 Hello, welcome to the session on precision clampers. Learning outcomes are at the end of session students will be able to explain the working of clamper circuits using operational amplifier as well as they can sketch the output waveforms for the key 1 input signal. Contents are like this before moving towards the positive clamper circuit you must know what are the clamper circuit and why the precision clamper circuits are required. So, clamper circuits are nothing but the circuit which clamps nothing but shifts the DC level of the signal towards the positive or negative side with reference to the signal applied to it that is reference voltage. In the usual clamper circuit here we need to use the capacitor and diode and with these two components we can obtain the clamping action. But while doing that diodes are practical diodes whose cutting voltage is 0.7 volt we cannot amplify or we cannot rectify the signals below that cutting voltage value. Therefore, we can use the operational amplifier and the diode when the diode will act as an ideal diode. So, when we connect the diode in the feedback path of the operational amplifier circuit then then the diode will act as ideal diode. So, this combination can be utilized to obtain the clamper circuit called as precision clamper circuit which will precisely gives you the shifting in the DC level. So, as you know that cutting voltage of diode divided by its open loop gain which will eliminate the cutting voltage and therefore it can able to rectify the signal in the range of milli volt or micro volts also. So, as you know this will give you the predetermined DC level and signal swing or the total amplitude will remain same simply it will give you the shift in the DC level. So, the need of precision clampers is to precisely shift the DC level and to preserve the signal or the output voltage over here. Figure 2 shows the circuit diagram for positive clamper using operational amplifier where capacitor plays the very important role to clamp the signal. So, this is capacitor CI or C1 and this is the register which should be of very large value in such a way that not provide the path to discharge the capacitor. Input signal is applied to the inverting terminal of opamp and reference signal is applied to the non-inverting input of opamp and diode is connected in the feedback path to provide precise clamping action. To analyze the circuit we will apply the concept of superposition theorem and then we will calculate the total output voltage. So, first we assume input signal is 0 and at that time what will be the output voltage as the V reference is connected to VCC via this part we will obtain the positive reference voltage at the output because we are connecting the reference voltage to the non-inverting input of opamp. So, this is the first case where output voltage is equal to reference voltage. Now the second case we assume reference voltage is not applied and input signal is applied. Now, when we consider the input signal then there will be the two assumptions positive half and the negative half. Therefore, when reference voltage is 0 and input signal is greater than 0 nothing, but it is positive half cycle. During positive half cycle what happens over here this inverting terminal will give you change in the phase. So, as positive signal is here, but now negative signal will be appeared at the output of opamp and therefore diode will not be in the conducting state. So, diode will be off. So, what will be the output here? So, we cannot obtain any output with this, but an output voltage will now be the negative. So, in the negative half diode will not conduct. So, it will be act as a open circuit over here. At that time output voltage will not be there for the first positive half cycle, but for the further half cycles you will get the output voltage. Now, the second condition where input voltage is less than 0 nothing, but negative half. Negative half is at the inverting input which will change the signal to the positive half due to the inverting terminal and then this positive half will give you the conduction of diode and as the diode is conducting which will charge the capacitor up to its peak value that is Vm. So, in the negative half cycle diode will be conducting will retain capacitor will retain its charge up to Vm and therefore you will obtain the output voltage equation for this case that is when V reference is 0 and input voltage is applied. Therefore, output voltage will be equal to voltage across capacitor plus input voltage because capacitor is connected in series with the input signal, but here as capacitor is charged up to its peak value. So, V0 can be Vm plus Vin. So, we can add these two values for getting the resultant voltage equation. So, this V0 equal to V reference plus Vm plus Vin. So, this is the analysis of positive clamper using operational amplifier. So, keep in mind reference voltage can be positive or negative and based on that you will get the clamping action in the positive or in the negative direction. Now, look at the waveforms for the positive reference voltage by putting the various values of input voltage in the equation of V0 you can obtain this clamping action in the positive direction and you will get the clamping with respect to Vm as capacitor gets charged up to Vm. Therefore, the total swing remains same, but only the shifting will be with respect to V reference and this is the waveform for the negative reference voltage. Now, think and write what should be the values of capacitor and resistor for the precision clamping and that can be obtained for the period of input signal waveform which is equal to 0.4 seconds. So, when we say input signal time period is 0.4 second that is positive as well as negative half. So, you have to find what are the conditions for charging of capacitor as charging of capacitor will be done at the peak value. So, we can just observe or just concentrate on the half of the period because as you know it is a sinusoidal waveform. So, in the t by 2 time period you will get the charging of capacitor. So, pause the video and select the proper component value for the precision clamping action. Yes and the answer is option E because here when you multiply this R and C you will get the time period which is less than this 0.4 by 2 where 0.4 is the time period of the input signal. So, where this precision clamping circuits are used? So, as you know wave shaping circuits are used in the digital computers and the communication systems and for these are the examples like TV and the FM receivers and this wave shaping can be done with the help of clipping and clamping circuits where the predetermined DC level is inserted in the output. Therefore, this circuit is also called as DC restorer. When a current lower points has to be fixed at a specific value then these clamping circuits are used and it is very useful in the testing equipments, radar systems, electronic, major systems and solar systems. References are like this. Thank you.