Question: Why blocking diodes and not MOSFETs on solar panels? - Page 1
Forum section: electronics / general

author: Dave92F1
author_rank: Contributor
timestamp: January 25, 2025, 05:15:53 pm
content: Forgive me if this has been asked before.In my experience, solar panels are universally connected with blocking diodes, to prevent power loss from current flowing backward thru the panel when it's not fully illuminated.Why a blocking diode, which loses power due to the forward voltage drop, and not a MOSFET circuit?Using a MOSFET to stop the backward flow of current would seem to work just as well, and the loss would only be due to the RDSon, probably losing less than 20% as much power.There must be a good reason. MOSFETs are cheap. Why?

author: uer166
author_rank: Super Contributor
timestamp: January 25, 2025, 07:00:20 pm
content: Since when do they have series blocking diodes? Citation required about the "universally have blocking diodes" part.Almost all large solar panels that I've worked with have parallel bypass diodes for sections of it, to prevent partial shading from reducing string power to zero. I don't remember there being series diodes though.If you do see series output diodes, it would actually make sense to not make them FETs due to complexity and low loss that diodes already have. Say for a 80V panel, a 0.5V schottkys will waste 0.6% of available power, why do you care to reduce that to 0.2% with a FET?

author: Siwastaja
author_rank: Super Contributor
timestamp: January 25, 2025, 07:38:32 pm
content: Those parallel bypass diodes of course could be ideal MOSFET diodes but what OP is missing is control of the MOSFET. MOSFET itself isn't magically an ideal diode. Remember to consider leakage current and risk of accidental activation. And the cost of providing a power supply for the controller, which needs to supply gate voltage for the MOSFET.Any gain would be completely marginal. It doesn't make sense to put extra money into extracting some 0.1% more power under partial shading condition, which should be <10% of the time so 0.01% in total.Adding a few more of those bypass diodes would be a simpler and cheaper optimization. Like, for a 60-cell panel you can have three (one bypassing each 20 cells). Or you can add a fourth so that the three still bypass 20-cell segments and the fourth bypasses all 60 cells, so that if all three segments are shaded then the voltage loss over the panel is smaller (because the current from other panels only needs to pass one diode, not three).Maybethis fourth diode is worth it, but just barely so. Any more complicated than that and it won't make any sense.

author: coppice
author_rank: Super Contributor
timestamp: January 25, 2025, 07:43:27 pm
content: There are special devices based on MOSFETs made specifically for that application. I think TI make some, but they are not alone. I've never really looked at how they work, but I assume they use more than 2 terminals, so get enough voltage to hard switch the FET.

author: nctnico
author_rank: Super Contributor
timestamp: January 25, 2025, 10:30:58 pm
content: A MOSFET is not going to work as it conducts in both directions when 'on'. So when a shading condition stops, the MOSFET will short the cells it is connected across. IOW: you'd need a very complicated control circuit which isn't going to work better or more reliable compared to a 'dumb' diode.

author: coppice
author_rank: Super Contributor
timestamp: January 25, 2025, 10:44:10 pm
content: Quote from: nctnico on January 25, 2025, 10:30:58 pmOf course it requires more the a MOSFET. The whole point of using a switch is to switch it. The purpose it to cut the voltage drop well below what a simple diode, or even a Schottky diode, would drop, just as you do in synchronous rectifier applications.

author: langwadt
author_rank: Super Contributor
timestamp: January 25, 2025, 11:38:35 pm
content: Quote from: nctnico on January 25, 2025, 10:30:58 pmhttps://robotics.ong.id.au/2014/07/30/raspberry-pi-b-power-protection-circuit/

author: bdunham7
author_rank: Super Contributor
timestamp: January 25, 2025, 11:52:48 pm
content: Quote from: Dave92F1 on January 25, 2025, 05:15:53 pmOnly multiple strings in parallel would need such blocking diodes and then you'd be looking at a Vfof 0.7-1.2V in a string of 300-1000V. That's pretty small and shouldn't really happen anyway during high-power production unless the system is badly designed. Such diodes are cheap, simple and very durable.The bypass diodes internal to the panel are different, but again you'd be talking about a pretty small improvement and only during shading or malfunctions.

author: nctnico
author_rank: Super Contributor
timestamp: January 26, 2025, 12:37:41 am
content: Quote from: langwadt on January 25, 2025, 11:38:35 pmThat circuit only works reliable (in volume production, over a wide temperature range and with room for external interference) with a large amount of RDSon in the MOSFET, a high enough current flow and extremely tight matching of the transistors making the comparator. IOW: it doesn't really work in a practicle situation. It is a waste of components as D5 and the fuse already provide a reverse polarity protection.

author: langwadt
author_rank: Super Contributor
timestamp: January 26, 2025, 01:27:33 am
content: Quote from: nctnico on January 26, 2025, 12:37:41 amit prevents back feeding the USB if you you power the 5V from somewhere else

author: Siwastaja
author_rank: Super Contributor
timestamp: January 26, 2025, 06:45:46 am
content: Quote from: nctnico on January 25, 2025, 10:30:58 pmIdeal diode circuit of a MOSFET wires the MOSFET such that the body diode direction matches the desired current direction. Then a control circuit monitors the voltage across the MOSFET to emulate a diode, somehow measuring the current to keep the MOSFET on (bypassing the Vf=0.7V body diode) during forward current, but quickly turn the MOSFET off if the direction of current changes - just like a diode would do. Most obvious way to measure the direction of current is to measure Vds - to maintain a small voltage drop over the MOSFET - just like a diode has Vf. The more accurate the control circuit is, the lower this drop can be, but drop is still needed.So now we have a complex, more expensive MOSFET circuit, which needs external supply voltage, and whichstillhas Vf drop - maybe just 0.1V instead of 0.3V that of a schottky diode.This drive complexity, requirement of external power supply, and minuscule gain to be had, are the reasons why this is not being done on solar panels.

author: max_torque
author_rank: Super Contributor
timestamp: February 01, 2025, 12:12:09 pm
content: Because "perfect" is the enemy of "good enough"!!!Today, with our modern tools, processes, simulation and manufacturing we really can just about do anything, so in most cases the limiting factor, or more accurately the level of "optimisation" required for any design is actually set primarily by economics and not physics.So yes, you could make a solar panel with slightly lower loss, but would the customer benefit, or again, more accurately speaking, are they willing to pay for that benefit?