Biasing of Differential Amplifiers

Constant Current Bias:

In the dc analysis of differential amplifier, we have seen that the emitter current I_E depends upon the value of b_dc. To make operating point stable I_E current should be constant irrespective value ofb_dc.

For constant I_E, R_E should be very large. This also increases the value of CMRR but if R_E value is increased to very large value, I_E (quiescent operating current) decreases. To maintain same value of I_E, the emitter supply V_EE must be increased. To get very high value of resistance R_E and constant I_E, current, current bias is used.

Figure 5.1

Fig. 1, shows the dual input balanced output differential amplifier using a constant current bias. The resistance R_E is replace by constant current transistor Q₃. The dc collector current in Q₃ is established by R₁, R₂, & R_E.

Applying the voltage divider rule, the voltage at the base of Q₃ is

Because the two halves of the differential amplifiers are symmetrical, each has half of the current I_C3.

The collector current, I_C3 in transistor Q₃ is fixed because no signal is injected into either the emitter or the base of Q₃.

Besides supplying constant emitter current, the constant current bias also provides a very high source resistance since the ac equivalent or the dc source is ideally an open circuit. Therefore, all the performance equations obtained for differential amplifier using emitter bias are also valid.

As seen in I_E expressions, the current depends upon V_BE3. If temperature changes, V_BE changes and current I_E also changes. To improve thermal stability, a diode is placed in series with resistance R₁as shown in fig. 2.

Fig. 2

This helps to hold the current I_E3 constant even though the temperature changes. Applying KVL to the base circuit of Q₃.

Therefore, the current I_E3 is constant and independent of temperature because of the added diode D. Without D the current would vary with temperature because V_BE3 decreases approximately by 2mV/° C. The diode has same temperature dependence and hence the two variations cancel each other and I_E3 does not vary appreciably with temperature. Since the cut – in voltage V_D of diode approximately the same value as the base to emitter voltage V_BE3 of a transistor the above condition cannot be satisfied with one diode. Hence two diodes are used in series for V_D. In this case the common mode gain reduces to zero.

Some times zener diode may be used in place of diodes and resistance as shown in fig. 3. Zeners are available over a wide range of voltages and can have matching temperature coefficient The voltage at the base of transistor QB is

Fig. 3

The value of R₂ is selected so that I₂ » 1.2 I_Z(min) where I_Z is the minimum current required to cause the zener diode to conduct in the reverse region, that is to block the rated voltage V_Z.

Current Mirror:

The circuit in which the output current is forced to equal the input current is said to be a current mirror circuit. Thus in a current mirror circuit, the output current is a mirror image of the input current. The current mirror circuit is shown in fig. 4.

Fig. 4

Once the current I₂ is set up, the current I_C3 is automatically established to be nearly equal to I₂. The current mirror is a special case of constant current bias and the current mirror bias requires of constant current bias and therefore can be used to set up currents in differential amplifier stages. The current mirror bias requires fewer components than constant current bias circuits.

Since Q3 and Q4 are identical transistors the current and voltage are approximately same

For satisfactory operation two identical transistors are necessary.

Example - 1

Design a zener constant current bias circuit as shown in fig. 5 according to the following specifications.     (a). Emitter current -IE = 5 mA     (b). Zener diode with Vz = 4.7 V and Iz = 53 mA.     (c). βac = βdc = 100, VBE = 0.715V     (d). Supply voltage - VEE = - 9 V. Solution: From fig. 6 using KVL we get	Fig. 5
Practically we use RE = 820 kΩ Practically we use R2 = 68 Ω The designed component values are:                 RE = 860 Ω                 R2 = 68 Ω	Fig. 6

Example - 2

Design the dual-input balanced output differential amplifier using the diode constant current bias to meet the following specifications.

1. supply voltage = ± 12 V.
2. Emitter current I_E in each differential amplifier transistor = 1.5 mA.
3. Voltage gain ≤ 60.

Solution: The voltage at the base of transistor Q3 is Assuming that the transistor Q3 has the same characteristics as diode D1 and D2 that is VD = VBE3, then Practically we take RE = 240 Ω. Practically we take R2 = 3.6 kΩ. To obtain the differential gain of 60, the required value of the collector resistor is The following fig. 7 shows the dual input, balanced output differential amplifier with the designed component values as RC = 1K, RE = 240 Ω, and R2 = 3.6KΩ.

Fig. 7