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I Straight line plot of bipolar transistor - temperature?

  1. Nov 23, 2017 #1
    Hey everyone,

    I'm trying to plot a straight line for a bipolar junction transistor to find the room temperature, T, using my experimental results for the associated base-emitter voltage, ##{V}_{BE}## and collector current, ##I_C##. Here's the equation that I'm using:

    $$ I_C = α_F {I}_{EO} [exp(e{V}_{BE} / kT) - 1] + {I}_{CBO} $$


    There is a few things that I'm unsure of though:
    1. I assume that to create the straight line plot, I have to somehow take the logarithm of both sides and then convert it to the straight line equation, y = mx + c, however the issue is I'm not sure how to convert it and then deal with the extra terms.
    2. I believe I have to assume some of the variables that I'm missing such as ##α_F## and ##{I}_{EO}##, I know ##α_F## to be the forward current transfer ratio, which for each I can estimate as 1. However I haven't a clue what to do with ##{I}_{EO}##, which I believe to be the reverse bias saturation current (not completely sure though).

    I'll attach my results in excel, in case anybody wants to have a look.

    Kind Regards,
    Strides
     
  2. jcsd
  3. Nov 23, 2017 #2
    Here's my attached results
     

    Attached Files:

  4. Nov 23, 2017 #3
    Hey everyone,

    I'm trying to form a straight line plot, using the following equation:

    $$ I_C = α_F {I}_{EO} [exp(e{V}_{BE} / kT) - 1] + {I}_{CBO} $$

    My aim is to find the temperature, T, by plotting a graph using the associated base-emitter voltage, ##{V}_{BE}## and collector current, ##I_C##.

    I've tried taking the logaritihims of both sides of the equation, but I'm not getting anywhere. Does anybody have a method I can use create a straight line plot.

    Thanks for all the help
     
  5. Nov 23, 2017 #4

    berkeman

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    Staff: Mentor

    Could you also print out your results as a PDF and upload that file? You can use PrimoPDF or some other free PDF writer to do that. Many people (like me) will be hesitant to open an Excel spreadsheet, which can contain macros and other problematic structures. Thanks.

    Also, which part of the transistor characteristic are you wanting to use for this temperature calculation? One good way to measure temperature using a PN junction is to measure the forward voltage drop at two different test currents...
     
  6. Nov 23, 2017 #5
    I'm measuring the collector current flowing into the transistor, as a function of the voltage at the base, while ensuring that no current flows through the base. I'm hoping that the temperature of the transistor across my results should be around room temperature (but it's most likely not, due to some unforeseen error).

    Here's the results in pdf format:
     

    Attached Files:

  7. Nov 23, 2017 #6

    robphy

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    Gold Member

    Before taking logs, algebraically isolate the exp() term. Then, take the natural log.
     
  8. Nov 23, 2017 #7
    Is there a reason why you think you can transform it into a straight line?
     
  9. Nov 23, 2017 #8
    Assuming you mean

    [itex]I_C = \alpha _F \cdot I_{EO} \cdot \left[ {\exp \left( {\frac{{e \cdot V_{BE} }}{{k \cdot T}}} \right) - 1} \right] + I_{CBO}[/itex]

    I would try it with

    [itex]\ln \left( {\frac{{I_C - I_{CBO} }}{{\alpha _F \cdot I_{EO} }} + 1} \right) = \frac{{e \cdot V_{BE} }}{{k \cdot T}}[/itex]

    Edit:

    Another linearization without logarithm but with numerical integration is

    [itex]\int {z\;d} V_{BE} = \frac{{k \cdot T}}{e} \cdot \left( {z - z_0 } \right)[/itex]

    with

    [itex]z: = \frac{{I_C - I_{CBO} }}{{\alpha _F \cdot I_{EO} }} + 1[/itex]

    Maybe there are other solutions. It will depend on your data which method works best.
     
    Last edited: Nov 23, 2017
  10. Nov 23, 2017 #9
    That's not a problem for the equation. The question is if it works with real data. That will depend on the noise.
     
  11. Nov 23, 2017 #10

    FactChecker

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    It's not clear to me what you want, but
    $$T=\frac{{V_{\mathit{BE}}} e}{k\, \ln{\left( -\frac{{I_{\mathit{CBO}}}}{{I_{\mathit{EO}}} {{\alpha }_F}}+\frac{{I_C}}{{I_{\mathit{EO}}} {{\alpha }_F}}+1\right) }}$$ is a linear function of ##V_{BE}## but not of ##I_C##.
     
  12. Nov 23, 2017 #11

    berkeman

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    Staff: Mentor

    << Mentor Note -- Two threads in different forumjs merged >>
     
  13. Nov 24, 2017 #12
    Hey, thanks for all the help everyone. I've just fitted my data and I've got T = 300K from the gradient of the graph, which matches my theoretical temperature quite nicely. However I'm unsure of what the y-intercept stands for in this scenario, which is around 1000, and how does it fit into my data?
     
  14. Nov 24, 2017 #13
    I got ##T = 296\;K## and ##\alpha _F I_{EO} = 1.95 \cdot 10^{ - 8} \mu A## with a non-linear fit. I also tried to get both parameters with linearisations but that didn't work.
     
  15. Nov 26, 2017 #14

    vanhees71

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    If I understand it right, you want ##I_C## as a function of ##V_{\text{BE}}##. Then you have to make a log-plot of ##I_C-I_{CBO}##. Given data, you should make a linear fit of ##\ln [(I_C-I_{CBO})/I_{EO}]## to get the corresponding constants in the resulting equation, linear in ##V_{BE}##.
     
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