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Amplifier and step function

  1. Apr 23, 2005 #1
    I am given an ideal OP-AMP like this (R-resistor, Cap-capacitor):
    Code (Text):

              ____+Cap-_
             |          |
     ___R____|__|\______|_____ +
    |          -|/
    +         |
    V in      |                  V out
    -         |
    |         |                
    |_________|______________ -

     
    and V in is given as a step function like so:
    Code (Text):

    /\ (volts)
    |
    |____1____2_______> (t, sec)
    |____| (-2)                  
    |
    |
     
    that is V(in) is -2 on [0,1] and then 0 further. I need to find V(out) at 2sec.
    So far I think I got the expression I need to work with.
    V(out) = -Vc
    Vc = 1/(RC) * integral(V in dt)
    But my problem is how I deal with a step function in this case of integration. Any help is appreciated.
    Thanks.
     
    Last edited: Apr 23, 2005
  2. jcsd
  3. Apr 23, 2005 #2
    [tex]Vc = \frac{1}{C} \int_{0}^{t} i_c dt [/tex]

    In your circuit [tex]i_c = V_{in}/R[/tex]

    so your form,

    [tex]Vc = \frac{1}{RC} \int_{0}^{t} V_{in} dt[/tex]

    looks right. So, assuming Vc started at zero at To with Vin = -2V, you have,

    [tex]Vc = \left[\frac{1}{RC} \times -2(t)\right]_{0}^{1}[/tex]

    When you switch from Vin = -2V to Vin = 0V the output will stop ramping because the current goes to zero.

    Does that make sense?
     
  4. Apr 24, 2005 #3
    Thanks for reply.
    So, to find V(out) at t = 2, I need to add integral from 0 to 1s and then from 1 to 2s? That is my question.
     
  5. Apr 24, 2005 #4

    SGT

    User Avatar

    Not really! When the input voltage switchs to zero the capacitor is charged. So it will start discharging towards zero
    [tex]Vc(t) = Vc(1) + \frac{1}{RC} \int_{1}^{t} V_{in} dt[/tex]
    Where [tex]V_{in} = 2.u(t-1)[/tex]
     
  6. Apr 24, 2005 #5
    SGT,

    Ignore the math for a second and check the circuit. When Vin is at zero the current through R is zero. Since there is no current flow at the virtual node (which is always at zero volts - the reference - in this circuit) the voltage across the cap as measured at Vout will hold at whatever voltage is across it.
     
  7. Apr 24, 2005 #6
    EvLer,

    Yes. Evaluating, you will have:


    [tex]Vc = \left[\frac{1}{RC} \times -2(t)\right]_{0}^{1} \plus+ \left[\frac{1}{RC} \times 0(t)\right]_{1}^{2}[/tex]

    You can see that the second term is zero forever with Vin = 0.
     
  8. Apr 25, 2005 #7

    SGT

    User Avatar

    Nope! The capacitor is negatively charged and since there is a path to ground through R, the capacitor will discharge.
     
  9. Apr 25, 2005 #8
    Nope. The voltage at the RC node is always zero (this is a virtual gnd). With Vin at zero, the circuit looks like this:

    Code (Text):

                          _ Vout
                         |
                         | -
             I           C
            <---         | +
    gnd ____ R __________|
                     |
                   Vgnd
     
    I = zero, and no current can flow into or out of a virtual gnd, therefore the cap has no discharge path and holds at whatever the last ramp voltage was.

    Think about it. This is an ideal integrator. Assume R and C are 1 so it sums -Vin x (t).

    t....Vin....Vout
    --------------
    0.....0.......0
    1.....1......-1
    2.....1......-2
    3.....0......-2
    4.....0......-2
    5.....1......-3
    6.....1......-4
    7.....0......-4

    Looks like this:

    Code (Text):

     0 _
    -1  \
    -2   \____
    -3        \
    -4         \_
     
    If it discharged at Vin = 0, it would be useless as an integrator.
     
    Last edited: Apr 25, 2005
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