basics of PID control

[PG1995]

Thank you, Steve.

Let's freeze this discussion about the block diagram for some time. We can come back to it later. At the moment I need to discuss something more important. I believe that I don't really have good intuitive understanding of MPPT which I came to to realize after post #15 above.

The reply of NorthGuy to Q4 in this post made things really easier for me. After his reply, I started to picture the capacitor, Cin, as playing the role of input voltage source for the buck converter and at the same time also functioning as dummy load or resistance to the panel. I was comfortable with this concept and you also okayed the way I was thinking of Cin. Then, you had to come along and destroy everything for me! As a matter of fact, I had already known that there was a flaw in my thinking but I simply didn't think about it much.

Let's discuss few simple questions first which directly and indirectly relate to maximum power tracking.

Q1: A 100-watt incandescent bulb operated at 220V would take 0.5A current. It means that its resistance is 440Ω. What would happen if such an incandescent bulb is directly connected to this solar panel which can function at Vmp and Imp. Would we get open circuit voltage of 22.7V across the bulb with no current through it?

Q2: I think the words "heavy load" and "high resistance" are opposite of each other. A heavy load means something with low electrical resistance which take heavy current. For example, an electric heater is a heavy load. On the other hand, something with high resistance takes little current and can be called light load. Do I have it correct?

Regards
PG

 

[steveB]

Thank you, Steve.

Let's freeze this discussion about the block diagram for some time. We can come back to it later. At the moment I need to discuss something more important. I believe that I don't really have good intuitive understanding of MPPT which I came to to realize after post #15 above.

The reply of NorthGuy to Q4 in this post made things really easier for me. After his reply, I started to picture the capacitor, Cin, as playing the role of input voltage source for the buck converter and at the same time also functioning as dummy load or resistance to the panel. I was comfortable with this concept and you also okayed the way I was thinking of Cin. Then, you had to come along and destroy everything for me! As a matter of fact, I had already known that there was a flaw in my thinking but I simply didn't think about it much.

Let's discuss few simple questions first which directly and indirectly relate to maximum power tracking.

Q1: A 100-watt incandescent bulb operated at 220V would take 0.5A current. It means that its resistance is 440Ω. What would happen if such an incandescent bulb is directly connected to this solar panel which can function at Vmp and Imp. Would we get open circuit voltage of 22.7V across the bulb with no current through it?

Q2: I think the words "heavy load" and "high resistance" are opposite of each other. A heavy load means something with low electrical resistance which take heavy current. For example, an electric heater is a heavy load. On the other hand, something with high resistance takes little current and can be called light load. Do I have it correct?

Regards
PG

Ok, we can come back to the block diagram later.

I did "OK" your Cin viewpoint because it seemed correct if I understood you accurately. Still, I also tried to offer another viewpoint based on input resistance which is less confusing, in my opinion. There is always more than one way to get an intuitive understanding of something.

Q1: If we forget about the fact that a light bulb resistance is temperature dependent which makes it dependent on current, we can treat it as a 440 Ohm resistance. In fact lets talk about a 440 Ohm resistance instead of a lightbulb. If you say you would get zero current and open circuit voltage, you are saying this 440 Ohm resistor is behaving as an open circuit. That makes no sense at all. The solar cell will operate at a point of the curve where Vmp/Imp=440 Ohms. It's as simple as that.

Q2: You are correct.

 

[PG1995]

Thank you.

I did "OK" your Cin viewpoint because it seemed correct if I understood you accurately. Still, I also tried to offer another viewpoint based on input resistance which is less confusing, in my opinion. There is always more than one way to get an intuitive understanding of something.

Yes, you are right and the only one who is to be blamed is myself.

Q1: If we forget about the fact that a light bulb resistance is temperature dependent which makes it dependent on current, we can treat it as a 440 Ohm resistance. In fact lets talk about a 440 Ohm resistance instead of a lightbulb. If you say you would get zero current and open circuit voltage, you are saying this 440 Ohm resistor is behaving as an open circuit. That makes no sense at all. The solar cell will operate at a point of the curve where Vmp/Imp=440 Ohms. It's as simple as that.

