I would like to understand exactly what happens when a PN doped crystalline silicon cell is shaded.Lets assume we have ten solar cells wired together in a series string.That there is plenty of sunlight on the first nine cells but cell ten is completely shaded.Lets say the forward bias of each solar cell is .5 volts lets say the conventional current is moving from anode to cathode and Electron flow is cathode to anode.This would make the anode more positive {more holes} than the cathode {more electrons}When current is flowing in the first nine solar cells at a forward bias how does the tenth cell reverse in bias {now the anode is less positive and the cathode is more positive} ? Do the other nine cells which are getting plenty of photons from the sun also go into reverse bias or just the shaded cell? Is the current flow of electrons and holes in the tenth cell the shaded one going in the opposite direction of the other nine ? What causes the shaded cell to create extreme heat?Lets assume we have no by pass diodes in this ten celled string.lets assume this collection of solar cells is a solar module Thank you ecvolt.
Current through each cell will be limited by the number of photons that strike it. So the output of several solar cells in series is limited by illumination of the most shaded cell. If you have 9 cells receiving light and the 10th one does not, you will not get any current flowing. You'd get the same effect if you just put a diode "wrong way" after 9 cells. The 9 illuminated cells will set up a voltage that reverse-bias the 10th, and no current will flow.
Thank You.Are you saying that cells 1-9 are still in a forward bias state and the shaded cell is in a reverse bias state? I am not sure of what you mean "the nine illuminated cells will set up a voltage that reverse-bias the 10th and no current will flow"To me the fact that the 10th cell is shaded means that the depletion zone at the PN junction of cell 10 has widened and no current will flow thru cell 10 Thanks ecvolt
Cells 1-9 are going to be near the diode forward voltage, so yes, they are going to behave as forward-biased. The depletion zone will be widened due to applied voltage from other cells. But yes, otherwise, that's what happens.
If the cells are connected neg to positive neg to positive etc. would that be like connecting diodes cathode to anode cathode to anode etc.?what would be the voltage between cell nine and cell 10? Thank You ecvolt
There would be no voltage between cells. Just across cells. Here is an image showing equivalent cell. You can picture this without the resistors. Just a current source and diode in parallel. For the 9 illuminated cells, I_{L} will be non-zero. For the last cell, it will be zero. Since the current can't flow through the last cell with I_{L}=0, there is no current flowing through that cell. That means, for every other cell, I_{L} = I_{D}. That means voltage across each cell will be very close to V_{d} for the equivalent diode. Assuming the circuit is closed, since there is no current across the load, the voltage across cell 10 is minus the sum of voltages across the 9 other cells. Note that in all of this, I am ignoring the possibility of a breakdown. With 9 cells, you might actually have enough voltage for breakdown at the 10th cell.
Would the anode of cell nine be connected to the cathode of cell ten?And would the voltage at the anode of cell 9 be 4.5 volts and seeing that the cathode of cell ten is connected to cell 9 would the voltage also be 4.5 volts at the cathode of cell 10? then what would be the voltage at the anode of cell 10?Thanks ecvolt
When I take a dc multi meter and read a voltage open circuit of say of 37 volts DC on a solar module comprised of 60 solar cells wired in series am i reading the additive voltage drop of all 60 cells ? so the voltage drop across each cell would approximately, . 616 volts Thank you ecvolt
Yes. Well, in one possible arrangement, depending which side you count 1-10 from. 4.5V relative to what? Voltage only makes sense as differential. That depends entirely on how you close the circuit. If you leave it open, voltage across cell ten will be zero. If you close the circuit across a resistive load, since there is no current, the voltage across cell ten must cancel voltage across all other cells. Yes.
