Page 2 of 2 FirstFirst 12
Results 11 to 13 of 13

Thread: Power my setup. Questions on Inverters!

  1. #11
    Who am I? HiJackZX1's Avatar
    Join Date
    Jul 2007
    Location
    Miami
    Posts
    6,603
    Quote Originally Posted by OldSpark View Post
    PS - Maybe a hacked PC supply for 5V @ 20A or more?
    PPS - though at the price of PaulF's link... IMO 10A for $32 is good.

    Geez HiJack, you reckon my answers are complex - what about your questions?! (They actually scope a lot! Though PaulF summed beautifully - with a great tip too!)


    As to "numbers", you want the INPUT power for all loads, and the OUTPUT power your supply is capable of.
    Conversions have inefficiencies. And these vary. I usually add 20% or maybe 30% to conversion max (continuous) outputs if I don't have their max input figures.
    Or I use their (input) fuse rating as a guide.
    But both can vary wildly from reality if used for multiple devices.

    EG - a 10A fuse means 80W is designed for minimum 8V operation, 138W@13.8V; 140W@14.4V or 160W if designed for a max of 16V.
    BUT - that 10A fuse may be for a 80W "constant power" supply (eg; HIDs) which is normally ~5.5A near normal 14V. But then the load is always (say) 80W so you don't have worry about its voltage/current relationship. (Ah - they beauty of dealing with "constant power" devices. Except when figuring what fuse and power cable to use!)
    EG - power conversion - an 80% efficient amplifier or dc-ac converter, or a 95% efficient dc-dc converter; or a 40mW resistor-fed LED that "draws" nearly 300mW (ie, 15% efficient)?

    Remember - the inefficiency is usually heat. Its power is P=VI where V or I are the "lost" quantities.
    Cooling prolongs life and cooling can improve efficiency (because of less resistance etc).
    If cooling prevents damage, then it is required (eg, heatsinks, fans, car radiators).
    Since cooling reduces wire etc resistance, there should be less heat (P = heat = IxIxR from P=VI & that V=IR hence V=I/R).
    But cooling can cause other inefficiencies, though I doubt that in this type of situation - ie, I can't think of anything other than battery issues)


    As to the rest - PaulF's excellent reply sums it up.
    But since I started this before he replied ( !!)
    So, included for reference and bit buckets...

    This will be long, but I'm trying to break it all down, cover some basics (to be certain), so please bear with me. (And yes - I still love you, but I specifically wrote "bear" with me, not "bare with me".)
    Also, I'll try "exec mode" of writing - first "heading" line is the basic answer or summary.
    Later lines become less important wrt giving "the answer", but provide further explanation or detail. Or they can provide examples & trivia.
    The rule: Read first heading line(s) only for short answer. Read on if not understood, or if extra (pedantic) info is sought.
    GO TO NEXT heading when too complex or enough is enough!
    And [ square brackets ] usually mean "real pedantic, but I'm preventing later objectors that know otherwise" etc, though sometimes they enclose nested brackets [(maybe (too) many (other brackets) get confusing)].
    Normally I'd use more inter-line breaks (between indirect sentences) and double line breaks between "real blocks", but to save space it's been crushed up. (Yes - I often use a hierarchy - full-stop & continue, full stop & new line; full-stop and blank line - three levels grouping.)


    Switching Regulator: switches dc on/off to convert voltage.
    More detail: Only AC transforms, hence convert DC to on-off DC which is AC (by definition and behavior).
    SMPS (Switched Mode Power Supplies) are Switching....
    SMPS (switching) is used in all step-up DC-DC converters, and now in many step-down DC and AC-DC converters.
    [Ped: An ac-ac converter is a transformer. Ignoring (minor) inefficiency, power IN equals power OUT. So if 120V 10A in and 12V out, that's 100A out. (ie 1200W in = 1200W out. P = VI; Watts - Volts time Amps.]
    The alternative to (step down) switching is Linear - eg, resistors, transistor circuits etc. But with them, at best current in = current out & hence inefficient. EG 120V 10A in = 12V 10A out; so 1200W in = 120W out. The Linear converter generates (1200-120=) 1080W heat because it "drops" or burns 108V at 10A. (Burns (C)PaulF 2010.)

