Saturday, 17 October 2015

Circuit with adjustable source and a diode


I'm struggling a bit with diode circuits.


schematic


simulate this circuit – Schematic created using CircuitLab


The source is adjustable.



  1. Which value the source has to be to the Diode start conducting?


  2. What's the value of entrance's current and the diode's current when the source is 10V?


Any explanation on why/how the diode will start conducting at certain value will help!



Answer




  • if the left side is 2/3 of V+ and the right side is 1/3 of V+


  • then the diode drop is now = Vf = (2/3 - 1/3) * V+





  • So what V+ starts conduction for Vf=0.6V?




@arthurg et al... Next question.




  • When diode conducts what happens to over all current from source.




  • How much does the network resistance change? from what to what?





Answer:



  • Using parallel (//) equivalents below and above threshold

  • Diode OFF , R_tot= (1+2) // (2+1) = 3//3 = 1.5 [kΩ]

  • then assume Diode Resistance much less or << 1 kΩ ( like a short circuit )

  • Diode ON , R_tot= (1//2) + (2//1)= 2/3 + 2/3 = 4/3 [kΩ]


So impedance change is 1.5 to 1.33 [kΩ] for V+ > 1.8V



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