I want to thank you for helping me.Perhaps you can check this post tomorrow I am gone to post a new scenario .It was a big help when you said there is no voltage between cells just across cells Thank You ecvolt
Does a diode follow Ohms law? How can you calculate the resistance of a solar cell in the forward bias ? If the forward voltage bias of a solar cell is .616 and the curent runnig thru the solar cell is 5 amps would the forward bias resistance be .1232 ohms. Thanks ecvolt
Not at all. While you can define effective resistance, you don't really need it. It's simply not how you would compute current through a cell. Lets say you have an ideal solar cell that gives you 0.5V. You connect two of them in series, and attach a 100 Ohm load. Lets also say that the two cells are not evenly illuminated. Say first cell gets enough light for I_{L}=17mA, and second cell I_{L}=12mA. (See diagram in previous post.) First of all, what is the current through the load? Well, the two cells apply 1V combined, so the current through the load will be 10mA. That means that for both cells, I_{L}-I_{D} = 10mA. This tells you that I_{D} will be 7mA and 2mA respectively for the two cells. Notice that from outside this just looks like a battery supplying 10mA at 0.5V. The fact that this is a solar cell has not become relevant yet. Now, suppose the illumination on first cell decreased, and now the I_{L}=7mA. For second cell, we still have I_{L}=12mA. The total current can no longer be 10mA, because that would require negative value for I_{D} for the first cell. Because it's a diode, you'll have I_{D}=0 instead. That means you'll have total current of 7mA through both cells and the load. At 100 Ohm, that puts total voltage at 0.7V. How do these 0.7V distribute between the two cells? Well, I_{D} for second cell is now 5mA. That means the voltage across the diode is V_{D}, which is the same 0.5V. So the second cell is still at 0.5V, and that leaves first cell at 0.2V. Notice that if illumination drops further, the voltage across first cell can change sign, and start opposing voltage across the second cell. When the amount of light that hits the cell is insufficient for the current you are trying to draw, the voltage across the cell drops. It behaves as if it has high internal resistance, not unlike a rundown battery. All of this assumed ideal cell, however, which can be visualized as just a current source and a diode. A real photovoltaic cell will also have sources of Ohmic resistance. See the diagram. In a good cell, resistor in parallel with diode will have high resistance, and resistor in series will have low resistance.
OK are you saying that cell 1 gets a light generated current of 17 ma and cell 2 gets a light generated current of 12ma? How did you calculate the current through the load of 10ma? I can see the 1volt part of the equation{.5volts plus .5volts=1 volt but not sure how you got the 10ma . Does Il mean light current generated or current of the load? Is Id mean the current through the diode?I rely appreciate your Patience !!!! I am a electrician with limited electronics knowledge I am getting this but I have to move slow Thanks ecvolt
Yes, I_{D} is current through the diode. It's opposite in direction to light generated current. You can think of it as sort of a relief valve that prevents the voltage building up in the cell past whatever output voltage it's meant to have. 10mA is the current through the load. V/R. Nothing tricky. But the current flowing in and out of every junction must be the same. That means that 10mA flowing through the load are the same 10mA flowing out of the second solar cell, which must be equal to I_{L}-I_{D}. That means current between the two cells is also 10mA. And that means that the same relation is true for the first cell. Which finally allows you to match the same 10mA flowing out of the load and back into the first cell. Basically, all of these are just applications of Kirchoff Laws. There is nothing terribly strange going on with solar cells with the exception of there being a constant current source, which you might not be used to. Once you just accept that it's there and use Kirchoff Laws systematically, you can work out the rest.
OK due to Kirchoff laws of current the current in a series circuit is the same for solar cell one and solar cell 2 I got that part.Is it the resistance of the load that determines the current flow in the circuit or the lowest iradiance of the individual cells that determines the amount of current flowing .Lower iradiance means less photons which mean less electron hole pairs which mean less current flow through the cell I think? Thanks ecvolt
When you said " Id is current through the diode.It's opposite in direction to light generated current" Are you referring to the fact that hole movement in a pn junction is in one direction and electron movement is in the opposite direction. Or are you referring to something else thanks ecvolt
when you said "that means that for both cells , Il-Id=10 ma" are you saying Il minus Id= 10ma. are you saying Il and Id =10ma thanks
No, the actual conventional current is in the opposite direction. Light pushes electrons from anode to the cathode. But if the voltage across is high enough, electrons will start flowing back from cathode to anode, as they would with an ordinary diode. The conventional current is opposite, of course. The light current is from cathode to anode, and diode current is from anode to cathode. Both. Basically, whichever of the two is the limiting factor. If the total current will be limited by the light current in the least illuminated cell, then the voltage of your assembly will drop so that the V=IR still works for the load. Otherwise, V is just sum of V_{D} of individual cells, and the limiting factor is the current through the load due to that voltage. This one, yes. The difference of the two currents must be equal to total current. The reason it's difference and not the sum is because I_{L} and I_{D} are assumed to have opposite direction. Just a convention thing.
So we have two currents light generated current and diode current that are involved in the circuit ?Is the light current generated by the photons and is the diode current derived from voltage drop over the solar cell which is acting like a diode
Correct. Keep in mind, though, that this is a description from equivalent circuit, where the current source and the diode are treated as two separate elements. In reality, both processes are happening within the same diode that makes up the photovoltaic cell. That means that you won't really have two separate currents flowing two separate paths, but rather a net current that looks like the difference of the two.