# What is the correct way to block current going into a solar panel from battery

I have a 6v 750mA solar cell that I am using to charge a couple of panasonic 18650 3400mAh 3.7v batteries. I have the positive from the solar going through a 1N4007 diode (cathode side toward the battery) then to the positive side of the batteries. I have the negative side going to the negative side of the batteries. I do not want any of the current from the batteries to go back to the solar panel from the batteries at night so I installed the 1N4007. However I just read the 1N4007 is only good for about 1A. Is this an ok diode to use of should I use something else.

• There is no one "correct" way. There are different ways that have different trade-offs. The simplest way is a diode. The down-side is that a diode has a forward voltage drop. Schottky diodes have lower forward voltage drop, but greater reverse leakage current. Power MOSFET's can be used, but they require additional circuitry. So, there are options and trade-offs. Commented Apr 19, 2021 at 22:52
• A second point I would like to make. I hope you are not intending to charge your battery directly from a solar panel. You need a battery charging circuit for that. Without one, you risk damaging your battery, or perhaps explosion. Commented Apr 19, 2021 at 22:56
• No I am making sure the battery does not over or under charge. How much voltage drop is there for a 1N4007 and how much reverse leakage is there? Thanks Commented Apr 19, 2021 at 23:05
• The forward voltage drop for all silicon diodes (except Schottky diodes) is generally between 0.6V and 0.7V, but for the 1N4007, according to the datasheet it can rise to about 1V when a 1 amp pulse is passing through. The reverse leakage current depends upon the temperature, and reverse voltage. It will probably be less than 1 uV in your situation. Commented Apr 19, 2021 at 23:16
• @user1114881 The 1N4007 isn't designed for this.Aren't you better off with "blocking" schottky diodes that are actually selected for this function with solar panels. Just look them up on Amazon. Cheap and readily able to handle 15 A, just for an example.
– jonk
Commented Apr 19, 2021 at 23:16

Your proposed standard diode is rated for 1000V .You do not need more than 10V for your 2 series Li cells. The power wasted is very significant in % terms due to your low voltages. 1 volt drop is 16% of 6V so efficiency could never top 90% .Si Schottky diodes come in lower reverse voltage ratings and have about half of the voltage drop .The 20V types drop less than the 200V types in the same series. This is simple and easy to understand .Mosfets can do even better but are slightly more complicated.

• Your answer would be correct in many cases. In this case : Both cells cannot be in series as Vpv is far too low. So they must be in parallel or charged one at a time. Vbat max is 4.2V. PV rated V and I are usually and hopefully at Vmp so he has lots of headroom and diode drop will not be an issue. Cell will load panel down to current battery voltage under charge and Ichg in max sun will be slightly more than Imp as loaded cells I increases towards Isc as load is above max power point (which only applies for optimised load). SO in this case any 1N400x will be OK. Commented Apr 20, 2021 at 3:04
• Unless the OP has a buck converter, the efficiency will be the same regardless of "how" the voltage is dropped. Some of it could be dropped across a diode, some of it could be dropped across an LDO linear regulator. But, whatever the Vsolar - Vbat is, without a buck, it is wasted. Also, the reverse leakage current is based upon the % of rated reverse voltage. So, 1000V 1N400x series will have lower reverse leakage than 1N4001. In this case, it looks like less than 1uA (which could be 1000x less than with a Schottky -- which again, without a buck, gives no advantage.) Commented Apr 20, 2021 at 3:34

In some cases, the forward voltage drop of a diode is an important consideration. If one has a buck converter this may be the case. On the other hand, if voltage is lowered by some linear device, such as a LDO regulator, or a typical battery charging circuit, there is no efficiency advantage to using Schottky diode with a lower forward voltage drop. The voltage between the Vsolar and the Vbat is going to be dropped one way or another.

In some cases, the reverse leakage of a diode may be a consideration. The reverse leakage of the 1N4007 will probably be less than 1 uA, according to the charts in the datasheet. The lower voltage diodes in the 1N400x series will have higher reverse leakage, because the leakage is proportional what percentage the reverse voltage is, to the rated reverse voltage. Thus, the higher rated diodes have less leakage at the same voltage. However, Even if you assume a Schottky diode with 1000 times the reverse leakage current, or 1mA, after 12 hours of darkness, the battery would have lost a mere 12mAh. For a battery with 3400mAh, that is not alot. So, the 1N4007 might be marginally better than some other diodes, but not by much.

One slight possible advantage that a Schottky, with its lower forward voltage drop might be that will give the LDO or battery charging circuit very slightly more headroom. There would be a narrow band of lighting levels where the Schottky could allow the battery to charge, but the 1N4007 would not. This could be a band of 0.3-0.4V. What percentage of the daylight hours would drive the solar cells at a voltage within this band? Probably not a whole lot, but I can't say. However, if the batteries can be charged in the remaining daylight hours, once again, it won't really make any difference to the task of charging the batteries daily. If one really wants to find an advantage, it will be that on some days, where lighting is just right, the battery might not be able to be charged with a 0.7V voltage drop, but could with a 0.4V voltage drop. If one wants to design for that, go for it.

Long story short, if one is not using a switch-mode buck converter, it probably doesn't make a great deal of difference which diode is chosen.

• For interest: PV cells act reasonably like an insolation controlled current source down to write low light levels. Insolation probably is under 5% max before voltage falls much. There is some energy they're but not vast. Even in overcast conditions lux level is usually above sayv5% and often surprisingly high. Commented Apr 20, 2021 at 7:35