 Sudhakar Barbade, Assistant Professor, Electronics and Telecommunication Engineering, Walchand Institute of Technology, Solapur. Today, we will discuss event counting using IC-74C926. Learning outcome. At the end of this session, friends will be able to describe event counting using IC-74C926. Contents. We will study event counting by its block diagram first. Then, we will study and design input interface required for all the measurements whatsoever we have done. Event counting. Here, as usual, we have 74C926 and its associated display circuit connected as shown. And here, we require the clock input to be given to the IC-74C926 and reset to be given when counting is over. And this display select is connected to ground so that we can display the data. Here, the events are converted into electrical signal and this electrical signal is given to event processing circuit, which converts the input electrical signal into the appropriate digital signal which is required to be given to the AND gate input. And another input coming from the external trigger input through this control signal generation circuitry block. So, whenever we want to start event counting, an external trigger is to be given. This trigger generates the logic one voltage on this and whatever the event signal is passed to the clock input of the 74C926 and it starts counting. Whenever counting is over, we can give another external trigger so that this output becomes zero events are now blocked at this AND gate. And whenever we want to start a new counting process, this reset signal we can generate by using this control signal circuit. So, this is what a block diagram of event counting. Now, let us see the input interface present in all of the IC-74C related circuits. Here, this is what the block diagram of input interface circuit which consists of amplifier and shemit trigger. Amplifier amplifies the input sinusoidal signal or analog signal we can say which is a very weak one. So, we obtain an amplified signal here and that amplified signal is given to the shemit trigger which then generates the pure digital output which can be given to the input of the measurement. As we see here, it consists of mainly amplifier and shemit trigger. So, here the circuit is built using opamp and here opamp is used as an amplifier. If you look at the input signal which is an analog one is passed through this potentiometer. We can adjust this input voltage quite for calibration and this is passed through the capacitor which blocks any DC components present in the signal to be given to the opamp input. So, only AC signal will be passed and this will be given to the opamp non-inverting input terminal. Whereas, inverting input terminal is connected through these resistors and the gain can be adjusted by using this feedback resistor and this input resistor. So, this is what amplification of any analog input then amplified analog input is given to the shemit trigger input. This is what opamp used as a shemit trigger. Here we give the input voltage to the inverting input of the opamp and non-inverting input is connected to input reference voltage that is through R2 and R1 and this is the feedback resistor and we get the output. So, we will see later on the details of this shemit trigger. Let us see the analysis of the previous circuit. In this analysis we are assuming this R1, R2, R3 of 10 K value and V reference is equal to 5 volt. When this is the case now let us assume that now the opamp output is 5 volt. Now when V0 is equal to 5 volt then what will happen is just to look at when V0 is equal to 5 volt what will happen? V reference is also 5 volt, V0 is also 5 volt that means this point and this point are tied together and means this R2 and R3 come in parallel with each other and R1 resistance is connected through ground that means whatever the potential across this we are finding this is a point A that potential at that point we are finding. So potential at point A that is for example this the equivalent circuit is here drawn V reference is connected to 5 volt already we have assumed and V0 is 5 volt our first assumption that means R2 and R3 come in parallel and this R1 is connected through ground so this is point A. So voltage at this point VA will be equal to since this is a common terminal and there we have 5 volt. So V reference divided by total resistance total resistance is a R1 plus parallel combination of R2 and R3 that is why V reference divided by R1 plus R2 in parallel with R3 into resistance R1 that is why I multiplied by R1. So if you put these values R1, R2, R3 is equal to 10 K we get VA is equal to 3.33 volt. So this is named as VTH higher trigger point and when V0 is equal to 0 for example we have assumed that this V0 is 0 that means this terminal connected to ground that means R1 and R3 appear in parallel and this point is this connected through R2 to V reference. So equivalent circuit will look like this R1 and R3 will come in parallel this is a ground V0 is also ground that means these two points are as good as connected together and this point A is connected to V reference through R2. So voltage at point A is found VA equal to R2 plus parallel combination of these two. So R2 plus parallel combination of R1 and R3 and the reference voltage is V5 volt as we know. So V reference divided by this total effective resistance multiplied by the resistance across at point A. So resistance at point A will be R1 in parallel with R3. So multiplied by R1 in parallel with R3. So with this we get VA is equal to 1.66 volt which is known as VTL lower trigger point. So this way we get different voltages at point A when the op amp is at positive saturation or at ground. So what type of output we get in the above circuit? So the answer should be in terms of whether the output is in phase or inverted. You pause the video and answer the question. See the question was what type of output we get for the circuit given. This is what the circuit given to us and this for example this is what the analog input and this is what the output. So if you look at here when the input signal is going positive and when it reaches this higher trigger point the output is going to logic 0. And when this signal is coming down and when it reaches lower trigger point at that point the output is becoming high. That means what we are getting here is inverted output. Now let us see how we can understand the Schmitt trigger using this diagram. On x axis we have V in and on y axis we have V0. So if you look at when the input voltage is going in the positive direction and we assume at that time the output is logic high. This output is logic high or 5 volt we can say. So here if it is a 5 volt here this terminal already we have seen this is a 3.33 volt. Potential at point A at this point is 3.33 volt. So as the input increases and it approaches this Vth known as higher trigger point that is 3.33 volt the output of the Schmitt trigger changes to the other state called 0. And it remains 0 see look at here when the output was high when Vth reaches when the input voltage reaches to Vth the transition takes place in the output. Output becomes 0 and it remains 0 till the input voltage drops below this Vtl that is lower trigger point. So if you look at this output remain the output will remain 0 till this input reaches to Vtl. So whenever input reaches Vtl again the output of the op-amp changes its states from low to high and this process continues. References. Electronic system designed by Vibhav T. Tarate. Electrotech publication Satara. And one more website I have taken for preparing these slides that is given here. Thank you very much.