I knew it didn't make much sense but still I wanted to hear your reply before going any further. So, we have a 440 ohm resistor connected to a solar panel directly and the incident irradiance is 1000W/m^2 and ambient temperature is 25 Celsius. The panel is 220 watt (i.e. Vmp x Imp=220W at 1000W/m^2 and 25C). Could you please tell me from this figure that at what voltage and current the panel will operate? Just a rough idea. Thank you.

Regards
PG

 

[steveB]

The operating point will be at roughly 37 V and 84 mA.

Note that 37V/84mA = 0.44 kOhms

 

[steveB]

By the way, it's clear that the power at that point is only 3.1 W, which is far short of the maximum point of 220 W

 

[steveB]

And to complete the thought, a resistance of 4 Ohms would put you a lot closer to the maximum power point.

 

[PG1995]

Thank you.

If you notice carefully, it looks like there is a little mistake when it comes to the grading of voltage scale between 35V and 40V. What value for Voc did you use? 38V?

 

[steveB]

Yes, I noticed that before, but you asked only for a rough estimate, so it does not matter. But, as I mentioned above, I estimated it at 37 V , not 38 V.

 

[PG1995]

Now let's proceed with that example of 440 ohm resistor using a new setup. We will use this schematic.

Let me repeat some details. We are operating at 1000 W/m^2 at temperature of 25 C. The open circuit voltage, Voc, is almost 37V and Isc is about 8A. The maximum power values Vmp and Imp are 30V and 7.8A respectively. It means that to get maximum power we need to operate the panel in the knee area or close to maximum power point. The value of variable resistor is set at 440 ohm. We have seen above that we are only getting 3.1 W of power.

(This is where I will interrupt my question and would like to ask you something first. You can simply say "yes" or "no". Your answer will decide how I should frame rest of my question.)

Just assume that the system has somehow reached the point of maximum power after it is turned on. Would then there be a need to turn the transistor on and off to keep the system running at maximum power point? Or, can it just be kept in on position and maximum power will still be available? Thank you.

Regards
PG

 

[steveB]

Now let's proceed with that example of 440 ohm resistor using a new setup. We will use this schematic.

Q1: I think it wouldn't a right choice to put the current sensor between Cin and the MOSFET. Simple "yes" would do if I'm correct.

I can't say you are necessarily correct. I think I know why you are saying that, and I would say that I would also try that as a first attempt. The problem here is that neither one of us has done a detailed control analysis of this system. After a control analysis, you might find that sensing at this other position and filtering is better. I'm not saying that is in fact the case, and I think your choice is probably correct. But, to say that it wouldn't be the right choice is premature, - for me at least.

Q2: Let me repeat some details. We are operating at 1000 W/m^2 at temperature of 25 C. The open circuit voltage, Voc, is almost 37V and Isc is about 8A. The maximum power values Vmp and Imp are 30V and 7.8A respectively. It means that to get maximum power we need to operate the panel in the knee area or close to maximum power point. The value of variable resistor is set at 440 ohm. We have seen above that we are only getting 3.1 W of power.

(This is where I will interrupt my Q2 and would like to ask you something first. You can simply say "yes" or "no". Your answer will decide how I should frame my question.)

Just assume that the system has somehow reached the point of maximum power after it is turned on. Would then there be a need to turn the transistor on and off to keep the system running at maximum power point? Or, can it just be kept in on position and maximum power will still be available? Thank you.

Yes, in general there would be a need to turn the transistor on and off to keep running at the maximum power point.

 

[PG1995]

Thank you.

I wasn't able to edit my last post in time. If you see now, I have removed Q1 from the post because it had occurred to me that I should ask it later.

Yes, in general there would be a need to turn the transistor on and off to keep running at the maximum power point.

It would mean Cin gets discharged when the transistor is turned on, and to maintain Vmp the transistor should be turned off so that the charge can accumulate and the voltage returns to Vmp. Actually I just wanted to confirm that. Now you have told me that it really does, I can at least start thinking along the right path. Thanks.

