Week 9

Speakers

1. Measure the resistance of the speaker. Compare this value with the value you would find online.

All microspeakers from CUI are 8 Ohms +/- 15%. Our resistance value is 8.3 Ohms.

2. Build the following circuit using a function generator setting the amplitude to 5V (0V offset). What happens when you change the frequency? (video) Figure 1: Test setup for the speaker. Fill the following table. Discuss your results. 

The higher the frequency of the function generator, the higher the pitch of the speaker:
Table1
Table 1: Our group’s description of the sound from the speaker at different frequencies

3. Add one resistor to the circuit in series with the speaker (first 47 Ω, then 820 Ω). Measure the voltage across the speaker. Briefly explain your observations. Discuss your results.

Table2
Table 2: The oscilloscope amplitude comparison when using different valued resistors
The higher the value of the resistor, the less amplitude the oscilloscope reads. This makes sense, as the higher resistor would absorb more power from the source than the lower value one would.

4. Build the following circuit. Add a resistor in series to the speaker to have an equivalent resistance of 100 Ω. Note that this circuit is a high pass filter. Set the amplitude of the input signal to 8 V. Change the frequency from low to high to observe the speaker sound. You should not hear anything at the beginning and start hearing the sound after a certain frequency. Use 22 nF for the capacitor.

Screenshot 2016-03-14 at 8.51.51 AM.png

a. Explain the operation. (video)

The voltage source is connected to a capacitor and is grounded. The capacitor is connected to the output and the load. The load in this case is the speaker and resistor in series. The capacitor and resistor work to reduce the amplitude(and thus the power) of frequencies above approximately 67 kHz.

b. Fill out the following table by adding enough (10-15 data points) frequency measurements. Vout is measured with the DMM, thus it will be rms value.

 Table3

Table 3: The root-mean-square voltage across the load, and the ratio of Vout to Vin

c. Draw Vout/Vin with respect to frequency using Excel.

W9Q4c

A graph of Vout/Vin vs Frequency with an exponential x axis and a linear y axis

d. What is the cut off frequency by looking at the plot in b?

The cutoff frequency is meant to be approximately 67kHz, however from our experimental observations it is actually around 3.5 MHz.

5. Design the circuit in 4 to act as a low pass filter and show its operation. Where would you put the speaker? Repeat 4a-g using the new designed circuit (e, f, g are for blogI).

a. Explain the operation. (video)

 

The voltage source is connected to the load and is grounded. The load is connected to the output and the capacitor in parallel. The load in this case is the speaker and resistor in series. The capacitor and resistor work to reduce the amplitude(and thus the power) of frequencies below approximately 67 kHz.

b. Fill out the following table by adding enough (10-15 data points) frequency measurements. Vout is measured with the DMM, thus it will be rms value.

 Table4.PNG

Table 4: The root-mean-square voltage across the load, and the ratio of Vout to Vin

c. Draw Vout/Vin with respect to frequency using Excel.

W9Q5c

A graph of Vout/Vin vs Frequency with an exponential x axis and a linear y axis

d. What is the cut off frequency by looking at the plot in b?

The cutoff frequency is meant to be approximately 67kHz, however from our experimental observations it is actually around 5.6 kHz.
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2 thoughts on “Week 9

  1. I like format of the blog. Everything looks good for the graphs and I am wondering if I should have taken more points at the right areas to visually see the purpose of each filter.

    Like

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