Exercise 2: AC Voltage and Power Gains

Exercise 2: AC Voltage and Power Gains an oscilloscope. Signals of equal magnitude but opposite polarity are needed for each transistor (Q1 and Q2). Center-tapped input transformer T1 is used as a phase splitter that develops two signals to Q1 and Q2 that are equal in amplitude and a. 180º out of phase. b. in phase. The transistor phase-splitter circuit used in the previous unit could also have been used to provide the 180º out-of-phase inputs to Q1 and Q2. Festo Didactic 91566-P0 139

The ac output signal (V o a. in phase with the input signal (V i ). b. out of phase with the input signal (V i ). Do transistors Q1 and Q2 conduct during the same half of the ac input signal (V i ) at the sine wave generator? a. yes b. no Transistors Q1 and Q2 conduct during different alternations of the input signal (V i ). 140 Festo Didactic 91566-P0

Transistor Q1 conducts during the a. positive half of V i. b. negative half of V i. Transistor Q2 conducts during the a. positive half of V i. b. negative half of V i. The 180º signal outputs from Q1 and Q2 are combined by output transformer T2 to form V o (360º). Output transformer T2 also acts as an impedance matcher between the high impedance of each transistor circuit and the low impedance of the load. Festo Didactic 91566-P0 141

When the ac input signal (V i ) is 0, no ac base current (I B1 or I B2 As V i becomes more positive, the ac voltage adds to the small dc bias,and I B2 to conduct. During the positive half of V i, Q1 is a. cut off. b. in saturation. As V i becomes negative, Q2 begins to move into cutoff, and Q1 is biased into ac conduction. 142 Festo Didactic 91566-P0

I B1 B2 becomes a. 0. b. negative. Voltage gain is not very high, but current gain is high; as a result, power gain, the product of voltage and current gains, is a. high. b. low. You must use rms values to calculate power. The formula below converts peak-to-peak voltage to rms voltage. V rms = (V pk-pk Input power is measured across the T1 primary coil and resistor R1. The equation below calculates input power (P i ). P i = I i(rms) V i(rms) Festo Didactic 91566-P0 143

I i(rms) is calculated from the voltage drop across R1. Output power is measured across resistor R6 in the T2 secondary coil circuit. The equation below calculates output power (P o ). P o = I o(rms) V o(rms) = V o(rms) 2 Crossover distortion is a type of waveform distortion in which the output signal becomes distorted near the zero crossing point. To prevent crossover distortion, transistors Q1 and Q2 have a Q-point a. slightly above the cutoff point. b. at the cutoff point. 144 Festo Didactic 91566-P0

Transistors Q1 and Q2 conduct for more than 180º but less than 360º of the input signal, allowing Festo Didactic 91566-P0 145

a. b. c. o ) from the 146 Festo Didactic 91566-P0

Locate the PUSH-PULL POWER AMPLIFIER circuit block, and connect the circuit shown. Adjust the positive variable supply (V A ) to 9.0 Vdc. While observing the signal on oscilloscope channel 1, adjust the generator for a 3.0 V pk-pk, 1 khz input signal (V i ). Festo Didactic 91566-P0 147

Connect the channel 2 oscilloscope probe to the base of Q1. Is the signal to Q1 in phase or out of phase with the input signal (V i ) at the generator? a. in phase b. out of phase Connect the channel 2 probe to the base of Q2. Is the signal to Q2 in phase or out of phase with V i? a. in phase b. out of phase 148 Festo Didactic 91566-P0

Connect the channel 1 oscilloscope probe to the base of Q1. Is the signal to Q1 in phase or 180º out of phase with the signal to the Q2 base on channel 2? a. in phase b. 180º out of phase Connect the oscilloscope channel 2 probe to the emitter of Q1. Transistor Q1 conducts during which alternation of the Q1 base signal? a. positive b. negative Festo Didactic 91566-P0 149

