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BJT Lab help(2)

Discussion in 'General Electronics Discussion' started by go2255, Mar 5, 2014.

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  1. go2255

    go2255

    9
    0
    Oct 28, 2013
    [​IMG]
    This is the result from a lab,but I am not sure the results are correct.

    I guess the combination of R1=10k and R2=10k is under linear region,but how come the calculated Ib is zero.

    For combination of R1=10k and R2=1k,Idon't know why the Ie is negative.Negative value means error?Is it the cut off mode?

    And for combination of R1=1k and R2=10k,it should be in saturation?
    I guess either the measured or calculated result is wrong.What is the problem?

    Also,there are many differences(highlighted).Is my calculation wrong?
     

    Attached Files:

  2. Arouse1973

    Arouse1973 Adam

    5,165
    1,087
    Dec 18, 2013
    Don't assume 0.7V Vbe, it will be less than that. It's about 6V with such small collector current. Vbe will be 0.7V at about 6mA IC. So your base current is about 2uA. The current is not the same in the collector as the emitter. This is where your figures are incorrect. You need to know the gain of the transistor also. But this is a bit woolly and changes with temperature and collector current and Vce. It's best to use the info from the datasheet to get you close and then make actual measurements and adjust if needed.
    Adam
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,448
    2,809
    Jan 21, 2010
    You show measured results of 2 resistors, but not all 4. In addition these measured values seem rather too accurate (or your meter has not very much resolution. I would have measured these with the most accurate multimeter I could find. Do you have access to a 4 1/2 digit meter? Or better?

    The problem with not knowing these values more accurately is that the voltages across them cannot be translated easily into currents.

    Another issue that then arises is that measuring the voltages causes errors! (But you can calculate the effect of this to some extent)

    The calculated Ib will be zero where the accumulated error, or the lack of resolution in your measurements combined with its small magnitude cause it to be lost in the noise or not resolvable.

    You show 2 sets of results. One lot claim to be measures, and you have calculations for the others.

    Did you actually measure everything you have in the table of measurements? If not, don't label them as measurements.

    I note you show a measured Ve in your table. I assume this was measured directly. Why do you use something else to calculate Ie? If all your voltages (and resistances) are positive, you can't end up with a negative current.

    Can this circuit ever place the transistor into either saturation or cutoff? Examine very closely the action of Re. Google it. Look it up in your text. This is *VERY* important.

    You'll need to answer my questions in respect of calculated vs. measured for me to answer these.

    Also, can you see how Vbe changes?

    I'm a little suspicious of your very exact readings for Vb. Are you sure you wrote down *exactly* what the multimeter said? 1.000, 8.50, 5.00 (?) Recording *exactly* what the meter says is *VERY* important. (just as important as knowing the value of the resistors). Again, if I had access to a mete with 4 1/2 digits, or better, that's what I'd use (a high Z input would also be a great benefit)
     
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