 Let's now talk about other application of p-engension. So there are so many applications of p-engension itself, so we'll take it one by one. So right now we have taken one application of p-engension that is rectifier. The second application of p-engension is from a very special purpose p-engension diode that is Zener diode. So right now I am going to talk about an application of a special purpose diode. It is called Zener diode, fine. Now Zener diode first of all is represented by this symbol. This is how you represent a Zener diode, okay? A unique thing about Zener diode is that once breakdown happens for the Zener diode, you can reuse it again. Usually what happens when some breakdown happens, the current becomes so high that the diode burns off, okay? A lot of heat gets generated and after that the diode becomes useless. But Zener diode is not like that. You can reuse it, fine? So I hope you remember the characteristic graph of a diode. Zener diode characteristic graph is also very similar to just like any other diode. The only difference is it can be reused after the breakdown, okay? So we are going to use Zener diode as voltage regulator or voltage stabilizer, fine? Write down Zener diode as voltage stabilizer, voltage regulator is better word, as voltage regulator. So it regulates the voltage and it does not let the voltage increase beyond a point, okay? So let me first draw the circuit diagram of the voltage regulator using Zener diode. Then we can discuss it in greater detail. This is how it is. So this is simple representation of the voltage regulator. What happens is that we are connecting load. Load is your device, okay? You are connecting a device parallel to the Zener diode, fine? So when an unregulated voltage comes as an input, okay? Then if you have like a huge amount of reverse bias voltage that is unregulated, then what will happen? Zener diode will break down, okay? And if Zener diode breaks down, lot of current will flow through the Zener diode, okay? And one more unique thing happens that across a Zener diode, the voltage becomes constant. So beyond this voltage, across a Zener diode, the voltage won't go beyond this, fine? So if you connect a load resistance parallel to the Zener diode, then after break down it can become a constant voltage across the load, fine? So that is how it works and there is a safety criteria also here. Safety criteria is like just to ensure that both load and Zener diode, they are safe. The safety criteria is the current in the Zener diode should be 5 times the current in the load. This actually ensures that Zener diode doesn't have such huge amount of current that even Zener diode burns off and also it ensures that whatever is excess current that doesn't goes through the load but only Zener diode captures it, okay? So I will again repeat here, we are right now exploiting the fact that after break down the potential difference across the Zener diode becomes constant. We are exploiting that property and how will you exploit that? You connect a load parallel to the Zener diode, okay? So whatever Zener diode potential is, which becomes constant after a while, the load will feel that resistance. So the load will feel that voltage, fine? So beyond a certain point, the voltage becomes constant no matter what is the input voltage, fine? So this RS which you connect here, this is called series resistance in a voltage regulator. This series resistance is required otherwise the current can grow very, very quickly, okay? Current can be very large. So in order to work safely, we have a series resistance that decreases the current little bit and also do not forget this safety criteria, okay? Fine, so let us solve a numerical based on the voltage regulator. By the way, do you have any doubts? Anyone? No, sir. No, sir. Okay. Fine. So here is a question. In a Zener regulated power supply, Zener diode has VZ to be equal to 6 volt. What does it mean? VZ is 6 volt. This is the breakdown potential, fine? VZ is 6 volt. The load current, which is IL, load current is given as 4 milliampere, okay? And unregulated voltage, that is the input voltage is given as 10 volt, okay? You need to find out the value of series resistance, RS should be equal to what? Okay? Keep in mind the safety criteria also. So this, both of you? So is it 0.6? 0.6 ohms? Yes, sir. Oh, no, no, no. I need to have converted. No, that's not correct. Okay. Ammo, did you? Okay. 300 ohms. Ammo is saying, no, that's not correct. So that's 600. No. So is it 250? Let me solve this. See, unregulated voltage is 10 volt, fine? Voltage across the diode will be 6 volt, okay? Because the Zener diode, once the breakdown happens, which will happen in this scenario, because external voltage is 10 volt and Zener diode operates in the reverse bias. So this will be equal to 6 volt, okay? Now the remaining voltage will feature here. This much will be how many volts, 4 volts, isn't it? So we have load current to be equal to 4 into 10 raise to the power minus 3 ampere. So the Zener current should be equal to what? Zener current should be equal to 5 times the load current, safety criteria. So that is 20 into 10 raise to the power minus 3 ampere, fine? So now Zener current is this and the load current is this. So the current through RS is what? Current through RS is sum of load current and Zener current, fine? This becomes 24 into 10 raise to the power minus 3 ampere, isn't it? You can apply the junction rule here. Current incoming should be equal to current this side plus current that side. This is I L and this is I Z, fine? So this is the current across RS and voltage across RS is 4 volt. So resistance RS will be equal to what? That will be equal to 4 divided by current V by I, isn't it? So it will come out to be 4 divided by 2.4 into 10 raise to the power minus 3. So it will be around 167 ohms. All of you clear? Yes, sir. Where you made an error? So I didn't take the voltage properly. I took it as the entire unregulated voltage and then I tried using Kirchoff's law. You can use Kirchoff's loop rule, that is correct. You can't use Ohm's law and also see the reason why you guys are not getting it right is because this is different, okay? This, your first step, you are analyzing a circuit with a diode in it, okay? So that automatically gives you a lower confidence when you solve this particular question, okay? But then when you look at the solution, you know, it's straightforward in a way, okay? So that hesitation or that lower confidence will go only when you practice a lot of questions.