bleed resistor

Discussion in 'Electronic Design' started by Paul Taylor, Jan 5, 2006.

  1. Paul Taylor

    Paul Taylor Guest

    hi all
    I need to place a bleed resistor across a power supply reservoir capacitor
    of 15000uf to discharge it on turn off is there any rule of thumb to
    calculate the right value for shortest turn off and wattage required.

    thanks
    Paul Taylor, Jan 5, 2006
    #1
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  2. Paul Taylor

    Pooh Bear Guest

    Paul Taylor wrote:

    > hi all
    > I need to place a bleed resistor across a power supply reservoir capacitor
    > of 15000uf to discharge it on turn off is there any rule of thumb to
    > calculate the right value for shortest turn off and wattage required.


    Choose a time constant that you like.

    The bleed resistor will dissipate under normal running conditions. I normally
    choose one ( in a high power amplifier ) to be ~ 1W dissipation.

    Graham
    Pooh Bear, Jan 5, 2006
    #2
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  3. Paul Taylor

    Jim Thompson Guest

    On Thu, 5 Jan 2006 12:23:11 -0000, "Paul Taylor"
    <> wrote:

    >hi all
    >I need to place a bleed resistor across a power supply reservoir capacitor
    >of 15000uf to discharge it on turn off is there any rule of thumb to
    >calculate the right value for shortest turn off and wattage required.
    >
    >thanks
    >


    For mathematical amusement, why don't you calculate it ?:)

    ...Jim Thompson
    --
    | James E.Thompson, P.E. | mens |
    | Analog Innovations, Inc. | et |
    | Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
    | Phoenix, Arizona Voice:(480)460-2350 | |
    | E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
    | http://www.analog-innovations.com | 1962 |

    I love to cook with wine. Sometimes I even put it in the food.
    Jim Thompson, Jan 5, 2006
    #3
  4. Paul Taylor

    Tam/WB2TT Guest

    "Paul Taylor" <> wrote in message
    news:dpj33e$t0h$...
    > hi all
    > I need to place a bleed resistor across a power supply reservoir capacitor
    > of 15000uf to discharge it on turn off is there any rule of thumb to
    > calculate the right value for shortest turn off and wattage required.
    >
    > thanks
    >
    >

    Pick a wattage, and then calculate the resistor value. If you only have room
    for a 1/8W resistor, the turnoff time will be longer than for a 5W resistor.
    I would pick a wattage less than 1% of the PS rating. Also, you will want to
    pull enough bleeder current to keep the supply in regulation when the
    external load is removed.

    Tam
    Tam/WB2TT, Jan 5, 2006
    #4
  5. Pooh Bear wrote...
    >
    > Paul Taylor wrote:
    >
    >> I need to place a bleed resistor across a power supply reservoir capacitor
    >> of 15000uf to discharge it on turn off is there any rule of thumb to
    >> calculate the right value for shortest turn off and wattage required.

    >
    > Choose a time constant that you like.
    >
    > The bleed resistor will dissipate under normal running conditions. I
    > normally choose one ( in a high power amplifier ) to be ~ 1W dissipation.


    These two criteria are often at odds with each other. If the voltage
    is high enough to be dangerous, and the instrument will sometimes need
    working on, the time-constant is best to be no more than a few seconds.
    Since that usually means excessive dissipation, my approach is to pick
    the appropriately-timed resistor, with its wattage sized to handle the
    discharge energy from the capacitors, and switch it with relay contacts.
    When unpowered the relay contacts should be closed. When AC power is
    applied the contacts will open an instant later, and the ac transformer/
    diode-bridge/resistor simply handle the high current flow for twenty or
    so ms. If the relay fails to open, the fuse should blow after a second
    or two (make sure the resistor is small enough for this to happen).


    --
    Thanks,
    - Win
    Winfield Hill, Jan 5, 2006
    #5
  6. Paul Taylor

    Ian Bell Guest

    Paul Taylor wrote:

    > hi all
    > I need to place a bleed resistor across a power supply reservoir capacitor
    > of 15000uf to discharge it on turn off is there any rule of thumb to
    > calculate the right value for shortest turn off and wattage required.
    >
    > thanks


    The amount of energy stored in the capacitor is given by:

    0.5CV^2 joules

    A joule dissipated over a second is one watt so an 1/8th watt resistor
    should not be expected to dissipate more than one joule every 8 seconds, a
    quarter watt resistor in 4 seconds, half watt in two seconds and one watt
    resistor in one second. Work out your joules and decide how quickly you
    want to dissipate the energy and that will give you the resistor wattage.
    Remember that this resistor will dissipate this power when the supply is on
    as well so it would be better to over rate it e.g. use a 1watt type where
    the dissipation is half a watt.

