Sunday, November 21, 2010

Transient Circuits lab discussion and related final practice problem

For your last lab, which is the week after Thanksgiving,  everyone will do the "Transient Circuit Analysis" lab*. (Please ignore that it says not offered in fall 2010 in your lab book.) Please do the "Transient Circuits" prelab!!!

This lab is more aligned with what we have been emphasizing in class and will allow you to see the time-dependent behavior of circuits. It is very important to notice and understand that the duration of each square wave is much longer than the characteristic time scale associated with your circuits in this lab. Thus each square wave functions as essentially a DC voltage; the beginning of each square wave "on-cycle" is essentially equivalent to the closing of a switch and the application of a DC voltage. You will look at the time-dependent response after the switch is closed.

*In this context, transient circuit analysis means analysis of the time-dependent response of the circuit to a sudden change in voltage, which we have discussed quite a bit in class and on HW.

Please feel free to post your comments and questions here.

added Friday, Dec 3
Here is a  new practice problem:

P.S. What part of the problem does not really make sense after you replace the capacitor by an inductor?  Why?
Posted by at

9 comments:

  1. UnknownDecember 2, 2010 at 11:45 AM

    Zach,
    I did well on the midterm but feel a little overwhelmed for the amount of info on the final. You had posted some problems for the midterm that were especially important and that helped a lot. Is there any way you could be somewhat specific about what concepts to really study and what problems are good indicators of understanding the concepts? Thanks!

    ReplyDelete
  2. Zack SchlesingerDecember 2, 2010 at 12:19 PM

    yes. i will get to work on that. Thanks.

    ReplyDelete
  3. Kelly A StanleyDecember 3, 2010 at 10:53 PM

    This comment has been removed by the author.

    ReplyDelete
  4. Kelly A StanleyDecember 4, 2010 at 5:10 PM

    For part b are you asking us to graph Va-Vd once gate 2 is closed? So there would be no battery source and current would decrease. So this is not the same as in our lab because the voltage is added back again correct? So it should be an exponentially decreasing graph?

    ReplyDelete
  5. Zack SchlesingerDecember 4, 2010 at 7:10 PM

    but when the gate is closed there is a path between a and d with R=0. That implies that Va=Vd whenever the gate #2 is closed, doesn't it?

    ReplyDelete
  6. Kelly A StanleyDecember 4, 2010 at 8:44 PM

    so you are saying that there is no voltage change at all between Va-Vd? They should both have a voltage of zero?

    ReplyDelete
  7. Zack SchlesingerDecember 4, 2010 at 9:18 PM

    when gate#2 is closed: yes. definitely. Va-Vd = 0.

    ReplyDelete
  8. UnknownDecember 5, 2010 at 10:49 AM

    For part a with the inductor, we don't need to calculate Q(t), right?

    ReplyDelete
  9. Zack SchlesingerDecember 5, 2010 at 10:51 AM

    good call.

    ReplyDelete

Subscribe to: Post Comments (Atom)