Peculiar precision full-wave rectifier needs no matched resistors

A classic analog application is the precision active full-wave rectifier. Many different implementations exist of this theme, each with its own supposed advantages. However, one circuit element needed by (almost) all active full-wave rectifier designs is an inverter with matched resistors to set its gain to an accurate -1.0. In such topologies, symmetry of rectification relies upon and can be no better than the accuracy of this resistor match. For an example, see a well known (veritable classic!) design in Figure 1 with op-amp U1b acting as the inverter and R1 and R2 as its matched gainset resistors. Unless R1 = R2, rectifier output for negative Vin excursions are (very) unlikely to equal output for positive Vin excursions. 

Figure 1 Conventional precision rectifier design with R1 and R2 matched symmetry-resistors.

For positive Vin inputs, D1 turns off and D2 conducts, establishing non-inverting unity gain for the circuit that’s unaffected by resistor values: Vout/Vin = +1.

For negative inputs, D1 conducts, D2 turns off and U1b becomes an inverter with gain Vout/Vin = –R2/R1 = -1 only if R2 = R1. Otherwise, not, creating poor rectification symmetry.

Figure 2 shows another (less conventional) design. But unconventional or not, here are Q2 and Q3 acting as the inverter and matched gainset symmetry-resistors R1 and R2 performing just as in Figure 1.

Figure 2 Unconventional rectifier with discrete circuit inverter still uses symmetry-setting resistors: R1 and R2.

But now, just to break the monotony, regard Figure 3. Note the (shocking) absence of matched resistors. Here’s how this nonconformist works.

Figure 3 Unconventional precision rectifier design without matched symmetry-resistors.

Q1 and Q2 provide simple cross-over compensation to cancel the Vbe drops of Q3 and Q4. Consequently, negative Vin excursions are inverted by A1 and output by Q4 to filter R3C3. Meanwhile, positive Vin excursions turn Q3 on, causing C2 to integrate their time and current product: charge. The accumulated charge is stored as voltage on C2 which is added to subsequent opposite polarity half-cycles with Q3 and Q4 acting as a simple full-wave charge pump. The net result: 

Vout = Avg(Abs(Vin)) R3 / R2 / R1.

Accurate rectification symmetry is therefore inherent as long as transistor Vbe’s match reasonably well which, being the same type and operating in similar contexts, they will.

Stephen Woodward’s relationship with EDN’s DI column goes back quite a long way. Over 100 submissions have been accepted since his first contribution back in 1974.

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