Capacitor and diode under swithcing and constant votage input.

Discussion in 'General Electronics Chat' started by supermankid, May 10, 2014.

  1. supermankid

    Thread Starter Member

    May 26, 2013
    36
    1
    I was trying to figure out how the following circuit worked!!
    I have 5V input that charges the capacitor connected to switching voltage (0-2V).
    I cannot clearly understand why is the voltage at VA shifted up (5-7V)(also switching).
    How is the point VA, charged to 7V.
    One last thing. If I want to do some math, how can I model a diode. (a 0.7V source??)

    DC shift.JPG
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,421
    3,355
    When V2 is at 0V, capacitor C1 gets charged by V1 through D1 to 5V.
    D1 acts as a one-way current switch. Current can flow only in one direction.

    When V2 is raised to 2V, C1 retains its 5V. Hence VA becomes 5V + 2V = 7V.
     
  3. supermankid

    Thread Starter Member

    May 26, 2013
    36
    1
    Mr. Chips. Thank you for your answer. The second part of your answer doesnot ring a bell. What does capacitor retain?? previous voltage(V1)..at which node....and how does V2 effect....on which node.....how do they add up..??
     
  4. MrChips

    Moderator

    Oct 2, 2009
    12,421
    3,355
    Think of the capacitor as a battery or voltage source.
    The voltage across the capacitor is 5V.
    When you raise the bottom end (node Vout) of the capacitor to 2V, the voltage across the capacitor has to remain at 5V because the diode D1 is non-conducting.
     
    supermankid likes this.
  5. supermankid

    Thread Starter Member

    May 26, 2013
    36
    1
    :)
    Many many thanks.....fruitful weekend
     
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