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Data sheet for transistors.
Data sheet for transistors.


[[Media:2N3904.pdf]][[Media:2N3906.pdf]]
[[Media:2N3904.pdf]]   [[Media:2N3906.pdf]]


[[File:2N3904_PinOuts.png | 150 px]][[File:2N3906_PinOuts.png | 150 px]]
[[File:2N3904_PinOuts.png | 150 px]][[File:2N3906_PinOuts.png | 150 px]]
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=Transistor circuit=
=Transistor circuit=
== Identify the type (n-p-n or p-n-p) of transistor you are using and fill in the following specifications. ==
'''1.) Identify the type (n-p-n or p-n-p) of transistor you are using and fill in the following specifications. '''




I am going to use n-p-n transistor '''2N3904'''. Below are some specifications from data shits for this type of transistor:
I am going to use '''n-p-n''' transistor '''2N3904'''. Below are some specifications from data shits for this type of transistor:


{| border="1" cellspacing="0"  cellpadding="10" style="text-align: center"
{| border="1" cellspacing="0"  cellpadding="10" style="text-align: center"
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|}
|}


== Construct the circuit below according to the type of transistor you have. ==
 
<br><br><br><br><br><br>
'''2.) Construct the circuit below according to the type of transistor you have. '''


[[File:TF_EIM_Lab13a_Circuit.png | 300 px]][[File:TF_EIM_Lab13_Circuit.png | 300 px]]
[[File:TF_EIM_Lab13a_Circuit.png | 300 px]][[File:TF_EIM_Lab13_Circuit.png | 300 px]]
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== Measure the emitter current <math>I_E</math> for several values of <math>V_{CE}</math> by changing <math>V_{CC}</math> such that the base current <math>I_B = 2 \mu</math> A is  constant. <math>I_B \approx \frac{V_{BB}-V_{BE}}{R_B}</math> ==
'''3.) Measure the emitter current <math>I_E</math> for several values of <math>V_{CE}</math> by changing <math>V_{CC}</math> such that the base current <math>I_B = 2 \mu</math> A is  constant. <math>I_B \approx \frac{V_{BB}-V_{BE}}{R_B}</math> '''
 
 
I used:
<math>R_1 = (199.5 \pm 1.0)\ k\Omega </math>
<math>R_1 = (198.7 \pm 1.0)\ k\Omega </math>
<math>R_1 = (100.0 \pm 1.0)\ k\Omega </math>
<math>R_B = (R_1 + R_2 + R_3) = (498.2 \pm 1.7)\ k\Omega </math>
and
<math>R_E = (100.0 \pm 1.0)\ \Omega </math>
 
 
Below is the table with my measurements:
 
[[File:Table 2uA 01 corrected.png | 700 px]]
 
 
And below is my currents and power calculation:
 
Here:
 
<math>I_{E} = \frac{V_E}{R_E}</math>
<math>I_{B} \approx \frac{V_{BB}-V_{BE}}{R_B}</math>
<math>P_{max} = I_C \cdot V_{CE} = (I_E - I_B) \cdot V_{CE} </math>
 
and I have used approximation <math>V_{E} \approx o</math>
 
 
[[File:Table 2uA 02 corrected.png | 700 px]]
 
 
'''4a.) Repeat the previous measurements for <math>I_B \approx 5\ \mu A</math>.  Remember to keep <math>I_CV_{CE} < P_{max}</math> so the transistor doesn't burn out'''
 
 
I used:
<math>R_B = (199.5 \pm 1.0)\ k\Omega </math>
and
<math>R_E = (100.0 \pm 1.0)\ \Omega </math>
 
 
Below is the table with my measurements:
 
[[File:Table 5uA 01 corrected.png | 700 px]]
 
 
And below is my currents and power calculation:
 
Here:
 
<math>I_{E} = \frac{V_E}{R_E}</math>
<math>I_{B} \approx \frac{V_{BB}-V_{BE}}{R_B}</math>
<math>P_{max} = I_C \cdot V_{CE} = (I_E - I_B) \cdot V_{CE} </math>
 
and I have used approximation <math>V_{E} \approx o</math>
 
 
[[File:Table 5uA 02 corrected.png | 700 px]]
 
 
 
