Can I do this…? V A & W

Discussion in 'General Electronics Chat' started by ranatungawk, Jan 4, 2015.

  1. ranatungawk

    Thread Starter Senior Member

    Oct 30, 2008
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    HI friends,

    I need to know whether the following assumption is correct......:cool: LED driver.jpg


    I have got a small LED driving circuit (please see the attachment) with power-rating :230V-12V/5W.

    As per the above rating : I can get 12V with 0.41A output. (5W/12V=0.41A)

    However if I modified the circuit with “7805 Regulator” to get 5V output , will I be able to get nearly 1A current ? (5W/5V =1A)
     
  2. ericgibbs

    Senior Member

    Jan 29, 2010
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    hi,
    Sorry, it doesn't work like that.
    Consider even with a 7805 at 1A, the 12V source driving the 7805 would have to be providing 1Amp.. that's 12 * 1 =12Watts
    E
     
  3. ranatungawk

    Thread Starter Senior Member

    Oct 30, 2008
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    thanks for replying ...can you please be more specific....so with 5v output ...what would be the current ??
     
  4. ScottWang

    Moderator

    Aug 23, 2012
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    V_7805 = 12V-5V = 7V.
    W_7805 = 7V*0.416A = 2.9W
    Wo = Wi - W_7805 = 5W-2.9W=2.1W
    I_7805 = 2.1W/5V= 0.42A, so you can see the 7805 only can get the original current.

    According to the V=I*R, when the current is not enough for the load then the voltage will be draw down naturally, I'm not sure what's your load and what voltage it can afford, sometimes we can using a 12V/0.416A power for a 5V/1A load, and no need the 7805.

    Edit: But there is need to concerned is that the internal design of switching power, does it allow to do this?
     
    Last edited: Jan 4, 2015
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  5. WBahn

    Moderator

    Mar 31, 2012
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    The 7805 is a linear regulator which basically means that the current out of it is the current into it except at a lower voltage (with the remaining voltage dropped across the regulator).
     
  6. wayneh

    Expert

    Sep 9, 2010
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    An ideal power converter could do this in theory, but that would require 100% power efficiency. There is no such converter. A buck DC-DC converter could give you ~85% I believe. These are commonly used in cars to plug into the accessory jack and provide USB 5V power. I recommend one of those for your project - they're cheap and plentiful.

    As you've been told, the linear regulator can give you the voltage you want, but it does that by just burning off the excess as waste heat.
     
  7. ranatungawk

    Thread Starter Senior Member

    Oct 30, 2008
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    *** Thanks for replying! for further clarification.. if i use a 7805 to get 5V out will "max-output-current" drop below 0.41A???? (such as : input current 0.41A * 85% = 0.34A)
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Assuming that you can actually get 0.41A out of the buck regulator, then you will be able to get nearly all of that out of the 7805. There is some current that flows out the third pin, but it is very small. One thing to keep in mind is that you have to get rid of the head dissipated by the 7805 which will be nearly 3W. That's quite a bit of heat and it will need to have a suitable heatsink.
     
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  9. wayneh

    Expert

    Sep 9, 2010
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    Current is conserved, like energy, and cannot accumulate in a circuit except in specialized devices like batteries and capacitors. So if it comes into the 7805, it has to leave and cannot accumulate there. As noted above, a small current is sent to ground within the 7805, so the output is not exactly as large as the input. Voltage, on the other hand, is not conserved. It, and the associated energy, is converted to heat and dissipated by the 7805.
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    I think this has the potential to be very confusing for some people. After all, what about a transformer? It is commonly the case that the current in many times different than the current out -- as seen by many folks. The same for that buck converter he is starting with. If you look at all of the terminals, then current is conserved just as it is for batteries and capacitors. Charge accumulation and or chemical conversion does not violate conservation of charge/current.
     
  11. wayneh

    Expert

    Sep 9, 2010
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    Yes, I was alluding to Kirchoff's law as a useful way to think about the OP's question (quoted in #9) but it probably requires more explanation.
     
  12. profbuxton

    Member

    Feb 21, 2014
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    If your source only supplies .41A that basically all you can get regardless of how low you regulate the volts
     
  13. ranatungawk

    Thread Starter Senior Member

    Oct 30, 2008
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    This can be done with "LM2575HV" successful than 7805, i hope ! isn't it ?

    LM2575HV-circuits.jpg
     
  14. ScottWang

    Moderator

    Aug 23, 2012
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    This ic is better than 7805, but you still need to increasing the input current more than what you want, otherwise that is useless.
     
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  15. WBahn

    Moderator

    Mar 31, 2012
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    I don't follow what you mean by still needing to "increasing the input current more than what you want".

    If he was going to power the original buck regulator with 12V, then he doesn't need that one at all and can just use the 2575. The spec sheet indicates a typical conversion efficiency of 77% at 12V input and 1A output (for the 5V output model). So his input current should be something close to

    <br />
I_{in} \; = \; \frac{I_{out} \( \frac{V_{out}}{V_{in}} \) }{\eta}<br />

    which would be ~550 mA.

    Not enough information was given in the first post to know what kind of input current he considers acceptable. In fact, I can't really tell what the output voltage of that little switcher is. I wouldn't think it is adjustable, so it is a fixed 12V output?
     
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  16. ScottWang

    Moderator

    Aug 23, 2012
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    What I mean that "more than what you want" is that he want the output current 1A, but he only provided 0.42A, so if he didn't given the enough current over 1A for the input then he can't get 1A from output.

    If the efficiency of dc-dc is 77%, that is too low for our standard spec of products, it should over 80% at least, a several months ago, I got a case from school, they asked for 3A ouput, but our skills is not enough to do that, so the engineer to bought a dc-dc from China, the efficiency over 95%, the current of dc-dc board still not enough, but engineer learned the skills of high efficiency and to completed the case, I can't tell the details about voltage and load.
     
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  17. ranatungawk

    Thread Starter Senior Member

    Oct 30, 2008
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    This small switcher is a fixed 12V\ 5W unit. with 77% efficiency 2575 can out put 5V 0.32A ?? correct ? (0.42A x 77% = 0.32A )
     
  18. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    If the input is 5W, then the 2575 would output roughly 5*0.77 = 3.85W, which gives 0.77A at 5V output voltage.
     
  19. crutschow

    Expert

    Mar 14, 2008
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    Efficiency is the ratio of the power out versus power in, not of the current.
     
  20. ScottWang

    Moderator

    Aug 23, 2012
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    Win = 5W, Vin=12V, Iin=0.42mA.
    Wo = Wi * η = 5W * 77% = 3.85W
    Io = Wo/Vo = 3.85W/5V = 0.77A
    So Iout = 770mA.
     
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