TRANSISTOR AMPLIFIER DESIGN

Good stability and sufficient amplification in Transistor based circuits depends on the proper selection of components and their layout. Gain of the transistor used as well as the current and voltage through it are the most important aspect in the working of a transistor amplifier. Here explains the design parameters of an one transistor amplifier

A PNP signal amplifier is shown in Fig.1. It is used to amplify the signals using a PNP transistor. Resistors are placed in their positions but values are not fixed. The resistors around T1 includes

1. R2-R1- A potential divider to provide base current to T1.
2. R3 – Emitter resistor of T1 that determines the Emitter voltage and current.
3. R4 –Collector resistor of T1 that determines the collector voltage and current and hence the strength of output signal.

Fig.1 Transistor Amplifier Layout

This circuit is designed to use with 9 volt DC power supply. The desirable collector current at the output is 1 mA. Thus we have to drop 1 volt across R3.Let us fix the value of R3 first.

Voltage drop required across R3 is 1 volt and desired current is 1 mA. Then 1V / 1mA = 1000 Ohms or IK. So IK resistor is selected as R3 so that 1 mA current appears in the collector since Emitter current = Collector current. Now let us go to the collector resistor R4.

Since IK resistor is used in the emitter, 1 mA current appears in collector resistor R4. For good stability of the circuit, the collector voltage should be equal to half of the supply voltage – emitter voltage. There fore
Collector voltage = 9-1 / 2 = 4 volts

Collector current as determined by 1K emitter resistor is 1mA and therefore the value of R4 is
4/1mA = 4000 Ohms or 4K. Nearest value is 3.9K. So 3.9K resistor is selected as R4.

Now we have to determine the values of R2 and R1, the potential divider that bias T1.

Value of R2 must be 10 times higher than that of R3.The value of R3 is 1K so 10K resistor is suitable for R2.

The base voltage of T1 is Vbe + emitter voltage. Here a Germanium transistor like SK100 is used. So Vbe is 0.2 volts. If you use a silicon transistor Vbe goes to 0.6 volts. Thus the base voltage of T1 is
0.2 + 1V =1.2 volts

Current passing through R1 equals the current passing through R2.

Current through R1 is 1.2V (Vbe)/10K (R2) = 120 mA.

R1 drops 9-1.2 volts. That is 7.8 volts at 120 mA.

Therefore the value of R1 is 7.8V / 120mA = 65K. Nearest value of R1 is 68K.

Power Rating of Resistors

After fixing the values of Resistors, it is important to assess the current flowing through them to select the power rating. Power of the resistor is calculated as

W = Amp x Amp x Ohm or Amp x Volt

A ¼ watt resistor can handle 50 mA or less. ½ w resistor handles 70 mA while 1w resistor handles 100 mA current. A 2 w resistor can handle up to 140 mA. If the power rating of resistor is not suitable, excess heat will be generated which causes power loss and the circuit’s functioning will be poor.

Here R3 and R4 passes only 1 mA current so ¼ w resistor is sufficient while R1 and R2 pass more current so that 1 w resistor is necessary.

The completed circuit of the transistor amplifier is shown in Fig.2

Transistor Amplifier Completed Circuit