Find Re to get Ie = 0.5mA

Discussion in 'Homework Help' started by Halim, May 11, 2011.

  1. Halim

    Thread Starter New Member

    May 7, 2011
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    Hi. I have to find Re so that Ie becomes equal to 0.5 mA. can anyone help ?
     
  2. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
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    Im looking for next weeks lottery numbers.. Any help?

    ;)

    Sorry, had to.

    You are looking for a resistance to achieve a particular current.. At what voltage?
     
  3. Halim

    Thread Starter New Member

    May 7, 2011
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    Nice one :p. well we have 2 DC voltage sources + if we remove the AC voltage source can't we find that resistor value ?
     
  4. retched

    AAC Fanatic!

    Dec 5, 2009
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    Have you learned ohms law yet?

    Resistance depends on Voltage and Current.

    R=I/V

    You need 2 to find the third.

    A 1 ohm resistor will give you 10A at 10V But 525A and 525V

    Resistors "convert" voltage into current per se.
     
  5. Halim

    Thread Starter New Member

    May 7, 2011
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    How can i find this Re with only these informations ?
     
  6. dcarson7

    New Member

    Apr 28, 2011
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    1
    Looks like a rather odd bias circuit there.
    From what I gather the TR only turns on when
    the AC signal is at a voltage that is 0.6V greater
    than the voltage at the emitter.

    Based on TR operating rules;
    Code ( (Unknown Language)):
    1.    ++
    2.    /
    3. +-{
    4.    \
    5.     -
    or as is here;
    Code ( (Unknown Language)):
    1.     +
    2.     /
    3. - -{
    4.     \
    5.    --
    I'm stumped as to why there are two negative
    DC supplies at -15V each.

    Ie at 0.5mA ?
    Ie includes Ib and Ic.
    In this case Ic also goes through RL.
    So Ie = Ib + Ic + Irl?
    IC is ~ Ie, usually and thus Ib is neglected.
    But Irl, that could be significant.

    Good luck with this one.
     
  7. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Since the two supplies are of equal and opposite polarity and connected in series, they may be replaced with a short.

    This circuit has no DC voltage applied to it. Thus there is no value of Re that will produce any current.
     
  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
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    Perhaps we could cut Halim a break and assume their orientation of the supply connections is incorrectly drawn. Also forget about any missing coupling capacitors & so forth.

    The emitter leg supply is -15V.

    Presumably Halim just wants the steady state (quiescent) bias conditions - so we could assume the AC signal input is 0V for the moment.

    The steady-state (quiescent) emitter current would be found from

    15-Vbe=IeRe+Ie\frac{Rb}{(1+\beta)}

    or

    Ie=\frac{(15-Vbe)}{Re+\frac{Rb}{(1+\beta)}}

    If Rb, Vbe, β are known and Ie=0.5mA one would re-arrange the equation to solve for Re. Which might be a useful exercise for Halim.
     
    Last edited: May 12, 2011
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  9. Halim

    Thread Starter New Member

    May 7, 2011
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    Thanks all but t_n_k gave the closest answer :p
     
  10. nbw

    Member

    May 8, 2011
    36
    10
    I'm still keen to find out the winning lottery numbers for the coming week... anyone? :)
     
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