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The objective Don John Purpose 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. Activity One Diagram of a npn BJT 2 N 390 Theory 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. Data Section Outline of procedures: 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. 5 mA and VCE = 5 Volts. 3) Determine &# 61538; DC. 4) Determine &# 61538; AC = &# 61508; IC/&# 61508; 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? 8) Determine the early voltage. Data Table / Calculations / Analysis 1) Completed in lab IC = 560 &# 61549; A VCE = 5 Volts IB is found to be 5 &# 61549; A according to curve tracer&# 61538; DC = IB / IC 560 &# 61549; A / 5 &# 61549; A = 11&# 61538; AC = &# 61508; IC/&# 61508; IB.
IC 1 = 420 &# 61549; A, IB 1 = 4 &# 61549; A, VCE = 5 Volts IC 2 = 680 &# 61549; A, IB 2 = 6 &# 61549; A, VCE = 5 Volts &# 61538; AC = IC 2 IC 1 /IB 2 IB 1 &# 61538; AC = (680 - 420) / (6 - 4) = 130 A diagram is attached explaining the origin of the values clearly point is 5 Volts Point Point B VCE = 1 Volt VCE = 9 Volts IC = 660 &# 61549; AIC = 700 &# 61549; A IB = 6 &# 61549; AIB = 6 &# 61549; A Ro = [ (700 &# 61549; A 660 &# 61549; A) / (9 - 1) ]- 1 = 200000 &# 61527; A diagram is attached explaining the origin of the values clearly. 6) Q 1 is 3 Volts Point Point B VCE = 1 Volt VCE = 5 Volts IC = 660 &# 61549; AIC = 680 &# 61549; A IB = 6 &# 61549; AIB = 6 &# 61549; A Ro = [ (680 &# 61549; A 660 &# 61549; A) / (5 - 1) ]- 1 = 200000 &# 61527; Q 2 is 7 Volts Point Point B VCE = 5 Volt VCE = 9 Volts IC = 680 &# 61549; AIC = 700 &# 61549; A IB = 6 &# 61549; AIB = 6 &# 61549; A Ro = [ (700 &# 61549; A 680 &# 61549; A) / (9 - 5) ]- 1 = 200000 &# 61527; 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. 7) Q 1 is 5 Volts Point Point B VCE = 1 Volt VCE = 9 Volts IC = 1. 16 m AIC = 1. 22 mA IB = 10 &# 61549; AIB = 10 &# 61549; A Ro = [ (1. 22 mA? 1. 16 mA) / (9 - 1) ]- 1 = 133333 &# 61527; Q 2 point is 5 Volts Point Point B VCE = 1 Volt VCE = 9 Volts IC = 660 &# 61549; AIC = 700 &# 61549; A IB = 6 &# 61549; AIB = 6 &# 61549; A Ro = [ (700 &# 61549; A 660 &# 61549; A) / (9 - 1) ]- 1 = 200000 &# 61527; 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 Early Voltage (VA) Ro = VA / IC - IC Ro = VA 560 &# 61549; A 200000 &# 61527; = 112 This value matches up with the value determined at the beginning of Activity 1 (3). Conclusion 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. Activity Two Diagram of a npn Bjtheory 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 &# 61537; and &# 61538; 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 &# 61537; from &# 61538; &# 61472; &# 61472; and &# 61538; from &# 61537; &# 61472; 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. Data Section Outline of procedures: ESTABLISHING DEVICE CURRENTS: 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. 4) Calculate VBE, IE, IB, IC 5) Calculate &# 61538; and &# 61537; from currents in part (4) 6) Calculate &# 61537; from &# 61538; 7) Calculate &# 61538; from &# 61537; IDENTIFYING THE CONTROLLING JUNCTION: 8) Set V+ = + 10 Volts and V- = - 5 Volts 9) Measure VBE, VE, VB, VC 10) Calculate all terminal currents, &# 61538; and &# 61537; 11) Set V+ = + 5 Volts and V- = - 5 Volts 12) Measure VBE, VE, VB, VC 13) Calculate all terminal currents, &# 61538; and &# 61537; 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 19) Verify VE, VB, VC. 20) Shunt Rb by another 10000 &# 61527; resistor, and measure VE, VB, VC. 21) Calculate all terminal currents, &# 61538; and &# 61537; 22) Remove the resistor, and shunt Rc by another 10000 &# 61527; resistor, and measure VE, VB, VC. 23) Calculate all terminal currents, &# 61538; and &# 61537; 24) Remove the resistor, and shunt RE by another 10000 &# 61527; resistor, and measure VE, VB, VC. 25) Calculate all terminal currents, &# 61538; and &# 61537; 26) Change V- to - 5 Volts.
Measure and calculate again. 27) Compile a neat table of all data. Data table 1) Completed in lab 2) Completed in lab 3) Attached 4) Attached 5) Attached 6) Attached 7) Attached 8) Completed in lab 9) Attached 10) Attached 11) Completed in lab 12) Attached 13) Attached 14) Attached 15) Attached 16) Attached 17) Attached 18) Completed in lab 19) Completed in lab 20) Attached 21) Attached 22) Attached 23) Attached 24) Attached 25) Attached 26) Attache Calculations are attached Conclusion 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. &# 61538; and &# 61537; remain nearly the same no matter what the conditions of the V+ and V- while in active mode.
When saturation occurs &# 61538; and &# 61537; are affected greatly. Activity Three Diagram of a pnp Bjtheory Refer to the theory statement listed in Activity Two. Data Section Outline of procedures: ESTABLISHING DEVICE CURRENTS: 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. 4) Calculate VBE, IE, IB, IC 5) Calculate &# 61538; and &# 61537; from currents in part (4) 6) Calculate &# 61537; from &# 61538; 7) Calculate &# 61538; from &# 61537; Data Table 1) Completed in lab. 2) Completed in lab. 3) Attached 4) Attached 5) Attached 6) Attached 7) Attached Calculations are attached Conclusion 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 &# 61537; from &# 61538; &# 61472; and &# 61538; from &# 61537; is caused by the fact that the resistor do not all have the same values, they are not completely matched.
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