 Good morning, dear friends, and welcome to this session on DC characteristics of op-amp. We are going to have rather two sessions on this. This is the first session on DC characteristics of op-amp, and Dr. Sachin R. Gengze, Professor and Head, Department of Electronics Engineering at Walshchand Institute of Technology, Seoul. What are the learning outcomes of this session? After completing this session, student will be able to explain difference between an ideal op-amp and practical op-amp. Also he can explain some of the input characteristics like input bias current, input offset current and input offset voltage of an op-amp. The contents of this session include, first we will try to understand difference between an ideal and practical op-amp, and then we will have a look at the different characteristic of op-amp like input bias current, input offset current and input offset voltage. What is an operational amplifier? We already know op-amp is a directly coupled hydrogen amplifier. It is available as a single chip or single IC. It is a versatile device, can be used for DC and AC applications and also it can be used for some of the mathematical operations like addition, subtraction, multiplication, integration. It can be used to design oscillators, active filter, comparator, regulator and many more other circuits. What are the characteristics of an ideal op-amp? Few of the characteristics of ideal op-amp includes, first it has infinite input impedance or resistance. So, R i is equal to infinity. It has a zero output resistance, R 0 equal to 0. Infinite open loop voltage gain, that is open loop voltage gain is A which is equal to infinite. It has an infinite bandwidth, it can it means that it can amplify all the signals, all frequency signal with the same voltage gain and it has a zero phase shape. It means that the output and input signal are in phase and then it has a zero offset voltage. However, when we are dealing with the practical op-amp, it is going to be different than the ideal op-amp. Let us have a look at an equivalent circuit, a very simplified equivalent circuit of a practical op-amp. Look at this, this is called as an input resistance. We know that in case of ideal op-amp, this input resistance is infinite. In other words, this should be an open circuit. But in case of practical op-amp, this R i is not infinite, but will have some finite value. Similarly, look at this, which is an R 0 or the output resistance. In case of ideal op-amp, we say that the R 0 equals to 0. It means that this part should be actually a short circuit. But in case of practical op-amp, the R 0 is not 0, but has some small finite value. This is what is the actual gain offered by the op-amp, which is nothing but A that is open loop gain multiplied by V i d, a differential voltage, which is nothing but a difference in a voltage between the non-inverting and inverting terminal. Similarly, this represents the voltage transfer curve of an op-amp, which is represented by the V i d. So, plus V i d, you can see that on excess, I am representing V i d. So, if it is a plus V i d, it means that V i d is nothing but V 1 minus V 2. So, if it is a plus V i d, then it simply means that it is a difference between non-inverting and inverting input. Non-inverting input is greater and then similarly, when it is a minus V i d, it simply means that the inverting input is inverting input voltage is greater than non-inverting input voltage. In this case, when the non-inverting input is greater, that this is a differential voltage and you can see that the output of the op-amp will rise very steeply and then it becomes into a saturation. Similarly, when the V i d is negative, it means that the inverting input is greater than also the output voltage will rise on Y axis. We have an output voltage which will rise very steeply and then it becomes into a saturation. This part of the op-amp is called this is where we can use op-amp in an active region. This is the part, this part and this part where we are using op-amp into a saturation region. This A represent the slope of this, this is represented by A, which is nothing but the open loop gain of an op-amp. In case of an ideal op-amp, this is going to be exactly vertical because A is going to be infinite. But in case of practical op-amp, the slope is very high or this line is very steep, but it will have some finite slope. Now, what is the difference between an ideal op-amp and a practical op-amp? The ideal op-amp do not take any current from source and its response is independent of temperature. That is what we assume the ideal op-amp as the input input resistance of an ideal op-amp is infinite. It is not taking any current from the source which is driving it and similarly, the working op-amp this op-amp is not depending on the temperature. However, for practical op-amp, there are three things which are happening. First, the practical op-amp do take some current from the source. Secondly, the two input that is inverting and non-inverting input respond differently to voltage and current due to the mismatch of the two transistors at the input stage. Later, we are going to look at there is a differential amplifier at the input stage of the op-amp and there are two transistors which are rather responding differently to the voltages and current and thirdly, the working of the practical op-amp depend upon the temperature. As you can see, this is an equivalent circuit of an op-amp. The first stage