1. To examine the different methods of setting the D.C. operating point of a transistor. This is usually referred to as the Q-point or quiescent operation point. That is the operation point for the transistor when no ac signal is applied to the system. The collector current , IC ,at this operating point is called, ICQ , and the collector to emitter voltage at this point is called, VCEQ. Providing the necessary voltages and currents to set this operating point is referred to as biasing the transistor.
2. To determine the most effective configuration for providing bias ensuring good stability with respect to changes in the Beta ( b ).
3. To understand how the change of
b
in a transistor
circuit can change the Q-point of the circuit.
Transistors: NPN - 2N5088 (
b
> 350)
NPN - 2N2222 (100 <
b
< 300)
NPN - 2N5550 (
b < 100)
Resistors: 2.2 kW, 1 kW, 470 kW and 100
W , plus other assorted
resistors as calculated.
+15 VDC power supply
Digital multimeter
Breadboard and wire
Hambley, Allan R., "Electronics," 2nd Ed. Prentice Hall, 2000, pp. 232-248
Serda and Smith, "Microelectronic Circuits," Oxford
University Press, New York, NY, 1991, pp. 191-220, or
Savant, Roden, and Carpenter, "Electronic Design," The Benjamin/Cummings
Publishing Company, Inc., Redwood City, CA, 1991, pp. 79-89, 213-226.
1. Build the circuit below with RB = 470 kW and use it to calculate the beta of each of the three transistors. This is done by first measuring the following node voltages: VCC, VC, and VB. Then calculate IC = (VCC - VC)/RC, IB = (VCC - VB)/RB, and finally b = IC/IB. The three transistors should have widely different values of b, One below 100, one greater than 300, and one somewhere near the middle of the range. Record which transistor has the medium value of beta. This one will be used for designing the four bias circuits to be used in the rest of the experiment.
Use RC = 1 kW, VCC = 15 V.
2. Small Signal Bias circuit. This is the same as the circuit used to find the b of transistors.
a) Using the medium value of b found in step 1 calculate the value of base current required to give a collector current, IC = 6 mA. Then calculate the value of resistor, RB, to give that base current using the value of VBE measured in step 1. Note: Since the emitter is grounded in this circuit the voltage from base to emitter, VBE, is the same as the base node voltage, VB.
b) Replace RB in the circuit with the standard resistor closest to the calculated value. Then insert the medium b transistor. Turn on the 15 V power and check to see that VC is between 8.5 V and 9.5 V. Note: If IC = 6 mA then VC should be 9 V. (VC = VCC - ICRC ) If VC is between 8.5 V and 9.5 V then IC is between 5.5 mA and 6.5 ma. This will be close enough for this experiment. If VC is not between 8.5 V and 9.5 V then calculate the % difference between the desired and actual value of IC and adjust the resistor value by the same % using two standard resistors in series to get close to the new value. Remember if the current is too large then the resistor must be increased and if the current is too small then the resistor must be decreased. Measure and record the actual value of each resistor or combination of resistors used in the circuit.
c) Once you have the circuit working at the correct current with the medium b transistor measure and record all the node voltages. Then replace the transistor with the high b transistor and low b transistor repeating the node voltage measurements in each case to see how the voltages and therefore the currents change as the b of the transistor changes.
3. Emitter Feedback Bias Circuit
Use RC = 1 kW and RE = 100 W, VCC = 15 V.
a) This circuit has a resistor between the emitter of the transistor and ground. The voltage drop across the resistor raises the voltage of the emitter above zero and therefore raises the base voltage since VBE is like the forward bias voltage on a diode is usually in the range of 0.6 to 0.7 V. Therefore the total voltage across RB is reduced. If IC is increased due to an increase in the b of the transistor then VE is increased causing VB to increase. This causes the voltage across RB to decrease and therefore causes IB to decrease. This in turn reduces the value of IC and brings it back closer to the current at the design value of b. This is a form of negative feedback used to stabilize a system variable against a disturbance in system parameters. You will have to calculate the value of RB to give IC = 6mA. Use the value of VBE found part 2 with the medium b transistor. Remember IE = IC + IB, where the transistor currents are the same as in the previous circuit. Write KVL around the loop containing the DC supply, RB, VBE, and RE. Then solve for the required value of RB, using the measured value of RE.
b) Build this circuit using the standard resistor closest to the calculated value RB. Then insert the medium b transistor. Turn on the 15 V power and check to see that VC is between 8.5 V and 9.5 V. If VC is not between 8.5 V and 9.5 V then calculate the % difference between the desired and actual value of IC and adjust the resistor value by the same % using two standard resistors in series to get close to the new value. Remember if the current is too large then the resistor must be increased and if the current is too small then the resistor must be decreased. Measure and record the actual value of each resistor or combination of resistors used in the circuit.
c) Once you have the circuit working at the correct current with the medium b transistor measure and record all the node voltages. There are 4 node voltages in this circuit not just 3 as in the previous circuit. Then repeating the node voltage measurements the high b transistor and low b transistor to see how the voltages and therefore the currents change as the b of the transistor changes.
