Ideal-Practical Opamp

Characteritic parameters

Differential Amplifier

Virtual Short in Opamp

Instrumentation Amplifier

Sqare Wave Generator

Schmitt trigger

741 Opamp PIN Diagram

Voltage follower- sample and hold

Lag and Lead Compensators

Bridge amplifier

Precision diode- Halfwave Rectifier

Peak and Zero Crossing Detector

Integrator-Differentiator

Log and Anti-Log Amplifiers

Inverting- Non inverting Amplifiers

Oscillators

**Logarithmic amplifier**

**Definition**

Logarithmic amplifier gives the output proportional to the logarithm of input signal.If V_{i} is the input signal applied to a differentiator then the output is V_{o} = K*ln(V_{i})+l where K is gain of logarithmic amplifier,l is constant.

## Logarithmic amplifier operation

The circuit diagram of logarithmic amplifier is as shown below

It is obvious from the circuit shown above that negative feedback is provided from output to inverting terminal.Using the concept of virtual short between the input terminals of an opamp the voltage at inverting terminal will be zero volts.(Since the non inverting terminal of opamp is at ground potential). The logarithmic circuit can be redrawn as follows

The current equation of diode is given as I_{d} = I_{do}*(exp (V/V_{t})-1) where I_{do} is reverse saturation current,

V is voltage applied across diode; V_{t} is the voltage equivalent of temperature. Hence applying KCL at inverting terminal of opamp, we get

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** **(0-V_{in})/R_{1} + I_{d} = 0 implies I_{d} = V_{in}/R_{1 }

Substituting the equation for current in the above equation we get I_{do}*(exp (V/V_{t})-1) = V_{in}/R_{1}. Assuming exp (V/V_{t}) >> 1 i.e. V>>V_{t}** **and V = – V_{o},** **we get I_{do}*exp (-V_{o} / V_{t}) = V_{in}/R_{1}. Applying Antilog on both sides we get

V_{o} = – V_{t} * ln (V_{in}/(R_{1}*I_{do})).

**Gain of logarithmic amplifier**

Gain of amplifier **K = -V _{t} **

**Anti log amplifier**

**Definition**

Anti log amplifier is one which provides output proportional to the anti log i.e. exponential to the input voltage.If V_{i} is the input signal applied to a Anti log amplifier then the output is V_{o}=K*exp(a*V_{i}) where K is proportionality constant, a is constant.

**Anti log amplifier operation**

A simple Anti log amplifier is shown below

It is obvious from the circuit shown above that negative feedback is provided from output to inverting terminal.Using the concept of virtual short between the input terminals of an opamp the voltage at inverting terminal will be zero volts.(Since the non inverting terminal of opamp is at ground potential). The anti log amplifier can be redrawn as follows

The current equation of diode is given as I_{d} = I_{do}*(exp (V/V_{t})-1) where I_{do} is reverse saturation current,V is voltage applied across diode; V_{t} is the voltage equivalent of temperature

Applying KCL at inverting node of opamp we get

I_{d} = (0-V_{o})/R = I_{o}*(exp (V_{in}/V_{t})) (assumed V_{in} /V_{t} >> 1)

Hence V_{o} = -I_{o}*R*(exp (V_{in}/V_{t})).

**Gain of Anti log amplifier**

Gain of Anti log amplifier **K= -Io*R**