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Homework Help: Voltage drop across voltage divider

  1. Feb 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A voltage divider with two resistors R1 and R2 is given. The output is connected to a transistor so VB is always equal to the base emitter voltage=0.7v.

    2. Relevant equations

    How can i explain physicaly (wath happens to the electron flow and such) why when R1 is verry high, VB is lower then 0.7V. I know all the math and theory behind that, but i want to be able to understand that on the verry basic scale. I mean like when you explain what heat is and you say that heat is the friction of the atoms caused by lets say a passing electron flow.
    3. The attempt at a solution
  2. jcsd
  3. Feb 16, 2016 #2
    good question. not obvious how tor reconcile physical effects [observations, measurements] with theory. Einstein was GREAT at that; he had visionary physical insights.

    You can think of the current flow in two parts.

    First, electrons bump along, one against the other in a continuous line. Individually they move rather slowly. See drift current in resistors and diffusion current for semiconductors.

    You can think of conduction electrons in a circuit bumping along akin to cars entering and leaving highway entrances and exits: when one is pushed free by an electrical potential, an emf, others are pushed also very quickly but less than the speed of light. When one electron enters a resistor, another exits almost immediately. They cannot build up in a resistor.

    q=it and i = E/R so q = Et/R so what happens when voltage E is increased....more charge [q] flows...electrons get pushed along faster! This assumes R is constant and that's an approximation since resistances can get hot...like on an electric stove top. What happens when more times passes at a fixed voltage...same thing, if your source is a battery, chemical changes deplete the battery.

    Conductors have loosely bound outer electrons, resistors have more tightly bound electrons, and in insulators electrons are REALLY tightly connected to their individual atoms. High resistances have conduction electrons more tightly bound than low resistances.
    Try here for additional detail: https://en.wikipedia.org/wiki/Valence_and_conduction_bands#Electrical_conductivity
  4. Feb 16, 2016 #3


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    It's sometimes worth taking things to extremes.... What happens if R1 is infinite eg removed from the circuit?
  5. Feb 17, 2016 #4
    Oops, sorry i write that twice, but i am new here and clicked the wrong thing the first time:-D. Thank you for youre answer, it realy helped me out :-)
  6. Feb 17, 2016 #5
    Good question, CWatters! Apperantly all resistors are in nature still conduct. In order to conduct, they must have loosely held electrons. As a resistance grows, it gets harder and harder for these electrons to break, untill atlast, in nonconductive materials there are no free charges. Current just cant flow through such material couse in order to flow, there must be free charges.
    At first, all voltage drops across R1, then you connect R2 and the voltage drops to 0,7v after R1. If R1 resistance is increased, VB falls beneath 0,7v and the transistor is in cut off mode. But what really is voltage drop?- in resistors the electron's velocity is reduced. That electron applies a repulsion force on the one behind him and a chain reaction occurs. If the current through R2 is I2, then that current applies the same repulsion force to the base current and maybe that slows it enough so that it cant leave the pn junction.. But do i realy need to go that deeply with the physics? (i study for engineer)
  7. Feb 17, 2016 #6
    Rereading this, I realize I should have also mentioned that with more energy, more emf, more electrons are ripped free from their atom, although it is mentioned in the second reference I posted:
    "As such, the electrical conductivity of a solid depends on its capability to flow electrons from valence band to conduction band."
    See also the comments there about " In solids, the ability of electrons to act as charge carriers depends on availability of vacant electronic states..."

    So when the basic math of electrical circuits was formulated, mostly likely people did NOT understand the underlying mechanisms, especially in semiconductors which not yet been invented. That's common in science where progress is made over time as theory, discoveries and scientific apparatus advances.

    You might also think about why resistors get hot.

    That's a good tip, especially when you are not sure about the incremental direction a change might make. Assuming infinite resistance [open circuit] or zero resistance [short circuit] usually points you in the right direction.....similar for mechanical situations, even relativity.
  8. Feb 17, 2016 #7
    Oooh, i think i get it now:-D. Not so long ago i searched the internet for explanation on how charhed particles get their charge.. only to find later that this wasnt yet explained by science. We live in a such era, that we use basic laws (Ohms law for exam) that were invented long ago, and something in them we understand both theoreticaly and physicaly, but something we still dont have answer for, but we continue using it, couse we know that the law tells us its correct.
  9. Feb 17, 2016 #8
    If you read about the Standard Model of Particle Physics, which contains our accumulated knowledge of all sub atomic particles you might even be discouraged: Despite all the great accumulated scientific knowledge that has been amassed, despite our ability to model and predict so many things, nobody knows just what the fundamental particles 'really are'. Some think that's not the role of physics; I think it is, and hopefully someday we'll understand 'why' the electron, for example, has the charge and mass we measure.
  10. Feb 17, 2016 #9


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    Perhaps you should post a circuit because I don't think this is correct.

    That is correct.

    A bipolar transistor is a device that manipulates current - so in general it is best to think about the base current rather than the base voltage.
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