 

[PG1995]

Hi

Q1: Just a quick and regular question. Why is a solar panel rated at 12 V when its peak voltage could be 17 V? The reason could be that it might be used to charge a 12V battery. But I don't really buy this reason. A 12V battery is charged at almost 14V and when it is complete discharged it sits at 10V. It could also mean that if sunlight reaching the panel is enough then most of the time it will be able to charge up a 12 V battery. Thanks.

Q2: You can see that both panels are rated at 12v but their power ratings differ. I think that the 100 W panel can deliver more power because its cells have been made in such a way that they contain generate more current for the same amount of sunlight compared to the 50 W model. Do I have it right? Please let me know.

Q3: By the way, did you notice that I'm not using any buck converter in this schematic which was used for my query in the post above? I hope you don't mind my asking this. Actually, you have used the phrase "in general" which might imply that we need to turn the transistor on and off in many other cases but it might not be applicable for the case shown in your schematic. Just wanted to make sure because your reply was really important.

Regards
PG

Useful links:
1: http://www.freesunpower.com/solarpanels.php
2: http://www.wholesalesolar.com/Information-SolarFolder/solar-panel-efficiency.html
3: **broken link removed**
4: **broken link removed**
5: **broken link removed**
6: http://www.amazon.com/RENOGY-Monocr...Charging/dp/B009Z6CW7O/ref=zg_bs_2236628011_7

 

[steveB]

I can only answer Q3. Hopefully someone else who knows more about solar cells can answer Q1 and Q2 for you.

Q3: Actually I did not notice that you left out the buck converter. When I say, "in general", I'm just trying to be cautious that there might be specific cases that are exceptions. In general, (there I did it again), rules have exceptions. As to whether there are or are not exceptions in this case, I haven't thought much about it. Remember that when I answer you questions, I'm giving quick answers off the top of my head. I know from experience that quick answers usually are not completely accurate and correct, hence, I try to phrase my answers with some caution to minimize damage from my oversights.

 

[PG1995]

Thank you.

Perhaps, NorthGuy could help me with those queries because he already knows what I'm into.

Actually I did not notice that you left out the buck converter.

So, it was good thing that I asked you about that. But your answer still holds up that there would be a need to turn the transistor on and off to keep running at the maximum power point even when the buck is excluded? Thanks.

 

[PG1995]

Hi

The Q3 from post #32 has already been answered. But I still need help with Q1 and Q2. I have noticed that the panels in Q2 differ in dimensions. It would mean that the 100 W panel uses cells with larger surface area and hence can generate larger current. Thanks.

 

[NorthGuy]

Q1/Q2: There are real ratings: Vmp is MPPT voltage at 25C and full insulation. Voc is open circuit voltage in these conditions.

The 12V nomination means that the panel is suitable for charging 12V battery with non-MPPT controller. Such panel will have an Vmp rating around 18V.

There are few places where voltage gets lost. Panels usually get real hot, so MPPT voltage point goes lower. It is also a little bit lower when it gets cloudy. Controller drops some voltage. Battery may need up to 16V for equalization charge (not the sealed one!).

Taken all together, it does mean that to charge 12V battery you need 18V panel.

With buck MPPT you need to obey this difference too, although higher differences (up to some point) will work too.

 

[KeepItSimpleStupid]

Q1: I worked in the research end of this industry. I suppose that a nominal 12V panel would work with a 12V "system". e.g. 12 V common batteries. Voc has to be know so the inverter will work. If you exceed it, the inverter won;t start. If Isc is exceeded, the inverter shuts down.

Panels should be specified at Am 1,5 Global spectrum, 25 deg C , and the power at Vmp and Isc. So, that's where your. 100 W comes from. Current generated is by unit area. So, the bigger the panel the higher the current assuming the same efficiency cells. Cell Interconnection within the panel also plays a role. So, you have a "nominal" voltage system which is usually your battery stack.

Now this could mean load balancing and a much higher battery stack like 400 V for certain installations.

