 Good morning student friends. In today's session we are going to discuss about one of the popular applications of the op-amp and this is called as the peak detectors. The learning outcomes of this session include after completing this session student will able to describe working of peak detector circuit and then they can comment about the components selection for the components which are used in designing peak detector circuits. The contain include we are going to discuss about in general what is a peak detector then we will discuss about positive peak detector and lastly we will have a look at what are the requirements for the components selection of designing the peak detector. As the name suggests peak detector is a circuit for which I am applying AC signal or a sinus signal or any other signal and the output of this is a DC signal which always represent the peak value of the signal. So, in this diagram as you can see the signal is varying then this is a peak value. So, the output is it remain at this value then for then the input goes down, but the output remain constant and when again the input rises to a new peak value the output again attains that peak value and remain constant. So, even though the input is going low the output remain constant. So, the output always represent the peak value of the input signal. There can be two types of the peak detectors one is a positive peak detector which always detects the positive of the input signal positive peak of the input signal and other one is a negative peak detector which detects or rather the output of which is the negative peak value of the input signal. This is a simple circuit which represent or which describes the positive peak detector circuit. The circuit is designed using of course operational amplifier and few components. As you can see the there is a operational amplifier 741 is used the output and then the components output component include there is one diode which is placed in the feedback path of the op amp diode D1 then there is one more diode D2 and of course there is a capacitor. Now if you know dear students peak detector passive peak detector can also be designed using simply diode a load resistance and a capacitor. However the passive peak detector design may suffer with what is called as the loading effect and to avoid the loading effect we can see that the using the similar component that is nothing but the load resistance then the capacitor and the diode we have designed a peak detector. But in this case as operational amplifier is being used it is going to avoid the loading effect. Let us have some more details about this circuit as you can see the op amp work op amp requires two type of the power supply plus VCC is of 15 volt minus VEE as of a minus 15 volt then this represent the load resistance the capacitor is being used which is actually detecting the peak voltage. Then interestingly you can see there is a resistance ROM now this resistance ROM as we know is used for the compensation of the input bias current. So in total this is a circuit which is acting as a peak detector. Now how this work for in order to analyze the working of this circuit we need to consider operational the form for this circuit we have to consider the input sinusoidal signal instead of sinusoidal signal I have considered a square wave signal over here. So as we know there are two half cycles a positive half cycle and a negative half cycle. So what happen during the positive half cycle and what happen during the negative half cycle we need to analyze. Let us first understand what happen during the positive half cycle ok. So during the positive half cycle as you can see when the input is positive as you can see then the output of the op amp is going to be positive when the output of the op amp is positive which makes the diode D1 on when the diode D1 is on the feedback loop is completed and op amp simply work as a non inverting voltage follower. So whatever the VIN is applied over here it appears at this point. Now when it is appearing at this point it also makes the capacitor charge. So as you can see the capacitor start charging and the capacitor charges to the maximum very rapidly the capacitor charges to the maximum value and that voltage will appear over here. So as you can see for this positive half cycle the output is nothing but the voltage across the capacitor which is which is nothing but the peak voltage of the input signal. During negative half cycle now when the input goes negative the output of this becomes negative now when output of this op amp becomes negative the diode D1 is no longer on it becomes open circuit and then the loop is not completed and the output of the op amp will go into the saturation. However what is the voltage which is delivered to the load resistance the voltage which is delivered to load resistance is nothing but the voltage across the capacitor and then the capacitor then the this voltage remain constant and that is being delivered across the load resistance. So we have seen that during positive half cycle during positive half cycle the capacitor get charge to the peak value and during negative half cycle. Now during negative half cycle you may one may think that why only the capacitor will try to deliver the voltage across the load resistance. However there may be two more paths for the voltage across the capacitor one of course the path is to the diode but as during the negative half cycle diode is open circuit the voltage will not reach the second