Month: May 2015

        The frequency and the sine wave was set by the function generator. The input, output wave forms were observed from the CRO. The conditions taken were  a 50KΩ , 1KΩ and a current source. The input was fed to the 2.1 speaker’s connector jack to produce the sound of the fed sine wave.

From the tabulation the maximum output 300mV is achieved when the input is set to a peak to peak value of 5.5V connected from the current source (PNP BJT’s collector junction ).

To determine the amplification,

the voltage at the base of the NPN transistors is ≈ 2.5V. now divided by the resistance of resistor R6 gives us the current.

 now, the obtained current value multiplied by R5

gives the amplified output ≈ 5V.

• We wanted to use the transistor as an amplifier.
• The input voltage at the base was 3v.
• Two resistors, one at the emiiter(R1) and the other at the collector(aR1) were connected.
• 9v supply was given at R1.

Constraints :

• We wanted head room of 2v.
• We wanted to obtain maximum gain.

Design :

• Subtracting 0.7v at the emitter from 3v at the base we get 2.3v.
• Current across R1 is

• As we needed 2v head room, the voltage drop across aR1 needed to be 4.7v(ie, 4.7v+2.3v=7v) which is 2v lesser then 9v.
• Current across aR1 is

• The maximum gain we can get is a gain of 2.

• Hence the resistor value at the collector should be two times more than the one at the emitter.

• We wanted to use a transistor as a current source.
• Two resistors, one at the emiiter(R1) and the other at the collector(R2) were connected.
• A voltage divider circuit with a supply of 9v was connected at the base that could supply certain voltage.

Constraints :

• We wanted 4.5mA of current to pass through.
• 9v supply was given to the resistor at the emitter.

Design :

• The resistor at the collector(R2) was chosen as 1k based upon the resistance of the mic.
• We knew that the diode would turn on at around 0.7v.
• The voltage drop across R1 should be some resistance that gives 4.5mA of current.
• Depending upon the availability of the resistors, 330 ohms was chosen for R1, keeping the 9v supply in mind and voltage at emiiter, it was calculated that 1.5v drop would give us 4.5mA of current.
• Now from the voltage divider, a certain amount of voltage needed to be supplied so that the drop across R1 is 1.5v (ie, 9v – 7.5v)
• So we wanted 6.8v from the voltage divider (ie, 6.8v + 0.7v = 7.5v)
• Hence, the voltage divider circuit was designed with the following formula.

• The ratio of the scaling factor was approximately 3 and depending upon the availability of the resistors, R3 was taken as 4.7k and R4 was taken as 14.5k.