Uncategorized So, welcome to this particular module. And in this module, we will be understanding what are the application of Op-Amp in terms of wave shaping circuits. So, what kind of wave shaping circuits you can use and what kind of circuits you can implement to shape the wave, alright. So, one of the circuit ah, that is used for shaping the signal ah is called clipper alright. So, clip; what does clip do? It will kind of cut right, clipping clipping. It will cut the signal or you can shape the wave So, either the signal; we are shaping the positive part of the signal is called positive clipper. If we are going to ah shape the negative part of the signal is called negative clipper So, very easy to understand, very easy to identify ah which kind of clippers we should use for shaping what kind of waveforms right So, ah how a operational amplifier can be used as a clipper circuit, alright. How operational amplifier can be used as a clipper circuit, ah? We will see and then, we will implement, will implement this clipper circuit using our favorite 3 equipment right. DC power supply function generator and oscilloscope alright 3 things and we have a simple thing called breadboard right. We have our operational amplifier ICs, we have some resistors, we have some ah ah we have some diodes and we will see that how we can implement this particular circuit So, ah let us see on the screen ah, the opamp circuits that using diodes, opamp circuits using diodes and that will act as a clipper, that will act as a clipper Now, we have seen the opamp circuits with diodes which act as a rectifiers right. We have used half wave rectifier; we have used full wave rectifier. Now, we are looking at the clippers alright. So, what exactly clipper means and ah how we can implement this circuit? So, this particular circuit that you see here right, what is it? This is minus 15 volts alright. Plus, ok. Now, we have V C C right And we have we are applying a reference voltage through to your anode of the diode. We are applying some voltage to the anode of diode Now, to conduct this diode, the anode should be more positive compared to cathode right The anode voltage should be higher than cathode voltage or the cathode voltage should be lower than anode voltage, right. That we already know So, here the reference voltage right. This voltage, reference voltage, this one is given through the potentiometer. What is it? Potentiometer p o t e n t i o m e t e r potentiometer, alright So, potentiometer is ah a component right, using which we can change the resistance, using which we can change the resistance We have seen the potentiometer in earlier lectures ah So, using a potentiometer , what i can do? I can set my reference voltage V reference I can set my reference voltage V reference This is my load resistor R L. This is my diode which is my P N junction diode. This is connected in ah connected in this particular way N P right. This is C, this is diode which is anode, this is cathode right is an diode is P right This anode is P cathode is N You have operational amplifier. You are applying signal through the non-inverting terminal of the operational amplifier, non-inverting terminal of the operational amplifier. Now, let us see if i have this circuit, how it will act how it will act hm. So, this is your positive clipper circuit, positive clipper alright. So, a diode can be a diode can also be used to clip off certain portion of input signal to obtain the desired output waveform That means, that if you see this particular signal, which is your input signal you see this signal which is your output signal. What we observe is, by using this circuit which is your positive clipper, you can clip off, you can clip off certain portion of your signal You see here, the this signal is clipped off right. That is what we are looking at. A diode, the diode can also be used to clip of certain  cutoff to drive diode in to cut off. Diode will not start. It will not conduct. Diode will not conduct right In this case, the opamp operates in open loop and the output voltage is V reference right When it does not conduct, opamp is like open loop because, is not connected. It is not connected. It is a open loop. Feedback is not connected because diode is not conducting Becomes a open loop circuit. When it becomes open loop circuit, the output is nothing but your V reference voltage. Output will be your V reference voltage So now, now our output is V reference excellent However, if V reference is made negative, then the entire output waveform V reference would be clipped off. That means, that if i have V reference right, V reference as negative like here right. If i make V reference equals to minus, let us say 5 volts and the signal is signal is 5 volts, then the entire signal would be clipped off right. This entire signal would be my output here in this particular case Let us see closely ok. It is very important point if right. Now, my V reference is right Now, my V reference is 1 volt alright. But, what if my V reference, what if my V reference is minus 5 volt for