In this lecture we will start a new chapter Logic gates. First we will discuss what is logic gate then we will move to different types of logic gates available to us. So what is a logic gate?
It is a physical device which performs logic operation on one or more logical inputs and produces a single logical output. So the first thing is physical device. Logic gate is a physical device. and it performs logic operation on one or more logical inputs and produces a single logical output.
Very clear. The logic operations like inversion, logical multiplication, logical sum etc. we can perform them by using logic gate. and we can have one or more logical inputs.
For example in inversion we have only one input and in logical multiplication we can have two or more inputs. So in logic gates we can have one or more logical inputs and it produces a single logical output. So this is a formal definition for logic gates.
Now we will categorize logic gates into three groups. So let's do it. The group number one is basic gates Basic gates in this group we have three gates first one is not gate second one is And gate and last one is our gate now why we call not and and our Basic gates because we can implement any logic by using not and and are By using these three gates only we can implement any digital circuit That's why we call them basic gates. The second one is Universal gates. And in this group we have two gates.
First one is NAND gate and second one is NOR gate. We call them Universal gates because by using Only NAND or only NOR we can implement any digital system. The last group is arithmetic gates. And in this group also we have two gates. First one is XOR and second one is XNOR.
We call them arithmetic gates because we use them in arithmetic operations. We will discuss all these gates one by one. and in this presentation we will start with NOT gate.
This is very important presentation because we are going to discuss some very important problems in this presentation because most of you already know what is NOT gate, how it is represented and what is the truth table for NOT gate but the important thing is the problem. I am going to discuss some problems which appeared in IES and gate exam and I will also give homework problems based on the concepts we will learn. So let's move to NOT gate. Not gate first we will see how we represent not gate symbol for not gate This is the symbol for not gate a is the input Y is the output and y is equal to a complement right we can also represent not gate like this Instead of putting bubble at the last we can put bubble in the front And the operation will remain the same. Y is equal to A complement.
Now we will move to truth table. Truth table. But first I want to discuss what is truth table.
Then we will see truth table for not gate. It is the table having outputs from all possible combination of input or inputs. So let me write this down. It is the table.
Table. having outputs from all the possible all the possible combinations of input or inputs right so this is the formal definition for truth table now we will make truth table for not gate there is only one input In case of NOT gate, let's say it is A. and y is the output and it is equal to a complement.
As there is only one input, we will have two combinations 0, 1. Complement of 0 is equal to 1 and complement of 1 is equal to 0. So this is the truth table for NOT gate. Now we will move to one question, question number 1 and this question appeared in IES 2010 and the same question appeared In gate 2000, gate 2010 also. So let's see what is the problem.
In this problem we have the circuit like this. These are the two not gates and we have feedback like this. Why is the output?
And we have to tell whether the circuit is buffer. B option is A stable multivibrator, A stable multivibrator. C option is Bi stable multivibrator, Bi stable multivibrator. And D option is square wave generator. Let's see how we can solve this.
First I will make the circuit. This is the feedback and y is the output. Before moving to the circuit, first I would like to explain buffer. What is buffer?
In buffer if we apply 0, we will have 0 and if we apply 1, we will have 1. So basically this is opposite of NOT gate and it is represented like this. If you compare the symbol of buffer, this is buffer. If you compare the symbol with NOT gate you will find there is no bubble in buffer. Because if the input is A, the output is also going to be A.
If the input is 0, output will be 0. If the input is 1, output will be 1. And if there is no input, no input, there is no output. So this is what we mean by buffer in digital electronics. Now we will analyze the given circuit.
Let's say here we have one. As 1 is the input to this NOT gate, output is going to be 0. This 0 is acting as input to this NOT gate, so output is going to be 1. So input was 1 and now we have output y equals to 1. Right? Again we have feedback and this feedback will give us 1 again because output is 1. And we have 1 feedback and because of this feedback we are again going to get 1. So again we will get 1. So output is going to be 1 always. If we have input equals to 1 and if I make input equals to 0, here we will have 1. This 1 is acting as input to this NOT gate.