4. Collector Feedback Bias Circuit
Use RC = 1 kW, VCC
= 15 V.
a) This circuit like the small signal bias circuit has the emitter of the transistor connected directly to the ground. Therefore VB = VBE. The current to the base of the transistor is supplied by a resistor, RF, connected to the collector of the transistor. As the collector current increases the voltage drop across the collector is resistor increased. This causes the voltage across RF to decrease and therefore causes IB to decrease. This in turn reduces the value of IC and brings it back closer to the current at the design value of b. This is another form of negative feedback. You will have to calculate the value of RF to give IC = 6mA. Use the value of VBE found part 2 with the medium b transistor. Write KVL around the loop containing the DC supply, RC, RF, and VBE. Then solve for the required value of RF, using the measured value of RC. Remember the transistor currents are the same as in the previous two circuits.
b) Build this circuit using the standard resistor closest to the calculated value RF. Then insert the medium b transistor. Turn on the 15 V power and check to see that VC is between 8.5 V and 9.5 V. If VC is not between 8.5 V and 9.5 V then adjust the resistor value as done in the previous two circuits. Measure and record the actual value of each resistor or combination of resistors used in the circuit.
c) Once you have the circuit working at the correct current with the medium b transistor measure and record all the node voltages. Back to just 3 nodes to measure in this circuit. Then repeating the node voltage measurements the high b transistor and low b transistor to see how the voltages and therefore the currents change as the b of the transistor changes.
5. Voltage Divider Bias
Use RC = 1 kW, RE = 100 W, R2= 2.2 kW, VCC = 15 V.
a) This circuit uses an emitter resistor to provide negative feedback along with a voltage divider to provide a nearly constant VB. The current to the base of the transistor is supplied by the voltage divider. The Thévenin equivalent resistance of the voltage divider is low, so the variation in base current does not cause the base voltage to change very much. This improves the negative feed back effect of the emitter resistor. You will have to calculate the value of R1 to give IC = 6mA. Use the value of VBE found part 2 with the medium b transistor. Design the voltage divider to provide the needed base current at a base voltage corresponding to VBE + the voltage drop across the emitter resistor using the given value for R2. Hint: The transistor currents and the emitter resistor the same as in part 3 , therefore the base voltage, VB, is the same. Use the base voltage to find IR2. Then write the node equation for the VB node and solve for R1.
b) Build this circuit using the standard resistor closest to the calculated value of RR1. Then insert the medium b transistor. Turn on the 15 V power and check to see that VC is between 8.5 V and 9.5 V. If VC is not between 8.5 V and 9.5 V then adjust the resistor value as done in the previous three circuits. Measure and record the actual value of each resistor or combination of resistors used in the circuit.
c) Once you have the circuit working at the correct current with the medium b transistor measure and record all the node voltages. Back to 4 nodes to measure in this circuit. Then repeating the node voltage measurements the high b transistor and low b transistor to see how the voltages and therefore the currents change as the b of the transistor changes.
1. Calculate, ICQ, the DC operating current, by dividing VRC = VCC - VC by the measured value of RC for each case.
2. Calculate IB = ( VCC - VB )/RB = VRB /RB for the small signal bias circuit used in step 2 and use it to recalculate the b of the transistors at the current used for the experiment.
3. Calculate the range of variation of ICQ for each circuit. Express this range as a percent of the value of ICQ for the medium b transistor.
%range = 100%*(ICQ high -ICQ low )/ICQ
medium
1. Rate the bias circuits in order from most stable (smallest variation in ICQ ) to the least stable.
2. What effect does adding an emitter resistor to the small-signal bias circuit have on the bias stability?
3. Derive the equations used to calculate the bias resistor values for each circuit by using circuit analysis and the definition of b.
4. Explain why biasing is needed in transistor amplifier circuits.
What can you say about bias stability for Bipolar Junction Transistors based on the results of this experiment?
This page last updated 10/30/2007