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Random Theory Thoughts: RMS and Smoothing

Discussion in 'General Electronics Discussion' started by Raven Luni, Feb 1, 2012.

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  1. Raven Luni

    Raven Luni

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    Oct 15, 2011
    I was just thinking - so much emphasis is placed on choosing the best capacitors for smoothing a rectified AC supply, with the ideal pure DC being unattainable - so why not take a higher starting voltage, smooth that, then use a zener / regulator / etc. to ensure that the output voltage is never more than the RMS of the input and the current drawn is never more than is available at the input. Surely that would give headroom for a flat DC supply?

    Edit: D'OH! I think most power supplys do exactly that. Sleep deprivation is a wonderful thing :eek: It's still an interesting subject for discussion though :D
     
    Last edited: Feb 1, 2012
  2. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Your second thought is a good one. Right, that's what a power supply usually does. Still there is reason to chose the smoothing capacitor wisely:
    The cap will charge to the top of the rectified AC which is Vcap=1.4*Vrms. The difference between this voltage and the output voltage will result in lost power in the regulator (approximately P=I*(Vcap-Vout), not accounting for the ripple). So you want to have a low voltage drop, meaning Vcap is as small ass possible.
    On the other hand, as you have rightly observed, Vcap will drop due to load current when the input voltage is lower than Vcap (rectifier is then nonconducting). Approximately DeltaV=Iout*DeltaT/C or C = Iout*DeltaT/DeltaV where
    DeltaV = voltage droop during time DeltaT and
    DeltaT = time between refreshing of the cap, that is 10 ms for a 50 Hz power line.
    In order for the regulator to operate properly, (Vcap-DeltaV)-Vregulator > Vout is required (Vregulator being the minimum Input-Output voltage differential of the regulator)

    Let us assume you have:
    Vrms = 10 V -> Vcap = 14V
    Iout = 1 A
    Vregulator = 1 V (there are better ones and worse ones)
    Vout = 9 V
    Frequency = 50 Hz -> DeltaT = 10 ms

    Putting all this together gives:
    (Vcap-DeltaV)-Vregulator > Vout -> (14 V - DeltaV) - 1 V > 9 V -> DeltaV < 4 V
    C = Iout*DeltaT/DeltaV -> C = 1 A*10 ms/4 V -> C ~2.5 mF (one would chose at least 4700µF to account for tolerances and aging).
    To estimate the power lost in the regulator let us assume from the above values a mean voltage of (14 V + 10 V)/2 = 12 V at the input of the regulator. The mean voltage dropped across the regulator is then 12 V - 9 V = 3 V which results in a mean power of 3 W lost in the regulator.

    Now let us assume we use a much bigger capacitor, say 10000µF, the same regulator, output voltage and current.
    From DeltaV=Iout*DeltaT/C we get DeltaV = 1 A * 10 ms / 10000µF = 1 V
    From (Vcap-DeltaV)-Vregulator > Vout we then get Vcap > Vout + Vregulator + DeltaV = 9 V + 1 V + 1 V = 11 V
    To estimate the power lost in the regulator let us assume from the above values a mean voltage of (11 V + 10 V)/2 = 10.5 V at the input of the regulator. The mean voltage dropped across the regulator is then 10.5 V - 9 V = 1.5 V which results in a mean power of 1.5 W lost in the regulator. This is half of what is used for heating with the first example.

    Regards,
    Harald
     
  3. Raven Luni

    Raven Luni

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    Oct 15, 2011
    Very interesting - thankyou :) I'll put that knowledge to use at some point (got a few future projects that will need custom power supplies) :)
     
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