    "sine wave inverters have little or no noise at all". There output should be pure sine which - like DC - has no (other) "noise".
    But like most power conversion, outputs have varying noise in practice - from negligible to disgusting. And that can depend on the load characteristics (size, inductive, SMPS etc).
    Inverters are - in simple terms - a dc-dc converter without the output rectification (ac-dc).

    Trivial: Because dc-dc is switching (AC) there is noise. But this is "internal" and might not come out of the output (ie, pure DC or sinewave) and should be shielded from radiating out (EMI/EMC etc)). So all switching converters can cause (both types of) noise.

    [ The "non-sinwave" inverters are "pseudo-sinewave". Instead of a pure (smooth) sinewave, they approximate it using 2 stepped voltage levels (either side of zero volts 0V).
    This "blocky" sharp-edged sinewave can be unfriendly to loads. It can overheat load input components (in their PSU).
    Technically, the squarish output is a pure sinewave with lots of harmonics - sinewaves at higher frequencies for which the 50Hz or 60Hz load-PSU may not be designed for. Filter out those harmonics and they'll be ok. Harmonics are multiples of the "fundumental" frequency - eg, 2x60, 3x60, 4x60 etc = 120, 180, 240Hz etc. (Hence those noise frequencies...) ]

    "will provide regulated power": All converters usually provide regulated power - ie, designed to maintain a certain voltage output irrespective of load current, load type, and input voltage.
    How well they do this depends on their design (specs).
    dc-dc & dc-ac can both suffer regulation problems.
    dc-ac inverters are more prone to regulation problems due to bigger load variations - ie, phase angles, harmonics, inductive spikes (motors etc)....
    dc outputs merely have to regulate voltage (as it changes with current load) since there are no output/load inductance, capacitance or frequencies involved. In theory that is...

    "inverters (are) very inefficient: No.... they can be 90% or better, but most are only about 85% at best.
    And that is at a certain output load.
    Higher or lower than that and the efficiency drops.
    And inverters (like dc-dc converters) always have some "idling current" - they still consume power just sitting there with no load. (Because they have whatever control electronics - circuit or uPCs or PIC etc.)
    But idling power should be very low - maybe 1W or less - thanks to modern low power electronics. Some use to be tens of Watts.
    Idling power may be insignificant when powering a 100W load etc, but if sitting on a battery in standby mode for hours....
    Hence "standby" consumption - the power/current a dc-dc or dc-ac device consumes just sitting and waiting.
    Standby power might be much less than "idling" power/current. VIZ - a low power supervisory pr monitoring circuit senses when conversion is needed and it starts up the "high power" converter...

    "inverters (are) very inefficient - Part 2: No - they just add an extra inefficiency that is missing in the dc-dc converter.
    They both have the same dc-ac inefficiency [ped: in principle, but that depends on design...] but the internal dc-dc converters ac-dc inefficiency is much less than the inverters external ac-dc efficiencies (eh, plug packs).


    Did anyone read this line? (LOL)
    (Maybe next time I'll add pre-planning to the exec mode!)
    Thanx for the post. I do know one thing for sure, the switching regulators will not be powered at all when the car is off. Basically I like to tie everything to PC1. That way when the OPUS powers up, I know everything is on. Also when I power down, I know its still on untill the PC fully shuts down. In saying that, there will be relays through out the car, all controlled by the OPUS.

    LOL.... I know my questions are complex..... I never said that about yours, I said they are too technical and you use terminalogy sometimes that I have no clue about, all though I understood this post perfectly.

    So do switching regulators generate much heat? Will I have to prepare to make a cooling system for them?

    Quote Originally Posted by PaulF View Post
    Haha oldspark, I read the whole response. I enjoyed the copyright in my name too.
    I know you're always there when crazy details are needed.

    Hijack,
    Slow down there... I doubt you need 30 amps.

    First, why do you need both an access point and an ethernet switch? Why not a wireless router with a 4 port switch built in? That would save space, wiring, and power. Unless of course you need more ports...

    Regardless, the adapter may say 2.6 amps. I highly doubt the device is using that though. Check it with a multimeter. It may have been cheaper for them to just throw in a generic 2.6 amp adapter than the 1.0 that was necessary. Or they may have wanted to give some extra room for any additional overhead.

    Really, check the given INPUT values on the devices themselves. Check them with a multimeter on the 5v side if necessary.

