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.

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.

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.