'''4a.) Repeat the previous measurements for <math>I_B \approx\ 10 \mu A</math>.  Remember to keep <math>I_CV_{CE} < P_{max}</math> so the transistor doesn't burn out'''
 
 
 
I used:
<math>R_B = (100.0 \pm 1.0)\ k\Omega </math>
and
<math>R_E = (100.0 \pm 1.0)\ \Omega </math>
 
 
Below is the table with my measurements:
 
[[File:Table 10uA 01 corrected.png | 700 px]]
 
 
And below is my currents and power calculation:
 
Here:
 
<math>I_{E} = \frac{V_E}{R_E}</math>
<math>I_{B} \approx \frac{V_{BB}-V_{BE}}{R_B}</math>
<math>P_{max} = I_C \cdot V_{CE} = (I_E - I_B) \cdot V_{CE} </math>
 
and I have used approximation <math>V_{E} \approx o</math>
 


[[File:Table 10uA 02 corrected.png | 700 px]]




{| border="1" cellspacing="0"  cellpadding="10" style="text-align: center"
!scope="row" |<math>V_{CC}</math>
!scope="row" |<math>V_{B}</math>
!scope="row" |<math>V_{BB}</math>
!scope="row" |<math>V_{EC}</math>
!scope="row" |<math>V_{E}</math>
!scope="row" |<math>R_{E}</math>
!scope="row" |<math>R_{B}</math>
!scope="row" |<math>I_{E} = \frac{V_E}{R_E}</math>
!scope="row" |<math>I_{B} = \frac{V_{BB}-V_B}{R_B}</math>
!scope="row" |<math>P_{max} = \frac{V_{BB}-V_B}{R_B}</math>
|-
|mV || mV || V || mV || mV || <math>\Omega</math> || k<math>\Omega</math>|| mA|| <math>\mu A</math>
|-
|<math>41.5\pm 0.5</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>0.0\pm 1</math>||<math>40\pm 2</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||0.4±0.02 ||2.0±0.18
|-
|<math>106.7\pm 0.5</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>4.0\pm 1</math>||<math>100\pm 5</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||0.4±0.02 ||2.0±0.18
|-
|<math>142.0\pm 0.5</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>10.0\pm 1</math>||<math>140\pm 5</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>170.8\pm 0.5</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>16.0\pm 1</math>||<math>170\pm 5</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>204.9\pm 0.5</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>22.0\pm 1</math>||<math>200\pm 5</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>233.0\pm 0.5</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>26.0\pm 1</math>||<math>240\pm 10</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>266.2\pm 0.5</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>28.0\pm 1</math>||<math>260\pm 10</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>296.1\pm 0.5</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>29.0\pm 1</math>||<math>300\pm 10</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>338.0\pm 0.5</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>29.0\pm 1</math>||<math>340\pm 10</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>406.0\pm 2.0</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>29.0\pm 1</math>||<math>400\pm 10</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>554.0\pm 2.0</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>29.0\pm 1</math>||<math>560\pm 20</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>809.0\pm 2.0</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>30.0\pm 1</math>||<math>800\pm 20</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|-
|<math>1041.0\pm 2.0</math>||<math>600\pm 50</math>||<math>1.6\pm 0.05</math>||<math>30.0\pm 1</math>||<math>1000\pm 50</math>||<math>100\pm 0.5</math>||<math>494.7\pm 0.5</math>||1.4±0.05 ||2.0±0.18
|}


== Repeat the previous measurements for <math>I_B \approx 5 \mbox{ and } 10 \mu</math> A.  Remember to keep <math>I_CV_{CE} < P_{max}</math> so the transistor doesn't burn out==


{| border="1"  |cellpadding="20" cellspacing="0 ==
'''5.) Graph <math>I_C</math> -vs- <math>V_{CE}</math> for each value of <math>I_B</math> and <math>V_{CC}</math> above. (40 pnts)'''
|-
 
|V_{CC} || V_B || V_{BB} || V_ {EC} || V_ E || R_E || R_B || I_E || I_B
Bellow is my plot for the case of <math>I_B = 2 \mu A</math>
|-
   
|mV || mV || V || mV || mV || <math>\Omega</math> || k<math>\Omega</math>|| mA|| \muA
[[File:plot 2uA.png | 1000 px ]]
|-
 
| || || || || || || || ||
<br><br><br><br><br><br><br><br>
Bellow is my plot for the case of <math>I_B = 5 \mu A</math>
 
