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Diode Analysis and Thevenins Theorem

  1. Mar 17, 2017 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations


    3. The attempt at a solution
    6dcdc0497a.jpg
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    The correct answer is 0.25V. I'm getting 0.3335V
     
  2. jcsd
  3. Mar 17, 2017 #2

    gneill

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    Staff: Mentor

    Check your Thevenin voltage derivation.
     
  4. Mar 17, 2017 #3
    ce48b4b3b8.jpg

    I tried it again, same answer.
     
  5. Mar 17, 2017 #4

    gneill

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    Staff: Mentor

    Still incorrect. You're mucking up after the first stage when you apply the voltage division with the 500 Ohm resistor. Take the circuit one stage at a time:
    upload_2017-3-17_21-29-56.png

    First break the circuit at (1) and find the Thevenin equivalent. What do you find?
     
  6. Mar 17, 2017 #5
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  7. Mar 17, 2017 #6

    gneill

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    No. There are three resistors involved and you haven't dealt with the voltage divider. A Thevenin model consists of a voltage source with a series resistance and does not form a closed loop!
     
  8. Mar 17, 2017 #7
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    my apologies, is this fine
     
  9. Mar 17, 2017 #8

    gneill

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    No. How do you arrive at 2 kΩ? How come the original 2 V hasn't been divided by the voltage divider consisting of the two 1 kΩ resistors? This is the circuit that you are starting with for this first part:

    upload_2017-3-17_21-53-34.png

    Apply the procedure to find its Thevenin equivalent.
     
  10. Mar 17, 2017 #9
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  11. Mar 17, 2017 #10

    gneill

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    Right!

    Now tack on the next stage and repeat:

    upload_2017-3-17_22-10-46.png
    What's the Thevenin equivalent for the terminals at (2)?
     
  12. Mar 17, 2017 #11
    Ah. I seem to have gotten it now. Thank you, not entirely sure why my method that I did initially to find Vth is wrong.
     
  13. Mar 17, 2017 #12

    gneill

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    Great.

    You were doing something strange with 1.5 kΩ that I couldn't see the reasoning behind. It's best to either break the circuit up into discrete stages and proceed methodically, or apply something like mesh or nodal analysis when you want to do it all at once.
     
  14. Mar 17, 2017 #13
    9a0db1136a.jpg

    Okay, i've written in detail to what my approach was. I have a feeling that I'm neglecting that extension of A-B at the right and assuming no current flows? But isn't that what I'm meant to do. I mean Vth is the open circuit voltage?
     
  15. Mar 17, 2017 #14
    I think I see my mistake. I'm applying voltage division even though the current through the resistors is different.
     
  16. Mar 17, 2017 #15

    gneill

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    The problem is that you started finding the Thevenin equivalent for the first section of the circuit and did not complete the operation before you applied the partial result to the next section. Your paragraph labelled (2) assumed that the 1 V first Thevenin voltage would be expressed across the first vertical 1 k resistor and hence would be applied to the following section's voltage divider consisting of the 500 Ω resistor and 1k Ω resistor. That was an error. The 1 V should be behind a 1 kΩ series resistor (the Thevenin resistance of the first stage) that includes the 500 Ω resistor.
     
  17. Mar 26, 2017 #16
    Thanks!
     
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