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Electro ripple current hint

Discussion in 'Electronic Design' started by Phil Allison, Jul 5, 2013.

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  1. Phil Allison

    Phil Allison Guest

    ** Hi,

    electro cap makers specify the maximum, safe "ripple current" for each
    category and size. Normally, two figures are given - one at AC supply
    frequencies and another at a much higher frequency - reflecting the
    differing ESR values at the two frequencies. In order to measure this
    current, a low value resistor needs to be installed in series with the
    particular electro and the resulting voltage measured with a true RMS volt
    meter or maybe a DSO that can do the same task.

    Finding a suitable resistor and fitting it in series could prove very
    tedious (or even unsafe) in practice. However, there is a simple way around
    this that works for low and AC supply frequency measurements.

    The classic formula " I = C dv/dt " shows the current flowing in an electro
    is proportional to the dv/dt of the voltage across the terminals any point
    in time - so if you derive a current proportional to dv/dt, then its RMS
    value is a measure of the ripple current.

    A series RC network does the job, long as its time constant is short in
    relation to the high frequency components in the current wave. For AC supply
    frequencies a value of 100uS proves to be OK.

    All you need then is a 100nF film cap and a 1kohm resistor, connected in
    series across the electro and simply measure the RMS voltage appearing
    across the resistor (Vrms). The voltage wave replicates the electro current
    wave almost exactly.

    The necessary scale factor is simply the electro's value in uF multiplied by
    10,000 ( as a 1mA current in the RC network requires a dv/dt of 10,000 V/S.)

    Electro ripple amps = C x Vrms x 10,000

    BTW:

    While ripple current in electros operating at AC supply frequencies is not
    regularly an issue - it can become one when the supply impedance is
    unusually low, as with direct mains operation or with large transformers and
    small value caps.

    In these cases, the peak charging current can exceed the average load
    current by a factor of more than 10:1 , exaggerating the RMS current value
    considerably.

    Larger than usual ripple voltage percentages also increase peak and RMS
    ripple currents, particularly if combined with the above.


    ..... Phil
     
  2. Tim Williams

    Tim Williams Guest

    If you guess the time constant correctly, you already know (that was the
    point).

    Works for inductive shunt resistors at high frequency, too:

    Iin
    o
    | R
    +---RRR--+---o out
    | |
    R Rs |
    R |
    | === C
    L Ls |
    L |
    | |
    +--------+---o GND
    |
    o
    Iout

    R*C = Ls/Rs

    Tim
     
  3. Tim Williams

    Tim Williams Guest

    Hold that thought.....

    Unfortunately, copper traces have an excellent Q and make better inductors
    than resistors at, oh, just a few kHz really. Copper is hopelessly
    inductive at SMPS frequencies. You'd have to know Rs and Ls very well
    indeed to compensate that trace well enough to know anything. Circular
    reasoning. :)

    Anyways, you'd have to be a pretty awful PCB designer to leave any length
    between rectifier and cap. :)

    Tim
     
  4. P E Schoen

    P E Schoen Guest

    "John Larkin" wrote in message
    I tried this using LTSpice, for a FWB with a 560 uF capacitor and 120 VRMS,
    with loads of 50 and 500 ohms, and capacitor ESR of 0.018 and 0.18 ohms, and
    it seems to work well. However, with an ESR of 1.8 ohms, it has an error of
    60%. But that is an extreme case at which there is 8W dissipation in the
    capacitor. Also, if you use a 1 uF capacitor instead of 100 nF, the result
    then becomes much closer. But then there is a large error at lower ESR
    values.

