller circuit to see
    f - over the range 3.3v to 2.4v the current is constant.

    Regards
    Joe
     
  11. Sounds correct.
    No, because you discharge wit ha _constant_ current, not in a resistor.
     
  12. John Fields

    John Fields Guest

     
  13. Eeyore

    Eeyore Guest

    Introducing a strawman is so totally predictable for you.

    Do you have any other tricks you'd like to perform for us ?

    Graham
     
  14. Fred Bloggs

    Fred Bloggs Guest

    Right, if it's a CMOS digital then I=Cp*F*Vdd where Cp is the so-called
    equivalent power dissipation capacitance. This makes CpF the load
    conductance or an equivalent discharge resistance of Re=1/(CpF). So the
    hold up time is derived from 2.4=3.3*exp(-t/CRe) or t=ReC*Ln(3.3/2.4) or
    t=C*Ln(3.3/2.4)/CpF, which means it will hold up longer than the
    constant current drain model, so he should stay with his original method
    if he doesn't mind a bit oversized component.
     
  15. Tom Bruhns

    Tom Bruhns Guest

    If it's op amps, the current probably will be very nearly constant,
    assuming the signal levels stay the same. If it's the input to a
    regulator, the output current will presumably be constant so long as
    the input doesn't drop below what's required to hold up the output
    voltage. But if it's a whole mess of cmos gates operating directly
    off the supply at a constant frequency, the current certainly will
    depend on the supply voltage. HOWEVER, the OP DID STATE that the
    current is (assumed) constant.

    Assuming the current is constant will adequetely cover the case where
    current decreases with voltage, of course, and unless space is very
    tight or the production volumes are huge, it's cheap insurance to just
    assume the current is constant. But there's one case where you could
    get burned: if that input voltage goes to a switching regulated
    supply, the current will INCREASE as the input voltage drops. You can
    get fancy and look at the efficiency of the particular supply as a
    function of input voltage for the particular output, or just assume
    constant input power and probably be OK.

    Cheers,
    Tom
     
  16. Andrew Edge

    Andrew Edge Guest

    The only way way you can get a current out of a capacitor is by using
    a resistor.It may have a small value but a resistor it is. All
    currents and voltages in an RC circuit
    have the form
    i= io*exp -t/Req*C
    v=voexp-t/Req*C

    Andy
     
  17. (snip)

    You might want to rethink that generality a bit. The
    universe is more complicated than that.
     
  18. No way, 'current' equals electrons.
    There are many ways other then 'resistors'.
    Look up vacuum tube, transistor, semiconductor, and in this context
    especially 'current source'.

    To make it a bit more clear, in a _resistor_ the current decreases linear
    with the increase in resistance.

    In something that allows control of the current, say a transistor, or tube,
    the current does _not_ rise proportionally to the applied voltage.

    In a (junction) transistor the Ic becomes greatly independent above some minimum
    voltage, and then only depends on Ib.
    Same for a 'penthode' tube.
    Many integrated circuits use current sources and current mirrors, and as such
    Isupply may depend little on Usupply.

    This is one for [sci].electronics.basics.
     
  19. Andrew Edge

    Andrew Edge Guest

    Ouch ... That one has a damn large resistance ... Infinity!
    ..


    You said it.

    Andy
     
  20. Andrew Edge

    Andrew Edge Guest


    It sure is. I am always willing to learn ... so give me an example.

    Andy
     
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