[[File:plot 5uA.png | 1000 px ]]
 
<br><br><br><br>
Bellow is my plot for the case of <math>I_B = 10 \mu A</math>
 
[[File:plot 10uA.png | 1000 px ]]
 
 
 
 
'''6.) Overlay points from the transistor's data sheet on the graph in part 5.).(10 pnts)'''
 
I can not really do it because there are not good points to compare from data sheet. I can take for example this data from data sheet:
 
[[File:Table sheet 01.png | 800 px]]
 
And from the table above I can only extract the range of <math>I_B</math> like
 
1) <math>I_C = 0.1\ mA,\ V_{CE} = 1.0\ V \rightarrow I_B = \frac{I_C}{\beta} = \frac{0.1}{40..300} = (2.5 - 0.33)\ \mu A</math>
 
2) <math>I_C = 1.0\ mA,\ V_{CE} = 1.0\ V \rightarrow I_B = \frac{I_C}{\beta} = \frac{0.1}{70..300} = (14.2 - 3.3)\ \mu A</math>
 
And my measurements looks like:
 
  1) <math>I_B = 2\ \mu A,\ I_C = 0.3\ mA,\ V_{CE} = 1.0\ V \rightarrow \beta = \frac{I_C}{I_B} = \frac{0.3\ mA}{2\ uA} = 150</math>


|}
2) <math>I_B = 5\ \mu A,\ I_C = 0.72\ mA,\ V_{CE} = 1.0\ V \rightarrow \beta = \frac{I_C}{I_B} = \frac{0.72\ mA}{5\ uA} = 144</math>


3) <math>I_B = 10\ \mu A,\ I_C = 1.40\ mA,\ V_{CE} = 1.0\ V \rightarrow \beta = \frac{I_C}{I_B} = \frac{1.40\ mA}{10\ uA} = 140</math>


First point to note that <math>I_C</math> from my measurements and from data sheet are different so I can not really overlay them and compare.
Second, I have specific values of <math>I_B</math>, otherwise from data sheet I have the range of <math>I_B</math>.
And third, I can't really plot this range of <math>I_B</math> from data sheet on my plots just because my plot doesn't have the <math>I_B</math> axis.


5.) Graph <math>I_C</math> -vs- <math>V_{CE}</math> for each value of <math>I_B</math> and <math>V_{CC}</math> above. (40 pnts)


6.) Overlay points from the transistor's data sheet on the graph in part 5.).(10 pnts)
All I can do here is to say that my measured <math>\beta</math> are inside the range of data sheet <math>\beta</math>.


=Questions=
=Questions=


#Compare your measured value of <math>h_{FE}</math> or <math>\beta</math> for the transistor to the spec sheet? (10 pnts)
1) Compare your measured value of <math>h_{FE}</math> or <math>\beta</math> for the transistor to the spec sheet? (10 pnts)
#What is <math>\alpha</math> for the transistor?(10 pnts)
 
#The base must always be more _________(________) than the emitter for a npn (pnp)transistor to conduct I_C.(10 pnts)
 
#For a transistor to conduct I_C the base-emitter  junction must be ___________ biased.(10 pnts)
I will calculate my <math>\beta</math> from my measurements above in active region like:
#For a transistor to conduct I_C the collector-base  junction must be ___________ biased.(10 pnts)
 
1)<math>I_B = 2\ \mu A</math>:  <math>\beta = \frac{I_C}{I_B} = \frac{(0.298 \pm 0.010) mA}{(1.967 \pm 0.108) uA} = (151 \pm 9) </math>
 
2)<math>I_B = 5\ \mu A</math>:  <math>\beta = \frac{I_C}{I_B} = \frac{(0.725 \pm 0.021) mA}{(4.862 \pm 0.121) uA} = (149 \pm 6) </math>
 
3)<math>I_B = 10\ \mu A</math>:  <math>\beta = \frac{I_C}{I_B} = \frac{(1.391 \pm 0.052) mA}{(9.200 \pm 0.372) uA} = (151 \pm 8) </math>
 
 
And above values of <math>\beta</math> are in agreement with range of <math>\beta</math> from the spec sheet which is from 30 to 300. But I can not say nothing more because 1) my <math>I_C</math> current doesn't correspond to published in data sheet. 2) My <math>\beta</math> calculation is for specific value of <math>I_B</math> current. But in the data sheet the range of <math>\beta</math> is reported for specific values of <math>I_C</math> and <math>V_{CE}</math>.
 
