An Enhanced Op-Amp Balanced Amplifier

Dec. 16, 1994
An improved version of the classic op amp balanced amplifier circuit computes the resistor values differently. It equalizes the impedance of both inputs by considering the op amp's active participation.

The classic version of the single op-amp balance amplifier has been accepted and used extensively throughout the industry (Fig. 1). It works well in low source impedance (bridging) configurations, but loses its punch in higher source impedance applications because of the varying input impedance of each input referred to ground. Consequently, the ability to reject outside signal injection is reduced, negating the design’s original purpose.

1. The classic single op-amp balanced circuit works well in low-source-impedance configurations, but tends to struggle in higher-source-impedance applications because of the varying input impedance of the inputs referred to ground.

An improvement on that classic design uses a different set of formulas to determine the resistor values (Fig. 2). The cost to implement the enhanced design is the same as the original version, but the new design equalizes the impedance of both inputs by taking into account the active participation of the op amp.

2. A modified version of the classic op-amp configuration of Figure 1 employs a different set of formulas to determine the resistor values. It equalizes the impedance of both inputs by considering the op amp’s active participation.

For convenience, both figures are set up for a gain of 1 and an input impedance of 600 Ω. If a 2-V differential signal is applied to each with –1 V at the inverting input and +1 V at the non-inverting input, the resulting output from each circuit is +2 V. in analyzing the first circuit, it can be seen that the actual input impedance of the positive input is 300 Ω to ground, as desired. However, the input impedance of the inverting input isn’t the same. It also isn’t the 150 Ω that some may think it is by virtue of the investing node.

The positive input to the op amp sits at 0.5 V due to the division of the positive voltage input by two equal resistors. Because the op amp is active and closed loop, the investing node at the op amp also must be +0.5 V. This means that across R1, there’s a differential voltage of 1.5 V, indicating that the source voltage to the positive input of the balanced amplifier must supply a full 10 mA. If this current requirement is referenced to ground using the 1-V input, 1 V/10 mA = 100 Ω to the load while the positive input presents 300 Ω. This is not functionally balanced. Similar analysis at higher grains reveal even worse imbalances. Obviously, this isn’t an optimized method. By changing R1 and R2 to the prescribed values calculated from the formulas in Figure 2, R1 becomes 450 Ω with the same 1.5 V across it, requiring 3.33 mA from the load. However, that’s identical to the requirement for the positive input (1 V/300 Ω). With the current referenced to ground as before, 1 V/3.33 mA = 300 Ω. And both inputs reflect the same impedance back to the source. The impedance back to the source. The advantage is that induced interference from outside sources will now have equal impedances to work into, which improves the balanced input amplifier’s ability to reject them.

The method illustrated in Figure 2 more closely approximates the ideal of “balanced” operation using a single op amp. As always, 1% values and possible trimming will maximize the common-mode rejection. Figure 1 works well if the source impedance is very low. However, if the source impedance is that low, a balanced receiver may not be necessary. Figure 1 won’t provide full rejection of a common-mode signal induced through the wire pair itself.

The questions are valid for gains from under 1 to over 1000, but higher gains (>30) become impractical without trimming adjustments. Even without trimming, the author achieved a 20-dB cable-induced noise improvement in an inexpensive low-impedance mike mixer by simply changing two resistors in each preamplifier as described previously.

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