Line up the signals on the oscilloscope screen. Compare the Q1 base signal on channel 1 with the Q1 emitter signal on channel 2. Measure the number of degrees of the base signal that Q1 is conducting. Number of degrees that Q1 is conducting = degrees (Recall Value 1) Connect the channel 1 oscilloscope probe to the base of Q2. Connect the channel 2 oscilloscope probe to the emitter of Q2. Transistor Q2 conducts during which alternation of the Q2 base signal? a. positive b. negative 150 Festo Didactic 91566-P0

On the oscilloscope screen, line up the signals, and compare the Q2 base signal on channel 1 with the Q2 emitter signal on channel 2. Measure the number of degrees of the base signal that Q1 is conducting. Number of degrees that Q2 is conducting = degrees (Recall Value 2) Each transistor is operating as a class a. b. c. Connect the channel 1 oscilloscope probe to the emitter of Q1. Ensure that the channel 2 probe is still connected to the emitter of Q2. Compare the Q1 emitter signal and the Q2 emitter signal (channel 2). i ) cycle? a. no b. yes Festo Didactic 91566-P0 151

Connect the channel 1 oscilloscope probe to the collector of Q1, and connect the channel 2 oscilloscope probe to the collector of Q2. Compare the Q1 and Q2 collector signals. How do the collector signals compare in terms of magnitude and phase? a. equal in magnitude and 180º out of phase b. equal in magnitude and in phase c. unequal in magnitude and out of phase Connect the channel 1 oscilloscope probe to the input at the sine wave generator, and across R6. Compare the input and output signals. The signals are a. in phase. b. out of phase. 152 Festo Didactic 91566-P0

Measure the output signal across R6 (V o ) on channel 2. V o = V pk-pk (Recall Value 3) Calculate the voltage gain (A v of V o at V pk-pk (Step 14, Recall Value 3). V i is set at 3.0 V pk-pk. A v = V o i A v = ( [Step 14, Recall Value 3 Input Power A v = (Recall Value 4) i ) to the rms voltage. Next, Then multiply the rms values of input voltage and current to obtain input power. Festo Didactic 91566-P0 153

Convert the peak-to-peak input voltage (V i at 3.0 V pk-pk ) to rms voltage. V i(rms) = (0.707 V pk-pk V i(rms) V i(rms) = V rms (Recall Value 5) Connect the channel 2 oscilloscope probe across R1. Measure the voltage drop across R1 (V R1 ). V R1 = mv pk-pk (Recall Value 6) resistor R1. I R1 = V R1 I R1 = ( mv pk-pk [Step 18, Recall Value 6 I R1 = pk-pk (Recall Value 7) 154 Festo Didactic 91566-P0

Convert the peak-to-peak input current (I R1 ) to rms current. I R1(rms) = (0.707 I pk-pk I R1(rms) = {0.707 [ A pk-pk (Step 19, Recall Value 7 I R1(rms) = rms (Recall Value 8) Calculate the input power (P i ). Your value of I R1 is A rms (Step 20, Recall Value 8) and your value of V i (rms) is V rms. (Step 17, Recall Value 5) P i = I R1(rms) V i(rms) P i = [ A rms (Step 20, Recall Value 8)] [ V rms (Step 17, Recall Value 5)] P i = Recall Value 9) o = V o(rms)2 directly use output current. Convert your measured V o ( rms value. V o(rms) = (0.707 V pk-pk V pk-pk [Step 14, Recall Value 3]) across R6 to the V o(rms) = {0.707 [ V pk-pk (Step 14, Recall Value 3 V o(rms) = V rms (Recall Value 10) Festo Didactic 91566-P0 155

P o = V o(rms)2 P o = ( 2 [Step 23, Recall Value 10 P o = mw (Recall Value 11) Calculate the power gain (A p of P o at ( mw [Step 24, Recall Value 11]) and P i at ( W [Step 21, Recall Value 9]). A p = P o i A p = ( mw [Step 24, Recall Value 11 W [Step 21, Recall Value 9]) A p = (Recall Value 12) Your calculated voltage and power gains were the following values: A v = ( [Step 14, Recall Value 4]) and A p = ( [Step 25, Recall Value 12]). Is the power gain considerably greater than the voltage gain? a. yes b. no The current gain was not calculated. However, because the power gain is considerably greater than the voltage gain, you can infer that the current gain is very a. low. b. high. 156 Festo Didactic 91566-P0

