The differential amplifier, abbreviated as DIFF AMP, is the basic stage of an integrated OP AMP with differential input. Its design is, therefore, mainly related to IC fabrication techniques. However, employing discrete components it is also used in some circuits.
Generally, the function of a differential amplifier is to amplify the difference of two signals. Fig. 4.7(a) shows a linear active device Fig. 4.7(a) with two input signals V1, V2 and one output signal Vout, all measured with respect to ground. Fig. 4.7(b) represents the basic differential amplifier circuit, the two transistors Q1 and Q2, of which have identical characteristics with a common emitter resistor RE. The collector load resistors are also made equal, i.e., RL1 = RL2 and the inputs are identical, i.e., R1 = R2 and V1 = V2.
In an ideal differential amplifier the output signal is proportional to the difference between the two input signals.
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This can be expressed as-
Equation (4.21), as stated, is for an ideal case. For a practical DIFF AMP equation (4.21) is not considered, in general, as the output depends not only upon the difference signal Vd but also on the average level, known as the common mode signal. This common mode signal can be represented by Vc, where-
ii. When the two input signals are equal in amplitude but 180° out of phase, we have-
Therefore, in such differential amplifier the output signal becomes twice the gain times of the input signal.
The Subtracting Amplifier:
In the circuit shown in Fig. 4.8 neither of the operational amplifier input terminals is connected directly to ground, so it cannot be assumed that they are both at ground potential. Therefore, in writing the current flow equations, a value V3 is assigned to the voltage level of both the inverting and non-inverting inputs, measured with respect to the ground. That both can be considered to be at the same potential is a consequence of the fact that any value of V4 can be produced by a negligibly small value of p.d. between the two input terminals.
As no current flows into the amplifier input terminals, we can write-
RC Active Filters:
Active filters are filters that employ passive elements, usually resistors and capacitors in conjunction with active elements, like OP AMP, to obtain characteristics similar to those of LCR, passive filters. Since, RC active circuits contain no inductors, it is possible to integrate them.
Consider the circuit in Fig. 4.9. It consists of an OP AMP connected as a voltage follower so that it has a gain of unity. Two resistors and two capacitors are required to make the filter second order. We can write two simultaneous equations for this circuit.
The current in R1 equals the sum of those in R2 and C1:
Note that these equations are written assuming that the same voltage V3 appears at both the non-inverting input and the output of the OP AMP. This is justified, since the amplifier is connected as a voltage follower. Equations (4.25) and (4.26) can be solved to give-
A Constant-Amplitude Phase Shifter:
The circuit is most easily analyzed by using operational calculus in which 1/p represents integration. Assuming this we can write the relationship for a capacitor-
It can be shown that equation (4.30) describes an all-pass transfer function, which means that if V1 is held constant in magnitude as frequency is changed, V3 will also remain constant and equal to V1 in magnitude while the phase of V3 with respect to V1 changes. This transfer function is independent not only of the operational amplifier characteristics but also of R. It can be shown also that if r is varied from 0 to ∞ at a particular frequency, V3 will shift in phase 180° with respect to V1 while its magnitude remains constant.
Equation (4.30) is the transfer function of a low-pass Butterworth filter with a 3 dB cut-off frequency of 1/2π Hz, from elementary filter theory. The filter is maximally flat because the coefficient of the p term in the denominator is √2.
Analysis of the Differential Amplifier:
The basic circuit used to provide gain in the OP AMP is as shown in Fig. 4.11. The emitters of the two transistors are joined and connected to a constant current source.
With V1 = V2 the collector currents are IC1 = IC2 ≈ 1/2 (neglecting the base currents).
With V2 fixed, an increase in V1 will divert a larger fraction of the fixed current I into T1. Hence VC1 will fall in, V1 will divert I into T1. Hence VC1 will fall in, V1 will divert a large part of I into T2.
Hence the gain, g = ∆VC1/∆V1 from the input of T1 to the collector of T1 will be negative (inverting).
While the gain ∆VC2/∆V2 will be positive (non-inverting). Corresponding comments apply to the gain of a signal applied to the base of T2.
As long as the current source I is precisely fixed, the change in current in one transistor must be equal and opposite of the change in current in the other transistor. In this case, the various gains must be equal or equal and opposite; i.e.-
Let then that starting from an arbitrary initial condition V1 and V2 are changed by arbitrary increments ∆V1 and ∆V2. Then the change in the output at the collector of T1 would be-
It thus appears that if ∆V1 and ∆V2 are equal, i.e., common-mode signal is applied to both inputs, the outputs ∆VC1, and ∆VC2 will be zero. This feature is described by saying that the amplifier rejects a common- mode signal or by saying that the common-mode gain is zero.
On the other hand, when a difference develops between ∆V1 and ∆V2, this difference is amplified. For this reason the circuit is often referred to as a differential amplifier.
To calculate the gain g of the differential amplifier we replace the transistors by the equivalent representation shown in Fig. 4.12. This equivalent circuit is a simplified form of the h-parameter circuit.
When the simplified h-parameter equivalent circuit transistor shown in Fig. 4.12 is used, the incremental equivalent circuit of the difference amplifier appears as in Fig. 4.13.
Circuit to ground and the current source in the emitter has been replaced by an open circuit.
The sum of the currents entering node p or p’ must equal to zero.
Hence,
Consequently, the current ∆IP = 0 and the lead from P’ to P may be removed.
Finally, the equivalent circuit for the purpose of calculating the currents through the collector resistors RC is as shown in Fig. 4.14.
Here the two current sources in series, each carrying the same current, have been replaced by a single current source hFE∆lBI, where,
So the input impedance becomes 50 kΩ. However, higher input impedances are possible through the use of Darlington input circuits and FET inputs.
Common-Mode Rejection Ratio:
Let the input signal to the OP AMP be V1 and V2 as in Fig. 4.15.
Instead of specifying the inputs directly by giving V1 and V2, we may equivalently specify the inputs in terms of a difference signal input Vd and a common-mode input Vc defined in terms of V1 and V2 by-
If V1 and V2 are equal and opposite, then Vc = 0; if V1 and V2 are equal, Vd = 0. The signals Vd and Vc. measure respectively the difference and average value of the input signal. The signals V1 and V2 are uniquely determined by the equations,
Now, let us suppose that as in an ideal amplifier, the gain A1 measured with respect to input 1 and the gain A2 measured with respect to input 2 are equal and opposite. Then if V1 and V2 are equal, Vd = 0 and the output will be zero even if Vc ≠ 0.
In this case we thus find that there is not only a gain Ad = 1/2(A1 – A2) for the difference of the input signal Vd but also a gain Ac = A1 + A2 to the common-mode signal Vc .