 
2) What is <math>\alpha</math> for the transistor? <math>\alpha = \frac {I_{C}}{I_{E}}</math> (10 pnts)
 
3) The base must always be more <u>positive</u> (<u>negative</u>) than the emitter for a npn (pnp) transistor to conduct I_C.(10 pnts)
 
4) For a transistor to conduct <math>I_{C}</math> the base-emitter  junction must be <u>forward</u> biased.(10 pnts)
 
5) For a transistor to conduct <math>I_{C}</math> the collector-base  junction must be <u>reversed</u> biased.(10 pnts)


=Extra credit=
=Extra credit=


Measure the Base-Emmiter breakdown voltage. (10 pnts)
Measure the Base-Emitter breakdown voltage. (10 pnts)




I expect to see a graph <math>(I_{B} -vs- V_{BE} )</math> and a linear fit which is similar to the forward biased diode curves.  Compare your result to what is reported in the data sheet.
I expect to see a graph <math>(I_{B} -vs- V_{BE} )</math> and a linear fit which is similar to the forward biased diode curves.  Compare your result to what is reported in the data sheet.


I used:
<math>R_B = (199.5 \pm 1.0)\ k\Omega </math>
<math>R_E = (100.0 \pm 1.0)\ \Omega </math>
<math>V_{CC} = (840 \pm 20)\ mV </math>
Below is the table with my measurements and current calculations:
Here I have used exact formula to calculate <math>I_B</math>:
<math>I_{B} = \frac{V_{BB}-V_B}{R_B}</math>
[[File:Table extra.png | 800 px]]
And bellow is my plot for the Base-Emitter breakdown voltage
[[File:Plot extra fitted.png | 1000 px ]]
The fitting line is <math>I_B\ (\mu A) = (111.7 \pm 10.61) + (0.1809 \pm 0.01634)\ (mV) </math>. The intersection this line with x-axis gives the forward turn on voltage:
<math>V_{BE} = \frac{p_0}{p_1} = \frac{111.7 \pm 10.61}{0.1809 \pm 0.01634} = (617.46 \pm 80.93)\ mV</math>
Actually what we are measuring here is better to call the forward turn on voltage for base-emitter junction (Base-Emitter breakdown voltage is for reverse current measurement). From the data sheet this point (called the base-emitter saturation voltage) is 0.65 V and this point is inside my predicted values <math>(617.46 \pm 80.93)\ mV</math>






[https://wiki.iac.isu.edu/index.php/Electronics_RS Go Back to All Lab Reports] [[Forest_Electronic_Instrumentation_and_Measurement]]
[https://wiki.iac.isu.edu/index.php/Electronics_RS Go Back to All Lab Reports] [[Forest_Electronic_Instrumentation_and_Measurement]]

Latest revision as of 04:07, 22 March 2011

Go Back to All Lab Reports


DC Bipolar Transistor Curves

Data sheet for transistors.

Media:2N3904.pdf Media:2N3906.pdf


Using 2N3904 is more srtaight forward in this lab.

Transistor circuit

1.) Identify the type (n-p-n or p-n-p) of transistor you are using and fill in the following specifications.


I am going to use n-p-n transistor 2N3904. Below are some specifications from data shits for this type of transistor:

