Common Emitter Output characteristics graph

Discussion in 'Homework Help' started by u01sgs4, May 11, 2012.

  1. u01sgs4

    Thread Starter New Member

    May 11, 2012
    3
    0
    Hi, First off let me apologise if this has been covered elsewhere ( the search does not seem to be functioning at the moment).
    The questions i have are in relation to this Question from a past paper for my upcoming electronics exam.

    [​IMG]

    First off i'm a bit embarrassed by this but how does this graph work? do i select Ib in relation to the current output that is required so for this question it would be 3 micro amps?

    So starting from the start. a) am i correct in assuming that this is β=Ic/Ib so using Ic = 1mA and Ib =3μA
    =1mA /3μA
    = 1000

    b) if that's the case would i then calculate the internal resistance (ΔV/ΔI) and divide Ic by it to get the Early Voltage?

    Would be grateful for any help on this?
     
  2. mlog

    Member

    Feb 11, 2012
    276
    36
    Beta is approximately IC/IB. As you can see from the plot, it can vary a little depending on the specific value of VCE.

    In order to calculate the Early Voltage, find the equation of a line of the form y=mx+b. If y is IC and x is VCE, then you can find the Early Voltage by solving the resulting equation for IC=0.

    Graphically you're looking for the intercept of the extended IB line(s). You should find it far out on the left side of the plot, i.e. some negative VCE.

    Let us know what you find.
     
  3. u01sgs4

    Thread Starter New Member

    May 11, 2012
    3
    0
    Cheers for that,
    so picking two sets of co-ordintes of the 3μA line
    at (2,1) and (1,0.95)

    i get y=0.05x+0.9
    when y=0. 0.05x+9=0
    x=-9/0.05
    that ends up being = -18v for my Early Voltage?

    Am i on the right tracks with that?
     
  4. mlog

    Member

    Feb 11, 2012
    276
    36
    You're on the right track, except I got a different value because I used the top line (Ib=6uA). I chose that line because it had better resolution. Here's what I did.

    I said the slope was (2.0 - 1.8) / (5 - 1) = 0.05

    y = mx+b = (0.05)(x)+b

    Solving for b, I chose x=5 and y=2 and got b=1.75.

    y= mx+b = (0.05)(x) + 1.75

    Now when I solve for y=0, I get an Early Voltage of -35 volts.

    I think the secret is in choosing a line with points that gives you good graphical estimates with the least error.
     
  5. u01sgs4

    Thread Starter New Member

    May 11, 2012
    3
    0
    fantastic thank you so much,that's really helped.

    just want to get something straight in my head though. does it not matter which Ib line i choose. for this question my current mirror is producing a 1ma current so i was assuming all calculations would need to be done on the 3μA line since that crosses the 1ma in its stable area.

    Does that factor in at all?
     
  6. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    It should not matter which Ib line you choose when calculating the Early voltage. However, as mlog has indicated, the line Ib = 6uA has a more pronounced slope and so choosing it gives a more realistic estimation of the value of Ve.

    hgmjr
     
  7. crutschow

    Expert

    Mar 14, 2008
    12,991
    3,227
    Your calculator needs new batteries. :rolleyes:
     
  8. mlog

    Member

    Feb 11, 2012
    276
    36
    Or a larger hammer.
     
  9. djsfantasi

    AAC Fanatic!

    Apr 11, 2010
    2,800
    831
    This is correct for very large values of 1 and very small values of 3...
    (old math joke :D)
     
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