minus 5 volt? Let us say, this is my minus 5 volt. If my signal if my signal is 5 volts, my V reference is minus 5 volts right. Then, what will i have the entire this waveform? The entire input signal right will be clipped off; that means, output will be nothing will be clipped off That means that, by changing V reference, by changing V reference you can you can control the shaping of the signal. You can control the shaping of the signal; you can control the clipping. When you clip it, it should it will look different. You see shaping. What is shaping? Shaping means, what does shaping means? Shaping of the signal signal is nothing but, you see if i have the signal right, i can clip it like this or i can clip it like this or i can clip it like this, right That means that, the signal which is your input signal. My output signal can be vary It can be of different ah shape, can be of different shape. That is what i mean by clipping and that is what clipper does, that is what is a function of a clipper ok. So, easy. Now, this is the positive clipper; that means, it will only act on the positive. What about i want to clip just negative? Only negative right. You are the input i have interested only in this part of the input, i want to shape the negative part of the signal. How can i design a circuit which can act for negative part of the signal So, in this case, what we will do in this case? What we will do ? Let us see, let us see the negative. So, opamp circuits using diodes which is a clipper and this is your negative clipper. This is your negative clipper Now, if you see the diode D 1 in the negative clipper and if you see diode D 1 in positive clipper ok. So, i am going back 1 slide, you see what is the main difference, what is the main difference? The main difference is that, when you compare diode D 1 ah in the in the in the in the negative clipper circuit and when you compare diode in the positive clipper circuit, the the the direction is opposite. The direction is opposite You see here, here see direction is opposite So, negative clipper you connect the diode, the anode of the diode to the output of the opamp and the cathode to the reference voltage, anode to the output cathode to the reference voltage P to the output N to the reference voltage P N junction. Now, it is in P N configuration right. So, rest of the circuit is similar Rest of circuit is exactly similar minus V or V E E. We have V C C, we can have V reference here, V reference right. This is your load resistor R L. This is you are output is connected to the ah cathode of the diode; that means,  to this 2 right. If i say, A and this is B, then A is not not connected to B. Why because, my diode is not conducting, my diode is not connecting When my diode conducts, then i can write like this When my diode is not conducting, i cannot draw like right; that means, a A and B is not connected. If A and B is not connected; that means, my opamp would be in open loop when my A and B is not connected; that means, my opamp can be or it will come into open loop situation or open loop mode and the voltage output voltage is nothing but your reference voltage So, if my opamp is open loop mode , then what will happen? If my opamp is open loop, then this is not connected; that means, whatever my output voltage is there, is nothing but here my reference voltage that i am creating here right. Super easy, extremely easy. There cannot be easier things than this ah this kind of circuits So, easy. Just you understand the operation of diode and rest becomes a history right Very easy diode. How diodes acts positive? This is negative if anode is greater than cathode, then it will start conducting. If anode is less than cathode, then ah it will not start conducting. When i say anode is greater, that means, the voltage at the anode is less than voltage at the cathode alright ah. Then, the diode will not start conducting and if you know this, then you can understand the clipper. You can understand the clipper alright So, however, if V reference is made positive and then the entire waveform entire waveform will clip. If my V reference is made positive, you see if right. Now, V reference is here, then you can see the entire waveform. This is my input signal But you see the output, see the output , almost entire waveform is gone except the voltage which was above the V reference voltage which was above the V reference voltage. But, if i have reference voltage V reference voltage, if i have this V reference voltage somewhere set here, then what i will get? The output i will get. The output would be this. Everything will be clipped off. Everything, whole signal, whole signal will be clipped off. That is what it says that if V reference is made positive, then the entire output waveform below V ref will be clipped off. Easy, super easy. Super easy Let us quickly quickly see, quickly see ah your clipper. Let us quickly see your clipper and let us see if i can ah understand what we have done Let us again see very quickly the positive clipper. Positive clipper diode is in this