Topic: Analog Circuits
For which of the following conditions, the circuit shown below will function as precision full wave rectifier?
(The circuit is a standard precision full wave rectifier using two op-amps, two diodes $D_1$ and $D_2$, and resistors $R_1, R_2, R_3, R$.)
- A. $R_{1}=R_{2}=R$
- B. $R_{1}=R_{3}=R$
- C. $R_{2}=2R_{1}$
- D. $R_{1}=R_{2}=R_{3}$
Answer: C. $R_{2}=2R_{1}$
Explanation: For $V_{in} > 0$, the diode $D_{1}$ is forward biased and diode $D_{2}$ is reverse biased. Applying KCL at the input of the second op-amp yields $\frac{V_{o}}{R_{3}} = \frac{V_{in}}{R_{1}} - \frac{V_{in}}{R_{2}}$.
For it to act as a full-wave rectifier, the gain magnitude must be proportional, e.g., using condition (c) $R_{2}=2R_{1}$:
$\frac{V_{o}}{R_{3}} = \frac{V_{in}}{R_{1}} - \frac{V_{in}}{2R_{1}} = \frac{V_{in}}{2R_{1}}$.
$\Rightarrow V_{o} = \frac{R_{3}}{2R_{1}}V_{in}$.
Because $V_{o} \propto V_{in}$ in a linear fashion for this half cycle matching the other, this condition is correct.