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power supply issue

Discussion in 'Electronic Basics' started by Anders Nesheim Vinje, May 3, 2005.

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  1. Hi.
    I have built this dual power supply.
    Transformer rating: 2 x 24 V 60VA
    Pretty standard with a bridge rectifier and ground connected to the center
    tap. So i got + 24 and -24.
    The i have two 4700uF 50v caps, one connected at each powerline to provide
    filtering. Then there is a 7815 and and a 7915 as regulators.
    Ok here is the issue. When i measure the voltage over the filtercaps i see
    +37,5V and - 37,5V! How is this remotly possible?. Even the peak to peak
    voltage isent that high. The 7815 and 7915 is only rated up to 35V so i am
    scared they might blow or something.

    Another problem is that since there is a voltage drop 37.5 -15 = 22 V over
    the regulators. It cant provide much current before they shut down because
    of the heat. I know from the datasheet they can handle up to 10W with a
    decent heatsink and a max current of 1,5A. I dont need that much current
    anyway since the transformer can only deliver 30VA / 24 = 1,25A per
    secondery anyway.
    I have another question too. How much current goes through the primary side
    of the transformer with max load? I was thinking of having a fuse there
    instead of buying a fuse to each of the secondery outputs.

    Thanks in advance

    Anders Vinje
     
  2. I took a new measurement and found out the the ac value was acutally 27V not
    24. So when i muliply with the square of two i roughly get 37,5V peak to
    peak.
    But why do i see the peak to peak voltage after it has been rectified???
    Even with load 40ohms load.

    Anders
     
  3. John Fields

    John Fields Guest

    ---
    With good reason.

    Since each of the supplies is connected to the transformer as a
    full-wave center-tapped supply, the rectifiers will only drop about
    0.7V in front of the filters, so with 27VRMS into the rectifiers the
    peak voltage (the voltage across each filter cap) will be:


    Vp = (Vrms * sqrt(2)) - 0.7V = 37.7V

    Which is pretty close to what you got.
    ---

    ---
    Since a transformer transfers power and the secondary can deliver:


    P = IE = 1.25A * 24V = 30 watts


    into a resistive load, then the primary will need to take 30 watts
    from the mains in order to transfer that into the load.

    Then, since:


    P= IE


    we can rearrange to solve for the primary current:

    P 30W
    I = --- = ------ = 0.25A
    E 120V

    for 120VRMS mains.

    Assuming your transformer is 80% efficient means that the mains will
    have to supply extra current to compensate for the transformer's
    losses, so that means you'll need to draw about

    0.25A
    It = ------- ~ 0.31 amperes
    0.8

    from 120V mains.
    ---


    ---
    Unless you're measuring the difference in voltage between the filter
    caps you're not measuring peak-to-peak, you're measuring peak.

    Since each of the supplies is connected to the transformer as a
    full-wave center-tapped supply, the rectifiers will only drop about
    0.7V in front of the filters, so with 27VRMS into the rectifiers the
    peak voltage (the voltage across each filter cap) will be:


    Vp = (Vrms * sqrt(2)) - 0.7V = 37.7V


    Which is pretty close to what you got.

    You don't say whether the load is across the filter cap or across the
    output of the regulator but, assuming it's across the filter cap, the
    current through the resistor will be

    E 37.7V
    I = --- = ------- ~ 0.94 amperes
    R 40R

    This will cause the voltage across the filter to vary as it charges up
    through the rectifier and discharge when the voltage out of the
    rectifier falls below the voltage on the cap. This voltage change is
    called 'ripple', and its amplitude varies with the load current, like
    this:

    I dT
    dV = ------
    C

    where dV is the ripple voltage in volts,
    I is the load current in amperes,
    dt is the ripple period in seconds, and
    C is the capacitance in farads

    So,

    0.94A * 8.3E-3s
    dV = ---------------- = 1.66 volts
    4.7E-3F


    That means that that the caps could charge up to 37.7V, but that that
    voltage would fall to about 36V when the rectifier output fell to
    lower than that. Your DC voltmeter would probably read the average
    of those two voltages, about 36.8V, so you probably wouldn't even
    notice the difference between loaded and unloaded unless you were
    intent on noticing it.

    If the 40 ohm load was connected across the output of a regulator, the
    current would be:


    15V
    I = ----- = 0.375A
    40R


    and the ripple voltage would fall to:


    0.375A * 8.3E-3s
    dV = ------------------ ~ 0.66 volts
    4.7E-3F

    so you'd probably be even less likely to notice that difference.

    There are a few more problems to consider, but in order to do that we
    need to know what your maximum load current is going to look like.
     
  4. jgreimer

    jgreimer Guest

    Why are you seeing 27 V instead of 24 V at the output of the transformer?
    Is your line voltage 27/24 * 120 = 135 V?
     
  5. Thank you for all answers.

    I see i need another transformer. Something like 2X 15V instead.
    The reason i see 27 V is under no load. I think its 24 V when under max
    load.
    I live in Norway and we got 220V 50HZ mains here.

    Anders
     
  6. Ban

    Ban Guest

    That is strange, I think Norway has 230V like almost all west-European
    countries. If the transformer is rated for 220V this will explain half the
    overvoltage. The secondary voltage is indeed rated at max. AC-current. This
    will explain the other half. Bigger transformers(M85b) have only 106% output
    voltage at no load.
     
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