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How do solar cells generate current from pn-junctions?

  1. Jul 17, 2017 #1
    Looking for a practical description. I have a feeling I am grossly misunderstanding something fundamental about electric circuits and apologize if my questions are confusing because of it. I have these two sources:

    acs


    hyperphysics

    My (probably naive) understanding of the process is as follows:

    1. p-type has no free electrons; n-type has free electrons.
    2. chemical potential drives n-type electrons to diffuse into the p-type region, leaving the n-type semiconductor positive and the p-type semiconductor negative, forming a dipole between the two types across a region over the junction called the depletion zone.
    3. When the electric potential of the dipole is equal to the chemical potential driving diffusion, diffusion stops and the system is in electrical/entropic equilibrium.
    4. Sunlight excites electrons in the p-type semiconductor. These electrons move to the LUMO of the n-type semiconductor (conduction band), because the band gap has been decreased across the two semiconductors as a result of the diffusion. The p-type's homo has been increased in energy as a result of the negative charge in its valence band, while the lumo of the n-type has been decreased in energy as a result of the positive charge in its valence band.
    5. Once the p-type valence band electron has been excited into the n-type conduction band, it moves to a copper wire to generate current? Here's where I'm stuck.

    The acs source says in the very last sentence:

    "If you connect the n-type and p-type layers with a metallic wire, the electrons will travel from the n-type layer to the p-type layer by crossing the depletion zone and then go through the external wire back of the n-type layer, creating a flow of electricity."

    This reads to me like the electrons "double back." They go from n -> p -> n -> copper wire. I thought a circuit needed to flow in one direction. In fact, in order for the copper wire to attract n-type electrons, the copper wire end attached to the n-type must be positively charged like an anode. However, that would mean the cathode/anode ends of the copper wire would be oppositely charged to the dipole orientation at the junction. As I understand, this would constitute a reverse-bias circuit, working against the chemical potential of diffusion by acting against the accumulation of an opposing electric potential within:

    Reverse-Bias Circuit
    upload_2017-7-18_0-36-33.gif

    I would think the potential should be in the opposite direction and consistent with the charge distribution of the depletion region. Electrons should flow from the negative p-type into the copper wire's anode, around the wire to the cathode, and into the positively charged n-type, replacing electrons at the depletion zone that had moved into holes resulting from outflowing electrons in the p-type. That would be a forward-biased circuit as shown:

    Forward-Bias Circuit
    upload_2017-7-17_23-56-15.gif

    This however contradicts the ACS source which explicitly depicts that in a solar cell, the electrons flow from the n-type into the wire and reenter at the p-type, as seen in the top right of this diagram:

    Solar Cell Circuit
    upload_2017-7-18_0-2-18.jpeg

    1.
    Could someone please explain what are the correct electric signs at each end of the copper wire, and how these come to be? My goal is to understand how electrons flow from the n-type into the copper wire and back into the p-type after electronic photoexcitation, and why this can proceed in the opposite direction of the depletion zone's dipole. I don't understand how the cathode/anode ends of the copper wire can coexist with an oppositely oriented dipole in the depletion field. I expect the two electric potentials to work in opposition.

    2.
    Why is sunlight even necessary? If the chemical potential of a pn-junction is already sufficient to drive the formation of a potential, can't we just attach a copper wire, whose electrons are driven by the potential from diffusion, and already have a functioning circuit?

    Thank you for your time.
     

    Attached Files:

  2. jcsd
  3. Jul 17, 2017 #2

    scottdave

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    Homework Helper

    Normally, the electrons which fill a valence shell would tend to stay with the atom, even with the electric field attracting the charge to move. Because the atoms are in a structure, the atoms cannot move. With no light present, the voltage difference will be small, and the electric field is limited to the depletion region.
    In semiconductors, the electrons need additional energy (as in a photon, for example) to leave their atom and travel. When an electron near the depletion region has left the atom, the electric field will force it "up" into the N-type material.

    Here is the way that I interpret the last paragraph. When the cell is connected to a circuit, the electrons leave the cell out of the n-type material and go into the circuit. When electrons come into the cell from the circuit (into the p-type) they then move through the cell material (jumping from hole to hole) and crossing the forward biased depletion region, into the n-type material again.

