Solar panels & blocking diodes

[Rigger Mortice]

Just a discussion point..........

I know that common wisdom says that blocking diodes are necessary on solar panels but it's not what I have found.

Nigel Calder says that power losses through a diode during the day can be greater than power losses due to current flowing through a panel at night when no diode is in place. In other words it is sometimes better to not have a diode. He goes on to discuss the differences between 36, 33 and 30 cell panels.

When my (10watt) panel is covered up and no diode is in place the current loss is only 25 milliamps. I'm not sure how much power is lost through the diode during the day but, depending on the type of diode, there can be a voltage drop of up to 0.7 volts. For the last couple of years I have removed the diode during the summer months and have found that my battery seems to get a better top up as a result.

Thoughts?

 

[Chalker]

Any chance thst you've measured the charge current with and without the diode?

 

[noelex]

Yes blocking diodes usually waste more power than they save. Very few modern panels have them.
Most solar controllers (if you have one) will disconnect the array anyway at night, making a blocking diode unnessary,but even without a controller it is often more efficient to omit any blocking diode.

Don't confuse the blocking diodes with bypass diodes, which serve a different function and are imprortant.
Do not remove diodes from a panel without knowing what they are. They are much more likely to be bypass diodes. If you remove these you risk damage to the solar cells.

 

[VicMallows]

Don't confuse the blocking diodes with bypass diodes, which serve a different function and are imprortant.

Agree totally. How many times have I heard "my panel has a diode".

However, surely the bypass diode is only relevant if you are connecting panels in series?

 

[VicS]

However, surely the bypass diode is only relevant if you are connecting panels in series?

I thought they were ( also ?) used in large panels made up of several banks of cells so that they are not put completely out of action by being partially shaded.

TTBOMK my little panel simply has a blocking diode built into the sealed terminal box.

 

[Rigger Mortice]

Any chance thst you've measured the charge current with and without the diode?

To be honest I took a few readings but they weren't conclusive. Some seemed to be better without the diode, some seemed to make little difference.
What did seem to be better was that the state of the battery after a week away always seems to be better.

 

[noelex]

Agree totally. How many times have I heard "my panel has a diode".

However, surely the bypass diode is only relevant if you are connecting panels in series?

Bypass diodes are present mainly to reduce the the heating and damage within a solar panel as parts of the panel are shaded. At least every 10-15 cells require a bypass diode to prevent damage from overheating when localised shading occurs. (this means the minimum, nominal 12v panel will require at least 2 bypass diodes). The better panels tend to have more bypass diodes fitted). More bypass diodes also tend to increase the production (fractionally ) when the the panels are shaded.

When fitting panels in series an additional bypass diode, to bypass the whole panel, can be beneficial.

 

[William_H]

This thread has wandered into discussion of bypass diodes. These will on smaller panels if fitted be in the panel itself.

OP asked about blocking diodes and power loss.
if the panel is connected directly to the battery as I would expect for 10w panel ......
The thing about a solar panel is that it generates something like 20 volts no load. When you connect to a 12v battery the voltage is effectively pulled down to that of the battery. The panel is described as having an internal resistance. Now a blocking diode will drop .7 volt if it is a silicon diode. For a 10w panel at .5 amp (strong sunlight) the diode will waste .035 watt. However at the same time if the battery is at 13volts then the panel itself will waste 3 watts in its internal resistance.
Only when sunlight is low such that panel voltage falls to near the battery voltage will the diode volt drop become significant.
So if OP fits an amp meter between the panel and the battery he hopefully will see in normal sun .3 to .5 amp into the battery.
Now short the series diode and I think he will find very little increase in charge current.
If this is a concern perhaps at low light conditions then he can fit a schotky diode of suitable current rating. This will approximately half the volt drop of the diode.
To me 25 miliamps lost from the battery overnight is far more significant. So unless you have a controller regulator leave the diode in. good luck olewill

 

[noelex]

A Voc of 20v is only true at unreliastic panel temperatures.
When you add a blocking diode you effectivly remove one cell. So the panel is the equivalent of a 35 cell rather than a 36 cell panel.
Lower cell counts have been tried on panels, but they are rare these days. It has been found from experience that heat, shade mean that a 36 cell tends to be the most effective.
To accurately model the power loss of a blocking diode is not easy, but a simple calculation looking at the power loss in the diode comes close even when looking at a non MPPT regulator (or no regulator at all).

One of the big advantages of a blocking diode for a panel used to maintain the batteries while away is that if if the panel becomes damaged, or permanently shaded by something falling on it will not have a net power drain. Generally however in normal conditions the loss from the blocking diode will exceed the small reduction in night time discharge.

