Exercise 1: Phase Splitter DC Operation

Transkript

1 Exercise 1: DC Operation When you have completed this exercise, you will be able to measure dc operating voltages and currents by using a typical transistor phase splitter circuit. You will verify your results with a multimeter and an oscilloscope. The phase splitter circuit has a single NPN transistor, Q1. How is the transistor (Q1) connected? a. common base circuit b. common collector circuit c. common emitter circuit The ac input is capacitor-coupled (C1) to the transistor base (B). Output A is capacitor-coupled (C2) to the collector terminal (C). 82 Festo Didactic P0

2 Output B is capacitor-coupled (C3) to the emitter terminal (E). Why are the ac input and outputs capacitor-coupled? a. to isolate the transistor dc voltages from the ac input and outputs b. to produce A and B output signals that are out of phase c. to produce A and B output signals that have equal amplitudes At the base, R1 and R2 form a voltage divider network that sets a base voltage {V B = V A The base voltage (V B ) forward biases the Q1 base-emitter junction for all ac input signals. If the Q1 base voltage (V B ) is about 4.7 Vdc, what is the emitter voltage (V E )? a. 3.6 Vdc b. 4.1 Vdc c. 4.4 Vdc Festo Didactic P0 83

3 Emitter current (I E ) can be calculated from Ohm's law. I E = V E E Based on the circuit shown, what is the equation for the emitter current (I E )? a. I E = V R4 b. I E = V R4 c. I E =V R3 In transistor circuits, the emitter (I E ) and collector (I C ) currents are essentially equal because the base current (I B ) is very small. The emitter current (I E ) is related to the collector (I C ) and base (I B ) currents by which relationship? a. I E = I C I B b. I E = I C + I B c. I E = I C + 2(I B ) In a transistor phase splitter circuit, the collector resistor (R3) and the emitter resistor (R4) have a. different values. b. the same value. 84 Festo Didactic P0

4 Because emitter and collector currents are essentially equal, what is the voltage drop across R3? a. about equal to the R4 voltage drop b. twice the R4 voltage drop c. half the R4 voltage drop in the emitter (V R4 ) and collector (V R3 ) circuits are equal. As a result, the peak-to-peak voltages of the ac waveforms at output A and output B are a. different. b. equal. Because the transistor phase splitter is biased for class A operation, what does the ac input signal do? a. increases the total circuit current (I T ) b. does not affect I T c. decreases I T Festo Didactic P0 85

5 Locate the PHASE SPLITTER circuit block on the TRANSISTOR POWER AMPLIFIERS circuit board, and connect the circuit shown. While observing the signal on oscilloscope channel 1, adjust the sine wave generator for a 1.0 V pk-pk, 1 khz ac input signal (V i ) at the base of Q1. What is the total dc circuit current (I T(signal) ) with an ac input signal? I T(signal) = ma (Recall Value 1) 86 Festo Didactic P0

6 Disconnect the generator. What is the total dc circuit current (I T(no-signal) ) with no ac input signal? I T(no-signal) = ma (Recall Value 2) Do the total current measurements with and without an ac input signal demonstrate that the T(signal) = ma [Step 3, Recall Value 1] and I T(no-signal) = ma [Step 4, Recall Value 2])? a. yes b. no Replace the ammeter with a two-post connector in the phase splitter circuit. Measure the dc supply voltage (V A ). V A = Vdc (Recall Value 3) Festo Didactic P0 87

7 Measure the following dc voltages at the transistor (Q1). V C = Vdc (Recall Value 4) V B = Vdc (Recall Value 5) V E = Vdc (Recall Value 6) 88 Festo Didactic P0

8 Calculate V BE. V C = Vdc (Step 7, Recall Value 4) V B = Vdc (Step 7, Recall Value 5) V E = Vdc (Step 7, Recall Value 6) V BE = V B V E V BE = Vdc (Recall Value 7) V BE is Vdc (Step 8, Recall Value 7). The transistor (Q1) is turned a. on. b. off. The voltage drop across R4 is the value of V E ( is the voltage drop across R3? Vdc [Step 7, Recall Value 6]). What V A = Vdc (Step 6, Recall Value 3) V C = Vdc (Step 7, Recall Value 4) V R3 = V A V C V R3 = Vdc (Recall Value 8) Festo Didactic P0 89

9 Are V R3 ( Vdc [Step 10, Recall Value 8]) and V R4 ( Vdc [Step 7, Recall Value 6]) essentially equal? a. yes b. no Collector resistor R3 is 10 k, and V R3 is ( Vdc [Step 10, Recall Value 8]). Calculate collector current from Ohm's law. I C = V R3 I C = ma (Recall Value 9) Emitter resistor R4 is 10 k, and V R4 is ( Vdc [Step 7, Recall Value 6]). Calculate emitter current from Ohm's law. I E = V R4 I E = ma (Recall Value 10) Is I E ( ma [Step 13, Recall Value 10]) essentially equal to I C ( ma [Step 12, Recall Value 9])? a. yes b. no 90 Festo Didactic P0

10 Place CM switch 9 in the ON position. Measure the Q1 collector voltage (V C ). By taking this measurement, you will determine the effect of CM 9. V C = Vdc (Recall Value 11) With CM 9 activated, is Q1 turned on or off? a. on b. off With CM 9 activated, measure V B and V E. V B = Vdc (Recall Value 12) V E = Vdc (Recall Value 13) Festo Didactic P0 91

11 Did CM 9 cause a short or an open between the base and emitter? a. short b. open Make sure all CMs are cleared (turned off) before proceeding to the next section. The voltage divider circuit of the phase splitter sets the base voltage, which forward biases the baseemitter junction; the base-emitter junction voltage drop is approximately 0.6 Vdc. The total circuit current is not affected by the ac input signal; the phase splitter operates as a The voltage drops across the collector resistor and the emitter resistor of the phase splitter are essentially equal. Because the collector and emitter resistor values are equal, the collector and emitter currents are essentially equal. A short between the base and emitter turns off the transistor. 92 Festo Didactic P0

12 1. Locate the PHASE SPLITTER circuit block, and connect the circuit shown. Place CM switch 8 in the ON position. Measure the voltage drops across R3 and R4. V R3 = Vdc (Recall Value 1) V R4 = Vdc (Recall Value 2) The voltage drop across R3 a. equals V R4. b. is less than V R4. c. is greater than V R4. d. is opposite in polarity to V R4. Festo Didactic P0 93

13 2. Based on V R4 ( Vdc [Step 1, Recall Value 2]), calculate the emitter current (I E = V R4 E and I C are essentially equal. CM 8 changed the value of collector resistor R3 to about a b. 4.7 k. c. 1 k. d. 100 k. 3. The phase splitter transistor circuit is in the a. b. c. d. 4. For a phase splitter to function properly, the base-emitter always has to be forward biased, and the collector resistor (R3) value has to be a. equal to the emitter resistor (R4) value. b. twice the value of R4. c. at least 10 k. d. half the value of R4. 94 Festo Didactic P0

14 5. If an ac input signal were applied to the circuit, the total dc circuit current (I T ) would a. double. b. halve. c. remain the same. d. be indeterminable without more information. Make sure all CMs are cleared (turned off) before proceeding to the next section. Festo Didactic P0 95