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Ripple filter

Discussion in 'Electronic Design' started by John Doe, Jan 13, 2006.

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  1. John Doe

    John Doe Guest

    I would like to use a brushless 12v (0.07a) computer fan to vent a small
    area in my RV. I tried this, but the battery charger cooked the fan after a
    while due to ripple- it could not handle the peak voltage I guess. I
    replaced the fan but placed a 2 ohm (5w) resistor in series with it, and a
    0.1mfd cap and 15v (3w) zener in parallel with it to filter and hopefully
    protect the fan- but now the 15v zener gets too hot as it "clamps off the
    ripple" (I only want it to conduct for transient protection). Fan still
    going ok though.

    A transistor regulator would not be good as- correct me if wrong- it would
    show a 0.7 Vce drop. This would cut the fan performance and wasting current
    for regulation is not desirable when on battery power. Would a bigger
    capacitor provide enough filtering to prevent the zener from conducting
    (0.1mfd was a guess)? The resistor needs to be as smaller (ohms) if
    possible for faster fan. Can anyone improve on this? I am pushing my
    design abilities already. Thanks for any ideas!
  2. Guest

    A saturated transistor drops only 0.3v on load, Germaniums are even
    lower. A transistor regulator is the way to go.
  3. John,

    If the 15v zener is getting hot, then your battery charger is outputting
    something greater than 15V so a 12V regulator would still have at least
    a 3v drop to work with. Your zener is having to dissipate 15V times the
    current output of the battery charger minus the fan current so it is no
    wonder that it is getting hot. A voltage regulator would only need to
    dissipate whatever the drop across it actually turns out to be times the
    fan current. Likely a simple LM7812 would work fine.

    Hope this helps.
  4. kell

    kell Guest

    Don't assume the butler (ripple) did it.
    This is 60 cycles from line power we're talking about, isn't it?
    If your battery charger is connected to the RV's battery, then the
    battery should absorb all the 60 Hz ripple and whatever runs off the
    battery will be running off smooth DC, and I'm not kidding you about
    If ripple is getting into stuff there's something wrong with the way
    you are using the battery charger, or your battery is shot, or
    by the way 0.1 uF is WAAAYYY too small to do anything about ripple, if
    indeed you have a ripple problem. Think 1000 uF. At 120 Hz (full wave
    rectification), a load of .07 amps would need a smoothing cap of 560 uF
    to get ripple to 1 volt.

    You may just have too high a voltage. If your zener is dissipating
    more than 3 watts then you have at least .2 amps flowing through it.
    With that plus the fan current, say a total of .3 amps, going through
    the resistor, the voltage across the resistor would be .6 volts.
    Meaning 15.6 volts DC or more coming in. Considering that the zener
    voltage will rise because of the high current going through it, this is
    a conservative number.

    Have you put a voltmeter on the battery while you are using the

    There's no way to give you a definitive answer based on very limited
    information in your post.
  5. Rich Grise

    Rich Grise Guest

    I have an RV and its charger output is almost raw unfiltered pulsating
    DC, unless I hook the battery to it, in which case the 13.6 or whatever
    it is is as smooth as silk. I can see the difference on my 12V TV. :)

    Lead-acid batteries love a float charge. :)

  6. John Fields

    John Fields Guest

    For your application, here's what I'd do: (View in Courier)

    Vin Vout
    / /
    +| | +|
    [ ] | [FAN]
    | | |

    Assuming that your charger output is unsmoothed and outputs 16V
    peaks, then the cap will charge up to about 15.3V.

    Since the 7812 will output 12VDC for your fan, and the cap will be
    charged up to 16V, the differential voltage across the regulator
    will be:

    dV = Vin - Vout = 16V - 12V = 4V

    with a fan current of 0.7A, the power dissipated by the regulator
    will be:

    P = dVI = 4V * 0.07A = 0.28 watts

    With a dropout voltage of 2.5V for the 7812, a 70mA load, and a
    full-wave rectified, unsmoothed output from the charger at 120Hz,
    the value of the capacitor will be:

    I dt
    C = ------

    where C is the capacitance of the capacitor in farads
    I is the load current in amperes
    dt is the period of the ripple
    dV is the allowable ripple in volts.

