9.3. Differential Amplifier Balancing Circuit

The principle of the balancing of the differential-amplifier stage (e.g., in an opamp) is often based on a circuit similar to that in Fig. 9.2, as shown in Fig. 9.5. Imbalance could be due, for example, to V_tpo1 ≠ V_tpo2 and k_p1 ≠ k_p2. Balancing is implemented by adjusting R₁ ≠ R₂, where R₁ = R_S1//R_x and R₂ = R_S2||R_y.

The relationship among the currents, parameters, and resistors is obtained by writing the loop equation around the source resistors and the gate – source terminals, which is

Equation 9.11

A solution for V_SD2 can be obtained from (9.11) for a given set of parameters, resistors, and bias variables; on the other hand, R_x and R_y are adjusted to obtain a certain V_SD2.

For the special case of I_D1 = I_D2 = I_D, the difference between the resistors in (9.11) is

Equation 9.12

For a numerical example, suppose the goal is to set . By design, in (9.12),

Equation 9.13

Any slight possible difference in λ_p is neglected.

In a practical opamp with this balancing configuration, the source nodes are connected to external pins. These pins are connected through external resistors R_x and R_y to V_DD as shown in Fig. 9.5. For adequate adjustment sensitivity, R_x and R_y are greater than R_S1 ≈ R_S2 by at least a factor of 10. In practice, the external resistances are implemented with a potentiometer.

Assume that the transistor parameter values are V_tpo1 = 1.49 V, V_tpo2 = 1.51 V, k_p1 = 345 μA/V², k_p2 = 335 μA/V², and λ_p = 0.02 V^–1. Also assume that the bias current I_D = 100 μA. We select R_S1 = R_S2 = R_S = 3 kΩ (Voltage I_DR_s = 0.3 V) and we use R_x + R_y = R_pot = 25 kΩ; the two resistors are segments of a 25-kΩ potentiometer. The segments of the potentiometer are then determined from a solution to

Equation 9.14

where δR is obtained from (9.12) with V_SD2 = V_SG1 as obtained from (9.13). The resistor values are R_x = 16 kΩ and R_y = 9 kΩ to give R₁ = 2.53 kΩ and R₂ = 2.25 kΩ.

An evaluation of V_SD2 without the balancing circuit can be made by setting δR = 0 in (9.12) and for solving V_SD2 to obtain

Equation 9.15

V_SG1 is again obtained from (9.13).

With the numbers from the example above, V_SD2 = 7.87 V. We note that with λ_p = 0.01 V^–1, the solution is V_SD2 = 13.2 V and the circuit might well have exceeded the power-supply limits. If V_tpo1 = 1.45 V, V_tpo2 = 1.55 V, and λp = 0.02 V^–1 that is, for an extreme case of the difference of threshold voltage, the balancing circuit will still function but now R_x = 21.8 kΩ and R_y = 3.24 kΩ with a small R₂ = 1.56 kΩ.