Announcement

Collapse
No announcement yet.

Diodes and bypass diodes

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • longwolf
    replied
    Originally posted by Mike90250 View Post
    Yeah, the pictorial is not good, the gray wire shorts all the cells out. Tab wire goes from topside, to bottom side, not across the whole row.
    Well, that was a given. I guess I could try to shade the tabbing wires where they go under the next cell.

    Leave a comment:


  • Mike90250
    replied
    Yeah, the pictorial is not good, the gray wire shorts all the cells out. Tab wire goes from topside, to bottom side, not across the whole row.

    Leave a comment:


  • longwolf
    replied
    Originally posted by Mike90250 View Post
    .... that pictorial, it's not good,...
    Sorry Mike, you lost me again.
    The pictorial view is essentially the same as the second schematic image from 9th post in this thread.
    The new pic shows a bypass diode every 2 volts/ four cells.
    Maybe you're misreading the image. It shows thirty-six 6x6 cells. Four per row with nine rows.
    The gray lines represent the two tabbing wires across each cell and going to the buses on each side of the panel.

    I'd sure hate to give up on you and disappoint crxvfr

    Leave a comment:


  • Mike90250
    replied
    Originally posted by longwolf View Post
    All right!
    Thx again Mike.
    And, because I know you like them, here's a pictorial view of the solution.
    Sorry about the delay. that pictorial, it's not good, try a plain schematic, and go over it yourself, count up the volts as you go, 0V at the ground terminal, 6 series cells should be close to 3v (.5V per cell), and so on to your most positive point. then about every 2.5v, you could place a bypass diode.

    Leave a comment:


  • crxvfr
    replied
    Something that's different about this board that I like [i]sometimes[/i] is the way they don't give you an answer. I work in the motorcycle industry. If you ask a question on a motorcycle forum, you'll get tons of answers, and you have to figure out which one is the best. Here, they tend to make you show your work. Some people get completely turned off by that notion but I kind of like it, sometimes.

    Longwolf asks a question. Mike pretty much says no, take another guess.

    ...and he does. Eventually the problem is resolved and the work is there to show for it, ...so when shleps like me happen along this thread, I can follow the same, learning why and why not. The more I do this, the more I find out how much Mike knows about solar panels and stuff.

    Ya got yerself a stocker Lonewolf. Ha, that sounds so western.

    I've been reading the encapsulating thread you started in the other area, with interest.

    I'm going to be doing the same thing, kinda, but I'm a hillbilly.

    I want to try to make three 4x5 320 watt panels.

    Leave a comment:


  • longwolf
    replied
    All right!
    Thx again Mike.
    And, because I know you like them, here's a pictorial view of the solution.
    Attached Files

    Leave a comment:


  • Mike90250
    replied
    Great, the 2nd one is the better one. So, each PV cell, is .5V. A diode needs 1 volt to energize. And you won't need a zillion diodes. So about every 2.5V of PV, bypass that, with a diode. If your panel goes into bypass, the first step down from 18V would be 15.5 which is the bare minimum to charge a battery with. It's sort of why a "12V" panel produces power at 18V and measures 24V no load. And it's easier with a larger array, in series, then if you had a 40V array powering a 24v bank, and some shade came along, you still can keep charging

    Leave a comment:


  • longwolf
    replied
    Originally posted by Mike90250 View Post
    You have to re-draw it as a stack of batteries, with the diodes.
    Here's an attempt to do it as batteries in a strait line.
    Each four cell, horizontal row is replaced by one battery.

    The 1st image would show the circuit as it would be in the image from my previous post.
    But I suspect the correct way would be shown in the second image I'm uploading.
    Attached Files

    Leave a comment:


  • Mike90250
    replied
    You have to re-draw it as a stack of batteries, with the diodes. The pictorial view, while nifty is confusing because the whole thing is shorted with traces going side to side.

    + --[ |-- - one cell

    + --[ |----[ |----[ |----[ |----[ |----[ |----[ |----[ |----[ |----[ |-- - stack of cells

    Leave a comment:


  • longwolf
    replied
    I was an electrician and have a good handle on amps/volts/watts/ohms/parallel/series, wire sizing, etc

    I won't be leaving a ladder up there. But I am living out of my van and doing handyman work when I can find it. My biggest needs are running my laptop (to help find work) and to run fans (too hot here to sleep without them).
    I've already killed one set of batteries (the van's a diesel and uses two) so the first panel is a priority!