    Paul

    Version 4
    SHEET 1 1368 680
    WIRE 304 128 112 128
    WIRE 320 128 304 128
    WIRE 432 128 384 128
    WIRE 496 128 432 128
    WIRE 528 128 496 128
    WIRE 576 128 528 128
    WIRE 672 128 576 128
    WIRE 672 144 672 128
    WIRE 112 160 112 128
    WIRE 320 208 272 208
    WIRE 432 208 432 128
    WIRE 432 208 384 208
    WIRE 576 208 576 128
    WIRE 496 224 496 128
    WIRE 672 256 672 208
    WIRE 112 272 112 240
    WIRE 272 272 272 208
    WIRE 272 272 112 272
    WIRE 320 272 272 272
    WIRE 416 272 384 272
    WIRE 672 288 672 256
    WIRE 304 368 304 128
    WIRE 320 368 304 368
    WIRE 416 368 416 272
    WIRE 416 368 384 368
    WIRE 496 368 496 288
    WIRE 496 368 416 368
    WIRE 576 368 576 288
    WIRE 576 368 496 368
    WIRE 592 368 576 368
    WIRE 672 368 592 368
    WIRE 592 432 592 368
    FLAG 592 432 0
    FLAG 528 128 out
    FLAG 672 256 ripple
    SYMBOL voltage 112 144 R0
    WINDOW 123 0 0 Left 2
    WINDOW 39 -30 182 Left 2
    WINDOW 0 -15 57 Left 2
    WINDOW 3 -85 151 Left 2
    SYMATTR SpiceLine Rser=50m
    SYMATTR InstName V1
    SYMATTR Value SINE(0 175 60 0 0 0 100)
    SYMBOL diode 384 288 M270
    WINDOW 0 32 32 VTop 2
    WINDOW 3 0 32 VBottom 2
    SYMATTR InstName D2
    SYMATTR Value MUR460
    SYMBOL diode 320 224 R270
    WINDOW 0 32 32 VTop 2
    WINDOW 3 0 32 VBottom 2
    SYMATTR InstName D3
    SYMATTR Value MUR460
    SYMBOL polcap 480 224 R0
    WINDOW 3 24 64 Left 2
    SYMATTR Value 560u
    SYMATTR InstName C1
    SYMATTR Description Capacitor
    SYMATTR Type cap
    SYMATTR SpiceLine V=600 Irms=2.9 Rser=1.8 Lser=0
    SYMBOL res 560 192 R0
    SYMATTR InstName R1
    SYMATTR Value 50
    SYMBOL diode 320 144 R270
    WINDOW 0 32 32 VTop 2
    WINDOW 3 0 32 VBottom 2
    SYMATTR InstName D1
    SYMATTR Value MUR460
    SYMBOL diode 384 384 M270
    WINDOW 0 32 32 VTop 2
    WINDOW 3 0 32 VBottom 2
    SYMATTR InstName D4
    SYMATTR Value MUR460
    SYMBOL res 656 272 R0
    SYMATTR InstName R2
    SYMATTR Value 1k
    SYMBOL cap 656 144 R0
    SYMATTR InstName C2
    SYMATTR Value 100n
    TEXT 64 360 Left 2 !.tran 200m startup
    TEXT 112 424 Left 2 ;Electro ripple amps = C x Vrms x 10,000
    TEXT 752 112 Left 2 ;560u * 1.075 * 10,000 = 6.02
    TEXT 752 144 Left 2 ;for R1 = 50, I(C1) = 6.19 A RMS
    TEXT 440 312 Left 2 ;0.018 ohms
    TEXT 752 360 Left 2 ;for R1 = 500, I(C1) = 1.21 A RMS
    TEXT 752 328 Left 2 ;560u * 0.199 * 10,000 = 1.114
    TEXT 752 432 Left 2 ;for R1 = 500, ESR(C1)=0.18, I(C1) = 1.102 A RMS
    TEXT 752 400 Left 2 ;560u * 0.196 * 10,000 = 1.098
    TEXT 752 176 Left 2 ;560u * 1.070 * 10,000 = 5.99
    TEXT 752 208 Left 2 ;for R1 = 50, ESR(C1)=0.18, I(C1) = 5.99 A RMS
    TEXT 752 272 Left 2 ;for R1 = 50, ESR(C1)=1.8, I(C1) = 5.33 A RMS
    TEXT 752 240 Left 2 ;560u * 1.394 * 10,000 = 7.81
     
  5. Phil Allison

    Phil Allison Guest

    "P E Schoen"
    "Phil Allison"

    I tried this using LTSpice, for a FWB with a 560 uF capacitor and 120 VRMS,
    with loads of 50 and 500 ohms, and capacitor ESR of 0.018 and 0.18 ohms, and
    it seems to work well. However, with an ESR of 1.8 ohms, it has an error of
    60%.

    ** Simulations need to use realistic data or they are just plain stupid.

    Obviously, you did not look up the ESR or ripple current figures of any
    *real* 560 uF, 250V electro.

    A figure of 0.25 ohms (100/120Hz ) is about right at room temp for standard
    grade, 560uF, 250V electros - dropping sharply with increasing temp. But a
    figure of 1.8 ohms indicates a faulty cap.

    Also, your 50ohm load draws over 3 amps resulting in circa 35V p-p ripple -
    which is rather high.

    Was your source resistance for 120VAC zero ?

    If so, that was crazy too.


    But that is an extreme case at which there is 8W dissipation in the
    capacitor.

    ** That cap would have exploded in minutes - so is utterly irrelevant.

    You need to do some REAL tests with REAL electros.

    Cos simulations are mostly bullshit.



    .... Phil
     
  6. Guest

    Those inductance figures seems to be off by one order of magnitude.

    For narrow PCB tracks 1 nH/mm would be typical, so 0.4 nH/mm for a
    wide trace would be believable (10 nH/inch).
     
  7. Phil Allison

    Phil Allison Guest

    "P E Schoen"

    I tried this using LTSpice, for a FWB with a 560 uF capacitor and 120 VRMS,
    with loads of 50 and 500 ohms, and capacitor ESR of 0.018 and 0.18 ohms, and
    it seems to work well. However, with an ESR of 1.8 ohms, it has an error of
    60%. But that is an extreme case at which there is 8W dissipation in the
    capacitor.

    ** That 8W figure cannot be right - cap ripple current always exceeds DC
    load current.

    With a 3.1 amp load, the power must exceed 17 watts.

    To get the real answer I used the following:

    240/120V at 800VA step-down tranny

    470uF, 350V electro ( tested as 530 uF )

    1.2kW, 240V heater for the 50 ohms load.

    1.8 ohm, 50 watt metal clad resistor.

    Results:

    I ripple rms = 4.3 amps, power loss = 33watts

    BTW:

    The ripple voltage wave showed a big bump at the end of the charging
    period - the usual sign of an electro that is completely worn out.


    .... Phil
     
  8. P E Schoen

    P E Schoen Guest

    "Phil Allison" wrote in message
    Yup, I ran it again and I got 32.6 watts, with 4.35 A for I(C) and 3.02 A
    I(R).

    I thought it seemed a bit "off". Even 0.18 ohms gives 13 watts.

    But that's not quite right either. I(C1) is 6 amps, and thus the power
    should be 36*0.18 = 6.48W.

    The problem seems to be due to the way LTSpice computes power in a capacitor
    and you must set the sample window to an exact multiple of half cycles,
    because otherwise the reactive VA dominates the equation. Hmm. Even that
    does not seem to work...

    Thanks for the heads up.

    Paul
     
  9. John S

    John S Guest

    No, I don't think so. Try this, Paul:

    ..tran 0 .5 .4 10u

    This will give you a whole number of cycles to work with and will drop
    the initial surge.

    Then you can use the nice power and average functions available to do
    what you want.
    It is.
     
  10. Phil Allison

    Phil Allison Guest

    "John S"
    ** Not for long.

    As usual, reality has gone missing.

    Can I remind you:

    1. Electro ESR has a large negative tempco.

    2. A 560uF, 250V electro with a 6A ripple rating ( if they exist) does not
    have a 0.18 ohm ESR, but something a lot less.

    Recall what I said about simulations ?

    It's a mug's game.


    ..... Phil
     
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