Lab 13 RS: Difference between revisions
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Below is the table with my measurements: | Below is the table with my measurements: | ||
[[File:Table 2uA 01.png | | [[File:Table 2uA 01 corrected.png | 700 px]] | ||
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Here: | Here: | ||
<math>I_{E} = \frac{V_E}{R_E}</math> | <math>I_{E} = \frac{V_E}{R_E}</math> | ||
<math>I_{B} | <math>I_{B} \approx \frac{V_{BB}-V_{BE}}{R_B}</math> | ||
<math>P_{max} = I_C \cdot V_{ | <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.png | | [[File:Table 2uA 02 corrected.png | 700 px]] | ||
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Below is the table with my measurements: | Below is the table with my measurements: | ||
[[File:Table 5uA 01.png | | [[File:Table 5uA 01 corrected.png | 700 px]] | ||
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Here: | Here: | ||
<math>I_{E} = \frac{V_E}{R_E}</math> | <math>I_{E} = \frac{V_E}{R_E}</math> | ||
<math>I_{B} | <math>I_{B} \approx \frac{V_{BB}-V_{BE}}{R_B}</math> | ||
<math>P_{max} = I_C \cdot V_{ | <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.png | | [[File:Table 5uA 02 corrected.png | 700 px]] | ||
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Below is the table with my measurements: | Below is the table with my measurements: | ||
[[File:Table 10uA 01.png | | [[File:Table 10uA 01 corrected.png | 700 px]] | ||
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Here: | Here: | ||
<math>I_{E} = \frac{V_E}{R_E}</math> | <math>I_{E} = \frac{V_E}{R_E}</math> | ||
<math>I_{B} | <math>I_{B} \approx \frac{V_{BB}-V_{BE}}{R_B}</math> | ||
<math>P_{max} = I_C \cdot V_{ | <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.png | | [[File:Table 10uA 02 corrected.png | 700 px]] | ||
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And from the table above I can only extract the range of <math>I_B</math> like | 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}{ | 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}{ | 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: | And my measurements looks like: | ||
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Below is the table with my measurements and current calculations: | Below is the table with my measurements and current calculations: | ||
Here: | Here I have used exact formula to calculate <math>I_B</math>: | ||
<math>I_{B} = \frac{V_{BB}-V_B}{R_B}</math> | <math>I_{B} = \frac{V_{BB}-V_B}{R_B}</math> | ||
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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: | 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> | <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> | 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
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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