Q2: For the same type and efficiency cells, the same nominal voltage, the larger current is the one with the larger area.


Aside:
We had a discussion once as to what's "better". In space, weight is important. For a home, it might be size. Meaning, easy to carry up a ladder. For a solar farm, it might be total cost. Then for a repeater on a mountain top, a road sign, landscape lighting etc. Each application has it's own constraints.

Also watch funky terms, Isc (Amps) or per pabel short circuit current or Jsc or (Amps/area) e.f. Amps/sqcm or Amps per cm squared. The Amps per unit area is a metric. Fill factor is a metric. Efficiency is a metric. All of these are independent of area.

 

[PG1995]

Thank you, everyone.

It might look like that there are too many queries but almost all the queries are simple. You can simply say "yes" if you don't find any major flaw with my thinking. I don't need to be exactly correct. I wanted to do simplified models first to create an intuitive understanding of the system. Once I get green signal from you then I can take my 'analogy' to next stage where I will use a microcontroller. Thanks.

Let me repeat some details. We are operating the panel at 1000 W/m^2 at temperature of 25 C. The open circuit voltage, Voc, is almost 37V and Isc is about 8A. The maximum power values, Vmp and Imp, are 30V and 7.8A respectively. It means that to get maximum power we need to operate the panel in the knee area or close to maximum power point. You can have a look on this figure. We will be using the curve for 1000 W/m^2.

Q1: When the switch is in position B, the Cin is getting charged up and its voltage increases. The solar panel, which is a complex type of current source, sees it as a resistance. Up to the MPP, the solar panel can supply a constant amount of current to Cin. The voltage of Cin keeps increasing, or in other words, its resistance keeps increasing. The solar panel won't really be able to supply constant current to Cin once MPP is crossed. Do you agree?

Now let's try to analyze some simple configurations.

Q2: The switch is in position B and the voltage is increasing. At this stage, we need to see how voltage is increased. The solar panel is constantly generating current. Just for the sake of this discussion think that the Cin and panel are connected by some kind of pipe and the current generated by the panel consist of flexible and elastic electrons which can be compressed inside the pipe. More compression would mean more voltage or pressure. Assuming sunlight is constantly falling at the panel at the same rate, the panel is able to push more and more electrons into the pipe at the same rate and all the time electrons are getting more and more compressed. Then, there comes a point when the panel is no longer able to generate electrons at a constant rate (or, we can say able to push electrons into the pipe at a constant rate) and that's point is roughly MPP. Beyond this MPP point, even though the amount of sunlight falling on the panel hasn't changed, the rate of generation of electrons (or the number of electrons being pushed into the pipe) constantly falls but still the voltage or pressure will increase because the panel is still able to push some more electrons into pipe which increases the compression and hence the increase in voltage. But there will come a point, called Voc, when electrons inside the pipe has been compressed to such an extremely degree that the panel is longer able to push any more electrons into the pipe although amount of sunlight hasn't changed. Does my analogy sound fine to you?

Q3: At any instant, both the panel and Cin are sitting at the same voltage. Don't forget that the switch is still in position B. We can say that both Cin and panel are two voltage sources (or, you can say a current source and voltage source) sitting side by side just like two batteries assuming they have exactly same voltage can be connected in parallel. Compared to Cin, the panel is constantly pushing more electrons into the pipe and hence increase in voltage all the time. Right?

