Illumination

Discussion in 'General Electronics Chat' started by Shmurk, Apr 28, 2009.

  1. Shmurk

    Thread Starter Member

    Aug 13, 2007
    24
    0
    It's neither a question nor an answer, but something I wanted to share:

    I had an illumination today. I'm a newbie and I have a hard time understanding all this electronics thing, but I understood something today: the voltage divider is not some "magical" circuit form as I thought before, it's just a series circuit where you add two wires, one between two or more resistors to divide, and one for the ground.

    [​IMG]

    Well, I'm relieved, I'm not that stupid after all... :D
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Well done Schmurk - now you could try to synthesize (on paper at least) a voltage divider circuit with a given specification. Say you need to create a voltage divider that has input of 9V and an output of 1.5V. Could you specify the resistor ratios needed to do this? Are there standard resistor values that you could choose?
     
  3. mbohuntr

    Active Member

    Apr 6, 2009
    413
    32
    I still get that feelin after 4 years of studyin... This stuff is amazing! Keep going, more fun to come.
     
  4. Shmurk

    Thread Starter Member

    Aug 13, 2007
    24
    0
    [​IMG]
    Problem 1:
    Code ( (Unknown Language)):
    1. Vout = Vin * R2 / (R1+R2)
    2. 1.5 = 9 * R2 / (R1+R2)
    3. (R1+R2) / R2 = 6
    4. R1/R2 + R2/R2 = 6
    5.  
    6. R1/R2 = 5
    but, according to Wikipedia (http://en.wikipedia.org/wiki/Preferred_number), I can't do that with resistor values in the E series... Of course, we still can use resistors in series, for example:
    • R1 = 10 * "15 Ohms"
    • R2 = 3 * "10 Ohms"
    overkill? yes, maybe I could use some resistors in parallel. I'll try that later...

    Problem 2:
    I just thought of another problem to add: let's say I want 1.5V in output of my voltage divider, but with a precise value "Iout" of current going in the branch.

    We keep R1/R2 = 5
    Code ( (Unknown Language)):
    1. Ohm's law: Vin = (R1+R2) I
    2. I = 9 / (R1+R2)
    3.  
    4. let I2 be the current going through R2:
    5. I = Iout + I2
    6. Ohm's law: Vout = R2 * I2
    7. I2 = Vout / R2 = 1.5 / R2
    8.  
    9. Iout = I - I2 = 9 / (R1+R2) - 1.5 / R2
    10. Iout = (9*R2 - 1.5*(R1+R2)) / (R2 * (R1+R2))
    again, is this result good? useful? I don't really know, it's just an idea I had this time.

    And again, thank you all for the support :)

    Edit: My second problem is stupid, I guess. I know Vout, I just need to put a precise resistor value to adjust the current going through the branch, right? Or so I guess...
     
    Last edited: Apr 28, 2009
  5. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    2,039
    287
    My first illumination was when I discovered how to use imaginary numbers. I've been having "mini-illuminations" pretty much every year since then. :)

    eric
     
  6. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    Well done, Shmurk!
     
  7. Shmurk

    Thread Starter Member

    Aug 13, 2007
    24
    0
    OK, I won't disturb you anymore with my illuminations, but I had a last one today and wanted to share it with all the newbies who struggle like me :D

    I was reading the chapter 1 of the book (again...) and wondered "how can a voltmeter measure anything in an opened circuit because by definition, electrons are nowhere to be seen?" Of course, electrons "come" from the battery, and they are pushed one way (or another in conventional notation :p ) if the circuit is closed (or so I thought...)
    [​IMG]
    And there I was struck (by a lightning, or maybe it was the sugar in my soda...) : Electrons are already there, everywhere! They are just not moving!
    [​IMG]
    That's what current is all about: it's not the electrons, it's the flow of those stupid things! And that explains how a voltmeter can sense the tension between two pieces of wires in an open circuit. Wow, my ego is ten feet tall now...
     
  8. mbohuntr

    Active Member

    Apr 6, 2009
    413
    32
    Here's a tip for you. In a series circuit, current must flow from the neg. to the pos., Each resistor will get all of the current. Total current is found by V/R1+R2+R3 etc. or V/(R total). Then I(total current)/ Individual resistor = Voltage across that resistor.
    Parallel is different, just because current has different paths to reach V+ . you still have to find R-total. R-total = (all resistors in parallel) 1/ 1/ r1 +1/r2 +1/r3 etc. Use a calculator to find the inverse of each resistor, add them together, then take the inverse again. (the total value will be less than the smallest resistor in parallel.) V/R-total = total current. Each parallel resistor branch in the circuit gets full voltage, but partial current. :cool: (theres a couple of calculators on the net...)
     
    Last edited: Apr 29, 2009
  9. PRS

    Well-Known Member

    Aug 24, 2008
    989
    35
    I'm impressed with you Shmurk. You think! LOL. Here's something else to think about. Electrons are moving at the speed of light, right? But that is not the case between terminals on your battery. They are actually colliding and going all over the place but with a drift from positive to negative terminals (I'm a Franklinite). The drift is called drift current and it is much slower than the speed of light.
     
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