fashion right and cathode is connected to the output of opamp . Anode is connected to the V reference. V reference is the voltage across here. Your plus minus ah V C C and V is already given signal is given to the non-inverting terminal right ah. And then, what we have the ah the if you see the see if you see the waveform, you can see that for all the signals V , then V reference the the signal is ah the signal is clipped off in the output in the output the signal is clipped off Now, we have for the for first case, when your input voltage is less than reference voltage, input voltage is less than reference voltage right. In this case, my diode will conduct my diode will conduct and the opamp works as a voltage follower If my diode works, conducts, then my opamp works as a voltage follower when the case when my V I is greater than V ref my diode will not conduct and diode when does not conduct This becomes open loop and becomes open loop My output voltage is nothing but V reference voltage. If my if i keep my V reference voltage negative, then the entire waveform entire waveform would be clipped off. Now, you go to the negative clipper. When you go to the negative clipper, the diode is in this particular fashion; that means, the anode is connected to the output of the opamp, the cathode is connected to the reference voltage. Your reference voltage is here, reference voltage right And now, if i have a first condition when my input voltage is greater than reference voltage, my input voltage is greater than reference voltage, my diode will conduct When my diode conducts, my opamp becomes nothing but a voltage follower . My input voltage is less than my reference voltage, then this diode will not conduct because, this because the voltage across the cathode is more than voltage across the anode. And that is why, when it does not conduct, the opamp becomes open loop. When it becomes open loop, then my voltage at the output is nothing but my V reference voltage right And then, if i make my V reference voltage ah positive, then i can entirely clip off my signal; that means, that in both the cases the reference voltage that we generate with the help of potentiometer is extremely important factor to control the clipping of the signal, good Now, let us move to the experiment. Experiment right. We have to see the experiment for the ah clipper. First, we will see positive clipper and then we will see negative clipper ok So, first if ah, if you want to make a positive clipper, we all now know we all are expert now right. Because, all now know how diode works ah. By by understanding how diode works, we can understand how the clipper works right So, we draw the circuit as shown in figure 5. We show we have we have we can draw the circuit as shown in figure 5. We apply voltage at the non-inverting terminal of the operational amplifiers. How much voltage we will apply? We will apply a 5 volts, peak to peak sine wave at 1 kilohertz to the V 1 From where you will apply 5 volts peak to peak? Now you should all tell or say the same thing which i am saying. Before i give the answer, you should be able to give the answer alright. This 5 volts peak to peak 1 kilohertz frequency can be given to the input of the non invertary amplifier using frequency generator, using the function generator right. And the output voltage observe the output voltage V o and note down, it is peak to peak voltage This output voltage output voltage from where we will measure output voltage or using what we will measure output voltage? We will measure output voltage using oscilloscope. That is the correct answer, digital oscilloscope right Because we have DSO here, you can use a ah CRO also. You can use C R O cathode ray oscilloscope No problem. Absolutely no problem right. If you do not have oscilloscope, you can measure the voltage at the output with the help of digital multimeter . The only only disadvantage would be you will not be able to comment on the waveform. You will not be able to comment on the waveform, you will not able to see the waveforms. Otherwise, you can see the voltage. So, that is fine right So, here we will if we have if we have function generator , if we have DC power supply, if we have oscilloscope, if you have a breadboard, if you have opamp, if you have a resistors and if you have a diode and if ah if you know this circuit, then you can design this circuit and connect it as shown in figure and then apply 5 volts peak to peak, measure the output voltage V o and then comment on the shape of output signal Now, what we will be doing? We will adjust the potentiometer. We will adjust this potentiometer to vary the reference voltage and see how the output waveform changes. This is what this is my positive clipper, this is my positive clipper. So, ah let us perform ah this experiment T. A. Seetharam will use multisim for simulation So now, we will see the working of ah the clipper using multisim. Then, we will look into the experiment. So, let me construct the same circuit using multisim So, let me open a multisim. So, i will keep ah