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Panels and power loss during slight amount of shade
Former Community Member
I haven't been on here for a couple of years and looking to add a few more panels so I'm wanting to know what to look for. I've been using 3 45 watt sets I bought at Harbor freight about 15 years ago. They still work perfectly well and they only lose partial output when a power line throws a shadow across the panels. Two years ago I bought 4 100w panels from HQST. They put out fine also but lose approximately 50% output with the same line shadowing them. These are monocrystalline and I believe the origional panels are also. Two questions, what do I look for to match the performance of my origional panels and also what causes the newer panels to lose so much output?
The degree of droop from panels that are partly in shadow will depend on 2 things, firstly how well the inverter can match itself to the new reduced output voltage, and secondly how the panels are interconnected.
Consider the following very simple case. If you have 3 photo diodes in series, and one of them is in darkness, then it acts as a resistor, and is not only generating but also limiting the current available from the other two.
However, if you have them in parallel then the current is still nothing from the one in darkness, but the full output of the ones in the light is then available.
Each cell is just one (very flat, very open) photodiode and will generate 0.5 to 0.7 volt in any reasonable light level. However, the available current will vary pretty much linearly with the light level, as each photon that is correctly trapped will release one electron hole pair. An optimum system will try and match the load to the panel so that enough current is drawn that the voltage droops quite a bit from the no load case, but not droop too much, the aim is to optimise to the peak product of current times voltage for that exact light level. Many simpler systems do not do this adaptation very well.
For practical reasons the diodes in a large panel cannot be all connected in parallel - to generate 200A at half a volt is a non-starter. But there will be multiple strings of series photo diodes connected in parallel, and probably in parallel with ordinary diodes too, to prevent the effect of the full voltage of the panel being blocked by damage or shadows on a single cell.
It is not clear from your post if you have a low voltage DC 12 or 24V set-up or perhaps a higher voltage - this will affect the make up of the panels, and how they should best be interconnected, and what sort of regulator or inverter is used.
Common 100W panels may be as low as 12V, in which case there may be chains of 20-30 photodiodes to give an absolute maximum of perhaps 15-20V, which is then stepped down via some electronics to charge a 12V battery. However, at least in the UK, panels of higher voltage with larger numbers of smaller cells are often used when they are intended to be part of a larger array. That said a panel of more than 100w would normally be used.
However, in general depending on the shape of the shadow and the way the strings on the panel are wired, the loss will not be same for the same degree of shading, or even for varied (horizontal<>vertical )orientations of the same panel relative to the shadow.
So check your inverter voltages on the DC side during shadow and full light, and perhaps look more closely at how the panels are wired - and are they in series or parallel, and if in series, are the shunting diodes in place, or do you have many inverters, one per panel ?
The degree of droop from panels that are partly in shadow will depend on 2 things, firstly how well the inverter can match itself to the new reduced output voltage, and secondly how the panels are interconnected.
Consider the following very simple case. If you have 3 photo diodes in series, and one of them is in darkness, then it acts as a resistor, and is not only generating but also limiting the current available from the other two.
However, if you have them in parallel then the current is still nothing from the one in darkness, but the full output of the ones in the light is then available.
Each cell is just one (very flat, very open) photodiode and will generate 0.5 to 0.7 volt in any reasonable light level. However, the available current will vary pretty much linearly with the light level, as each photon that is correctly trapped will release one electron hole pair. An optimum system will try and match the load to the panel so that enough current is drawn that the voltage droops quite a bit from the no load case, but not droop too much, the aim is to optimise to the peak product of current times voltage for that exact light level. Many simpler systems do not do this adaptation very well.
For practical reasons the diodes in a large panel cannot be all connected in parallel - to generate 200A at half a volt is a non-starter. But there will be multiple strings of series photo diodes connected in parallel, and probably in parallel with ordinary diodes too, to prevent the effect of the full voltage of the panel being blocked by damage or shadows on a single cell.
It is not clear from your post if you have a low voltage DC 12 or 24V set-up or perhaps a higher voltage - this will affect the make up of the panels, and how they should best be interconnected, and what sort of regulator or inverter is used.
Common 100W panels may be as low as 12V, in which case there may be chains of 20-30 photodiodes to give an absolute maximum of perhaps 15-20V, which is then stepped down via some electronics to charge a 12V battery. However, at least in the UK, panels of higher voltage with larger numbers of smaller cells are often used when they are intended to be part of a larger array. That said a panel of more than 100w would normally be used.
However, in general depending on the shape of the shadow and the way the strings on the panel are wired, the loss will not be same for the same degree of shading, or even for varied (horizontal<>vertical )orientations of the same panel relative to the shadow.
So check your inverter voltages on the DC side during shadow and full light, and perhaps look more closely at how the panels are wired - and are they in series or parallel, and if in series, are the shunting diodes in place, or do you have many inverters, one per panel ?
