 Hello, welcome to the session on log amplifiers using opam. Learning outcomes are at the end of session students will be able to explain working of logarithmic amplifiers as well as they can analyze equation for the output voltage. Contains are like this figure one shows the circuit diagram for basic log amplifier where p n junction diode is used in the feedback path and thus we obtain the output voltage which is directly proportional to the log of input voltage signal. And register R 1 is used to minimize the offset voltage loss. When we assume opam as ideal then we can apply the concept of virtual ground means when non-inverting voltage is grounded then voltage at inverting terminal will also be treated as grounded. Because output voltage is equal to gain times the difference between input voltage signal and when we assume the gain is very very large then we assume here gain that is Vid equal to 0. And when we assume that we get the inverting terminal voltage equal to 0. And therefore, we can say here current flowing through register R 1 is equal to the current flowing through diode that is IF. Because as you know the resistance at the input terminals of opam is infinite or very very large therefore I 1 is equal to IF. So, we can write the equation for I 1 with respect to the input voltage terminal. So, that is V in divided by R 1 and here another end of the circuit that is V out is nothing but voltage across this diode p n junction diode. So, it is connected in the reverse fashion with respect to the output terminal therefore, V out equal to minus V f. So, we can analyze the output voltage equation by observing the current flowing through the diode. So, current flowing through diode can be written with the diode equation that is IF equal to I 0 times exponential of V f by eta V t minus 1, where I 0 is the leakage current V f is the forward or reverse voltage across this diode. Eta is the ideality factor which generally depends on the type of material use that is silicon or germanium and V t is the temperature equivalent voltage that is K into T, where T is in degree Kelvin that is temperature in degree Kelvin. So, which approximately comes equal to 26 millivolt. So, here we will put this value in the equation of IF and so we can get the ratio of IF by I 0 nothing but I 1 by I 0 which is in terms of this voltage that is V in by R 1. So, see here the next step. So, we can say IF by I 0 that is your leakage current equal to exponential of V f by eta V t minus 1 and as you know 1 is very small as compared to the exponential of this value. Therefore, we will get here or we will just neglect the value 1 over here and then taking natural log on both sides. So, when we take natural log on both sides we will get the equation which is in terms of V f by eta V t times this log. So, V f is nothing but eta V t times log to the base E IF by I 0, but we know here IF is equal to I 1 that is equal to V in by R 1 as well as we also know output voltage is opposite of the feedback voltage. So, V out is equal to minus V f by putting all these values in the equation of V out we will get V out equal to minus V f. Therefore, we are writing this equation with the negative sign. So, minus eta V t log to the base E IF is replaced by V in by R 1 and I 0 as it is. So, thus we can say output voltage is directly proportional to the logarithmic of the input voltage signal and this log amplifiers are widely used to get the signal processing of the voltage signal over here. So, log amplifiers are used in digital computers. So, we will see here how we can obtain the log amplifiers using transistor because if you see here this eta plays the important role in the output voltage equation to minimize this effect that is effect of ideality factor we can replace this diode with the transistor circuit. With this we can increase the wide range or we can increase the capability of log amplifier to handle the wide range of the signal. So, here P N junction diode is replaced by a N P N transistor here base terminal is grounded and as we again apply the previous concept of virtual ground. So, as non-inverting input is grounded, so voltage at this terminal is again 0. So, I 1 is equal to IC that is current flowing through the collector it same will flow through the emitter because here base is grounded. So, we can analyze that with the help of again equation of IC and as we know here it acts as a P N junction diode. So, same equation we can apply to obtain the relation between V out and the input voltage. So, as base is grounded IC is approximately equal to I E and this IC can be returned in terms of current that is leakage current that is I 0 into exponential of V f by eta V t. So, here also we will get the log on both sides. So, log of IC upon I equal to V f by eta V t and here we will get the V out is minus V f nothing, but minus V be voltage between meter and base terminals and IC is V in by R 1. So, here we will get the output voltage equation which is directly proportional to the log of input voltage signal. We can also obtain the output voltage which is directly proportional to the exponential of the signal and these circuits are called as antilog amplifiers. Just by replacing the position of resistor and the diode we can obtain that circuit. So, here input is applied to the P N junction diode and we can find the equation of I f and how can we obtain V 0 in terms of exponential of V in. So, here as V in equal to minus V f why it is so? Because see here this is the direction of V in and you are taking the voltage across this diode. So, it can be the minus V f over here that is voltage across the diode and I f equal to I 0 times the exponential of V in that is signal applied to the diode divided by eta V t. So, it can be calculated like this I is nothing, but now minus V 0 upon R f run flowing through the feedback resistor. So, it will be minus V 0 by R f as I and I f both are of same value. So, we can equate the equations like this minus V 0 R f minus V 0 upon R f equal to I 0 into exponential of V in by we can get here output voltage which is minus I 0 R f times the exponential of input signal. Thus we can say here exponential of the signal can be obtained just by adding the diode at the input circuit and in the feedback path connecting the resistor. So, here if the input signal is negative here we have written the V in equal to minus V f because input signal is positive over here, but to process the negative input voltage and to obtain the exponential of that what should be the arrangement? Yes, the arrangement should be just reversing the position of diode that is minus side here and plus side over. So, again now you think can we replace that in this antelog circuits? Yes, we can do that and now draw the circuit for antelog amplifier if input signal is negative. So, keep in mind you have to draw the antelog amplifier circuit for the negative input with the help of transistor. So, pause the video and draw the circuit. Yes, the answer is you have to use NPN transistor over here and in such a way that this emitter is connected to input of the signal that is NPN negative side is connected to V in over here and in this way we can obtain the antelog amplifier using transistor. Very important thing that is applications of log amplifier it can be used in the multiplication division to find the powers of the equation or powers of the root to calculate the 2 RMS conversion. So, all these processing of signals we can do with the help of log amplifiers and very important it will give you the value in decibels as we will get the log of the signal we can represent in the decimal and as well as we can have the signal compression of dynamic range of signal with the help of this log amplifiers. Log amplifiers are essential part of analog computers and with this we can get the simplification of equations just by having a simple clique. So, log amplifier plays very important role in the digital and analog circuits both references are like this. Thank you.