side by side. So, i will take an opamp So, i am taking an opamp . In this case, i am taking 3 terminal opamp ah. Let me flip it So, that it will looks similar to that what we have seen in the circuit 2 and and ah what we require? We require a diode. So, i am taking the diode here, but it should be in reverse It should be in different way. So, i am just flipping it and we need one 10 k resistor So, i will take 1 resistor which is nothing but a R L load resistance in this case and the value i am changing it to 10 k right right So, this is where that we have to provide a reference to the circuit. So, let me connect completely. So, this particular terminal has to be connected here and the diode the this has to be connected here and this portion has to be connect here So, this is the where we have to provide an input signal and this is where we have to ah provide an reference signal. So, if if you look into the circuit. So, we are providing a input reference voltage. Using a potentiometer in this case rather we can use a potentiometer or we can also use ah another voltage source So, for that, so, what i will be doing is that, i will take voltage source D C voltage in this case So, connecting from here to here ; connecting from ah positive terminal to ah this particular value and where as the other ah the terminal should be connected to the ground. So, let me connect it. So, what i will be doing is that input will be show connected to the positive terminal of an operational amplifier So, i will take ah AC voltage. I will be connecting from here to here and other terminal should be connected to the ground. Now, the circuit  the DC power supply to the opamp, so, we are applying the bias voltage through the opamp Now, you can see he has to turn on the opamp and we applied the bias voltage, the input signal. The input signal is 5 volts peak to peak, 5 volts peak to peak and the frequency is 1 kilohertz. So, if you see a function generator, what he is doing? He is adjusting or he has to adjust ah 1 kilohertz 5 volts peak to peak. So, 5 volts peak to peak is there. We have to change it to 1 kilo hertz ok So, 1 kilohertz frequency is set. We have 5 volts peak to peak which is also set. Now, we are looking at the output of the opamp, output of the circuit output of the positive clipper circuit. Let us see output voltage So, if you can focus on the if you can focus and you see he is changing the reference voltage, you see he is changing a beautiful. If he goes on changing the reference voltage, see now we have no clipping action no clipping action But, if i see slowly if i change the reference voltage, slowly if you change the reference voltage you will see see clipping action Excellent, he is changing the reference voltage with the help of potentiometer like you see and you see oscilloscope and an ah potentiometer These are we turning it. You see he is turning it. The potentiometer in a ah in a anticlockwise direction, in a anticlockwise direction and and you can see the change in the signal at the output volt at the oscilloscope Now, again he is changing. So, ah by changing the reference voltage can you can you just click positive ah yeah ok. So, in this case as you see right, we have completely ah cut or clip the positive ah of the of the signal So, signal that is you can see is 1 ah this is your 5 volts peak to peak ah which is your yellow signal, that is a input and the output is your blue signal and . You can see the positive part of the blue ah of the input signal is clipped is clipped completely is clipped completely with the help of the positive clipper right So, here you can very very easily understand how a positive clipper would be used to clip the signal and how you the reference voltage plays a part right to clip the signal correct So, you can see again . Since it is very interesting experiment, we have shown you 2 3 times or rather than we can see in live that how the clipping action occurs. So, the the circuit is in a perfect condition and this is how it should be demonstrated. So, ah this is guys for you ah how the clipping action would work particularly the positive clipper Now, another thing that we have learn is the negative clipper, is the negative clipper So, we all have studied we all have studied what is negative clipper. So, quickly if you want to tell me if in this circuit, if i want from positive clipper to negative clipper, what i will do? I will just change the direction of the diode, i just reverse the diode, i just reverse the diode alright. So, if i want to reverse the diode; that means, if i consider this particular circuit right you see. So, let this the diode here and the diode here right quickly see diode reversed or otherwise the circuit is same. See circuit is same positive clipper, negative clipper, positive clipper, negative clipper circuit is same; diode is reversed Alright, here is my V reference here is my input signal here is my output signal here is my output signal . Now what are the ah components here again we have the bias voltage that is using the DC regulated power