Connect the channel 2 oscilloscope probe across R6. Increase the input signal to 10 V pk-pk. Remove the two-post connector between R2 and R3 to remove the dc biasing. The output signal (V o ) at R6 on channel 2 exhibits a. crossover distortion. b. amplitude distortion. Without dc biasing, does the ac input signal cause Q1 and Q2 to conduct? a. yes b. no Install the two-post connector between R2 and R3 to restore the dc bias. Reduce the input signal (V i ) to 3.0 V pk-pk. If a fault occurred in the circuit to cause transistor Q1 to cut off (not conduct), would the circuit output across R6 measure 0 V pk-pk? a. yes b. no Festo Didactic 91566-P0 157

Place CM switch 16 in the ON position to open R4 and cause Q1 to stop conducting. The output signal across R6 exhibits a. crossover distortion. b. amplitude distortion. Each transistor (Q1 and Q2) forms its own circuit. With CM 16 activated, compare the input (V i ) and output (V o ) waveforms. For which half of the input does Q2 conduct? a. the positive alternation b. the negative alternation Make sure all CMs are cleared (turned off) before proceeding to the next section. phase splitter is used to conduct the input signal into two equal 180º out-of-phase signals. Each transistor conducts during opposite halves of the input signal for about 225º. A transformer combines transistor outputs into an output signal with no amplitude distortion. Even though voltage gain is less than 1.0, power gain is very high (about 500) because current gain is also very high. Crossover distortion occurs when the dc bias is removed. 1. Locate the PUSH-PULL POWER AMPLIFIER circuit block, and connect the circuit shown. Adjust the positive variable supply (V A ) to 9.0 Vdc. While observing the signal on oscilloscope channel 1, adjust the generator for a 3.0 V pk-pk, 1 khz input signal (V i ). 158 Festo Didactic 91566-P0

Place CM switch 15 in the ON position to change voltage divider resistor R3 from 100 to 50. The new R3 value causes the dc bias to change. Connect the channel 1 oscilloscope probe to the emitter of Q1, and connect the channel 2 oscilloscope probe to the emitter of Q2. Observe the Q1 and Q2 emitter signals. Each transistor is operating in class a. A. b. AB. c. B. d. C. Place CM switch 15 in the OFF position. Place CM switch 20 in the ON position to increase load resistor R6 from 8.2 to 18. Festo Didactic 91566-P0 159

In the procedure with R6 at 8.2, you calculated the output power (P o ) to be about 49.95 mw. Calculate P o with R6 changed to 18 by measuring the output peak-to-peak voltage, converting it to rms voltage, and calculating the output power. V o = V pk-pk (Recall Value 1) V o(rms) = (0.707 V pk-pk V o(rms) = V rms (Recall Value 2) P o = V o(rms) 2 P o = mw (Recall Value 3) note: Using 1000 converts watts (W) to milliwatts (mw). 2. With an 8.2 load resistor, V o was about 1.8 V pk-pk and P o was about 49.95 mw. With an 18 load resistor, V o is V pk-pk (Step 1, Recall Value 1) and P o is mw (Step 1, Recall Value 3). Increasing the value of output resistor R6 from 8.2 to 18 caused the a. output power (P o ) to increase. b. output power (P o ) to decrease. c. voltage gain (A v ) to decrease. d. current gain (A i ) to decrease. 3. The push-pull circuit needs two signals that are a. in phase and equal in magnitude. b. in phase and unequal in magnitude. c. 180º out of phase and equal in magnitude. d. 180º out of phase and unequal in magnitude. 4. a. b. is biased slightly above the cutoff point. c. conducts for less than 360º of the input signal. d. All of the above 160 Festo Didactic 91566-P0

5. a. an increase in the amplitude of the input signal. b. the removal of the dc bias of the circuit. c. an increase in total dc circuit current. d. biasing the transistors in the center of the load line. Make sure all CMs are cleared (turned off) before proceeding to the next section. Festo Didactic 91566-P0 161