Calculating time & amplitude

Discussion in 'Homework Help' started by colbygatte, Sep 5, 2011.

  1. colbygatte

    Thread Starter New Member

    Sep 5, 2011
    6
    0
    I am trying to find time & amplitude in a circuit with a diode and resistor. The teacher hasn't gone over very much, and everyone else in my class is confused right now.

    I'm not sure where to start, but I think the current is 9.8 mA?
     
    Last edited: Sep 5, 2011
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    When Vin is greater than 0.6V, the diode conducts and produces a voltage drop of 0.6V, as you point in your sheet. Consequently, in the output, the result will be an identical voltage as the input, but reduced by 0.6V.

    Anything less than 0.6V in the input, including negative voltages, will cause the diode to be cut off, and the input will be isolated by the output. As a result, no current will flow in the resistor and no voltage will be developped on it.

    Is that clear?
     
  3. colbygatte

    Thread Starter New Member

    Sep 5, 2011
    6
    0
    but what about the 20 ohm resistance in the diode? won't that cause a voltage drop?
     
  4. colbygatte

    Thread Starter New Member

    Sep 5, 2011
    6
    0
    I have that current is 9.8 mA, Vo is 9.2 volts, ω is 90 degrees, and
    T=\frac{ 2\pi }{ 90^{ \circ }\left( \frac{ \pi }{ 180^{\circ} }\right)} = 4
     
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    A value of 4 for T must be wrong, as you can see if you look at the original graph. The total cycle time is 4, so you should be looking for the half-cycle time, or more accurately the half-cycle less a couple of little angles to allow for the voltage being 0.6V minimum.

    What are the small angles? If the voltage reaches 10V in 1s, it must reach 0.6V in how long?

    The peak output amplitude is given by the original amplitude, less the 0.6V drop, then reduced by the potential division ratio given by 20Ω and 1kΩ.

    Remember: Vout = Vin*R2/(R1+R2)

    Edit: You appear to have been asked for A in terms of voltage. In any case the load current is less than 9.8mA
     
  6. colbygatte

    Thread Starter New Member

    Sep 5, 2011
    6
    0
    Okay, for current, I used an equation from the book:
    V_{in} = \frac{V_{in}-V_{D}}{R}
    and got 9.2 mA.

    Does this make more sense?
    Using your equation for voltage, I still get 9.2 volts for Vo, taking into consideration the .6 drop across the diode.

    I'm still confused on T, the period is less than 2, but how would I show that numerically?
     
    Last edited: Sep 6, 2011
  7. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Correct me if I 'm wrong, but typically isn't the period of the output signal exactly 4, as before? The time quantity that is a bit less that 2 is the diode conduction time, isn't it?
     
  8. colbygatte

    Thread Starter New Member

    Sep 5, 2011
    6
    0
    Using 2 linear equations, y=10x and y=-10x+20, I found that the diode period is 1.88.
     
  9. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    Actually the system is
    y=10x and
    y=0.6

    but your answer is correct, diode conduction time is 1.88sec.
     
Loading...