supply right. So, we have V E E, we have V C C right We have input signal , we have output signal, we have load resistor, we have potentiometer Easy finally, we have here operation amplifier operation amplifier Now, let us see if i connect this circuit as shown in figure 6 and if i want to measure the output voltage when i change, the when i apply input voltage of a 5 volts peak to peak sine wave at V I here right, i will observe my output voltage V o and note down it is peak to peak output voltage right What i will do? I will measure the input voltage and i will apply the input voltage 5 volts and i will measure the output voltage V o and note down it is peak to peak voltage peak to peak voltage. Suppose, this is input , what will be my output peak to peak voltage? I will measure at the output peak to peak voltage, i am applying at the input correct So, let us see and finally, what i will do? I will comment comment on the shape of the signal comment on the shape of the output signal . Again, we can use the potentiometer We can adjust the potentiometer to vary the reference voltage and see how the output voltage waveform changes, that is we are going to change the potentiometer and we will see corresponding change in the waveforms in the output Now, ah T. A. Seetharam will use multisim for simulation. Now, let us see how to analyze and how to design the negative clipper circuit using multisim. So, for that case let me open multisim right. So, let me construct the circuit So, what i will be doing is i will take an opamp let me clip it. So, it looks similar to what we have seen here and we need a diode So, but if we observe carefully, in in the previous case the the diode connections as well as ah for the negative clipper the diode, connections are different. As a result, it completely changes which portion of your input signal to be clipped off. In the previous case, which is of positive clipper right ah The this particular the cathode the cathode is connected to the output of your operational amplifier Whereas, in case of your ah negative clipper the anode is be connected to the output of your operational amplifier as a result, as a result ah ah it access a negative clipper; that means, that below the reference value which are the voltage that you have will be completely clipped off, but above the reference value will be there. So, in order to understand, what i will do is that, i will take a resistor need not to be connected to a resistor directly We can also ah we can connect ah to ah the voltage source directly at this 0.2. Now, i will take a DC voltage. I will be connecting the positive terminal to the 10 k input of your R 1 ah this R L. This case which is a nothing but R 1. In this case, resistive as a result, i will keep the magnitude as a negative So, we can understand what negative amplitude that we are applying as an input to ah input as a reference right and whereas, the negative inverting terminal should be connected in this way as a feedback now the input should ah the input voltage which is of sinusoidal as if we connected to the non (Refer Time:29:32) terminal. So, let me connect it other terminal to the ground. So, i will take the peak to peak value as ah 5 volt. So, the peak value should be 2.5 read one kilohertz. So, in order to understand how exactly the circuit works right. I will take a probes; one at the input side other one at the output So; that means, at which of the signal that we see represented using the green will be the input the sky sky blue represents the output signal. Now, if you if you compare the circuit, whatever we have seen in the ah presentation as well as we have seen in the multisim, there is a small difference That difference is that the input voltage We are connecting in this case using ah potentiometer But, where as in this case, we are connecting directly to the circuit. So, as a result whatever the reference value that you said about that reference will be remains and below the reference value will be completely removed off ok So, to analyze it ah, let me split the circuit and run the circuit. Now, what is happening here? We can see the green represents completely our input signal right. So, since the peak to peak is of 2.5. We can see peak to peak and the frequency of 1 kilo, but output output shows 12.5 . The reason is that the input whatever we have connected is of 12 volts So, since this shows 12 volts ah since the input is 12 volts and the reference values is of 12 volts and whereas, your input is 2.5 which is which ah which is very, very smaller than that. So, since there is no signal to be ah clipped off that is a reason the output will be whatever you applied So now, what i will be doing is to understand this, let me make it as 0 right. If you recall what we have seen in the positive clipper, when when the reference is 0 ah whichever the ah voltage ah input voltage below the this particular reference is completely clipped off. So, it looks similar to that of your ah half way rectifier is not it. But, whereas, in this case if you see that ah whichever  