Q4: Now assume that the system has somehow reached the point of maximum power just before the switch is put into the position "A". Please note that the switch has been put into position A. This is the simplified model. We will use the data given above. We were given that Vmp and Imp, are 30V and 7.8A respectively. What does it really mean? It means that if the resistor value were, R=Vmp/Imp=30/7.8, 3.5 ohm then we will be able to utilize maximum power from the panel and above all we can leave the switch in position A all the time assuming that the amount of sunlight remains constant. Let's look into it in little detail. Don't forget I'm assuming that we are using 3.5 ohm resistor. Let's further assume that the resistor represent an electric heater. Informally speaking, electrons enter a resistor at one end and exit the other end of it. While passing through a resistor electrons lose their energy and that energy is converted into some other form. In case of a heater, it will be converted into thermal energy. Further, the rating 3.5 ohm tells that the resistor will allow one coulomb of electrons to pass through it per second (1 A = 1 coulomb of charge per second) if they are compressed at one end of it with pressure or compression of 3.5 V. As we know that when the switch was put into position A, the compression of electrons or voltage was 30V. It means that the resistor would allow 7.8 coulomb of electrons to pass through it per second assuming the compression of electrons remains constant at its end. But we can see that the panel can happily push 7.8 coulomb of electrons into the pipe per second. Please note that when the switch was put into position A, the pipe network was extended a little because another pipe was connected to the system through the switch! We have a situation where the resistor lets 7.8 coulomb of electrons to enter per second and to hold the compression at a constant value the panel pushes more 7.8 coulomb of electrons into the pipe network per second. This means Cin won't be discharged and the switch can remain in position A assuming the amount of sunlight does not change. Do I make sense?

Q5: Now let's proceed with that example of 440 ohm resistor. We have seen above that we are only getting 3.1 W of power when 440 ohm resistor is connected directly to the panel. Again we will assume that just before the switch is put into position A, the Cin and panel were sitting at MPP point, i.e. Vmp and Imp, are 30V and 7.8A respectively. Further, amount of sunlight remains constant throughout this discussion. Again, 440 ohm rating means the resistor will allow one coulomb of electrons to pass through it per second if they are compressed at its entrance with a pressure of 440V. At the moment, when the switch has just been put into position A, the resistor will allow only 0.06818 coulomb of electrons to pass through it per second. At that instant the panel is pushing almost 7.8 A of electrons into the pipe network but only a small of them is able to exit through the resistor; more electrons are entering the pipe network than exiting. It means that the electrons will be compressed more tightly and hence increase in the voltage and which means the panel's capacity to push more electrons has started to decrease; see the "IV Curves" for 1000 W/m^2. On the other hand, as the pressure inside the pipe starts increasing, the electrons will be compressed more tightly at the entrance of the resistor so the number of electrons exiting the resistor per second will increase. But still the increase in number of electrons exiting the resistor is not enough to compete with the number of electrons the panel is pushing into the pipe network. But as the pressure keeps increasing and the panel's capacity to push more electrons into the pipe continuously decreases, there will come a time when number of electrons exiting the resistor and number of electrons being pushed into the pipe by the panel will match up and that number would be almost 84 mA. Do I make sense?

Q6: Now let's repeat the above procedure using a 2 ohm resistor now. The rating 2 ohm means that when electrons are compressed with pressure of 2V at the entrance of resistor, it will let one ampere (i.e. one coulomb per second) of electrons to pass through it. Again, just before the switch is put into position A, the system consisting of Cin and panel is sitting at MPP point, i.e. Vmp and Imp, are 30V and 7.8A respectively. It means that as soon as the switch gets into position A, the resistor lets, Vmp/R, 15 A of electrons to pass through it. You can have a look on "IV Curves" for 1000 W/m^2. You can see that the panel is only able to push 8 A of electrons into the pipe per second. So, the pressure inside the pipe network will decrease. Again, there will come a point when number of electrons being pushed into the pipe by the panel matches up with the number of electrons exiting the resistor and that point would be reached when the compression of electrons is reduced to almost 16V. Do you agree?

Thanks a lot for the help.

Regards
PG

 

[steveB]

Q1: When the switch is in position B, the Cin is getting charged up and its voltage increases. The solar panel, which is a complex type of current source, sees it as a resistance. Up to the MPP, the solar panel can supply a constant amount of current to Cin. The voltage of Cin keeps increasing, or in other words, its resistance keeps increasing. The solar panel won't really be able to supply constant current to Cin once MPP is crossed. Do you agree?

Of course, you already know that the current at the maximum power point is less than the short circuit current.

