**Differential Amplifier Circuit using Transistors:**

The Differential Amplifier Circuit using Transistors is widely applied in integrated circuitry, because it has both good bias stability and good voltage gain without the use of large bypass capacitors. Differential amplifiers can also be constructed as discrete component circuits.

Figure 12-31(a) shows that a basic Differential Amplifier Circuit using Transistors consists of two voltage divider bias circuits with a single emitter resistor. The circuit is also known as an **emitter-coupled amplifier**, because the transistors are coupled at the emitter terminals, If transistors Q_{1} and Q_{2} are assumed to be identical in all respects, and if V_{B1} = V_{B2}, then the emitter currents are equal, and the total emitter current is,

Like the emitter current in a single-transistor voltage divider bias circuit, I_{E} in the differential amplifier remains virtually constant regardless of the transistor h_{FE} value. This results in, I_{E1}, I_{E2}, I_{C1}, and I_{C2} all remaining substantially constant, and the constant collector current levels keep V_{C1 }and V_{C2} stable. So, the differential amplifier has the same excellent bias stability as a single-transistor voltage divider bias circuit.

The circuit of a Differential Amplifier Circuit using Transistors using a plus-minus supply is shown in Fig. 12-31(b). In this case, the voltage across the emitter resistor is (V_{EE} – V_{BE}), as illustrated,

The base resistors (R_{B1} and R_{B2}) are included to bias the transistor bases to ground while offering an acceptable input resistance to a signal applied to one of the bases. The transistor emitter currents (I_{E1} and I_{E2}) are exactly equal only if the devices are perfectly matched. To allow for some differences in transistor parameters, a small-value potentiometer (R_{EE}) is sometimes included between the emitters, ( see Fig. 12-32). Adjustment of R_{EE} increases the resistance in series with the emitter of one transistor, and reduces the emitter resistance for the other transistor. This reduces the I_{E} for one transistor and increases it for the other, while the total emitter current remains constant.

**AC Operation:**

Consider what happens when the at input voltage (v_{i}) at the base of Q_{1} is positive-going, as illustrated in Fig. 12-33. Q_{1} emitter current (I_{E1}) increases. Also, I_{E2} decreases, because the total emitter current (I_{E1} + I_{E2}) remains constant. This means that I_{C1 }increases and I_{C2}_{ }decreases, and consequently, V_{C1} falls and V_{C2} rises, as shown. So, the ac output voltage at Q_{1} collector is in anti-phase to v_{i} at Q_{1} base, and the output at Q_{2} collector is in phase with v_{i}.

**Voltage ****Gain:**

The voltage gain of a single-stage amplifier with an unbypassed emitter resistor and no external load is given by

Referring to Fig. 12-34, it is seen that the resistance looking into the emitter of Q_{2} is h_{ib}, so h_{ib}||R_{E} behaves like an unbypassed resistor in series with the emitter of Q_{1}. Neglecting R_{E} because it is very much larger than h_{ib}, the voltage gain from the base of Q_{1} to its collector is,

this reduces to,

Equation 12-24 gives the voltage gain from one input terminal to one output of a differential amplifier. It is seen to be half the voltage gain of a similar single-transistor CE amplifier with R_{E} bypassed; but note that the differential amplifier requires no bypass capacitor. This is an important advantage, because bypass capacitors are usually large and expensive.

Another way to contemplate the operation of the Differential Amplifier Circuit using Transistors is to think of the input voltage being equally divided between Q_{1} base-emitter and Q_{2} base-emitter. This is illustrated in Fig. 12-35 where it is seen that (for a positive-going input) v_{i}/2 is applied positive on the base of Q_{1}, while the other half of v_{i} appears positive on the emitter of Q_{2}. Thus, for v_{i} at Q_{1B}, transistor Q_{1} behaves as a common-emitter circuit, and because Q_{2} receives the input at its emitter, Q_{2} behaves as a common-base circuit. Consequently,

**Input and Output Impedances:**

The input impedance at the base of a CE circuit with an unbypassed emitter resistor is,

Referring to Fig. 12-34, the differential amplifier has. h_{ib}||R_{E} as an unbypassed resistor in series with the emitter of Q_{1}. Neglecting R_{E} (because R_{E} ≫ h_{ib}). The input resistance at Q_{1B} is,

This reduces to,

Note that there are usually bias resistors in parallel with Z_{b}, so that the circuit input impedance is

As in the case of CE and CB circuits, the output impedance at the transistor collector terminals is given by

**DC Amplification:**

When one transistor base is grounded in a Differential Amplifier Circuit using Transistors, and an input is applied to the other one, as already discussed, v_{i} is amplified to produce the outputs at the collector terminals. In this case v_{i} is the voltage difference between the two base terminals. Figure 12-36 shows a differential amplifier with dc input voltages V_{i1} and V_{i2} applied to the transistor bases. If the voltage, gain from the base to the collector is A_{v}, the dc voltage changes at the collectors are;

It is seen that the differential amplifier can be employed as a direct-coupled amplifier, or dc amplifier. The term difference amplifier is also used for this circuit.

**Design Calculations:**

Design procedures for a Differential Amplifier Circuit using Transistors are similar to those for voltage divider bias circuits. Because there is no bypass capacitor in a differential amplifier, one of the coupling capacitors determines the circuit lower cutoff frequency (f_{1}). The capacitor with the smallest resistance in series with it is normally the largest capacitor, and in the case of a differential amplifier this is usually the input coupling capacitor. So, the input coupling capacitor determines the circuit lower cut-off frequency.

Consider the capacitor-coupled differential amplifier in Fig. 12-37. The circuit uses a plus-minus supply, and a single collector resistor (R_{C}). No output is taken from Q_{1} collector, so there is no need for a collector resistor. R_{C} is selected in the usual way for a small-signal amplifier; R_{C} ≪ R_{L}. The collector-emitter voltage should be a minimum of 3 V, as always. Then, I_{C} is calculated from R_{C} and the selected voltage drop across R_{C}.

The total emitter current is determined as,

The base bias resistors are determined by,

As discussed, capacitor C_{1} sets the lower cutoff frequency. So.

and C** _{2}** is determined by,