Straight line plot of bipolar transistor - temperature?

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Discussion Overview

The discussion revolves around plotting a straight line for a bipolar junction transistor (BJT) to determine the room temperature, T, using experimental data related to the base-emitter voltage, ##{V}_{BE}##, and collector current, ##I_C##. Participants explore methods for transforming the relevant equation into a linear form suitable for graphing.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest taking the logarithm of both sides of the equation to create a linear relationship, but express uncertainty about how to handle additional terms.
  • Others propose isolating the exponential term before applying logarithms, indicating that this may simplify the transformation process.
  • One participant mentions the importance of measuring the forward voltage drop at two different test currents for temperature calculation.
  • There are discussions about the validity of transforming the equation into a straight line and whether it will work with real data, with some noting that noise could affect the results.
  • A later reply provides a specific transformation for the equation, suggesting a linear function of ##V_{BE}## but not of ##I_C##.
  • Some participants report fitting their data and obtaining temperature estimates, but express confusion regarding the interpretation of the y-intercept in their graphs.
  • There are mentions of different approaches to linearization, including numerical integration, and the effectiveness of these methods may depend on the data used.

Areas of Agreement / Disagreement

Participants express a range of methods for transforming the equation and fitting data, but there is no consensus on the best approach or the implications of the results. Some participants have successfully fitted their data to obtain temperature estimates, while others remain uncertain about the process and interpretation.

Contextual Notes

Participants acknowledge missing variables and assumptions, such as the values of ##α_F## and ##{I}_{EO}##, which may affect the analysis. The discussion also highlights the potential impact of noise on the accuracy of the temperature measurements derived from the data.

Strides
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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
 
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Here's my attached results
 

Attachments

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
 
Strides said:
Here's my attached results
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...
 
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:
 

Attachments

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

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

I would try it with

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

Edit:

Another linearization without logarithm but with numerical integration is

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

with

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

Maybe there are other solutions. It will depend on your data which method works best.
 
Last edited:
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rumborak said:
Is there a reason why you think you can transform it into a straight line?

That's not a problem for the equation. The question is if it works with real data. That will depend on the noise.
 
  • #10
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##.
 
  • #11
<< Mentor Note -- Two threads in different forumjs merged >>
 
  • #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?
 
  • #13
Strides said:
I've just fitted my data and I've got T = 300K from the gradient of the graph

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.
 
  • #14
Strides said:
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
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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