of op-amp consists of a differential amplifier formed because of these two input transistors Q1 and Q2. The non-inverting input is directly driving the base of the Q1. The inverting input is directly driving the base of the Q2 and if there is a mismatch in the fabrication of this Q1 and Q2 which is bound to happen, then the Q1 and Q2 and hence, the inverting terminal and non-inverting terminal are going to respond differently for the input voltages. Now, these non-ideal DC characteristics, what happened because of this? Because of this, it adds error to the working of an op-amp and what are these error or what are these, what happened because of this? Because of this, there are three, four things that may happen. The first is called as an input bias current. The second is called as an input offset current. Then there is an input offset voltage and then there is what is called as a thermal drip. So, let us have a look at these one by one. What is the input bias current? The first important DC characteristic of practical op-amp is nothing but an input bias current. Now, we know that there are two transistors at the input of the op-amp and they are either VJT or FET and they are biased in their linear region. Ideally, no current is drawn by the op-amp input. So, no current enters into the base of these two transistors but practically very small current is drawn by the basis of these two transistors. Now, these two current which is IB1 which is into a non-inverting terminal of an op-amp and then IB2 which is current into the inverting terminal of op-amp, they are ideally they should be 0 but they are not 0 to further complicate they are not equal also. So, these give rise to the first parameter of an op-amp which is called as an input bias current. So, as you can see, even though both the input terminals are connected to ground, some small current is entering and that into the base into the inverting and non-inverting terminal and that is called as an average of that is called as an input bias current. So, input bias current is the average of the current that flows into the inverting and non-inverting terminal of the op-amp. So, IB1 plus IB2 divided by 2 gives an IB. So, one of the way to specify the quality of the op-amp is depending upon how much is that IB. So, if for example, in case of 741C, the IB has a maximum value of plus or minus 500 nanoampere similarly and for a better op-amp like a precision 741C, this IB or input bias current is very less, it is equal to plus or minus 7 nanoampere. The second parameter is called as an input offset voltage. We already know that IB1 and IB2 are the two currents which are supposed to be 0, but they are non-zero, they are non-equal and then the difference between that, the absolute difference between the IB1 and IB2 is called as an input offset current. So, input offset current is the difference between the current that is entering into the inverting and non-inverting terminals of the op-amp and it is an absolute difference because we cannot predict whether the IB1 is greater or IB2 is greater. So, I am saying that the input offset current is an absolute difference between IB1 and IB2. Again, some of the specification, let us have look at the some of the specifications of a typical op-amp for 741C, the value of input offset current is 200 nanoampere and again for the precision 741C which is a better op-amp, this value is very less which is equal to 6 nanoampere. Now, you can pause the video and answer what is the input bias current and input offset current for an ideal op-amp? We have seen that there are some values for the practical op-amp, but what do you think should be the input bias current and input offset current of an ideal op-amp? So, please pause the video for some time and answer these questions. Lastly, we come to a parameter called as an output offset voltage. We know that when both the inputs of the op-amp are connected to the 0, that inverting and non-inverting are connected to ground, then there should not be any output voltage. However, even though both the inputs are connected to the ground, we get certain output voltage and that output offset output voltage is called as an output offset voltage. Now, in order to nullify this output offset voltage, we need to apply some small DC input voltage either to the inverting terminal or to the non-inverting terminal of an op-amp and that small voltage that we must apply it between the two input terminals of the op-amp, so as to make the output offset voltage equal to 0 is called as an input offset voltage. Again, for 741C, the input offset voltage is 6 milliholt and that for the precision 741C, the input offset voltage is 150 micro-olt. So, this is another important parameter of an op-amp called as an input offset voltage. With that, we come to the end of the today's session and I am leaving you with some points to be discussed. First is, is it possible to have ideal op-amp? If no, why? Secondly, even if real op-amp is available, will it be useful as an amplifier? The clue for discussion of this question is that please remember the transfer characteristic of an op-amp and think whether even if real op-amp is available, will it be useful as an amplifier? If no, why? References for today's session are again op-amp and linear integrated circuit by Ramakan Ghaikwad and the second book you have to refer is linear integrated circuit by Raj Choudhury and Shai Jai. Thank you, dear student for listening.