Lab 13 RS

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} = \frac{V_{BB}-V_B}{R_B} }[/math]
[math]\displaystyle{ P_{max} = I_C \cdot V_{EC} = (I_E - I_B) \cdot V_{EC}  }[/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} = \frac{V_{BB}-V_B}{R_B} }[/math]
[math]\displaystyle{ P_{max} = I_C \cdot V_{EC} = (I_E - I_B) \cdot V_{EC}  }[/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} = \frac{V_{BB}-V_B}{R_B} }[/math]
[math]\displaystyle{ P_{max} = I_C \cdot V_{EC} = (I_E - I_B) \cdot V_{EC}  }[/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 sheet:

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 Ic from my measurements and from data sheet are different so I can not really overplay them and compare. Second, I have specific values of Ib, otherwise from data sheet I have the range of Ib. And third, I can't really plot this range of Ib from data sheet on my plots just because my plot doesn't have the Ib axis.

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 saturation region:

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 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 [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:

[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]

Go Back to All Lab Reports Forest_Electronic_Instrumentation_and_Measurement