How to add gain to this op-amp based precision full wave rectifier?

Question

The design for the rectifier is rather clever, and comes from TI's TIDU030: Precision Full-Wave Rectifier, Dual-Supply.

It looks as follows:

^{simulate this circuit – Schematic created using CircuitLab}

R2=R3=R4, and this rectifier has gain of +/-1. The architecture is nice as it keeps both op-amps operating closed-loop, without saturation. D1 and D2 can be Schottky microwave RF diodes to lower crossover distortion.

Could it be made to have inherent gain greater than 1? It's not as simple as changing the R3/R2 ratio, since the gain for the positive input would still remain +1.

Answer 1

It turns out it's possible by adding R0 and R1 to control the non-inverting gain.

^{simulate this circuit – Schematic created using CircuitLab}

When the input is positive, D2 conducts, D1 is reverse-biased, and the equivalent DC circuit becomes:

^{simulate this circuit}

When the input is negative, D1 conducts, D2 is reverse-biased, and the equivalent DC circuit is:

^{simulate this circuit}

We want the gains in each case to have the same magnitude:

$$ A_V = 1+\frac{R_1+R_2+R_3}{R_0} = \left( 1 + {R_1 \over R_0} \right) \frac{R_3}{R_2} $$

We can substitute \$r_i=R_i/R_0\$, and

$$\begin{aligned} A_V &= 1+ r_1 + r_2 + r_3 \\ A_V &= (1+r_1)\frac{r_3}{r_2} \\ \\ A_V \frac{r_2}{r_3} &= 1 + r_1 \\ \\ A_V &= A_V \frac{r_2}{r_3} + r_2 + r_3 \\ A_V - r_3 &= \frac{r_2}{r_3} \left( A_V + r_3 \right) \\ \text{Let }k &= \boxed{ \frac{A_V-r_3}{A_V+r_3} } \\ r_2 &= \boxed{ r_3 \cdot k } \\ \\ r_1 &= A_V \frac{r_2}{r_3} - 1 \\ r_1 &= \boxed{ A_V \cdot k - 1 }. \\ \end{aligned}$$

Since \$r_1\$ must be non-negative, we have the constraint \$A_V \cdot k \ge 1\$. \$R_0\$ and \$r_3\$ are otherwise arbitrary.

Example: $$\begin{aligned} &\text{Let } A_V = \boxed{10}, \ r_3 = \boxed{5}, \text{ then} \\ \\ k &= \frac{A_V-r_3}{A_V+r_3} = \frac{10-5}{10+5} = \boxed{ \frac{1}{3} }\\ r_2 &= r_3 \cdot k = 5 \frac{1}{3} = \boxed{ \frac{5}{3} } \\ r_1 &= A_V \cdot k - 1 = 10 \frac{1}{3} - 1 = \frac{10}{3} - 1 = \boxed{ \frac{7}{3} }. \\ \end{aligned}$$

Letting \$R_0 = 1{\,\rm k\Omega}\$, we get

$$\begin{aligned} R_3 &= r_3 R_0 = 5 {\,\rm k\Omega} \\ R_2 &= r_2 R_0 = \frac{5}{3} {\,\rm k\Omega} \\ R_1 &= r_1 R_0 = \frac{7}{3} {\,\rm k\Omega} \\ \end{aligned}$$

A DC simulation confirms the derivation for this case:

^{simulate this circuit}

Answer 2

I have often used a similar, but simpler, precision rectifier circuit, which has the advantage of using a single supply and only 3 resistors, but it has a gain of 0.5. It also requires a low impedance source and a high impedance load. A second op-amp can provide as much gain as desired as well as driving a lower impedance load.