So here we will have 0. And as there is feedback, again we are going to get 0. So output is going to be 0. So if input is equal to 1, we are going to get output equal to 1. And if input is equal to 0, we are going to get output equals to 0. So this is how the circuit is working and now we have to find out correct option for the circuit. Initially it looks like this is buffer because I told you in buffer if input is 0, output is also 0 and if input is 1, output is also equal to 1. The same thing is happening here when input is equal to 1, output is 1, when input is equal to 0, output is 0. But this is not buffer because of this feedback. The feedback here, this is the feedback and because of this feedback the circuit is not acting as buffer. So option A is not correct.
Now we will move to option B, a stable multi vibrator. This is not a stable multi vibrator, because we have stable output, the output is stable. If it input is one, the output is going to be one until we change the input.
So output is one, which means it is stable. It is not astable because we have two stable states 1 and 0. So this is not astable but this is bistable. In bistable multivibrator we have two stable states and here we have 1 and 0, two stable states.
So the circuit is acting as bistable multivibrator. Option D is not correct because this is not a square wave generator. We have output y like this. This is 1. Or we have output y like this. This is 1, this is 0. And square wave looks something like this.
And we are not getting any square wave. So option C is correct. This question appeared in IES 2010 and GATE 2010. Now we will move to second question.
Question number 2. In the second question, first I will make the circuit, then we will analyze it. In this question, I was having two NOT gates, but now I have three NOT gates. And let's say the propagation delay for each NOT gate is equal to small t pd, where pd stands for propagation delay.
For each gate, we have propagation delay of t small pd. Now what is propagation delay? If we apply input to any digital system, let's say this is and a digital system and we are applying input to this digital system A, the output Y is not generated instantaneously. Because this circuit here will take its time to produce the output and this time is called as propagation delay.
So we have three NOT gates and each NOT gate is having its own propagation delay. Let me write this down TPD is equal to propagation delay of each NOT gate and we have to analyze the circuit. So let's do it. The output is Y and let's say Y is equal to 0 initially. Y is equal to 0. We have a feedback.
So here we are going to get 0. 0 is acting as the input to this NOT gate. So we will have 1 here, but still output is equal to 0. So for For TPD, because TPD is the propagation delay for this NOT gate, once we apply 0 to this NOT gate, after TPD we are going to get 1. So I'm making the timing diagram and after TPD also the output is equal to 0. Right? Now this 1 is acting as the input to this NOT gate.
So output is going to be 0. So for next TPD also, we will have y equals to 0. Right and the 0 is acting as the input to this NOT gate and we'll have 1. So after the third TPD the output will be 1. Very simple. And as the output is 1 and we have a feedback we'll have 1 here. So we have 0, 1 and 0. So after 3 TPD.
the output will again go to 0. So this is how it repeats itself. Let's try to find out time period. This is the time period and it is equal to this is 3 TPD.
This is 3 TPD again. So the time period is equal to 6 TPD. Right? Because the time period is the time after which function repeats itself.
We had 0, y equals to 0 and after 6 TPD we again have y equal to 0. So I can say capital T that is the time period is equal to 6 TPD or I can generalize this capital T equals to twice of small n TPD where this small n is the number of not gates. So if I have five not gates and TPD being their time delay, the time period is equal to two multiplied by five because five is equal to n, the number of not gates multiplied by TPD. So time period is equal to 10 TPD.
So this is how we can find out time period and you can easily calculate frequency if you have time period because frequency is equal to one by In the earlier problem, circuit was not square wave generator, but this circuit here with three not gates is a square wave generator. This circuit here is a square wave generator. And also this circuit is a stable multivibrator.
And this circuit is also clock generator. When you study sequential circuit, you will learn what is clock. and how it is used.
The circuit is acting as a clock generator and also the circuit is ring oscillator. Ring oscillator. So these are the different functionalities satisfied by this circuit.
Now we will move to homework problem. The two homework problems are very important. In first homework problem, in first homework problem we have a circuit having five NOT gates. and the propagation delay of each NOT gate is 100 picoseconds. So small tPD is equal to 100 picoseconds.
And we have to calculate the frequency of generated square wave. Very simple. Once you have time period, you can easily find out frequency as 1 by T.
These are the options. In problem number 2, the propagation delay of each NOT gate is 2 nanosecond. So tPD is equal to 2 nanoseconds.
And the time period of generated square wave we have to find. You have to find time period of generated square wave. But there is slight change in the circuit. The feedback is not from here.
The feedback is not from here. But the feedback is from here. So this problem, problem number two is very good problem and very important as well. So once you have your answers, post them in comment section.