    Here is another article, that you may enjoy reading. I did not have Java turned on, but was still able to read the article. http://micro.magnet.fsu.edu/primer/java/solarcell/
     
  4. Jul 17, 2017 #3

    Charles Link

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    Homework Helper

    In an open-circuit condition, there will be a potential created from light shining on the p-material as you described above, with the n side being more negative. You can measure this potential with a voltmeter, and you read a forward voltage from p to n. If you short circuit the photocell, you will get an external current flowing at zero voltage with the electrons on the n side going back into the valence band in the p material. Photodiodes used to measure light intensity are typically operated in this manner, where a current amplifier (op-amp circuit), that essentially has the inverting input at a virtual ground, puts out a voltage proportional to the current that flows when the photocell is short-circuited. ## \\ ## The photocurrent is a reverse current. Inside the photodiode, electrons travel from p material across the junction to the n side, and when short-circuited, a steady flow of electrons back into the p side from the external circuitry occurs, so the current through the diode is negative. The photodiode is very linear in its response, typically generating one photo-electron per incident photon with perhaps an 80% or better efficiency. When used as a solar cell, it generates a positive (forward) voltage at some negative current (the operating point is in the fourth quadrant on the I-V curve), and if you connect a resistor across the photocell, this makes a load line, basically superimposed on the I-V characteristic curve, that intersects the I-V characteristic curve of the photodiode in the fourth quadrant. There will be a positive (forward) voltage, but a negative current where the resistor load line (through the origin with slope ## I/V=-1/R ##. (Basically writing ## V_{diode}=-I_{diode} \, R ## where ## R ## is in series with the diode and ## I_{diode}=-I_{resistor} ##. Meanwhile, we also have the photodiode current is a function of the photodiode voltage, ## I_{diode}=I_{diode}(V_{diode}) ##, where this characteristic curve will be affected by how much light shines on the photodiode.) ## \\ ## Additional item: Without any light, the I-V curve for a photocell is that of a diode. With light incident on the photodiode, basically it causes the I-V curve of the photodiode to shift downward in the vertical direction, with the amount of shift proportional to the incident light intensity. The operating point is found in the 4th quadrant of the I-V curve. If the photodiode is open-circuited, the forward voltage can be read on the (positive) x-axis of the I-V curve for that particular light intensity. ## \\ ## Note: If you google Wikipedia "photodiode", you can essentially find the above load-line explanation with a diagram.
     
    Last edited: Jul 17, 2017
  5. Jul 17, 2017 #4
    Essentially what is happening is that photons, representing parcels of energy, are being absorbed by atoms of the semiconductor material.
    That causes electrical potential to change in the material.
    After that it's about efficiently extracting energy from the charged atoms, which would prefer to be in their normal ground state.
     
  6. Jul 18, 2017 at 2:56 PM #5
    Thank you very much for the responses. They all contributed in some way. The FSU link was also helpful for distilling the background I knew and providing some perspective on what I didn't know.

    After further looking around, I came to a few conclusions. A lot of it is superfluous information to what I asked, but writing it down helps me to reinforce the concepts, and maybe it will prove useful to someone else.

    On the interaction of the depletion zone with the reverse-bias potential:
    • The depletion zone is very localized and doesn't interact directly with broader effects. So a reverse bias does not oppose the formation of the depletion zone as I mentioned. Instead, it works in tandem and actually increases the size of the zone. The anode pulls electrons from an n-type semiconductor in the neutral region of the pn-junction, creating even more positive charge. A similar interaction with holes occurs in the p-side.This makes the junction act as an insulator, as the increased electric field under a reverse-bias even more strongly opposes forward current.
    Comparison of reverse-biased vs forward-biased junctions from hyperphysics. The width of each respective depletion region is what's relevant.

    Reverse-biased pn-junction

    upload_2017-7-18_21-8-37.gif

    Forward-biased pn-junction
    upload_2017-7-18_21-9-13.gif

    • If the reverse-bias becomes strong enough, breakdown occurs; the reverse potential overpowers the chemical potential at the junction and actually creates a current in reverse. For a reverse-bias below the breakdown voltage, all of the electron and hole movement is primarily in the valence band and no current is observed. Current flows in the conduction band. When breakdown happens to force current flow, an electron can hit another and knock it into the conduction band. This means a small current starting at a high voltage quickly knocks more and more electrons and cascades into an avalanche of electrons, and this giant current breaks many diodes. A Zener diode is specifically designed to have a low breakdown voltage and minimize the avalanche and remain stable. Because current only flows at a specific voltage threshold, the reverse-potential remains constant; any increase in it just results in more current to counteract the increase. This stability in output voltage from the current makes Zener diodes useful as reference diodes, taking in a varying voltage of current and outputting it at its reverse breakdown voltage.