    From the decided dissipation and the supply volts you can calculate the
    resistor value W = V^2/R so R = V^2/W

    HTH

    Ian
    Ian Bell, Jan 5, 2006
    #6
  7. Paul Taylor

    Jon Guest

    There are several approaches. Theoretically the capacitor will never
    discharge to zero volts.
    Approach 1)
    Decide what is a "safe" voltage (Vd)
    Decide what is the maximum acceptable time (t) to discharge to VD.
    Calculate R:
    Let Vc = the normal charged voltage.
    R = t*/{C*[ln(Vc) - ln(Vd)]}
    Where ln(x) is the natural logarithm of
    x.
    Calculate the wattage according to Paul Taylor's advice.
    ~
    Approach 2)
    Decide what is a "safe" voltage (Vd).
    Decide on the minimum value resistance that you can use, based on
    dissipation.
    Calculate the time required to discharge to Vd.
    t = R*C*[ln(Vc)-ln(Vd)]
    Jon, Jan 5, 2006
    #7
  8. Paul Taylor

    Tam/WB2TT Guest

    "Jon" <> wrote in message
    news:...
    > There are several approaches. Theoretically the capacitor will never
    > discharge to zero volts.
    > Approach 1)
    > Decide what is a "safe" voltage (Vd)
    > Decide what is the maximum acceptable time (t) to discharge to VD.
    > Calculate R:
    > Let Vc = the normal charged voltage.
    > R = t*/{C*[ln(Vc) - ln(Vd)]}
    > Where ln(x) is the natural logarithm of
    > x.
    > Calculate the wattage according to Paul Taylor's advice.
    > ~
    > Approach 2)
    > Decide what is a "safe" voltage (Vd).
    > Decide on the minimum value resistance that you can use, based on
    > dissipation.
    > Calculate the time required to discharge to Vd.
    > t = R*C*[ln(Vc)-ln(Vd)]
    >

    With 15000 uF, I assumed it was not a high voltage power supply; but then,
    why is he worried about it at all?

    Tam
    Tam/WB2TT, Jan 6, 2006
    #8
  9. Paul Taylor

    Paul Taylor Guest

    thanks for all your valuable information with this I have sort out the
    problem.

    thanks
    Paul Taylor, Jan 6, 2006
    #9
  10. On 2006-01-05, Winfield Hill <> wrote:

    > These two criteria are often at odds with each other. If the voltage
    > is high enough to be dangerous, and the instrument will sometimes need
    > working on, the time-constant is best to be no more than a few seconds.


    Years ago I worked in a (German) laser company in the power
    supply department. Once I asked the head engineer why he had
    littered the PS enclosure with more than a dozen screws. He told
    me that there was some safety regulation that required the SMPS
    primary filter caps to have discharged to a safe voltage within
    the time it takes to open the enclosure. Large bleed resistor --
    many screws. The relay approach is more elegant but probably
    entails more elaborate safety audits. Once some bureaucrat has
    figured out the average time it takes to remove one screw the
    math is easy.

    robert
    Robert Latest, Jan 7, 2006
    #10
  11. Paul Taylor

    Pooh Bear Guest

    Robert Latest wrote:

    > On 2006-01-05, Winfield Hill <> wrote:
    >
    > > These two criteria are often at odds with each other. If the voltage
    > > is high enough to be dangerous, and the instrument will sometimes need
    > > working on, the time-constant is best to be no more than a few seconds.

    >
    > Years ago I worked in a (German) laser company in the power
    > supply department. Once I asked the head engineer why he had
    > littered the PS enclosure with more than a dozen screws. He told
    > me that there was some safety regulation that required the SMPS
    > primary filter caps to have discharged to a safe voltage within
    > the time it takes to open the enclosure. Large bleed resistor --
    > many screws. The relay approach is more elegant but probably
    > entails more elaborate safety audits. Once some bureaucrat has
    > figured out the average time it takes to remove one screw the
    > math is easy.


    I'd like to know which IEC reg that was ! I suspect said engineer was taking
    an imaginative approach.

    Graham
    Pooh Bear, Jan 7, 2006
    #11
  12. In article <>,
    Robert Latest <> wrote:

    > Years ago I worked in a (German) laser company in the power
    > supply department. Once I asked the head engineer why he had
    > littered the PS enclosure with more than a dozen screws. He told
    > me that there was some safety regulation that required the SMPS
    > primary filter caps to have discharged to a safe voltage within
    > the time it takes to open the enclosure. Large bleed resistor --
    > many screws.


    Yes. I've had to work to similarly described specs.
    Plus requiring that the user should have to use a
    tool to gain access to the innards.... where a coin
    was specifically not classed as a tool, barring the
    use of such things as large-slotted Zeus fasteners
    for access panels.

    --
    Tony Williams.
    Tony Williams, Jan 7, 2006
    #12
  13. Paul Taylor

    Guest

    Paul Taylor wrote:
    > hi all
    > I need to place a bleed resistor across a power supply reservoir capacitor
    > of 15000uf to discharge it on turn off is there any rule of thumb to
    > calculate the right value for shortest turn off and wattage required.
    >
    > thanks



    I tried a clever trick when I was too young to be more sensible. Used a
    2 pole 2 way mains switch, and in the off position it switched the
    transf primary across the dc output. Thus the transformer discharged
    the reservoir.

    The problems with this are:
    1. if mains supply is lost but panel switch is on, output stays charged
    2. lack of adequate insulation reliability between switch contacts,
    creating a failure path from mains side to output.

    In principle this could be done with a relay to solve 1., but as
    someone mentioned a R would be safer!

    FWIW a constant current device would give better discharge time vs
    power waste, and a miniature lightbulb, whose R drops at lower V, would
    give even better result. These must be considerably underrun to ensure
    they last more or less indefinitely. Since you can solder miniature
    lamps straight into the pcb this becomes an option.


    NT
    , Jan 9, 2006
    #13
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