W1 - Lab1 Frequency Response-OG

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 = __1.81mA____________

3. Measure the AC current at the base. I b = __12.1uA____________ 4. Calculate β ac β ac = _Ic/Ib = 149.6___________ 5. Measure the DC current at the emitter I E = ___1.82mA___________ 6. Calculate r’e r’e = _25mV/Ie = 13.7 Ω 7. Measure the AC voltage at the Load. Vout =__298.8mV__________ 8. Measure the AC voltage at the Source. Vin =_____7.07mV________ 9. Calculate the mid-band gain. (Vout/Vin) A V = _42.3V____________ 10. Convert the mid-band gain to its dB value (20log(av)) A V(dB) = _32.5__________ 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 = 211.5hz b. Frequency due to the bypass capacitor. The source resistance in this case is in parallel. R 1 | | R 2 | | R S β ac r ' e + R E 1 +( ¿ ) ¿ R E 2 ∨ ¿¿ 2 π ¿ f c 2 = 1 ¿ Equation 2 c. Frequency due to the output capacitor. f c 3 = 1 2 π ( R c + R L ) C 3 Equation 3 = 1/2π(3900+5600) x .0001 = . 168hz 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 211.5 f C2 32.6 f C3 .168hz f CL(dom) 284.6hz