    Later, there are other devices I'll be using, up to 500 watts.
    It may be a while before I can afford an mppt, so I'm trying to keep the voltage low.

    Anyway, would this config work?
    what are the downsides?
    Attached Files

    Leave a comment:


  • Mike90250
    replied
    8A panels

    Those will need at least 12a diodes, larger if you can find them. This is why modern PV's are going to higher voltages, at the current density of 8 amps, your feed wires/tap/ribbon has to be large enough to carry 8A, and it begins to shade the cells underneath. Have you followed some of the other threads here about volts vs amps still gets you the same watts,
    http://www.solarpaneltalk.com/showthread.php?t=2200
    Each time you double the voltage, you improve the efficiency 400% for a given wire
    Bypass diodes only need to handle their own panel's amps, not the whole array.

    But if you are planning on putting them in series, your 25V OC panels, that jumps to 50V, and you should use 60-80V diodes.

    [B]Any[/B] shade is [B]bad[/B]. If you are leaving a ladder up, then just use a worm drive saw and slice off the bottom (shaded portion) of your array. It will then work the same as if it was shaded. [B]Poorly.[/B]

    Leave a comment:


  • longwolf
    replied
    Originally posted by Mike90250 View Post
    Almost. The bypass need to cover the entire panel, as you've drawn it, if either end gets shaded, the whole panel will be shut down.
    Ouch, not good. These will be going on a platform I made under my van's ladder rack. I occasionally do carry ladders and/or lumber up there. Also the ladder rack's own up-rights will cast shadows.


    Originally posted by Mike90250 View Post
    If you are going to use a charge controller, leave out the blocking diode, as modern controllers perform that function for you.
    For now, I'll only be using a cheap controller (all I can afford) with one panel . Latter I'll put the other two panels together and try for a mppt controller.


    Originally posted by Mike90250 View Post
    Amp of your bypass diodes should be 2x your panel max amps. (4 amp panel needs 8A diode, and heat sink, as all amps will flow thru diode and heat it up.) Try to use Schottky diodes, they have lower voltage loss. Diode voltage should be 1.5x your entire array Voc.
    The cells are rated at 0.5 volt/ 8amps and I'll be making 18 volt panels. Once I have all three panels, they will be in parallel. So that means the bypass diodes must handle at least 16 amps at 27 volts?
    Looking at eBay, the only ones I see in that range are the square, 3 pronged type. Will those work?


    Originally posted by Mike90250 View Post
    Also, try drawing it like a multi cell battery, and visualize 1 cell going bad, and how the power will flow around it, and what the voltage loss would be. Currently, with only 2 diodes, you will loose 50% of panel voltage, if shadow appears.
    I'm still not getting it.
    If I were to use 8 diodes, the way I added the two, that should work shouldn't it?
    Other than the price of the diodes, would there be any other downsides?


    Originally posted by Mike90250 View Post
    What do you use for your drawing program? Nice.
    Thx, it's an old copy of Jasc Paint shop Pro v7.01
    If you like those simple drawings, you should see some of the things I've done with TrueSpace, a free 3D program

    Leave a comment:


  • Mike90250
    replied
    Almost. The bypass need to cover the entire panel, as you've drawn it, if either end gets shaded, the whole panel will be shut down.

    If you are going to use a charge controller, leave out the blocking diode, as modern controllers perform that function for you.

    Amp of your bypass diodes should be 2x your panel max amps. (4 amp panel needs 8A diode, and heat sink, as all amps will flow thru diode and heat it up.) Try to use Schottky diodes, they have lower voltage loss. Diode voltage should be 1.5x your entire array Voc.

    Also, try drawing it like a multi cell battery, and visualize 1 cell going bad, and how the power will flow around it, and what the voltage loss would be. Currently, with only 2 diodes, you will loose 50% of panel voltage, if shadow appears.

    What do you use for your drawing program? Nice.

    Leave a comment:


  • longwolf
    replied
    Ok, so would this be the way? Two bypass and one blocking.
    Attached Files

    Leave a comment:


  • Mike90250
    replied
    Nope, those are series diodes. All current in your array must flow thru them, all the time. You will loose 1 volt for each diode you have drawn.

    Bypass diodes are normally reverse biased, and only get fwd biased when some cells are shaded.

    When a segment gets shaded, it goes to high resistance. The diode comes into play, and allows the power (reduced) to continue to flow
    Last edited by Mike90250; 06-29-2010, 05:54 PM.

    Leave a comment:

Working...
X