    I HIGHLY doubt the KVMP switches use 2.6 amps each! They're maybe a switch, buttons, LEDs and possibly a microcontroller. I doubt they even use 500mA!


    ----
    These switching regulators are actually bigger than the M4. But stacking them isn't a problem. They're enclosed. They don't generate much heat. I mean I wouldn't completely enclose them though haha.
    Dimensions (H x L x W): 1.5 in x 6.25 in x 3.8 in.
    I am going by the adapters themselves. Im guessing what they are putting on there is the max they can possibly use. I'd rather use that number so I dont have to worry incase the devices do happen to reach 2.6 amps at some point.

    Well, remember I was going to put 7 PC units in my car. Now though I have managed to get it down to two. One system is going to emulate 6 setups, then the other system is for all the high end stuff like gaming, etc. So in saying that, I bought everything before hand. I like the access point because the range is good. Its actually the device using the a lot also, 2 amps. All the hubs are rated at 1 amp.

    I have tried it without the plugs, they dont even turn on. They are KVMP switches, which do more. I know the 2.6 number is maybe a "could use up to" figure, but I feel more comfortable with it.

    Quote Originally Posted by soundman98 View Post


    all of the 2-4 computer kvm's i have worked with have been able to be powered through the usb/vga/ps2 port-so they definitely don't take more then about 500ma..

    though, for the usb hubs, i think you should still plan for at least 3a-- not all devices will consume the full 500ma usb ports are speced for, but it is better to have a little wiggle room..
    Yea I left wiggle room with the hubs also. Technically the install has a total of six 7 port hubs, but for 2 of them, the KVMPs are connected to them, and since the KVMP switches will be powered, I dont need to power the hubs. Then the last unpowered hub is only for data, for the arm rest ports. Three that are powered, one is for 5 usb soundcards. I tested without power and it seemed to work, but I am scared that once music is pumping through all of them at the same time, its going to eat up power and need more. The other is for the front center console, it will power my touchscreen, serial to USB, PPC, keyboard, Sony control dongle and stuff that I know will max out power. The final one is for the arm rest toggle system, it needs to be powered so if the user plugs in something that eats to much power. Basically it constantly powers the ports, so say you toggle to PC2, its still powering the ports, even though that hub sits on PC1.
    Nirwana Project, the Android/Win 7 hybrid system!

    1X Ainol Novo Flame Tab
    4X MK808b
    3x Perixx Touchpads
    3x 7 inch Screens
    1X 7 inch motorized Screen
    1x Win 7 PC

  2. #12
    FLAC
    Join Date
    Jan 2008
    Location
    Dartmouth, MA
    Posts
    916
    No, the power supplies are not a "could use up to" situation. They're complete BS.

    Say I'm Belkin... I just made a sick KVMP. Now I need to get the cheapest power adapter supplier for it. My device uses 1 amp. 1 amp supply from manufacturer A costs 40 cents, 2.6 amp supply from manufacturer B costs 36 cents. B gets the sale. Done.

    Seriously, measure the KVMP draw with a multimeter. I promise it probably won't even touch 1 amp.

    Switching regulators are not huge heat generators. They may get warm but never really hot.
    My Nearly Complete Car:
    http://www.mp3car.com/vbulletin/show...ed-car-pc.html

    Micro Control Center... Control Your Car Across the Internet
    http://www.mp3car.com/fusion-brain/1...-internet.html

    Website: (It's a work in progress, really. All my projects have taken me from ever really developing it.)
    http://paulfurtado.com/

  3. #13
    Raw Wave
    Join Date
    Nov 2009
    Posts
    2,113
    LOL! I am corrected - "too tech" is correct. Apols. But as for all - ask for clarification (if I keep reposting PSU = Pwr Spply Unit and other "basics" (LOL - if you know them!), it's space and boredom etc.

    But it gladens me that this time, the tech was "understood ... perfectly". (Moral - I should organise replies etc.)

    Again, PaulF sums it up...
    And agreed - DO NOT confuse "actual" power with "capability" - a 200A alternator or USB supply only delivers whatever is required. (The 200A capability can supply 200A, but otherwise it doesn't exist in real terms. If that makes sense.)
    And yes - PSUs are whatever is cheapest and common, and with enough headroom (aka wiggle for surges or transients that have insignificant heating effect.


