Efficency of power supply design

Discussion in 'The Projects Forum' started by AIN, May 12, 2011.

  1. AIN

    Thread Starter New Member

    Apr 11, 2011
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    Hello,

    I have this power supply block which supplys 3 other blocks.

    First block gets unregulated 24V out of the big cap of 5600 uF. The other two get 5V from the regulator.

    I want to calculate the efficiency AND power dissipated of the following:

    1- First block power

    2- Second and third block power.

    3- Overall power supply efficiency and power dissipated.


    Pls advise!
    Thanks
     
  2. wayneh

    Expert

    Sep 9, 2010
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    That's nice. What's stopping you? Perhaps a complete lack of data regarding current flow? Perhaps a definition of what a "block" is?
     
  3. gerty

    AAC Fanatic!

    Aug 30, 2007
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    Your fuse is placed across the primary of the transformer, if you apply power it will blow (if you are lucky). You need to put it in series with the transformer.
     
  4. AIN

    Thread Starter New Member

    Apr 11, 2011
    13
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    Block1 = Ri which draws 5 Amps
    Block2 and Block3 are represented by R4. The total current they draw is 400 mA.

    Now, I know efficiency = output power/input power.

    I just want to check if my calculation are correct.

    ===========
    Efficiency:
    Regarding block 1: eff = 5*24/8*24 = 62.5 % (since the transformer is rated with 24V/8Amps)

    Now for block 2 and 3: eff= (24V * I) / 5V * 400mA. Do I use I =7.5A which is provided from the datasheet. In the datasheet the regulator circuit is designed to give 5V/7.5A. Do I use I = 7.5A?

    I also want to compute the overall efficiency of the design. How would I combine these values to give the overall efficiency?

    ===========
    Dissipated Power = (Vin-Vout)*Iload

    blk 1 = (24-24)*5amps = 0 ? Is it possible to get zero dissipated power?

    blk 2 and 3 = (24-5)*0.4 = 7.6 watts

    And the overall dissipated power would be adding up the values.

    I'll fix the fuse. Fortunately, it is the only part that I have not soldered yet.


    Thanks gerty and wayneh,
     
  5. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Complete nonsense, that would be utilization of available power or something

    If block 1 is from the sine source up to Cr, without Ri which is another nonsense, then its eficinecy is (24*Iout-0.7*2*Iout)/(24*Iout)
    which boils down to about 94%.
    I have no idea how you want to divide the regulator into two blocks...
     
  6. AIN

    Thread Starter New Member

    Apr 11, 2011
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    Well, the output pin of the regulator will be connected to two blocks. Each block will draw the needed current. Ri and R4 are in the schematic just to simulate the blocks.
     
  7. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    I have no idea what youre talking about.
     
  8. AIN

    Thread Starter New Member

    Apr 11, 2011
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    OK. I don't understand what do you mean by dividing the regulator into two blocks.

    Basically, the output of the regulator is 5V. This 5V is connected to two load (R3 and R4). Now, R3 is drawing 200mA. Also, R4 is drawing 200 mA.
     
  9. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    So you want to calculate efficiency of the whole setup?
    Then you take current as constant, each load gets 5V*0.2A W

    VI loss on the bridge rectifier, that is 0.7V*2*0.2A W
    VI loss on the regulator - that is loss on R1 and R2 plus voltage difference*current

    I was wrong before, efficiency is Pin/(Pin + Ploss). Add the losses and calculate efficiency. Maybe you should take the regulator´s bias current into accout too.
     
  10. wayneh

    Expert

    Sep 9, 2010
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    To talk about efficiency you have to identify and analyze all your losses.

    One of the bigger losses in your circuit is transformer inefficiency, which is a black box so far. I think a typical transformer can be expected to give 85% efficiency but that's just a rule of thumb.

    Immediately after that you've got diode losses in your rectifier. You're probably dropping at least 2v across the rectifier, so that's nearly 10% right there.

    Then you've got Ri (why?) just burning off power. It's rippling DC, but you can get a good estimate of that power loss.

    I don't know that regulator but I'm guessing it's linear, meaning it drops voltage by burning off the excess. You can calculate that also, and it'll be larger than the power going to R4. The small current to operate the chip and provide the reference current are probably negligible, but are in the datasheet or can be calculated.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Your efficiency will be very poor; under 20%. You really should look at using a switching regulator, perhaps an LT1070. It's rated for 5A.
     
  12. Wendy

    Moderator

    Mar 24, 2008
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    Measuring loss of a power supply is very simple. Loss (in %) = Wattsout/Wattsin

    I agree, Ri is pure waste. I think it is intended to discharge the capacitor, but meanwhile it absorbs power and does nothing but eat watts.

    Digital regulators can be very efficient but noisy (90%), analog regulators can be very quiet but inefficient (no lower limit). Inefficiency in this case is heat, that is the waste byproduct.

    Transformers are all over the map efficiency wise. They can be the most efficient components made by man, or very poor. It all depends on their construction.

    Just repeating what everyone else has said, more or less.
     
  13. rwstowe

    New Member

    Mar 23, 2011
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    For a LT1084CP adjustable regulator per your drawing, the reference voltage per the datasheet is 1.25V. With the feedback divider you have shown, the output voltage would be 1.67V.

    I am not sure of exactly what you are trying to calculate efficiency for, but I offer you the following relations:

    Efficiency of a linear regulator is simply Vout/Vin.

    Generally speaking, efficiency is Pout/Pin = Pout/(Pout+Ploss) = (Vload*Iload)/(Vsource_rms*Isource_rms*power factor)

    I believe a typical power factor for a bridge rectifier might be about 0.6. I am not sure though and it would be worse with bigger filter caps or lighter loads.

    Use the principle of superposition (calculating each part, then summing the results for each part) to arrive at your answers for each efficiency.

    Data sheet ratings don't have an effect upon efficiency. They only tell you how much stress the part can take.

    Glad you fixed the fuse.

    See the following page for more information about power supply efficiency:

    http://www.truepowerresearch.com/2011/03/tutorial-power-supply-efficiency/
     
  14. wayneh

    Expert

    Sep 9, 2010
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    I'm not sure there is a "typical" factor for a rectifier. There's the voltage drop, but after that the power delivered to a purely resistive load is just V^2/R. Perhaps you were referring to the combined rectifier and filter. Power factors into capacitive loads is beyond my skill set! :cool:
     
  15. rwstowe

    New Member

    Mar 23, 2011
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    wayneh,

    Thanks for clarifying my explanation. Yes, I was referring a bridge rectifier with a capacitive filter.
     
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