Q.Explain with a neat circuit diagram working of a dual input balanced output differential amplifier.

In a dual input balanced output differential amplifier both bases are driven and the output is taken between the two collectors; its action is analysed by superposition of the two inputs.

[!ANSWER]

With $V_{i1}$ and $V_{i2}$ applied to the two bases, each input alone makes one transistor act as a CE stage and the other as a CB stage; by superposition the output taken between the two collectors is proportional to the difference of the inputs, with voltage gain $A_V = V_O/(V_{i1}\sim V_{i2})$.

[!TLDR]

Both inputs $V_{i1}$, $V_{i2}$ drive the two bases and the balanced output is taken between the two collectors; by superposition the circuit amplifies the difference $(V_{i1}\sim V_{i2})$ with gain $A_V = V_O/(V_{i1}\sim V_{i2})$.

Circuit: Two identical transistors $Q_1$ and $Q_2$ have equal collector resistors $R_{C1}=R_{C2}=R_C$ returned to $+V_{CC}$, and their emitters are joined and returned through a common resistor $R_E$ to $-V_{EE}$. Inputs $V_{i1}$ and $V_{i2}$ are applied to the bases of $Q_1$ and $Q_2$; outputs $V_{O1}$ and $V_{O2}$ are taken from the collectors and the balanced output $V_O$ is measured between the two collectors.

Working (by superposition):

1. Considering $V_{i1}$, grounding $V_{i2}$: $V_{i1}$ is the input to $Q_1$ at its base and the output $V_{O1}$ is taken at its collector, so $Q_1$ acts as a common-emitter (CE) amplifier and $V_{O1}$ is $180^{\circ}$ out of phase with $V_{i1}$. The same signal $V_{i1}$ appears at the common emitter and is fed to the emitter of $Q_2$, so $Q_2$ acts as a common-base (CB) amplifier; its collector output $V_{O2}$ is in phase with $V_{i1}$.

2. Considering $V_{i2}$, grounding $V_{i1}$: now $V_{i2}$ is the input to $Q_2$ which acts as a CE amplifier, giving $V_{O2}$ out of phase with $V_{i2}$; $Q_1$ acts as a CB amplifier, so $V_{O1}$ is in phase with $V_{i2}$.

When both signals are applied, the two effects add up. Because a balanced output is measured between the two collectors, the two collector voltages subtract, so the circuit responds to the difference of the inputs. The voltage gain is therefore

For a purely differential input ($V_{i1}\neq V_{i2}$) the differential-mode gain is $A_d = R_C/2r'_e$; for a common (equal) input the two collector signals are equal and, being taken differentially, cancel to give ideally zero common-mode output. This rejection of common signals (noise) while amplifying the difference is exactly why the differential amplifier is the input stage of every op-amp in the Karnataka 2nd PUC Electronics syllabus.

[!ANSWER]

With both bases driven, each input in turn makes one transistor a CE stage and the other a CB stage; taking the output between the two collectors makes the two collector voltages subtract, so the amplifier responds to the difference of the inputs with gain $A_V = V_O/(V_{i1}\sim V_{i2})$. Equal (common-mode) inputs give equal collector signals that cancel at the balanced output, so common-mode noise is rejected.