    Now, since we have 16V across the cap and the output of the
    regulator is at 12V, that gives us 4V of headroom. Since the
    regulator requires that its input never fall more than 2.5 above its
    output, that means that we must make sure that the voltage across
    the cap never go below:

    Vin(min) = Vout + Vdo = 12V + 2.5V = 14.5V

    Since we'll have 16V across the cap when it charges, somewhere on
    the rising edge of the charger's output peaks, we have to make sure
    that it never falls below 14.5V when the cap is dicharging into the
    load and the charger's output isn't charging the cap and driving the
    load. Since the difference between 16V and 14.5V is 1.5V, that
    means that the allowable ripple voltage out of the cap is 1.5V

    If we lower the stress on the cap and lower the ripple to 1V we can
    solve for the capacitance like this:

    I dt 0.07A * 8.3ms
    C = ------ = --------------- = 581µF
    dV 1V

    Most common small aluminum electrolytics have a capacitance
    tolerance of +/- 20%, so to make sure you get at least 581µF you
    need to increase its value to 697µF.

    The closest commonly available value is 820µF, which would be fine
    for full-wave rectified 60Hz.

    However, on the chance that the output of your charger is half-wave
    rectified 60Hz, the capacitance would need to be doubled for the
    same ripple, bringing the value to 697µF * 2 = 1394µF.

    A 1500µF/25V Panasonic EEU-FC1E152 would work, as would anything you
    might have lying around with at least that capacitance and a working
    voltage somewhat greater than the peak output voltage from your

    There's not much reason to worry about transients since the cap will
    soak them up, but if you want to you could put a Zener in parallel
    with the cap with a voltage rating equal to greater than the peak
    output of the charger but less than the working voltage of the cap.
  7. John Fields

    John Fields Guest


    The foregoing has some haste-makes-waste errors in it. Primarily,
    with 16V out of the charger, the input to the regulator will be
    15.3V which, with a dropout voltage of 2.5V for the regulator, means
    that the ripple out of the cap can't exceed 0.8V at 120Hz. That, in
    turn, means the cap has to have a capacitance of at least 813
    (820)µF with 120Hz ripple or 1626 (1800µF) with 60 Hz ripple.

    After all is said and done, I'd put a 2000µF/25V aluminum
    electrolytic in there and be done with it.
  8. Mark

    Mark Guest

    do you have the battery connected to the charger also or just the fan?

    connect the battery also, chargers do not put out DC...

    if you want to run the fan from AC, get a small DC power supply, not a
    battery charger...

  9. John Doe

    John Doe Guest

    ##snip ##

    THANKS for all replies!
    John I agree that that would be the easiest thing to try so will look for a
    large electrolytic capacitor and let you all know how it works. Thanks for
    doing the math- I was way off with my .1mfd. If that don't work will look
    look into the regulators mentioned. Do they still make germanium
    transistors (low voltage drop)? I looked for them once -not too hard- and
    all I could find was antiques.

    I wasn't too clear on the ripple source. The charger for the RV -called a
    converter- has 2 outputs, one for battery, the other powers the fuse bus for
    the RV accessories. I think they do this so charge rate won't be affected
    by accessory load. So I guess the battery might filter it's output but the
    fuse bus is not- wish I had a scope.
  10. ehsjr

    ehsjr Guest

    Just use John's design, as he designed it. If you use a filter cap
    and nothing else, the fan will still be exposed to too much voltage.


    Do they still make germanium
  11. John Fields

    John Fields Guest

    If you were blowing up fans before, and you connect up that 2000µF
    cap without using a regulator, you'll blow them up even quicker.

    You _must_ use a regulator or a series resistor if you expect your
    fan(s) to survive.
    You don't need a germanium transistor in there, you need a
    Do you have a multimeter?

    If you do, why don't you measure the DC voltage and the AC voltage
    on the fuse bus and post back with what you find?

    If you don't, why not buy a cheap one? They're handy...
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