    Trivial Detail and maybe a guide:
    "So do switching regulators generate much heat?". ROT (Rule Of Thumb) - assume 20% maybe 30% of MAX input is maximum heat. (Remember - from above - that's probably about 32 levels down in your "new details" brain-stack - not yet shifted to mull-brain and short-term memory.)
    BUT - as the load drops, there is less current, hence less heat. See [Ped] below where "nominal" heat is 1/3rd of full rated load heat.
    Hence too, don't "design" for the 200A supply if only 1A or 10A etc is being used. (The "exponential" size is explained(??) in [Ped] - ie 200A is (20x20=) 400 times the heat of 10A.... There is a BIG difference between a 1" heatsink and 400" heatsink!)

    Remember too that inefficiency is "100% minus efficiency%" [or 1-u if using "per unit" u=efficiency figures - eg 85% efficiency means u (mu; μ) = 0.85. Inefficiency = 1-μ = 1 - 0.85 = 0.15 (or 15 %)].


    Before I suggest "Brace Yourself!" (the Australian definition of foreplay), double-thanks to PaulF - a small one for his "crazy details" compliment (am I too naive? I know I often feel too unworthy! Yes - an Alice Cooper reference), and a BIG one for his link.
    Great seeing what seems to be quality gear and at reasonable prices, AND with a great spec sheet - ie, input voltage detail, efficiency guide, output (DC) noise and EMI/EMC noise both with notes. (Don't ask EMI/EMC rating's real meaning... me googles too!)


    Stop here, else Brace Yourselves!

    [ Ped: Converter efficiencies vary with load. As a fictitious example - maybe 50% at 20% load; 85% at nominal (say 70%) load; 80% at full 100% load.
    Max heat is at 100% load at 80% efficiency.
    As load drops, current decreases (so less heat) AND efficiency increases - a double win! IE - a much greater heat drop than current drop alone.
    Then below best efficiency (85%) at 70% load, efficiency increases BUT current is decreasing.

    Complication/Detail: heat is IIR = i-squared R = current times current - ignore Resistance R because let's assume that is constant for the wires and circuit that carries the current.
    So triple the current thru a wire generates 9x the heat [ I x I = 3I x 3I = (3x3)I = 9I ]
    EG - wire = 2R (1 Ohm). I = 10A. P = IIR (aka I^2R or I**2R) = 10 x 10 x 2 = 200 Watts. 3xI = 30A so it's now 30 x 30 x 2 = 1800W = (1800/200=) 9 time more - ie, 3x3 current more.
    That could make up for efficiency gain as efficiency increases but current drops.
    Hence heat at any load (even peak efficiency) could be higher than at max load.
    And to be honest, I have forgotten "trends", but hence why normally max rated load current is used - it's the highest IxI at less that best efficiency.

    { My sanity check: 100% I @ 80% efficiency = 1x1 x 0.2 = 0.2 "heat".
    70% I @ 85% efficiency = .7x.7 x .15 = 0.07 "heat". (Phew! )
    IE - 70% load generates nearly 1/3rd the heat of 100% load. It is very disproportionate. }

    FYI - I reckon the sizing of stuff (PSUs, cooling etc) is worse than replacing 2nd hand exhausts. And trust me, I hate the latter! Good Engineering as opposed to Over-Engineering is luck, or lucky under-knowledge, or trial and error, or damned good ROTs & experience, or freaks of nature. ]

    I think I'll go. For some reason I feel far from any ability to come.
    (To the party of course!.)

    Readers may now un-brace. And relax.

Page 2 of 2 FirstFirst 12

Similar Threads

  1. Power
    By ssteger in forum VaultWiki
    Replies: 0
    Last Post: 10-14-2009, 02:59 PM
  2. Hardware Review: M3-ATX Power Suply
    By Bimmerstyle in forum The Review Palace
    Replies: 8
    Last Post: 03-05-2008, 05:57 PM
  3. Newbie: questions about Power inverters and DC PSU's
    By HuseyinOzsut in forum Power Supplies
    Replies: 1
    Last Post: 02-22-2008, 04:28 PM
  4. del
    By hanzacra in forum Classified Archive
    Replies: 1
    Last Post: 08-07-2005, 10:32 PM

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •