Power supply filtering and bleeder resistor calcs

Discussion in 'Electronic Basics' started by seware, Jan 26, 2005.

  1. seware

    seware Guest

    I am wanting to build a power supply and need a bit o' help. I have an
    "Electronics for the hobbiest" type book that goes into fair detail on the
    subject, but where it talks of filtering and bleeder resistors it speaks of
    their use and suggests values for an example power supply but not how to
    calculate the values for your own. Their example uses a transformer with a
    12.6V 3A secondary wth a full-bridge rectifier. The bridge outputs have 3
    1000uF 35V caps in parallel and then a 1K 1/2W resistor in parallel with the
    caps for the bleeder. How do I calculate my needs for a transformer with a
    15V 5A secondary? A worked example would be nice but I can plug in numbers
    if someone will explain the reasoning. I understand that the size of the
    bleeder depends on how fast I want to discharge the caps, but I don't know
    what kind of time is reasonable. Should I reverse engineer the RC constant
    in the example and work forward from that or are their some better rules to
    follow? Thanks to all you professionals who sustain the questions of all of
    us wannabees.

    Steve
     
    seware, Jan 26, 2005
    #1
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  2. seware wrote:
    > I am wanting to build a power supply and need a bit o' help. I have an
    > "Electronics for the hobbiest" type book that goes into fair detail on the
    > subject, but where it talks of filtering and bleeder resistors it speaks of
    > their use and suggests values for an example power supply but not how to
    > calculate the values for your own. Their example uses a transformer with a
    > 12.6V 3A secondary wth a full-bridge rectifier. The bridge outputs have 3
    > 1000uF 35V caps in parallel and then a 1K 1/2W resistor in parallel with the
    > caps for the bleeder. How do I calculate my needs for a transformer with a
    > 15V 5A secondary? A worked example would be nice but I can plug in numbers
    > if someone will explain the reasoning. I understand that the size of the
    > bleeder depends on how fast I want to discharge the caps, but I don't know
    > what kind of time is reasonable. Should I reverse engineer the RC constant
    > in the example and work forward from that or are their some better rules to
    > follow? Thanks to all you professionals who sustain the questions of all of
    > us wannabees.
    >
    > Steve
    >
    >


    The way these things work is that the power line gives a pulse every
    1/60 seconds (or 1/120 seconds, if you are using a full bridge with 4
    diodes). So, the caps will be 'pumped up' to full voltage that often,
    and then power your supply between pulses.

    You will probably also use some kind of voltage regulator. If you were
    building a 5V supply, for example, you might use a 7805. These
    regulators will always require some 'headroom', meaning that the input
    voltage must not drop below some minimum for the regulator to maintain
    the output voltage. For a 7805, for example, this is usually stated as
    3V. Thus, you don't want the input voltage to drop below 8V.

    So, the caps get pumped up to 12.6V every 1/120 second, meaning that the
    drop from 12.6 to 8V cannot be faster than 1/120 seconds.

    Now, you need to decide how much current your maximum is. For a 7805,
    for example, the maximum current is about 1A. So,

    I = C * (dV/dt)

    I = 1A

    dV = (12.6-8)

    dt = (1/120)

    so C = 1A / (4.8*120) = 1812uF

    Usually, some fudge is thrown in, so if you use 2,000uF, you'll be safe.

    For a bleeder resistor, say you want the thing to discharge in 5
    seconds. The rule of thumb is that a resistor will discharge a cap in 5
    * RC. So RC in this case is 1. Thus, R = 1/2000uF = 500 ohms. Thus, 470
    ohm is a standard value which is near enough.

    --
    Regards,
    Robert Monsen

    "Your Highness, I have no need of this hypothesis."
    - Pierre Laplace (1749-1827), to Napoleon,
    on why his works on celestial mechanics make no mention of God.
     