Value Description
[math]\displaystyle{ V_{(BR)CEO} = 40\ V }[/math] Collector-Base breakdown voltage
[math]\displaystyle{ V_{(BR)EBO} = 6\ V }[/math] Emitter-Base Breakdown Voltage
[math]\displaystyle{ V_{(BR)CEO} = 40\ V }[/math] Maximum Collector-Emitter Voltage
[math]\displaystyle{ V_{(BR)CBO} = 60\ V }[/math] Maximum Collector-Emitter Voltage
[math]\displaystyle{ I_C = 200\ mA }[/math] Maximum Collector Current - Continuous
[math]\displaystyle{ P = 625\ mW }[/math] Transistor Power rating([math]\displaystyle{ P_{Max} }[/math])
[math]\displaystyle{ h_{FE}\ min \ }[/math] [math]\displaystyle{ h_{FE}\ max \ }[/math] [math]\displaystyle{ I_C }[/math], [math]\displaystyle{ V_{CE} }[/math]
40 300 [math]\displaystyle{ I_C=0.1\ mA }[/math], [math]\displaystyle{ V_{CE}=1.0\ V }[/math]
70 300 [math]\displaystyle{ I_C=1\ mA }[/math], [math]\displaystyle{ V_{CE}=1.0\ V }[/math]
100 300 [math]\displaystyle{ I_C=10\ mA }[/math], [math]\displaystyle{ V_{CE}=1.0\ V }[/math]
60 300 [math]\displaystyle{ I_C=50\ mA }[/math], [math]\displaystyle{ V_{CE}=1.0\ V }[/math]
30 300 [math]\displaystyle{ I_C=100\ mA }[/math], [math]\displaystyle{ V_{CE}=1.0\ V }[/math]








2.) Construct the circuit below according to the type of transistor you have.


Let [math]\displaystyle{ R_E = 100 \Omega }[/math].

[math]\displaystyle{ V_{CC} < 5 Volts }[/math] variable power supply

[math]\displaystyle{ V_{BE}= 1\ V }[/math].

Find the resistors you need to have

[math]\displaystyle{ I_B = 2 \mu A }[/math] , [math]\displaystyle{ 5 \mu A }[/math] , and [math]\displaystyle{ 10 \mu A }[/math]

By measurements I was able to find that [math]\displaystyle{ V_{BE}= 0.6\ V }[/math]. So I am going to use this value. Also let picks up [math]\displaystyle{ V_{BB}= 1.6\ V }[/math]. So my current [math]\displaystyle{ I_B = \frac{V_{BB} - V_{BE}}{R_B} = \frac{(1.6 - 0.6)\ V}{R_B} = \frac{1.0\ V}{R_B} }[/math]. 

Now to get [math]\displaystyle{ I_B = 2\ \mu A }[/math] I need to use [math]\displaystyle{ R_B = \frac{1.0\ V}{2\ \mu A} = 500\ k\Omega }[/math]
    To get [math]\displaystyle{ I_B = 5\ \mu A }[/math] I need to use [math]\displaystyle{ R_B = \frac{1.0\ V}{5\ \mu A} = 200\ k\Omega }[/math]
    To get [math]\displaystyle{ I_B = 10\ \mu A }[/math] I need to use [math]\displaystyle{ R_B = \frac{1.0\ V}{10\ \mu A} = 100\ k\Omega }[/math]



3.) Measure the emitter current [math]\displaystyle{ I_E }[/math] for several values of [math]\displaystyle{ V_{CE} }[/math] by changing [math]\displaystyle{ V_{CC} }[/math] such that the base current [math]\displaystyle{ I_B = 2 \mu }[/math] A is constant. [math]\displaystyle{ I_B \approx \frac{V_{BB}-V_{BE}}{R_B} }[/math]


I used: 

[math]\displaystyle{ R_1 = (199.5 \pm 1.0)\ k\Omega  }[/math]
[math]\displaystyle{ R_1 = (198.7 \pm 1.0)\ k\Omega  }[/math]
[math]\displaystyle{ R_1 = (100.0 \pm 1.0)\ k\Omega  }[/math]
[math]\displaystyle{ R_B = (R_1 + R_2 + R_3) = (498.2 \pm 1.7)\ k\Omega  }[/math]

and 

[math]\displaystyle{ R_E = (100.0 \pm 1.0)\ \Omega  }[/math]


Below is the table with my measurements:


And below is my currents and power calculation:

Here: 

[math]\displaystyle{ I_{E} = \frac{V_E}{R_E} }[/math]
[math]\displaystyle{ I_{B} \approx \frac{V_{BB}-V_{BE}}{R_B} }[/math]
[math]\displaystyle{ P_{max} = I_C \cdot V_{CE} = (I_E - I_B) \cdot V_{CE}  }[/math]

and I have used approximation [math]\displaystyle{ V_{E} \approx o }[/math]



4a.) Repeat the previous measurements for [math]\displaystyle{ I_B \approx 5\ \mu A }[/math]. Remember to keep [math]\displaystyle{ I_CV_{CE} < P_{max} }[/math] so the transistor doesn't burn out


I used: 

