Lab 13 RS

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DC Bipolar Transistor Curves

Data sheet for transistors.

Media:2N3904.pdfMedia:2N3906.pdf

Using 2N3904 is more srtaight forward in this lab.

Transistor circuit

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

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

Let [math]R_E = 100 \Omega[/math].

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

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

Find the resistors you need to have

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

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

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]

[math]V_{CC}[/math]	[math]V_{B}[/math]	[math]V_{BB}[/math]	[math]V_{EC}[/math]	[math]V_{E}[/math]	[math]R_{E}[/math]	[math]R_{B}[/math]	[math]I_{E}[/math]	[math]I_{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]
[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]
[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]
[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]
[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]
[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]
[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]
[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]
[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]
[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]
[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]
[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]
[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]

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

V_{CC}	V_B	V_{BB}	V_ {EC}	V_ E	R_E	R_B	I_E	I_B
mV	mV	V	mV	mV	[math]\Omega[/math]	k[math]\Omega[/math]	mA	\muA

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)

Questions

1. Compare your measured value of [math]h_{FE}[/math] or [math]\beta[/math] for the transistor to the spec sheet? (10 pnts)
2. What is [math]\alpha[/math] for the transistor?(10 pnts)
3. The base must always be more _________(________) 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 ___________ biased.(10 pnts)
5. For a transistor to conduct I_C the collector-base junction must be ___________ biased.(10 pnts)

Extra credit

Measure the Base-Emmiter 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.

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