W1 - Lab1 Frequency Response-OL (2)

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Jan 9, 2024

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Week 1 Lab Assignment: Frequency Response By: Deandre Wheelington 05/14/2023 Instructor: Professor J. Koscinski EET121
Experiment 1: Frequency Response Lab Objectives: After completing this experiment, you should be able to, 1. Measure β ac 2. Calculate the low frequency response for a common emitter amplifier 3. Measure the dominant frequency using a Bode Plotter. Procedure: 1. Build the circuit in Figure 1 using multisim. Figure 1 Common Emitter Amplifier Circuit In order to measure β ac you want the amplifier to be operating at mid-band frequencies and that the output isn’t distorted due to saturation or cutoff. Selecting 10 mVp amplitude at 5 kHz allows you to make this measurement. 2. Measure the AC current at the collector. I c = 66.8uA 3. Measure the AC current at the base. I b = 420nA
4. Calculate β ac β ac = Δ Ic Δ Ib = ¿ 158.04nA 5. Measure the DC current at the emitter I E = 1.80mA 6. Calculate r’e r’e = 14.32Ω 7. Measure the AC voltage at the Load. Vout = 844mV 8. Measure the AC voltage at the Source. Vin = 20mv 9. Calculate the mid-band gain. A V = A M = V out V ¿ = 42.2 10. Convert the mid-band gain to its dB value A V(dB) = 20log ( 42.2 ) = 32.51 dB 11. Calculate the low frequency cutoff values due to the 3 external capacitances. Show your work and record the values in Table 1. a. Frequency due to the input capacitor. The source resistance in this case is in series with Rin(tot) f c 1 = 1 2 π ( R S + ( R 1 | | R 2 | | ( β ac ( r ' e + R E 1 ) ) ) ) C 1 Equation 1 1 2 π ( 600 + ( R 1 | | R 2 | | ( 158 ( 14.32 + 32 ) ) ) ) C 1 158 ( 14.32 + 32 ) = 7318.6 1 ( 1 68000 + 1 22000 + 1 7318 ) = 5081.048166 1 2 π ( 5081 100 10 9 ) = 313. 2354715 Hz b. Frequency due to the bypass capacitor. The source resistance in this case is in parallel. f c 2 = 1 2 π ¿¿ Equation 2 2 π ( R E 2 ¿ ( 14.32 + 32 + 5081 158 ) ) C 2
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1 ¿¿ 579 158 = 3.66455 3.6 + 32 + 14.32 = 49.92 1 ( 1 1500 + 1 49.92 ) = 48.31217095 1 2 π ( 48.31 ) ( .0001 ) = 32.94451316 32.94 Hz c. Frequency due to the output capacitor. f c 3 = 1 2 π ( R c + R L ) C 3 Equation 3 1 2 π ( 3900 + 5600 ) ( .0001 ) = 167.53 Hz d. Which of the 3 low frequency response frequencies would be considered to be the dominant frequency? Include your answer in Table 1. Value Calculated (Hz) Measured (Hz) f C1 313.23 Hz f C2 32.94 Hz f C3 167.53 Hz f CL(dom) 313.23 Hz 317.573 Hz Table 1 Low Ferquency Response Values Measure the dominant frequency of the circuit using a Bode Plotter. A Bode Plotter is an instrument that measures the input and output of a circuit while is sweeps the frequency from low frequency to high frequency and plots the resulting gain as a function of the frequency. It can be found in the instruments in the instrument menu. 12. Connect the input of the Bode Plotter to the voltage source. It works like an oscilloscope so the second connection must be connected to ground. 13. Connect the output of the Bode Plotter to the load resistor.
14. Adjust the setting of the Bode Plotter by selecting Magnitude Mode. The Horizontal settings should be LOG with an upper frequency of 10 GHz and a lower frequency of 1 Hz. The Vertical settings should be LOG with an upper value of 40 dB and a lower setting of -20 dB 15. Run the simulation and measure the midband gain with the cursor. A V(dB) = 32.494 dB 16. How does this compare to the value measured in step 10? The measured gain is extremely close to the calculated gain of 32.51 dB. 17. In order to find the dominant frequency, move the cursor so that the gain has dropped by 3 dB on the lower side of the graph. Measure the value of the frequency at that point. f CL(dom) = 317.573 Hz . Add this value to Table 1. 18. Compare the two dominant frequencies. 19. Find the dominant high frequency value by finding the corresponding point on the high side of the graph. F CH(dom) = 30.844 MHz *********End of the experiment ********* SOURCES: Floyd, T. L. (2017). Electronic Devices (Conventional Current Version) (10th ed.). Pearson Education (US). https://ecpi.vitalsource.com/books/9780134414553