Q2: The switch is in position B and the voltage is increasing. At this stage, we need to see how voltage is increased. The solar panel is constantly generating current. Just for the sake of this discussion think that the Cin and panel are connected by some kind of pipe and the current generated by the panel consist of flexible and elastic electrons which can be compressed inside the pipe. More compression would mean more voltage or pressure. Assuming sunlight is constantly falling at the panel at the same rate, the panel is able to push more and more electrons into the pipe at the same rate and all the time electrons are getting more and more compressed. Then, there comes a point when the panel is no longer able to generate electrons at a constant rate (or, we can say able to push electrons into the pipe at a constant rate) and that's point is roughly MPP. Beyond this MPP point, even though the amount of sunlight falling on the panel hasn't changed, the rate of generation of electrons (or the number of electrons being pushed into the pipe) constantly falls but still the voltage or pressure will increase because the panel is still able to push some more electrons into pipe which increases the compression and hence the increase in voltage. But there will come a point, called Voc, when electrons inside the pipe has been compressed to such an extremely degree that the panel is longer able to push any more electrons into the pipe although amount of sunlight hasn't changed. Does my analogy sound fine to you?

I don't find any analogy needed here. We understand electricity and can describe it in the conventional way with less chance of confusion. Basic energy conservation is enough to get the gist of this anyway. If current could remain constant for any voltage, then you would reach a point where the output electrical energy was greater than the input photon energy. Hence, the maximum power point must have less than maximum current if things are changing continuously.

Q3: At any instant, both the panel and Cin are sitting at the same voltage. Don't forget that the switch is still in position B. We can say that both Cin and panel are two voltage sources (or, you can say a current source and voltage source) sitting side by side just like two batteries assuming they have exactly same voltage can be connected in parallel. Compared to Cin, the panel is constantly pushing more electrons into the pipe and hence increase in voltage all the time. Right?

Theoretically yes, but practically no. We know that eventually the current flow essentially stops. The capacitor voltage approaches the open circuit voltage of the solar cell and the solar cell current approaches the open circuit current, which is zero. At some point, the capacitor is effectively charged and the current is effectively zero.

Q4: Now assume that the system has somehow reached the point of maximum power just before the switch is put into the position "A". Please note that the switch has been put into position A. This is the simplified model. We will use the data given above. We were given that Vmp and Imp, are 30V and 7.8A respectively. What does it really mean? It means that if the resistor value were, R=Vmp/Imp=30/7.8, 3.5 ohm then we will be able to utilize maximum power from the panel and above all we can leave the switch in position A all the time assuming that the amount of sunlight remains constant. Let's look into it in little detail. Don't forget I'm assuming that we are using 3.5 ohm resistor. Let's further assume that the resistor represent an electric heater. Informally speaking, electrons enter a resistor at one end and exit the other end of it. While passing through a resistor electrons lose their energy and that energy is converted into some other form. In case of a heater, it will be converted into thermal energy. Further, the rating 3.5 ohm tells that the resistor will allow one coulomb of electrons to pass through it per second (1 A = 1 coulomb of charge per second) if they are compressed at one end of it with pressure or compression of 3.5 V. As we know that when the switch was put into position A, the compression of electrons or voltage was 30V. It means that the resistor would allow 7.8 coulomb of electrons to pass through it per second assuming the compression of electrons remains constant at its end. But we can see that the panel can happily push 7.8 coulomb of electrons into the pipe per second. Please note that when the switch was put into position A, the pipe network was extended a little because another pipe was connected to the system through the switch! We have a situation where the resistor lets 7.8 coulomb of electrons to enter per second and to hold the compression at a constant value the panel pushes more 7.8 coulomb of electrons into the pipe network per second. This means Cin won't be discharged and the switch can remain in position A assuming the amount of sunlight does not change. Do I make sense?

I'm not sure the full explanation makes sense because you are falling back on your previous analogy, but I think essentially your conclusion seems to make sense. I'm not sure why you need so many words to describe such a simple case. You have switched in a resistor that happens to load the cell at the maximum power point. Wait for the transients to settle out, and naturally you will have a steady state maximum power point condition.

