Problem drawing U(out) function

Discussion in 'Homework Help' started by Tobias Hildebrandt, Jan 24, 2016.

  1. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
    44
    0
    Capture43.PNG
    Hello,

    The task for all three exam questions is to draw U(out) into the diagram. (N O C A L C U L A T I O N S ! ! !)

    The Prof insisted on not calculating anything. So where do I start? Is there a recipe how to tackle such a problem?
     
  2. dannyf

    Well-Known Member

    Sep 13, 2015
    1,835
    367
    All you need to know is that diode opens at 0.7v. The answers are fairly simple.

    Or LTSpice, :)
     
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,963
    1,098
    For such a simple circuit you indeed do not need to do any calculations. But you need to understand how individual component works.
    For example for the first circuit ask yourself a question: what voltage at Vout node will turn-ON the D1. Then do the same for D1 diode.
    Sometime you may need to do some "side note" calculations if circuit is requires it.
     
  4. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
    44
    0
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,963
    1,098
    OMG you even do not understand how diode works. And you are wrongly interpreting the simulation results.

    Diode pass current only in one direction (from anode to cathode) if voltage at anode is 0.7V large then the voltage at cathode.
    111.PNG

    And if diode is ON the voltage drop across it is almost constant and equal to about 0.7V. And this is why you get +0.7V at point 1.
    But for Vin < 0.7V diode will be OFF. This means that we can remove the diode from the circuit. If so, for Vin<0.7V diode is OFF and the voltage at point 1 is equal to the Vin voltage. And this is why you get -10V max.
     
  6. dannyf

    Well-Known Member

    Sep 13, 2015
    1,835
    367
    I think the class you are taking is too much ahead of you, unfortunately.
     
  7. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
    44
    0
    Look, I know that I am struggling with electronics, but this is why I am here and ask questions. I graduated from the Gymnasium in 1995 and I started going to University two years ago. In the almost twenty years between school and university I never had to deal with electronics...
    When I picked this degree program I was under the impression, that the focus would be on renewables and not on the electronics part.

    Anyway, to cut a long story short. Thanks for your help. You explanation and me looking at the graph again cleared up my dilemma.
     
  8. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
    44
    0
    @danny sure as hell looks like it, but I have to pass this class on Thursday. I passed all the difficult classes like Math II (Differential Calculus for Multivariate Functions;Infinite Series, Taylor series and Fourier series,....) and I passed Physics II (Thermodynamics and Waves). These are the two most difficult courses for most of the students, but for me, it seems that ,electronics is a problem.
    This being said, I looked at all the old exams and he typically asks one question about low/high pass filters and emitter circuits. We can bring old exam papers, actually any kind of papers to the exam, so I think there is at least a chance that I pass.
     
  9. dannyf

    Well-Known Member

    Sep 13, 2015
    1,835
    367
    If your goal is to pass exams, you have wasted your time / money on an education.

    What you have shown so far is your willingness to answer questions, not your ability to learn. Others, including this forum, can help you but sooner or later you will have to stand on your own. Without a solid understanding of why, that isn't going to happen.

    with that said, I would pick the first question to help you.

    Diodes are essentially directional switches: when the voltage differential is greater than its threshold (0.7v for silicon diodes), it starts to conduct (=switch closed).

    In the first chart, that means if Vout is greater than V1 + 0.7v, it will stay at V1 + 0.7v.

    Conversely, if Vout is less than Vgnd - 0.7v = -0.7v, it will stay at Vgnd-0.7v=-0.7v.

    For anything in between the diodes are not closed and Vout = V2 as there is no current flow through R2.
     
Loading...