- Linear Integrated Circuits Applications
- Home
- Basics of Integrated Circuits Applications
- Basics Operational Amplifier
- Op-Amp applications
- Arithmetic Circuits
- Differentiator & Integrator
- Converters Of Electrical Quantities
- Comparators
- Log & Anti-Log Amplifiers
- Rectifiers
- Clippers
- Clampers
- Active Filters
- Sinusoidal Oscillators
- Waveform Generators
- 555 Timer
- Phase Locked Loop Ic
- Voltage Regulators
- Data Converters
- Digital to Analog Converters
- DAC Example Problem
- Direct Type ADCs
- Indirect Type ADC

- Useful Resources
- Quick Guide
- Useful Resources
- Discussion

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

**AC** and **DC** are two frequent terms that you encounter while studying the flow of electrical charge. **Alternating Current (AC)** has the property to change its state continuously. For example, if we consider a sine wave, the current flows in one direction for positive half cycle and in the opposite direction for negative half cycle. On the other hand, **Direct Current (DC)** flows only in one direction.

An electronic circuit, which produces either DC signal or a pulsated DC signal, when an AC signal is applied to it is called as a **rectifier**. This chapter discusses about op-amp based rectifiers in detail.

Rectifiers are classified into two types: **Half wave rectifier** and **Full wave rectifier**. This section discusses about these two types in detail.

A **half wave rectifier** is a rectifier that produces positive half cycles at the output for one half cycle of the input and zero output for the other half cycle of the input.

The **circuit diagram** of a half wave rectifier is shown in the following figure.

Observe that the circuit diagram of a half wave rectifier shown above looks like an inverting amplifier, with two diodes D_{1} and D_{2} in addition.

The **working** of the half wave rectifier circuit shown above is explained below

For the

**positive half cycle**of the sinusoidal input, the output of the op-amp will be negative. Hence, diode D_{1}will be forward biased.When diode D

_{1}is in forward bias, output voltage of the op-amp will be -0.7 V. So, diode D_{2}will be reverse biased. Hence, the**output voltage**of the above circuit is**zero**volts.Therefore, there is

**no (zero) output**of half wave rectifier for the positive half cycle of a sinusoidal input.For the

**negative half cycle**of sinusoidal input, the output of the op-amp will be positive. Hence, the diodes D_{1}and D_{2}will be reverse biased and forward biased respectively. So, the output voltage of above circuit will be −

$$V_0=-\left(\frac{R_f}{R_1}\right)V_1$$

Therefore, the output of a half wave rectifier will be a

**positive half cycle**for a negative half cycle of the sinusoidal input.

The **input** and **output waveforms** of a half wave rectifier are shown in the following figure

As you can see from the above graph, the half wave rectifier circuit diagram that we discussed will produce **positive half cycles** for negative half cycles of sinusoidal input and zero output for positive half cycles of sinusoidal input

A **full wave rectifier** produces positive half cycles at the output for both half cycles of the input.

The **circuit diagram** of a full wave rectifier is shown in the following figure −

The above circuit diagram consists of two op-amps, two diodes, D_{1} & D_{2} and five resistors, R_{1} to R_{5}. The **working** of the full wave rectifier circuit shown above is explained below −

For the

**positive half cycle**of a sinusoidal input, the output of the first op-amp will be negative. Hence, diodes D_{1}and D_{2}will be forward biased and reverse biased respectively.Then, the output voltage of the first op-amp will be −

$$V_{01}=-\left(\frac{R_2}{R_1}\right)V_i$$

Observe that the output of the first op-amp is connected to a resistor R

_{4}, which is connected to the inverting terminal of the second op-amp. The voltage present at the non-inverting terminal of second op-amp is 0 V. So, the second op-amp with resistors, R_{4}and R_{4}acts as an**inverting amplifier**.The output voltage of the second op-amp will be

$$V_0=-\left(\frac{R_5}{R_4}\right)V_{01}$$

**Substituting**the value of $V_{01}$ in the above equation, we get −$$=>V_{0}=-\left(\frac{R_5}{R_4}\right)\left \{ -\left(\frac{R_2}{R_1}\right)V_{i} \right \}$$

$$=>V_{0}=\left(\frac{R_2R_5}{R_1R_4}\right)V_{i}$$

Therefore, the output of a full wave rectifier will be a positive half cycle for the

**positive half cycle**of a sinusoidal input. In this case, the gain of the output is $\frac{R_2R_5}{R_1R_4}$. If we consider $R_{1}=R_{2}=R_{4}=R_{5}=R$, then the gain of the output will be one.For the

**negative half cycle**of a sinusoidal input, the output of the first op-amp will be positive. Hence, diodes D_{1}and D_{2}will be reverse biased and forward biased respectively.The output voltage of the first op-amp will be −

$$V_{01}=-\left(\frac{R_3}{R_1}\right)V_{i}$$

The output of the first op-amp is directly connected to the non-inverting terminal of the second op-amp. Now, the second op-amp with resistors, R

_{4}and R_{5}acts as a**non-inverting amplifier**.The output voltage of the second op-amp will be −

$$V_{0}=\left(1+\frac{R_5}{R_4}\right)V_{01}$$

**Substituting**the value of $V_{01}$ in the above equation, we get$$=>V_{0}=\left(1+\frac{R_5}{R_4}\right) \left\{-\left(\frac{R_3}{R_1}\right)V_{i}\right \} $$

$$=>V_{0}=-\left(\frac{R_3}{R_1}\right)\left(1+\frac{R_5}{R_4}\right)V_{i}$$

Therefore, the output of a full wave rectifier will be a

**positive half cycle**for the negative half cycle of sinusoidal input also. In this case, the magnitude of the gain of the output is $\left(\frac{R_3}{R_1}\right)\left(1+\frac{R_5}{R_4}\right)$. If we consider $R_{1}=2R_{3}=R_{4}=R_{5}=R$ then the gain of the output will be**one**.

The **input** and **output waveforms** of a full wave rectifier are shown in the following figure

As you can see in the above figure, the full wave rectifier circuit diagram that we considered will produce only **positive half cycles** for both positive and negative half cycles of a sinusoidal input.

Advertisements