EENG 4109 – Electronic Circuit Analysis II Laboratory
Lab 7
I-V characteristic of BJT
Name: ____________________________________________________________________ (50/______)
Purpose
In this lab, the characteristics of Bipolar Junction Transistor (BJT (PNP) will be investigated. These
include the characteristic curves (CE- Common Emitter) for the BJT.
Theory
The bipolar junction transistor (BJT) can be modeled as a current controlled current source. The circuit
symbol and the pin out for the actual device can be seen in Figure 3-1.
Figure 3-1. The Bipolar Junction Transistor.
The CE (Common Emitter) Collector characteristics constitute a family of static characteristics plots of
collector current versus collector-emitter voltage for several values of base current.
Figure 3-2. CE configuration circuit using NPN transistor.
Figure 3-3: Characteristic curves for the BJT transistor. (CE)
These curves can be used to calculate the large signal current gain βDC (or hFE) and the
small signal current gain, βAC (or hfe). These values are in general calculated for a given
bias point ICQ, VCEQ using the following equations:
βDC=ICQ/IBQ;
βAC = | ICQ - ICQ’ |/| IBQ- IBQ’|;
From this, one can see that a large signal gain depends only on the Q point and the small
signal gain depends only on small deviations around the Q point.
Equipment’s to be needed.
1) Transistor – 2N3904 (NPN)
2) Resistor – 1kΩ and 100 kΩ
3) Variable Power Supply
4) Hook up wires
5) Bread Board
6) Digital multimeter
Procedure
1) Take the measurement of BJT (2N3904) with the help of multimeter putting it in a
diode mode, before doing the circuit connection.
Table.1
Read the BJT before circuit connections
Junction Voltage Measured from DMM
1 B-E
2 B-C
3 C-E
4 E-C
Table.2
S.N. Resistor Value measured from DNN
1 R1 = 100 kΩ
2 R2 = 1kΩ
2) Connect the circuit as shown in figure below.
Figure 3-4. BJT (NPN) circuit diagram.
2) After connecting the circuit diagram, set the VBB as 1 volt and measure the
corresponding quantity as shown in the table.
Table 3.
VBB (V) VR1(V) IB (µA) VCC(V) VR2(V) VCE(V) VBE(V) IC (mA)
1 1
2
3
5
7
9
3) Repeat step 2 for VBB = 2V.
Table 4.
VBB (V) VR1(V) IB (µA) VCC(V) VR2(V) VCE(V) VBE(V) IC (mA)
2 1
2
3
5
7
9
4) Repeat step 2 for VBB =4V
Table 5.
VBB (V) VR1(V) IB (µA) VCC(V) VR2(V) VCE(V) VBE(V) IC (mA)
4 1
2
3
11
13
15
5) Repeat step 2 for VBB =6V
Table 6.
VBB (V) VR1(V) IB (µA) VCC(V) VR2(V) VCE(V) VBE(V) IC (mA)
6 1
2
3
11
13
15
6) Plot Ic VS VCE for different value of IB using Excel.
7) Calculate β (DC) and β (AC) for any value of VCEQ from the curve that you obtained from
Excel.
8) β (DC) = I CQ / I BQ = __________________.
9) Pick any two value for ICQ and corresponding IBQ for the VCQ that you take from the
graph.
Compute change in ICQ = ______________.
Compute change in IBQ = ______________.
10) β (AC) = change in ICQ / change in IBQ = __________________.
11) Calculate transconductance gm = ICQ / VT = ______________.
12) RΠ = β (AC) * VT / ICQ = ________________.