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So I've asked a similar question in the past, but I'm still curious about something.

On a fully sunny day, the solar panel tops out at 5.5v @ 260 mA. On overcast or cloudy days, the voltage stays just about the same, but the current drops to 50 mA.

Why does it do this? Does the intensity of the UV rays directly affect the amount of current in the solar panel? If so, why doesn't the voltage drop as drastically? Any way around this besides a new panel?

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  • \$\begingroup\$ What is the load? Maybe you are out of maximum power peak point. \$\endgroup\$ – Michal Podmanický Jun 9 at 18:28
  • \$\begingroup\$ The ideal load resistance, R is where the ratio of open circuit voltage to short circuit current Voc/Isc=R It acts as a high impedance current source from an open circuit voltage. The MPT will start at 85% of Voc and decline from there with Solar input. So choose your max battery voltage from 70 to 85% of Voc with an overvoltage load (LEDs) for battery protection if using it during the day or choose towards 100% Voc = max battery voltage (NiMh) electronics.stackexchange.com/questions/264225/… \$\endgroup\$ – Sunnyskyguy EE75 Jun 9 at 19:12
  • \$\begingroup\$ The load is a 12v dc hairdryer motor. I'm not an expect with electricity, actually quite a novice (which is why I'm here), but I'm trying to get it to power the motor, which it does at just about its peak output (5.5v). I'm testing the panels directly to the multimeter leads, not connected to the motor. \$\endgroup\$ – Couch Mango Jun 9 at 19:13
  • \$\begingroup\$ Actually, a simple model is a constant current generator proportional to the light intensity in parallel with a series chain of silicon diodes one for each cell in the array. Since the diodes have a logarithmic voltage vs current characteristic the array open circuit voltage has a logarithmic voltage vs light intensity characteristic. \$\endgroup\$ – Kevin White Jun 9 at 19:15
  • \$\begingroup\$ @Couch A motor is a poor load since they need 10x the current just to get started which reduces to 10% with no fan load and you only have 50% of the required voltage. \$\endgroup\$ – Sunnyskyguy EE75 Jun 9 at 19:17
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The generated power is, among other things, a function of the number of photons that strike the panel per square meter per second (the photon flux), AND the wavelengths (the "colors") of the photons that are striking the panel.

A photon's wavelength determines its energy. When a photon strikes the solar panel's surface (the silicon material), if the photon has the right amount of energy the collision knocks loose (it "ejects") an electron from its silicon atom. *Fun fact: Albert Einstein received his Nobel prize in physics for understanding and describing this photoelectric effect. The liberated electron (-1 charge) and the ionized silicon atom (+1 charge) now form a kind of battery. This is a GROSS oversimplification, but hopefully you get the idea.

Only certain photon wavelengths have the required energy to eject an electron from its silicon atom. Unfortunately, Earth's clouds are very effective at absorbing and reflecting those particular photon wavelengths. Therefore, when a cloud passes between the sun and the solar panel, fewer photons strike the panel per square meter per second (the photon flux reduces), and therefore there are fewer ejected electrons per square meter per second, and therefore the panel produces less power.

The panel's output power is the product of its output voltage and output current. So if the output voltage remains constant and the generated power drops due to reduced photon flux, then the output current must drop:

$$ P = I\,V \Rightarrow I = P/V $$

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  • \$\begingroup\$ This is a magnificent explanation to my question not only explained in major detail, but made simple enough for a novice like myself to understand. Thank you! \$\endgroup\$ – Couch Mango Jun 9 at 19:36
  • \$\begingroup\$ @CouchMango If so, do 'accept' this answer. It's the best way of saying Thank you \$\endgroup\$ – Huisman Jun 9 at 21:21
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The amount of energy hitting the panels is dissipated as it comes through the atmosphere, so clear days produce more power than cloudy days - you should have noticed this.

Try recording the weather each day and the power output each day and you will see...

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  • \$\begingroup\$ @MichalPodmanický I have no idea of the load... \$\endgroup\$ – Solar Mike Jun 9 at 18:28
  • \$\begingroup\$ I think OP is not questioning the reduced power, but the reduced current at constant voltage. \$\endgroup\$ – Huisman Jun 9 at 18:51
  • \$\begingroup\$ @Huisman so reduced current at constant voltage is reduced power : power being I*V ... \$\endgroup\$ – Solar Mike Jun 9 at 18:59
  • \$\begingroup\$ True. But it still does not explain why the current drops and the voltage remains the same. If the voltage would drop and the current remained the same, that would also be reduced power : power being I*V \$\endgroup\$ – Huisman Jun 9 at 19:07
  • \$\begingroup\$ Yes, exactly what you just said. \$\endgroup\$ – Couch Mango Jun 9 at 19:14
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Your solar panel is working totally normally.

Photons in the sunlight hit the panel. As they are absorbed, they knock electrons free, and those electrons form the current you measure. On a cloudy day, there is less light, so fewer photons, and so fewer electrons.

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  • \$\begingroup\$ This makes total sense actually. I figured it had something to do with the amount of direct light, but I didn't understand why it happened \$\endgroup\$ – Couch Mango Jun 9 at 19:15
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PV panels act like constant voltage sources. For example, if you have a 12V panel, it might put out 17V with no load, 12V up to about 0.3A then drop off rapidly as you increase the load above its max power point. The max power point depends on incident light and panel efficiency.

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  • \$\begingroup\$ No, it acts like a high impedance current source from an open circuit voltage created by the solar power input. \$\endgroup\$ – Sunnyskyguy EE75 Jun 9 at 18:55

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