Week 9

Week 9

Blog Week 9 - Group 2

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

We measured the resistance value of the speaker to be 8.9 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)

We found that when you increase the frequency, the pitch of the speaker goes way high.  We started the speaker off at 10K output with 5.00 Volts, and 1000KHz.  As we increase the frequency, the pitch of the speaker goes higher. 

Frequency	Observation
1 Hz	No sound
10 Hz	Very soft sound
0.1KHz	Vibrating
1kHz	Quite, high pitched tone
5kHz	Louder, higher pitched tone
10kHz	Very high pitched tone

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.

Resistor value	Oscilloscope output	Observation
47 Ω	1.5V peak to peak	As we increase the frequency from the function generator, the frequency across the speaker increase.
820 Ω	1.5V peak to peak	Peak to peak voltage across the speaker increases as amplitude of input voltage increases.

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.

a. Explain the operation. (video)

The function generator is connected to the capacitor, which is connected to a resistor and the speaker in series which has an equivalent resistance of 100 Ohms.  As we increase the frequency, the speaker pitch slowly - and I mean, very slowly - increases.  We didn't hear any sound from the speaker until 1.17 KHz.  But, this could be different by person as everyone's threshold of hearing is slightly different.

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.

Frequency( kHz)	Vout (V)	Vout/Vin( V)
0	.461	.082
10	.612	.109
20	1.22	.210
40	2.09	.375
60	2.78	.452
70	2.92	.508
80	3.08	.535
90	3.24	.558
100	3.34	.576
110	3.39	.588

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

d. What is the cut off frequency by looking at the plot in b?
the cut off frequency is going to be the nearest number to the 0.707V which comes from (1/sqrt2) which is 70 kHz

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. 

Frequency (kHz)	Vout( V)	Vout/Vin (V)
10	5.44	.954
20	5.19	.920
40	4.7	.818
50	4.3	.753
60	4.03	.689
70	3.64	.634
80	3.34	.557
90	3.12	.524
100	2.76	.503
110	2.54	.469

the cut off frequency is 90kHz here in the low pass filters based in the plot above.

6. Construct the following circuit and test the speaker with headsets. Connect the amplifier output directly to the headphone jack (without the potentiometer). Load is the headphone jack in the schematic. “Speculate” the operation of the circuit with a video