To achieve a specified amplifier gain in designing small signal amplifier circuits, we usuall use op-amp amplifier design to ease the gain setting. We can use transistor feedback amplifier as the base of the amplifier design when the gain accuracy is not critical. Following figure shows the circuit diagram :

feedback amplifier using transistors circuit diagram

feedback amplifier using transistors circuit diagram

For the transistor operation, without TR2, taking the TR1 emitter as the output then we can see an emitter follower circuit, the emitter voltage will follow the base voltage if the voltage variation if TR1 base. The mechanism of this is based on the face that the small base-emitter current will cause a large collector-emitter current. However this current cause the emitter voltage – the voltage across the emitter resistor – to rise and then this reaction will decrease the base-emitter since the current is the result of base-emitter voltage difference. The feedback mechanism in the circuit is based of this action and reaction. When we look at the transistor TR2, this transistor amplify the collector current of the TR1 and now the feedback source is supplied from the TR2 collector to the resistor R4. Since the feedback is now handled by TR2, the TR2 make the collector current of TR1 very small. And then the current drawn by the base of TR1 will become smaller and provides better amplifier impedance at the input.

For the Amplifier Gain, the feedback principle remain unchanged. Voltage at TR1 emitter is maintained to follow the base voltage. Since almost all current comes from R4, the voltage at the TR2 collector will have the equation as follows :

Vout=VR4+VR5;

Since VR5 follow the base voltage, and the voltage at R4 and R5 is proportional to their values, then:

VR4= (R4/R5) x VR5

The circuit is an ac amplifier, because the decoupling capacitor is used at the input and the output, so the offset error by TR1 base-emitter  drop can be neglected,  thus VR5=Vinput, so

Vout=VR4+Vinput;

Vout=((R4/R5)xVinput)+Vinput;

Vout=(1+(R4/R5))xVinput;

You get 1+(R4/R5) as the amplifier gain of this transistor circuit.

For the amplifier impedance, we can assume that the amplifier gain of the input only depends on R1 and R2. This is because the current drawn by TR1 can is very small and can be neglected. Therefore, the input impedance is R1 and R2 combination in parallel circuit, since the supply is modeled as a short circuit in AC analysis. The output impedance is not symmetric, thus you can drive the output instantly to charge the load capacitance since the transistor TR2 is not limited by any transistor. When the output have to fall down to zero, then the capacitor discharging process will be slower since it should flow through R4 and R5. For your worst case condition, the output impedance as the value of R4 and R5 combination in series circuit can be assumed.

For DC Amplification, the circuit can be used as DC amplifier by removing the C1 and C2 and the bias resistor R1 and R2, as long as the input voltage level is always above TR1 base-emitter bias voltage (around 0.7v). The output then is a valid product of the input and amplifier gain.

For the Amplifier Output Protection, a resistor can be inserted between the output tap and the TR2 collector. The output is tapped at the junction of the inserted resistor and the R4. When the output is shorted to ground, this will prevent the excessive current at TR2. This resistor insertion will limit the maximum output voltage of this amplifier circuit.  Select the optimal value between protection and the desired upper voltage limitation.

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