The objective of this lab is to introduce the Bipolar Junction Transistor (BJT). A BJT is a three terminal device composed of an emitter, base, and collector terminals. In this lab we will introduce two major types of BJT’s : npn and pnp. The first, npn, has an n-type emitter, a p-type base and a n-type collector. On the other hand the pnp has a p-type emitter, a n-type base, and a p-type collector. Also the transistor consists of two major pn junctions, the emitter-base junction (EBJ) and the collector-base junction (CBJ).

Depending on the bias condition of each of these junctions, there are different modes of operation. We will show that the basic principle of a BJT is the use of the voltage between two terminals on order to control the current in the third terminal.

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In this part of the lab we will use the curve tracer to display the common-emitter BJT family of curves. We will see the i-v characteristics of Ic vs.

Vce for steps of IB. The i-v characteristic showing Ic vs. Vce for different values of VBE are not linear. Thus we will see that the output resistance of the BJT change slightly with current. Then using the values of Ic and Ro, we can calculate the early voltage, Va. The important feature of this device is that the i-v characteristics are not perfect linear.

1) Use the curve tracer to display the common-emitter BJT family of curves (ic vs vCE for steps of iB).

2) Determine IB needed to set the Q-Point for Ic=0.5mA and VCE=5 Volts.

4) Determine AC = IC/IB.

5) Determine the output resistance, Ro, by measuring the slope of the i-v curve and taking the inverse of that.

6) Does the output resistance change with voltage on the same curve?

7) Does the output resistance change with current on different curves?

Data Table / Calculations / Analysis

IB is found to be 5A according to curve tracer.

4) AC = IC/IB.

IC1 = 420A , IB1 = 4A, VCE = 5 Volts

IC2 = 680A , IB2 = 6A, VCE = 5 Volts

A diagram is attached explaining the origin of the values clearly.

Ro = [(700A – 660A)/(9-1)]-1 = 200000

A diagram is attached explaining the origin of the values clearly.

Ro = [(680A – 660A)/(5-1)]-1 = 200000

Ro = [(700A – 680A)/(9-5)]-1 = 200000

The output resistance values determined with 2 different Q points along the same IB value (curve) shows that voltage does not change resistance.

A diagram is attached explaining the origin of the values clearly.

Ro = [(1.22mA – 1.16mA)/(9-1)]-1 = 133333

Ro = [(700A – 660A)/(9-1)]-1 = 200000

The output resistance values determined with the same Q point on two different IB values (different curves) shows as current increases, IC, resistance decreases.

A diagram is attached explaining the origin of the values clearly.

This value matches up with the value determined at the beginning of Activity 1 (3).

In conclusion the BJT characteristics were as expected. As current increased the output resistance decreased, and as voltage changes the output resistance did not change. Hence current change and not voltage change affect the output resistance.

In this part of the lab we will set the dc voltages to the terminals of the BJT and measure the corresponding voltages at the nodes. Then we will calculate the currents through the emitter, base and the collector terminals. Next, we will calculate and from these currents. We will see that even though the resistor values are not completely matched we will have some discrepancies in the currents. But for the most important part, we will show that when we will calculate from and from there will be a big change. In the second part, when we change the dc voltages we will show that the transistor current is more dependent on the emitter potential than the collector potential for both npn and pnp BJT’s.

1) Choose RC and RE to be well matched.

2) Adjust dc supplies to +10 Volts and –10 Volts.

3) Measure the dc voltages with the DVM at points E, B, C.

5) Calculate and from currents in part (4)

IDENTIFYING THE CONTROLLING JUNCTION:

8)Set V+ = +10 Volts and V- = -5 Volts

10) Calculate all terminal currents, and

11) Set V+ = +5 Volts and V- = -5 Volts

13) Calculate all terminal currents, and

14) Compare this data with the data found at + 10 Volts.

15) Do the transistor currents depend more on the conditions in the emitter of the collector?

16) Set up two-equations-in-two-unknowns and solve simultaneously for n and IS.

17) Are these values reasonable? Why or why not?

MEASURING EFFECTS OF CIRCUIT RESISTANCE

18) Set V+ = +10 Volts and V- = -10 Volts

20) Shunt Rb by another 10000 resistor, and measure VE, VB, VC.

21) Calculate all terminal currents, and

22) Remove the resistor, and shunt Rc by another 10000 resistor, and measure VE, VB, VC.

23) Calculate all terminal currents, and

24) Remove the resistor, and shunt RE by another 10000 resistor, and measure VE, VB, VC.

25) Calculate all terminal currents, and

26) Change V- to -5 Volts. Measure and calculate again.

27) Compile a neat table of all data.

In conclusion, the circuit worked as expected. VC changes according to the difference between V+ and V-. Since VB is grounded very little voltage is lost through the base collector so the voltage between the emitter and collector terminals remain almost the same while V+ and V- are equal but of opposite sign values. Also when VC is less than VE saturation occurs in the circuit, hence it is forward biased as opposed to being reversed biased in active mode. and remain nearly the same no matter what the conditions of the V+ and V- while in active mode. When saturation occurs and are affected greatly.

Refer to the theory statement listed in Activity Two.

1) Choose RC and RE to be well matched.

2) Adjust dc supplies to +10 Volts and –10 Volts.

3) Measure the dc voltages with the DVM at points E, B, C.

5) Calculate and from currents in part (4)

In conclusion from the results obtained in Activities Two and Three, the branch currents and node voltages of npn and pnp transistors, it can be said that the transistor currents depend more on the emitter potential than the collector potential. Also the error that is seen in calculating from and from is caused by the fact that the resistor do not all have the same values, they are not completely matched.