path is of course the path through the operational amplifier. But as we know the input impedance of the operational amplifier or input resistance of the operational amplifier is very high and that is why the only path for the capacitor is to is is to appear across the load resistance. So if we now look at the waveform across the load resistance it almost remain constant and that is nothing but the peak value of the input voltage. So this is how when the input is a square wave or a sinusoidal wave the output voltage across the load resistance remain almost constant I will not say it is constant but it is almost constant and it is equal to the peak voltage of the positive waveform. As you can see the output is is representing the positive peak voltage this is called as the positive peak detector. Now dear student please pause the video and answer this question how to design a negative peak detector. So I think you are ready with the answer see there are two diodes there is a diode D 1 and there is a diode D 2 the negative peak detector can be designed by just replacing the polarities or or reversing the polarities of the diode D 1 and D 2. So instead of this anode and cathode if I if we have if if I connect cathode at this point and anode at this point and this diode is also is also inverted then that will become a negative peak detector. By the way in this positive peak detector what is the what is the purpose of this diode D 2 is during negative half cycle when the output is negative this diode becomes on and which makes the and and and which and and when whatever the output saturation voltage for that this diode will act as a path. So that is about the positive peak detector. Now the ads we have used a capacitor and we have used a resistor and there is a load resistance we have to comment or we have to talk about what are the different types of the component that can be used and how this component can be selected. For a complete and quick charging of a capacitor the RC time constant associated with capacitor must be less than the time period of the input waveform as we can see we are applying a sinusoidal or a square wave input to the circuit and then there is a with every capacitor as we know there is RC time constant. Now in this circuit what is that RC time constant? So now during the charging as we know the capacitor charges through the diode the diode becomes on and that the the the resistance of the forward bias diodes is nothing but the R in this equation. So we can say that the C that is the capacitance of the capacitor multiplied by R d the C R d must be less than the time period. So for to be on the safer side we are we are just trying to see that this equation is satisfied the C R d is less than or equal to t by 10. So by by making this t by 10 we are ensuring that the RC time constant is much lesser than the time period of the waveform and as we know the C is nothing but the capacitor and R d is the resistance of the forward bias diode and that is that is that is equal to usually typically it is equal to 100 ohm. So this is about how the how the components are governed during the charging. Now during the discharging we can say that for the capacitor not to discharge RC time constant associated with the capacitor must be must be greater than time period of the input signal. Now this condition can be satisfied. So if we can if we can achieve C into R L now during when the capacitor is trying to discharge the capacitor is discharging the discharge for the capacitor is through the load resistance and then we can say that C into R L if you want that the capacitor should retain its charge for sufficient amount of time then the C R L must be greater than t and the t to be on the safer side we can say that C R L must be greater than equal to 10 t. So this ensures that ensures that the peak voltage is retained. So in this case the C is the capacitor and the discharge path is through the load resistance. So R L in the in in this case instead of for R we are using R L. Now this may this may raise one more question is that we are saying that we should have C R L should be greater than or equal to 10 t but please remember the R L is a load resistance on which we may not have a control. So if the value of R L is not sufficiently high then the then the peak voltage may not be retained. How to solve this problem? This problem can be solved by connecting one more op-amp circuit which is called as a non-inverting voltage follower between the capacitor and load resistance. So there will not be any gain offer by the operational amplifier but in this case it will act as a buffer and then it will it can solve this problem. Lastly we can say that a popular op-amp 741 can be used for a usual operations but if input frequency the signal frequency is very high then we need to use high speed precision op-amp operational amplifiers like 771 or 714. So lastly we will come to the discussion and I would like student you to have a discussion about what is going to be effect we have seen that there are two equations which need to be satisfied which are related to RC time constant. What happen if those equations are not satisfied? How the waveform is going to change if those equations are not satisfied? So that is what I am leaving for you discussion. The reference used for the today's session include operational amplifier and linear integrated circuit by Ramakant Gaikvar and linear integrated circuit by Roy Chaudhary and Jain. Thank you student for a patient listening.