**Single Input Balanced Output Differential Amplifier:**

In a single input balanced output differential amplifier an input signal is applied to either input, with the other input connected to ground. However, due to the common-emitter connection the input signal operates both transistors, resulting in output from both collectors.

Single input balanced output differential amplifier is depicted in Fig. 20.11. In this circuit, an input signal is applied to the base of transistor Q_{1}, with the other input connected to ground. Output is measured between the two collectors, which are at the same potential, and that is why the output is said to be **balanced output**.

**DC Analysis:**

The dc analysis process and the bias equations for this configuration are identical to those of the two previous configurations. This is because the dc equivalent circuit for all the three configurations is the same. Hence the quiescent collector current I_{CQ} and quiescent collector-emitter voltage V_{CEQ} are given by the expressions

**AC Analysis:**

Figure 20.12(a) depicts the ac equivalent circuit of this configuration with small signal T-equivalent models substituted for the transistors.

**1. Voltage Gain:** First of all let us discuss the important points about the circuit depicted in Fig. 20.12(a).

(i) During the positive half-cycle of the input signal v_{in1}, the V_{BE} for the transistor Q_{1} is positive and that of transistor Q_{2} is negative, as indicated in Fig. 20.11. Thus the collector current in transistor Q_{1} increases and that in transistor Q_{2} falls from the quiescent value I_{CQ}. This change in collector currents during the positive half cycle of the input signal is shown in Fig. 20.12(a) by indicating them flowing in the same direction. During the negative half-cycle of input signal v_{in1}, the the opposite action takes place i.e. i_{C1} falls and i_{C2} strengthens.

(ii) The voltage across the collector resistance of transistor Q_{2} is positive, and that across the transistor Q_{1} is negative with respect to ground. This is because the voltage across each resistor is consistent with the direction of current sources i_{C1} and i_{C2}_{.}

(iii) The output voltage v_{out }is equal to the difference of voltages at the two collectors C_{1} and C_{2 }i.e. v_{out} = v_{C1} -v_{C2}.

Applying KVL to loops I and II of the circuit given in Fig. 20.12 (a), we have

Usually R_{i1}/β_{ac }is negligibly small; therefore, neglecting it for simplicity, we have

Solving above equations for i_{e1} and i_{e2}, we have

The output voltage,

Substituting the values of i_{e1} and i_{e2} from Eqs. (20.30) and (20.31) respectively in Eq. (20.32), we have

Thus voltage gain,

The above equation reveals that the voltage gain of the single input balanced output differential amplifier is equal to that of dual-input, balanced-output differential amplifier [Eq. (20.15)]

The input and output waveforms for this configuration are shown in Fig. 20.12 (b).

**2. Differential Input Resistance:** The input resistance R_{in} seen from the input signal source is given by expression

Substituting the value of i_{e1} from Eq. (20.30) in above equation, we have

**3. Output Resistance:** The output resistance measured at collector C_{1}(R_{out1}) and the output resistance measured at the collector C_{2}(R_{out2}) are given by equation