Lab 1 - Speaker Circuit

The topmost picture is the actual circuit to implement the LM555 timer, while the one below is the schematic diagram for the same circuit. The following paragraphs describe how the resistance values were arrived at.



Given above is the part of our circuit where the transistor is placed. The aim was to produce a sound at a frequency which is decided by the parameters of the LM555 circuit. The signal from the LM555 circuit is received at the base of the transistor. The speaker volume depends on the current flowing through it and voltage drop across its terminals. Value R6 decides the current Ic while R5 will govern Ib.

• From the parameters defined for the speaker: Power = 0.5 W, Resistance = 7.3 Ώ (measured value). Using the equation P=I^2*R, we found the maximum current which can be passed through the speaker. Imax = 260 mA.

• Deciding upon a value of 100mA for current Ic, we calculated R6. R6 = (Vcc - Vce - 7.3*0.1 )/0.1 = 40 Ώ. Here Vcc = 5V, and Vce = 0.3V
  

• For the value of R5 we decided a current Ib of 15 mA. Then R5 = (Vbb – Vbe)/0.015. For a 74LS02 chip the typical output high voltage is 3.4V which is nothing but Vbb = 3.4V, while Vbe = 0.6V. Thus we got R5 = 187 Ohms.

• Selecting from the set of available resistances, we got R5 = 175 Ohms and R5 = 48 Ohms. The actual values of Ic and Ib were now 85 mA and 16 mA respectively.

• The LM555 produces a continuous waveform when operated in the astable mode for which the external circuit is given. The values of Ra, Rb and C decide the frequency of the waveform as well as its duty cycle.

• We were provided with a potentiometer of 10 k Ώ to be connected as Ra. As explained in class, it was necessary to include a fixed resistance in addition to the potentiometer. But we checked as to what happens without a fixed resistance, and found that the circuit did not work.

• The formulae for calculating frequency and duty cycle are:

Duty Cycle, D = Rb/(Ra+ 2Rb)
Frequency, F = 1.44/C/(Ra + 2Rb)

Here, Ra = R(pot) + R1
R(pot) varies from (0 – 10) k Ώ .

To get a frequency of 1000Hz when the potentiometer slide is at its mid portion

C = 0.1µF, R1 = 505 Ώ, Rb = 4.7k Ώ.

• Thus, when a potentiometer (Range = (0 – 10)k Ώ )was used in series with Rb, we got a range of frequencies from (723 – 1453)Hz, which corresponds approximately to an Octave.

• In order to get an Octave, one can stick to the equation R1 + 2Rb = 10000. 10000 is the maximum resistance given by the potentiometer.