Chat with us, powered by LiveChat The low frequency cutoff values due to the 3 external capacitances. ?Show your work and record the values in Table 1. ?? 16. How does this compare to the value measured i | Wridemy

The low frequency cutoff values due to the 3 external capacitances. ?Show your work and record the values in Table 1. ?? 16. How does this compare to the value measured i

The low frequency cutoff values due to the 3 external capacitances.  Show your work and record the values in Table 1.

  

16. How does this compare to the value measured in step 10?

  

18. Compare the two dominant frequencies.

Objectives:

After completing this experiment, you should be able to,

1. Measure βac

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 βac 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. Ic = 76.628 uA

3. Measure the AC current at the base. Ib = 809.134 nA

4. Calculate βac βac = 94.7

5. Measure the DC current at the emitter IE = 1.818 mA

6. Calculate r’e r’e = 13.75 ohms

7. Measure the AC voltage at the Load. Vout = 298.85 mV

8. Measure the AC voltage at the Source. Vin =7.071 mV

9. Calculate the mid-band gain. AV = 42.26

10. Convert the mid-band gain to its dB value AV(dB) = 32.52dB

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)

Equation 1

b. Frequency due to the bypass capacitor. The source resistance in this case is in parallel.

Equation 2

c. Frequency due to the output capacitor.

Equation 3

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)

fC1

 

 

fC2

 

 

fC3

 

 

fCL(dom)

 

 76.773 MHz

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. AV(dB) = 32.536 dB

16. How does this compare to the value measured in step 10?

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. fCL(dom) = 76.773 MHz. 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. FCH(dom) =12.652 Hz

*********End of the experiment *********

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