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The second equation we have here is what I showed you in the previous slide. It is simply an equation, which relates the diode voltage to the diode current. So, that is all that equation is. It relates diode voltage diode current, as I told you that comes from the basic behaviour of the charge carriers inside you know inside the semiconductors, and how they move across the junction, what is the flux going one way, what is the flux going the other way etcetera and then once you do all that calculation will arrive at that equation. So, that is that equation which we are not deriving which as I said you can look up in a book associated with semiconductors, but that is the one equation you have to look at. So, now if you look at it, what we can do is, we can substitute this equation into this here. Okay, so, I have an expression for Id, I am going to substitute it in equation 1 which is, let, let me call this equation 1. So, I am going to substitute equation 2 in equation 1 and do a little bit of rearrangement. So, a little bit of rearrangement. So, so that we can you know put something together. So, let just substitute it there and then we will take a look. So, if I want to write it in terms of say, let us say IL, Id is we have put this equation now. So, I just put that down there Iph minus I0 e power q Vd by gamma kT minus 1 equals IL ok. So, that is the thing we want to do. So, let me do some rearrangement here if I want an expression for the diode voltage. So, in terms of the currents that are out there. Supposing I want an expression for the diode voltage in terms of the currents that are present here, what would I have to do? I will rearrange this. So, we have Iph minus IL. So, I am just moving the IL to this side and moving the other currencies to the other side equals I0 to the power q Vd by kT, gamma kT minus 1. So, I have just rearranged. So, now, I will move the I0 this side and I will move the 1 also to the other. So, I will basically do that here, Iph minus IL by I0 and then I can, so, this is now one term and then I can move the 1 to this side. So, that will become a plus 1, equals e power q Vd by gamma kT, ok. So, now if we take natural logarithm on both sides, so, you basically and then do a little bit of rearrangement, you will get Vd equals, so, I will take the natural log here. So, I will put here lon of I ph minus IL by I0 plus 1. And this is q and this would have been a q by gamma k T would have been that next to this. So, here you would have had this q by gamma k T sitting here. So, I will shift that to the other side. So, I will have gamma k T by q. So, I have just. So, therefore, we have got an expression for the diode voltage, in terms of the currents that are present ok. So, that is the other one equation that we have arrived at, simply by looking at how, by simply substituting equation 2 in equation 1 which is simply the expression for how the diode current varies as a function of diode voltage into expression one, which talks of how the current is distributing as it comes off the solar cell into either the load circuit or into the diode that is internally present within the circuit. So, once you do that in terms of the current. So, now, we only have the load current the photocurrent and I0 that are present there and concerning that we have an expression for Vd. So, this is what we have put together. Now, let us look at equation 3. So, what is equation 3? We are simply saying that there is a voltage across this diode. So, that voltage is this voltage Vd. So, that voltage is this from here to here, what voltage you see, is the voltage Vd. The voltage that you see here across the load resistor, the voltage that you see here is Vlad V subscript L, that is the load resistor right and so, the voltage across the load resistor is VL. So, how do Vd and VL relate? We just had a relationship between you know Id and IL right, we came up with this equation 1 which is a relationship between Id and IL, which says that you have so much photocurrent something went into the diode remaining current goes into the load. Similarly, we can think of a relationship between the voltage across the diode and the voltage across the load. What is the relationship? Whatever is there across the diode will be equal to load because that is the source from where the, we know that is where the current is generating, that will be equal to whatever is there across the load plus the IR drop across this resistor RS right. This resistance, series resistor, resistor due to all the internal resistance is present in the diode and the contact resistances. So, if you include all that that is the resistor there. So, the IR drop across it is a voltage drop. So, you have generated a diode voltage, but some of them due to the current going through this series resistor there is an IR drop, the remaining voltage goes to the load ok. So, what is this I? This I is the same I that is coming here, this is the same as in this circuit you only have IL, once you cross this point you only have IL. So, this is IL, I subscript L is current, it is not just some I, it is I subscript L, it is a current that is going into the load, it is also the current that is going into the series resistor, I subscript L and the resistance there is this series resistor RS. So, therefore, the voltage of the on the load plus the IL RS together will equal the voltage across the diode. So, that is what we have here, the voltage across the diode is equal to the voltage across the load plus the IR drop that was there on that series resistor. So, Vd equals VL plus IL RS. So, now what we are simply going to do is, we are going to rearrange this. So, we are simply going to say therefore, VL equals Vd minus IL RS. So, this is what we are going to, we are simply rearranging that equation, and for Vd, we have an expression here, right.