- 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

The electronic circuits which perform the mathematical operations such as logarithm and anti-logarithm (exponential) with an amplification are called as **Logarithmic amplifier** and **Anti-Logarithmic amplifier** respectively.

This chapter discusses about the **Logarithmic amplifier** and **Anti-Logarithmic amplifier** in detail. Please note that these amplifiers fall under non-linear applications.

A **logarithmic amplifier**, or a **log amplifier**, is an electronic circuit that produces an output that is proportional to the logarithm of the applied input. This section discusses about the op-amp based logarithmic amplifier in detail.

An op-amp based logarithmic amplifier produces a voltage at the output, which is proportional to the logarithm of the voltage applied to the resistor connected to its inverting terminal. The **circuit diagram** of an op-amp based logarithmic amplifier is shown in the following figure −

In the above circuit, the non-inverting input terminal of the op-amp is connected to ground. That means zero volts is applied at the non-inverting input terminal of the op-amp.

According to the **virtual short concept**, the voltage at the inverting input terminal of an op-amp will be equal to the voltage at its non-inverting input terminal. So, the voltage at the inverting input terminal will be zero volts.

The **nodal equation** at the inverting input terminal’s node is −

$$\frac{0-V_i}{R_1}+I_{f}=0$$

$$=>I_{f}=\frac{V_i}{R_1}......Equation 1$$

The following is the **equation for current** flowing through a diode, when it is in forward bias −

$$I_{f}=I_{s} e^{(\frac{V_f}{nV_T})} ......Equation 2$$

where,

$I_{s}$ is the saturation current of the diode,

$V_{f}$ is the voltage drop across diode, when it is in forward bias,

$V_{T}$ is the diode’s thermal equivalent voltage.

The **KVL equation** around the feedback loop of the op amp will be −

$$0-V_{f}-V_{0}=0$$

$$=>V_{f}=-V_{0}$$

Substituting the value of $V_{f}$ in Equation 2, we get −

$$I_{f}=I_{s} e^{\left(\frac{-V_0}{nV_T}\right)} ......Equation 3$$

Observe that the left hand side terms of both equation 1 and equation 3 are same. Hence, equate the right hand side term of those two equations as shown below −

$$\frac{V_i}{R_1}=I_{s}e^{\left(\frac{-V_0}{nV_T}\right)}$$

$$\frac{V_i}{R_1I_s}= e^{\left(\frac{-V_0}{nV_T}\right)}$$

Applying **natural logarithm** on both sides, we get −

$$In\left(\frac{V_i}{R_1I_s}\right)= \frac{-V_0}{nV_T}$$

$$V_{0}=-{nV_T}In\left(\frac{V_i}{R_1I_s}\right)$$

Note that in the above equation, the parameters n, ${V_T}$ and $I_{s}$ are constants. So, the output voltage $V_{0}$ will be proportional to the **natural logarithm** of the input voltage $V_{i}$ for a fixed value of resistance $R_{1}$.

Therefore, the op-amp based logarithmic amplifier circuit discussed above will produce an output, which is proportional to the natural logarithm of the input voltage ${V_T}$, when ${R_1I_s}=1V$.

Observe that the output voltage $V_{0}$ has a **negative sign**, which indicates that there exists a 180^{0} phase difference between the input and the output.

An **anti-logarithmic amplifier**, or an **anti-log amplifier**, is an electronic circuit that produces an output that is proportional to the anti-logarithm of the applied input. This section discusses about the op-amp based anti-logarithmic amplifier in detail.

An op-amp based anti-logarithmic amplifier produces a voltage at the output, which is proportional to the anti-logarithm of the voltage that is applied to the diode connected to its inverting terminal.

The **circuit diagram** of an op-amp based anti-logarithmic amplifier is shown in the following figure −

In the circuit shown above, the non-inverting input terminal of the op-amp is connected to ground. It means zero volts is applied to its non-inverting input terminal.

According to the **virtual short concept**, the voltage at the inverting input terminal of op-amp will be equal to the voltage present at its non-inverting input terminal. So, the voltage at its inverting input terminal will be zero volts.

The **nodal equation** at the inverting input terminal’s node is −

$$-I_{f}+\frac{0-V_0}{R_f}=0$$

$$=>-\frac{V_0}{R_f}=I_{f}$$

$$=>V_{0}=-R_{f}I_{f}.........Equation 4$$

We know that the equation for the current flowing through a diode, when it is in forward bias, is as given below −

$$I_{f}=I_{s} e^{\left(\frac{V_f}{nV_T}\right)}$$

Substituting the value of $I_{f}$ in Equation 4, we get

$$V_{0}=-R_{f}\left \{{I_{s} e^{\left(\frac{V_f}{nV_T}\right)}}\right \}$$

$$V_{0}=-R_{f}{I_{s} e^{\left(\frac{V_f}{nV_T}\right)}}......Equation 5$$

The KVL equation at the input side of the inverting terminal of the op amp will be

$$V_{i}-V_{f}=0$$

$$V_{f}=V_{i}$$

Substituting, the value of 𝑉𝑓 in the Equation 5, we get −

$$V_{0}=-R_{f}{I_{s} e^{\left(\frac{V_i}{nV_T}\right)}}$$

Note that, in the above equation the parameters n, ${V_T}$ and $I_{s}$ are constants. So, the output voltage ${V_0}$ will be proportional to the **anti-natural logarithm** (exponential) of the input voltage ${V_i}$, for a fixed value of feedback resistance ${R_f}$.

Therefore, the op-amp based anti-logarithmic amplifier circuit discussed above will produce an output, which is proportional to the anti-natural logarithm (exponential) of the input voltage ${V_i}$ when, ${R_fI_s}= 1V$. Observe that the output voltage ${V_0}$ is having a **negative sign**, which indicates that there exists a 180^{0} phase difference between the input and the output.

Advertisements