Lab 13 RS

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

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

Let $R_E = 100 \Omega$.

$V_{CC} \lt 5 Volts$ variable power supply

$V_{BE}= 1\ V$.

Find the resistors you need to have

$I_B = 2 \mu A$ , $5 \mu A$ , and $10 \mu A$

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

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

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

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:

And below is my currents and power calculation:

Here:

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

4a.) Repeat the previous measurements for $I_B \approx 5\ \mu A$. Remember to keep $I_CV_{CE} \lt P_{max}$ 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:

And below is my currents and power calculation:

Here:

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

4a.) Repeat the previous measurements for $I_B \approx\ 10 \mu A$. Remember to keep $I_CV_{CE} \lt P_{max}$ 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:

And below is my currents and power calculation:

Here:

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

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

Bellow is my plot for the case of $I_B = 2 \mu A$

Bellow is my plot for the case of $I_B = 5 \mu A$

Bellow is my plot for the case of $I_B = 10 \mu A$

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

Questions

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

I will calculate my $\beta$ from my measurements above in saturation region:

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

2) What is $\alpha$ for the transistor? $\alpha = \frac {I_{C}}{I_{E}}$ (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 I_{C} the base-emitter junction must be forward biased.(10 pnts) 5) For a transistor to conduct I_{C} 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 $(I_{B} -vs- V_{BE} )$ 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:

[math]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 $I_B[\mu A] = (111.7 \pm 10.61) + (0.1809 \pm 0.01634)[mV]$. 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 $(617.46 \pm 80.93)[mV]$

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