    Current generated vs applied Voltage - Breakdown of a diode from Wikipedia
    upload_2017-7-18_21-14-36.png



    On solar cells with zero-bias and forward-bias:

    • In the absence of an external potential, the depletion region builds up as normal. However, it is localized and does not produce current.
    • If a forward-bias is applied, the opposite effect of a reverse-bias occurs and the depletion zone decreases. Electrons from the n-type semiconductor cannot diffuse too far into the p-type, as recombination (electron pairing with the hole) is favored. This is called the injection barrier.
    Example of injection barrier in a solar cell. In the diagram, a reverse-bias (negative voltage) generates no current, presumably because it's below the breakdown voltage.
    upload_2017-7-18_21-5-42.jpeg

    On recombination:
    • What might be an important concept here is the difference between a direct and indirect band gap. An indirect band gap results when the conduction band's minimum energy and valence band's maximum energy (the band gap) are states that have differing frequencies in the lattice. A direct band gap has the same frequency in the lattice for both bands at the band gap. The reason this is important is because it affects the efficiency of recombination. Recombination between an electron in the conducting band and a hole in the valence band implies a transition between two momentum states and two energy states. Energy and momentum must be conserved. Energy is conserved by releasing a photon. However, a photon has very little momentum, so it cannot compensate for an indirect band gap's difference in momentum. A phonon must also be released. A silicon pn-junction like in solar cells has an indirect band gap. This makes recombination a slower process in solar cells. The heating of the panel by the sun is also important for the absorption of a valence electron into the conducting band, and while I did not find a source on this, I presume is the reason why an electron in the n-type semiconductor does not favorably recombine with any holes in the depletion zone.
    • An LED is a pn-junction with a direct band gap. In LED, a forward-bias is applied, the injection barrier is surpassed to generate a current, and more importantly recombination is favored, occurring in the depletion zone. The photons from recombination fluoresce in all directions, so the system is placed in a reflective cup to send them toward the exit, where they light up our phone screens according to the energy taken from recombination over the band gap.
    Direct vs indirect band gaps

    Direct band gap
    upload_2017-7-18_21-26-56.jpeg


    Indirect band gap

    upload_2017-7-18_21-27-25.jpeg


    Absorption in an indirect band gap (Wikipedia).
    upload_2017-7-18_21-28-27.png


    The answer to my question mark in step number 5 of my first post I now believe to be:

    In a solar cell, when light excites a p-type electron into the n-type semiconductor, rather than recombine, it is actually favored to proceed toward the electrode, due to it being a faster process than recombination. I should emphasize I didn't find a source on that statement, but it was one of the obstacles to me understanding the process, so I ended up inventing this answer to overcome it (feel free to correct me if it's wrong). At any rate, crossing the depletion zone involves moving the electron toward a positively charged region, which means the electron gains energy, so a positive voltage is established over the n-type semiconductor. It runs through the external circuit, drops off its energy and returns. Thus, the pn junction is necessary to establish the voltage difference and prevent simply recombining after photoexcitation.

    So in summary: the electron is not favored to recombine; there is no applied bias, rather the reverse bias is a result of the electron gaining potential energy by moving toward a positive charge across the pn junction; and that leads to the reverse current.


    Note:

    The various articles I read almost all mentioned the electron is "swept toward" and becomes "stuck" on the n-type due to the electric field of the depletion zone, but this doesn't make intutitive sense to me as technically there should be a more powerful chemical potential to push it back -- that is after all why the electrons diffused into the p-type in the first place in spite of the accumulating electric field. I just think of it as the n-type conduction band being lower in energy and therefore unable to return to the p-type which has a higher-energy conduction band.

    There's a picture from Wiki on this:

    Excitation of p-type valence electron to n-type conduction band
    upload_2017-7-18_21-39-42.gif

    Source

    In addition to the URL links above and elsewhere in this thread, chapters 3 and 4 on this site (navigate via left sidebar) were also quite thorough and helpful.
     
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