Spatial Matrix

Discussion in 'General Electronics Chat' started by T.Jackson, Nov 22, 2011.

  1. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    • SM.zip
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  2. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
    328
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    Then you do this one ...

    :: OHM's Law Quiz ::

    Requirements:

    • 6 resistors ...
    • 5 specific Vrefs of: 1, 2.2, 3, 3.3, and 4.5VDC
    • IT = 20mA
    • Calculate value of R1 - R6
    Code ( (Unknown Language)):
    1.  
    2. [FONT=Courier New]              4.5V      3.3V      3V        2.2V      1V
    3.                |         |         |         |         |
    4.   5V o--/\/\/--+--/\/\/--+--/\/\/--+--/\/\/--+--/\/\/--+--/\/\/--o GND
    5.           R1        R2        R3        R4        R5        R6[/FONT]
    6.  
    Show all workings, natural voltages measurable from ground.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Is there a trick?

    R6=1/20mA=50Ω
    R5=(2.2-1)/20mA=60Ω
    R4=(3-2.2)/20mA=40Ω

    and so on ....

    R1=(5-4.5)/20mA=25Ω

    I'm reluctant to open a zip file for fear of viruses.
     
  4. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    That's wrong.

    No need to fear of virus. I don't do that.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Many thanks for the interesting challenge.

    If you mean my answer to your question is wrong then you might be kind enough to help me correct my errors.
     
  6. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    Let me get a calculator and type it all out.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
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    No, no calculators!

    You must use a slide rule.
     
  8. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    :: Problem Solution ::

    Code ( (Unknown Language)):
    1.  
    2.  
    3. [FONT=Courier New]Step 1.  Calculate RT …
    4.  
    5.             RT = V / IT
    6.                = 5 / 0.02
    7.                = 250Ω
    8.  
    9. Step 2. Calculate R1 – R5
    10.  
    11.            Using Formula: R = (V - Vref) / V x RT
    12.  
    13. Therefore ...
    14.  
    15.              R1 = V – Vref1
    16.                 = 5 - 4.5
    17.                 = 0.5 / V
    18.                 = 0.1 x RT
    19.                 = 25Ω
    20.  
    21.              R2 = Vref1 – Vref2
    22.                 = 4.5 - 3.3
    23.                 = 1.2 / V
    24.                 = 0.34 x RT
    25.                 = 60Ω
    26.  
    27.              R3 = Vref2 – Vref3
    28.                 = 4.5 - 3.0
    29.                 = 0.3 / V
    30.                 = 0.06 x RT
    31.                 = 15Ω
    32.  
    33.              R4 = Vref3 – Vref4
    34.                 = 3.0 - 2.2
    35.                 = 0.8 / V
    36.                 = 0.16 x RT
    37.                 = 40Ω
    38.  
    39.              R5 = Vref4 – Vref5
    40.                 = 2.2 - 1.0
    41.                 = 1.2 / V
    42.                 = 0.24 x RT
    43.                 = 60Ω
    44.  
    45.   Step 3. Calculate Trimming R
    46.  
    47.              R6 = RT –  (R1 + R2 + R3 + R4 + R5)
    48.                 = 250 – (25 + 60 + 15 + 40 + 60)
    49.                 = 50Ω[/FONT]
    50.  
     
  9. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    I think that's right. Been a while.
     
  10. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Many thanks for your work.

    I notice I got the same answers as you did despite being "wrong".
     
  11. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    Arh see when I just see numbers with no working I automatically assume it to be insanity and totally wrong.

    You need to explain how you arrive at the answer.
     
  12. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I understand. I'll keep that in mind. Cheers!
     
  13. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    How would you do it if the Vref was required across each resistor instead?
     
  14. Wendy

    Moderator

    Mar 24, 2008
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    Smell the smoke?
     
  15. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    Depends on how much current was needed at each voltage point.

    i.e. set I=1mA and you have a lot higher resistor values, set I=20 A and you have very low resistor values (and a lot of heat).
     
  16. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
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    He showed his work/formulas. The steps between could be done mentally instead of a calculator after making a few hundred voltage dividers for other circuits.
     
  17. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    Well he's extremely gifted if he can do all of that mentally. I sure as hell cannot.
     
  18. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    I actually remember it all as a 'protocol' (set of step-by-step rules) -- that I was formally taught.

    1. Decide on IT
    2. Find RT
    3. R = (the voltage that you want / the voltage that you have) x RT

    I have memorized many things such as resistor colour codes in the form of sequential colours.

    When you study, you write down on a piece of paper many times and you never forget.
     
    Last edited: Nov 23, 2011
  19. steveb

    Senior Member

    Jul 3, 2008
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    Actually, t_n_k is gifted at seeing easy solutions to complex problems. We've seen that here before. This problem might not be all that complex, but still a person might not see the simple approach right away. Often we take the hard way, out of habit.

    In general, problems have easy solutions and difficult solutions, so it is important to not just dive straight into a problem and do it by the standard methods. Often a little thinking reveals a viewpoint that makes the whole problem seem trivial.

    A computer program that was developed to solve very general resistor network problems, might not be able to find t_n_k's simplifying viewpoint on this problem, and it might write out a bunch of simultaneous equations and then solve them in a "brute-force" way. I'd be willing to bet that if students were given this problem on a test, some would dive straight in with simultaneous equations and waste 10 minutes that could be spent on the other problems. The calm guy (or girl) would take time to look carefully and then see the easy way.
     
  20. T.Jackson

    Thread Starter New Member

    Nov 22, 2011
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    It is hard to teach an old dog new tricks boss.
     
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