[math]\displaystyle{ R_B = (199.5 \pm 1.0)\ k\Omega  }[/math]

and 

[math]\displaystyle{ R_E = (100.0 \pm 1.0)\ \Omega  }[/math]


Below is the table with my measurements:


And below is my currents and power calculation:

Here: 

[math]\displaystyle{ I_{E} = \frac{V_E}{R_E} }[/math]
[math]\displaystyle{ I_{B} \approx \frac{V_{BB}-V_{BE}}{R_B} }[/math]
[math]\displaystyle{ P_{max} = I_C \cdot V_{CE} = (I_E - I_B) \cdot V_{CE}  }[/math]

and I have used approximation [math]\displaystyle{ V_{E} \approx o }[/math]



4a.) Repeat the previous measurements for [math]\displaystyle{ I_B \approx\ 10 \mu A }[/math]. Remember to keep [math]\displaystyle{ I_CV_{CE} < P_{max} }[/math] so the transistor doesn't burn out


I used: 

[math]\displaystyle{ R_B = (100.0 \pm 1.0)\ k\Omega  }[/math]

and 

[math]\displaystyle{ R_E = (100.0 \pm 1.0)\ \Omega  }[/math]


Below is the table with my measurements:


And below is my currents and power calculation:

Here: 

[math]\displaystyle{ I_{E} = \frac{V_E}{R_E} }[/math]
[math]\displaystyle{ I_{B} \approx \frac{V_{BB}-V_{BE}}{R_B} }[/math]
[math]\displaystyle{ P_{max} = I_C \cdot V_{CE} = (I_E - I_B) \cdot V_{CE}  }[/math]

and I have used approximation [math]\displaystyle{ V_{E} \approx o }[/math]




5.) Graph [math]\displaystyle{ I_C }[/math] -vs- [math]\displaystyle{ V_{CE} }[/math] for each value of [math]\displaystyle{ I_B }[/math] and [math]\displaystyle{ V_{CC} }[/math] above. (40 pnts)

Bellow is my plot for the case of [math]\displaystyle{ I_B = 2 \mu A }[/math]









Bellow is my plot for the case of [math]\displaystyle{ I_B = 5 \mu A }[/math]





Bellow is my plot for the case of [math]\displaystyle{ I_B = 10 \mu A }[/math]



6.) Overlay points from the transistor's data sheet on the graph in part 5.).(10 pnts)

I can not really do it because there are not good points to compare from data sheet. I can take for example this data from data sheet:

And from the table above I can only extract the range of [math]\displaystyle{ I_B }[/math] like 

1) [math]\displaystyle{ I_C = 0.1\ mA,\ V_{CE} = 1.0\ V \rightarrow I_B = \frac{I_C}{\beta} = \frac{0.1}{40..300} = (2.5 - 0.33)\ \mu A }[/math]

2) [math]\displaystyle{ I_C = 1.0\ mA,\ V_{CE} = 1.0\ V \rightarrow I_B = \frac{I_C}{\beta} = \frac{0.1}{70..300} = (14.2 - 3.3)\ \mu A }[/math]

And my measurements looks like: 

1) [math]\displaystyle{ I_B = 2\ \mu A,\ I_C = 0.3\ mA,\ V_{CE} = 1.0\ V \rightarrow \beta = \frac{I_C}{I_B} = \frac{0.3\ mA}{2\ uA} = 150 }[/math]

2) [math]\displaystyle{ I_B = 5\ \mu A,\ I_C = 0.72\ mA,\ V_{CE} = 1.0\ V \rightarrow \beta = \frac{I_C}{I_B} = \frac{0.72\ mA}{5\ uA} = 144 }[/math]

3) [math]\displaystyle{ I_B = 10\ \mu A,\ I_C = 1.40\ mA,\ V_{CE} = 1.0\ V \rightarrow \beta = \frac{I_C}{I_B} = \frac{1.40\ mA}{10\ uA} = 140 }[/math]

First point to note that [math]\displaystyle{ I_C }[/math] from my measurements and from data sheet are different so I can not really overlay them and compare. Second, I have specific values of [math]\displaystyle{ I_B }[/math], otherwise from data sheet I have the range of [math]\displaystyle{ I_B }[/math]. And third, I can't really plot this range of [math]\displaystyle{ I_B }[/math] from data sheet on my plots just because my plot doesn't have the [math]\displaystyle{ I_B }[/math] axis.