 

[Rigger Mortice]

. Now a blocking diode will drop .7 volt if it is a silicon diode. For a 10w panel at .5 amp (strong sunlight) the diode will waste .035 watt. However at the same time if the battery is at 13volts then the panel itself will waste 3 watts in its internal resistance.
Only when sunlight is low such that panel voltage falls to near the battery voltage will the diode volt drop become significant.
So if OP fits an amp meter between the panel and the battery he hopefully will see in normal sun .3 to .5 amp into the battery.
Now short the series diode and I think he will find very little increase in charge current.
If this is a concern perhaps at low light conditions then he can fit a schotky diode of suitable current rating. This will approximately half the volt drop of the diode.
To me 25 miliamps lost from the battery overnight is far more significant. So unless you have a controller regulator leave the diode in. good luck olewill

I'm struggling with your calculations. I make a 0.7 volt drop at .5 amp to be 0.35 watts (against your 0.035 watts).

The night time loss is 0.025 amps which at 13 volts will be .325 watts. In other words it will be slightly less than daytime loss due to having a diode in. I then factor in the amount of daylight at our latitude during the summer when it is dark for less than 8 hours a night. i.e. nighttime loss of 8hrs at .325 watts compares favourably with a daytime loss of 18hrs at .35 watts (though in reality I would expect it to be less than that).

I'm sure someone will pick up a flaw in my calculations but it does look as if a diode is not beneficial......I think!

 

[William_H]

I'm struggling with your calculations. I make a 0.7 volt drop at .5 amp to be 0.35 watts (against your 0.035 watts).

The night time loss is 0.025 amps which at 13 volts will be .325 watts. In other words it will be slightly less than daytime loss due to having a diode in. I then factor in the amount of daylight at our latitude during the summer when it is dark for less than 8 hours a night. i.e. nighttime loss of 8hrs at .325 watts compares favourably with a daytime loss of 18hrs at .35 watts (though in reality I would expect it to be less than that).

I'm sure someone will pick up a flaw in my calculations but it does look as if a diode is not beneficial......I think!

Yes sorry about my rusty maths. It is indeed .35watt wasted. And yes it does reduce the output by 2 cells worth. However my argument stands that under normal sunlight conditions with panel connected directly to the battery the power is mostly wasted in the internal resistance of the panel and power wasted in the diode is not significant. As I said try measuring actual current into the battery with and without diode. I think you will find very little difference. However under low light conditions it may make a difference.
The number of cells in a panel seem to have been determined historically by cost versus ability to charge in marginal sunshine and taking into account the blocking diode. More cells won't give more current just more voltage normally wasted but usable at dusk and dawn.
If however you fit an MPPT controller the diode will waste power which other wise could be used. You won't need a blocking diode anyway. Sadly however the inefficiencies of the MPPT controller may bring you nearer to original current anyway.
good luck olewill

 

[noelex]

Yes sorry about my rusty maths. It is indeed .35watt wasted. And yes it does reduce the output by 2 cells worth. However my argument stands that under normal sunlight conditions with panel connected directly to the battery the power is mostly wasted in the internal resistance of the panel and power wasted in the diode is not significant. As I said try measuring actual current into the battery with and without diode. I think you will find very little difference. However under low light conditions it may make a difference.
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You will see almost no difference in the output when the Vmp is above the battery voltage by a greater amount than the voltage loss of the diode. (If you don't have a MPPT regulator) A small but significant difference as the Vmp of the panel with the blocking diode falls below the battery Voltage.
There will be a total loss of output when the Voc of the panel with a blocking diode falls below the battery voltage.
The average loss will be dependent on circumstances, but as rough rule of thumb if you take the voltage loss in the diode and divide this by a Voc under real world conditions you will get close.
For a silicon diode the loss will be about 3%.

 

[nuuskamuikkunen]

I am still a bit troubled. I see this is an old thread, but it's no lesser applicable.

I have two panels in parallell at a summer house. And an active controller (PWM). One panel is quite old, about 20 years but still working. I have not checked how much power it delivers, I just know it works.

The panels are installed at different angles so during the morning and evening one will be in the shadows while the other one is in sunshine (most usually clouded).

To prevent leakage of current from the panel in the sun to the one in the shadows, I guess i should use a blocking diode on both panels? right?

Next question, I don't see any blocking diode inside the wireing box of the new panel, does it mean the panel has no blocking diode, or is it likely that a blocking diode is somewhere integrated and hidden inside the panels?

Does anyone have a good and easy test for checking if there is a diode or not inside the panel? Guess i could cover it, connect it directly to a 12v battery and then measure the reverse current. Or can i simply disconnect and measure the resistance in both directions? Any idea about what the values should be?

Both panels are for 12v system and about 100W each originally. (would be interesting to measure how much ompf is left in the old one.)
Greetings