    Robert Monsen, Jan 26, 2005
    #2
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  3. seware wrote:
    >
    > I am wanting to build a power supply and need a bit o' help. I have an
    > "Electronics for the hobbiest" type book that goes into fair detail on the
    > subject, but where it talks of filtering and bleeder resistors it speaks of
    > their use and suggests values for an example power supply but not how to
    > calculate the values for your own. Their example uses a transformer with a
    > 12.6V 3A secondary wth a full-bridge rectifier. The bridge outputs have 3
    > 1000uF 35V caps in parallel and then a 1K 1/2W resistor in parallel with the
    > caps for the bleeder. How do I calculate my needs for a transformer with a
    > 15V 5A secondary? A worked example would be nice but I can plug in numbers
    > if someone will explain the reasoning. I understand that the size of the
    > bleeder depends on how fast I want to discharge the caps, but I don't know
    > what kind of time is reasonable. Should I reverse engineer the RC constant
    > in the example and work forward from that or are their some better rules to
    > follow? Thanks to all you professionals who sustain the questions of all of
    > us wannabees.
    >
    > Steve


    There are no hard and fast rules for these things. More capacitance
    produces less ripple voltage at full load current but a bigger start
    up surge and a longer bleed down at turn off. Personally, I hate to
    wait for a lab supply to fade out after I turn it off, because I may
    do that dozens of times a day when working on circuit variations. So
    I like serious bleeders on a lab supply. The supply for an audio
    amplifier, I am not in such a hurry about, and there is some minimum
    load there that pulls the voltage down, anyway.

    But here are some formulas starting with the transformer.
    A capacitor input filter (cap directly to the bridge rectifier) and
    load heats the transformer more than a resistor that draws the same
    average current, connected directly to the transformer.

    The bigger the cap, the worse this gets, because the cap voltage tends
    to stay near the peak of the rectifier output wave, so all the current
    has to occur in a small blast right at the top of the wave, when the
    rectifiers forward bias. So don't expect more than about 75% of the
    transformer's current rating coming out as DC. Your transformer will
    supply about 4 amps continuously from a capacitor input filter.

    That peak voltage is almost 1.414 times the RMS AC voltage of the
    transformer, so your 15 volt transformer will pump an unloaded cap up
    to about 15*1.414=21.2 volts. How low the cap voltage sags to before
    another peak comes along depends on the ratio of capacitance to load
    current. If you want the output to be regulated to no more than about
    12 or 15 volts, you can afford a lot of ripple on the capacitor, but
    if you want to regulate something like 18 volts or use the supply
    unregulated, you will probably want to keep the full load ripple less
    than a volt or so.

    To simplify the math a bit, lets assume that the 60 hertz line
    frequency (if that is your line frequency) charges the cap
    instantaneously every half cycle or every 8.3 milliseconds. That
    means that the 4 amp output current is running entirely from the cap
    all that time. The formula that relates current to rate of change of
    voltage is I=C*(dv/dt) or current (in amperes) equals capacitance (in
    farads) times the rate of change of voltage (in volts per second). So
    to have a 1 volt sag before the next recharge, the 4 amps has to cause
    a rate of change of voltage of 1 volt per .0083 seconds or 120 volts
    per second. So the required C is 4/120 farads = 0.033 farads = 33,000
    microfarads. If you can tolerate 2 volts of ripple, half that
    capacitance will do. Or you can substitute your available capacitance
    and calculate the ripple voltage.

    Bleeders are really only essential for safety reasons (to dump the
    voltage to a safe level before you can get the case open). But you
    can decide what is a reasonable wait if you decide you want the supply
    to fade out on its own, even if it isn't a shock hazard. The time it
    takes to fade it to 37% of full voltage is R*C seconds where R is in
    ohms and C in in farads (or R in meg ohms and C in microfarads, etc.)

    After you get that all working to your satisfaction and you have load
    tested it with some pairs of 12 volt bulbs in series (or other 24 volt
    capable loads) to see if the ripple voltage is as expected and the
    transformer doesn't overheat, etc. you can start thinking about
    regulation.
    --
    John Popelish
     
    John Popelish, Jan 26, 2005
    #3
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