All I can do here is to say that my measured [math]\displaystyle{ \beta }[/math] are inside the range of data sheet [math]\displaystyle{ \beta }[/math].

Questions

1) Compare your measured value of [math]\displaystyle{ h_{FE} }[/math] or [math]\displaystyle{ \beta }[/math] for the transistor to the spec sheet? (10 pnts)


I will calculate my [math]\displaystyle{ \beta }[/math] from my measurements above in active region like: 

1)[math]\displaystyle{ I_B = 2\ \mu A }[/math]:  [math]\displaystyle{ \beta = \frac{I_C}{I_B} = \frac{(0.298 \pm 0.010) mA}{(1.967 \pm 0.108) uA} = (151 \pm 9)  }[/math] 

2)[math]\displaystyle{ I_B = 5\ \mu A }[/math]:  [math]\displaystyle{ \beta = \frac{I_C}{I_B} = \frac{(0.725 \pm 0.021) mA}{(4.862 \pm 0.121) uA} = (149 \pm 6)  }[/math] 

3)[math]\displaystyle{ I_B = 10\ \mu A }[/math]:  [math]\displaystyle{ \beta = \frac{I_C}{I_B} = \frac{(1.391 \pm 0.052) mA}{(9.200 \pm 0.372) uA} = (151 \pm 8)  }[/math]


And above values of [math]\displaystyle{ \beta }[/math] are in agreement with range of [math]\displaystyle{ \beta }[/math] from the spec sheet which is from 30 to 300. But I can not say nothing more because 1) my [math]\displaystyle{ I_C }[/math] current doesn't correspond to published in data sheet. 2) My [math]\displaystyle{ \beta }[/math] calculation is for specific value of [math]\displaystyle{ I_B }[/math] current. But in the data sheet the range of [math]\displaystyle{ \beta }[/math] is reported for specific values of [math]\displaystyle{ I_C }[/math] and [math]\displaystyle{ V_{CE} }[/math].


2) What is [math]\displaystyle{ \alpha }[/math] for the transistor? [math]\displaystyle{ \alpha = \frac {I_{C}}{I_{E}} }[/math] (10 pnts)

3) The base must always be more positive (negative) than the emitter for a npn (pnp) transistor to conduct I_C.(10 pnts)

4) For a transistor to conduct [math]\displaystyle{ I_{C} }[/math] the base-emitter junction must be forward biased.(10 pnts)

5) For a transistor to conduct [math]\displaystyle{ I_{C} }[/math] the collector-base junction must be reversed biased.(10 pnts)

Extra credit

Measure the Base-Emitter breakdown voltage. (10 pnts)


I expect to see a graph [math]\displaystyle{ (I_{B} -vs- V_{BE} ) }[/math] and a linear fit which is similar to the forward biased diode curves. Compare your result to what is reported in the data sheet.


I used: 

[math]\displaystyle{ R_B = (199.5 \pm 1.0)\ k\Omega  }[/math]
[math]\displaystyle{ R_E = (100.0 \pm 1.0)\ \Omega  }[/math]
[math]\displaystyle{ V_{CC} = (840 \pm 20)\ mV  }[/math]


Below is the table with my measurements and current calculations:

Here I have used exact formula to calculate [math]\displaystyle{ I_B }[/math]: 

[math]\displaystyle{ I_{B} = \frac{V_{BB}-V_B}{R_B} }[/math]


And bellow is my plot for the Base-Emitter breakdown voltage


The fitting line is [math]\displaystyle{ I_B\ (\mu A) = (111.7 \pm 10.61) + (0.1809 \pm 0.01634)\ (mV) }[/math]. The intersection this line with x-axis gives the forward turn on voltage: 

[math]\displaystyle{ V_{BE} = \frac{p_0}{p_1} = \frac{111.7 \pm 10.61}{0.1809 \pm 0.01634} = (617.46 \pm 80.93)\ mV }[/math]

Actually what we are measuring here is better to call the forward turn on voltage for base-emitter junction (Base-Emitter breakdown voltage is for reverse current measurement). From the data sheet this point (called the base-emitter saturation voltage) is 0.65 V and this point is inside my predicted values [math]\displaystyle{ (617.46 \pm 80.93)\ mV }[/math]


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