Q5: Now let's proceed with that example of 440 ohm resistor. We have seen above that we are only getting 3.1 W of power when 440 ohm resistor is connected directly to the panel. Again we will assume that just before the switch is put into position A, the Cin and panel were sitting at MPP point, i.e. Vmp and Imp, are 30V and 7.8A respectively. Further, amount of sunlight remains constant throughout this discussion. Again, 440 ohm rating means the resistor will allow one coulomb of electrons to pass through it per second if they are compressed at its entrance with a pressure of 440V. At the moment, when the switch has just been put into position A, the resistor will allow only 0.06818 coulomb of electrons to pass through it per second. At that instant the panel is pushing almost 7.8 A of electrons into the pipe network but only a small of them is able to exit through the resistor; more electrons are entering the pipe network than exiting. It means that the electrons will be compressed more tightly and hence increase in the voltage and which means the panel's capacity to push more electrons has started to decrease; see the "IV Curves" for 1000 W/m^2. On the other hand, as the pressure inside the pipe starts increasing, the electrons will be compressed more tightly at the entrance of the resistor so the number of electrons exiting the resistor per second will increase. But still the increase in number of electrons exiting the resistor is not enough to compete with the number of electrons the panel is pushing into the pipe network. But as the pressure keeps increasing and the panel's capacity to push more electrons into the pipe continuously decreases, there will come a time when number of electrons exiting the resistor and number of electrons being pushed into the pipe by the panel will match up and that number would be almost 84 mA. Do I make sense?

Again, I'm not following the analogy and all the unnecessary words. You switched in a 440 ohm resistor which gives only 3.1 W, and not the maximum power. Solving the system with the resistance load and solar cell VI curve gives the solution 84 mA as I already discussed above.

Q6: Now let's repeat the above procedure using a 2 ohm resistor now. The rating 2 ohm means that when electrons are compressed with pressure of 2V at the entrance of resistor, it will let one ampere (i.e. one coulomb per second) of electrons to pass through it. Again, just before the switch is put into position A, the system consisting of Cin and panel is sitting at MPP point, i.e. Vmp and Imp, are 30V and 7.8A respectively. It means that as soon as the switch gets into position A, the resistor lets, Vmp/R, 15 A of electrons to pass through it. You can have a look on "IV Curves" for 1000 W/m^2. You can see that the panel is only able to push 8 A of electrons into the pipe per second. So, the pressure inside the pipe network will decrease. Again, there will come a point when number of electrons being pushed into the pipe by the panel matches up with the number of electrons exiting the resistor and that point would be reached when the compression of electrons is reduced to almost 16V. Do you agree?

Again, you switched in another resistor which gives a different operating point with more power, but still not at the maximum power point.

I think you are making this terribly more complicated than it needs to be. The real control situation is a case where the switching is happening very fast and we only care about the average behavior of what that resistor looks like. The effective resistance seen in your examples would be a function of the duty cycle. The Cin would smooth out the fast changes and let the solar cell see the averages, effectively.

 

[PG1995]

Thank you.

My apologies for the redundancy and superfluity. As a matter of fact, I wanted to mention at the start of my previous post that there is a lot of redundancy and could have offered my apologies in advance had I not forgotten to do so. My analogy might not really make sense to you but it really helped me a little. Now I can look at 440 ohm and 2 ohm resistors from mppt point of view using a microcontroller.. But I need to confirm something before this.

In post #16 of yours, you said this:

You want to measure the average values of these two quantities so that you can calculate average power from the cell. The current in Cin is not relevant, nor is the current going into the converter input. However the sum of these two currents is the photocell current, which you do care about.

It would mean that we need two values from current sensor to measure average power - one value should be the one which is taken when input current into the converter is zero, i.e. transistor is not conducting, and the other value should be the one which is taken when there is a current into the converter, i.e. transistor is conducting. Then, the average of these two values should be taken to measure average power. Am I correct?

Don't we also need to use two values for voltage too just like current because voltage will also vary? Thanks.

Regards
PG