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Pinball machine - switch closure timing problem

Discussion in 'Electronic Design' started by frenchy, Mar 22, 2005.

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  1. Ken Taylor

    Ken Taylor Guest

    I'm sure you experimented and found quantifiable evidence that it was the
    appropriate value combination.

    Cheers.

    Ken
     
  2. Rich Grise

    Rich Grise Guest

    That is called "Empirical Design". Don't be ashamed of it - that's the
    way most people do it, and the rest lie about it, kinda like wanking. ;-)

    Cheers!
    Rich
     
  3. Rich Grise

    Rich Grise Guest

    I think he slapped it in and played the game, and tested different
    values until the game worked. To me, 4.7 sounds a little high, but hey,
    whatever works!

    Cheers!
    Rich
     
  4. frenchy

    frenchy Guest

    ..05 nonpolarized cap worked a little better than nothing, .1 a little
    better than that but still some hits still not registering. Since one
    pinball company once used 22 uf polarized and 100 ohm across some
    non-matrixed switches, I decided to try a much smaller cap 4.7
    polarized by itself and seemed 100% and no problems, then added a
    bigger resistor as a safety factor as per current, and to stretch the
    charge out, and still is registering 100%. A little better guesswork
    than grabbing two parts with my eyes closed but alas, precise it was
    not. I could probably keep experimenting with lower cap and bigger
    resistor till it starts to fail again. If anybody thinks my combo here
    could cause damage in a 5 volt system lemme know please, the 22uf setup
    was a 12 volt switch matrix....Frenchy
     
  5. Rich Grise

    Rich Grise Guest

    As I said: Empirical Design. :)
    Nah. You've not only done your homework, but you've reported it! This
    is WAY more than most do!

    I might be a wacko, and annoying, and politically incorrect, but when
    something works, it works, so what else is there?

    Do you put on both of your socks before you put on your shoes, or do
    you put on one sock, one shoe, then the other sock and the other shoe?

    Cheers!
    Rich
     
  6. frenchy

    frenchy Guest

    If it's Saturday? Neither!
     

  7. I have no choice. I have to wear surgical support hose all the time
    because of swelling in my legs so its both, then pants, and finally
    shoes, if they'll go on my feet that day.
     
  8. keith

    keith Guest

    Good grief! The socks get put on in the bedroom, while dressing. Shoes
    stay in the forier (or mud room) and never get close to the bedroom. What
    kind of a barn did you grow up in?
     
  9. Terry Given

    Terry Given Guest

    ?!

    If the [cap + Rdischarge] is across the switch, and the switch gets
    closed, the cap discharges thru Rdischarge. Next time the matrix gets
    around to the switch-in-parallel-with-RC the cap will begin to charge
    up, and the micro will read the switch as closed (even though its not).
    In practice you dont really need the charge resistor, re-drawn circuit
    becomes:


    |
    ----------|-------+-- [Row = switch to V+]
    | |
    | |
    | |
    | o--[+Cap]--+
    | / |
    | [switch] |
    | / |
    +---o----[Rdis]----+
    |
    |
    [Column = switch to 0V]


    Cheers
    Terry
     
  10. frenchy

    frenchy Guest

    I was told by somebody in the pinball newsgroup that an engineer
    indicated to them that a 4.7 cap sounded somewhat high for the
    particular circuit, I have changed it to a 1.0 with the same resistor
    and still seems to be working perfectly (crossing fingers). The
    engineer didn't really want to suggest any values, just that 4.7
    sounded high. Oh well.
    Ok so I think I get it now, the cap being charged causes the matrix
    read to still 'see' a closed switch even thought the actual switch is
    now open. Thanks....Frenchy
     
  11. Rich Grise

    Rich Grise Guest

    Sorry to hear about your condition. I hope you have ample opportunities
    to put them up, and relieve the pressure.

    Get Better! :)
    Rich
     
  12. Terry Given

    Terry Given Guest

    Hi Frenchy,

    Its been about 15 years since I last worked on videogames (since I
    graduated and got work as a power supply design engineer), but I
    regularly saw (and used) 4.7uF caps on switch inputs for videogames
    (including pinballs). I fiddled with values as high as 100uF (too big,
    they start missing short switch *releases* for large cap values, rather
    than missing closures for small cap values) and as low as 100nF. IME
    1uF-10uF was fine, whatever was available.

    Yep, you've got it exactly right. The cap charge time is long compared
    to its discharge time, and regardless of what circuitry is reading the
    switch, it is comparing the switch voltage to a threshold. As long as
    the cap voltage is below the threshold, the "switch" reads as closed.

    Cheers
    Terry
     
  13. Terry Given

    Terry Given Guest

    your empirical work is excellent. The highest voltage your cap can
    charge to is the supply voltage - 12V. The peak current flowing through
    the switch when it closes is the sum of the matrix current and the
    current coming from the cap as it discharges through Rdischarge. The
    discharge current is simply

    Idis_max = Vcap_max = 12V = 80mA peak
    ---------- --------
    Rdischarge 150 Ohms

    The actual discharge current will be a decaying exponential spike:

    Idis(t) = Idis_max*exp(-t/(Rdis*C))

    See how the cap value doesnt affect the peak - only the series resistor
    does.

    the area of this exponential decay curve (ie its integral) is exactly
    the same as the area of a rectangular pulse of the same (peak)
    amplitude, and width Tau_dis = Rdis*C (seconds).

    Tau_dis = 4.7uF*150R = 705us

    charge Q = C*V = I*t = 80mA*705us = 56.4uC

    double check: C*V = Q = 56.4uC so 56.4uC/12V = 4.7uF. as it should.

    Energy in cap E = 0.5*Q*V = 0.5*C*V^2 = 0.5*56.4uC*12V = 338.4uJ

    But see that the cap *does* control the charge and hence energy, which
    does not depend on the resistor.

    It takes 3 time constants to discharge to 5% of applied voltage.

    3*Tau_dis = 2.115ms, nice and short

    if we knew the pullup resistance we could look at the charge time
    constant. Guess 4.7k:

    Tau_chg = (4.7k + 150R)*4.7uF = 4.85k*4.7uF = 22.795ms

    3*Tau_chg = 68ms

    If you want to know the power dissipated in the discharge resistor, it
    depends on the frequency the switch is toggled at, in Hertz. In the case
    of a videogame, human reaction times are around 200ms/400ms for olympic
    athletes & normal people respectively, so 100ms is a pretty fast switch
    closure time. If we assume a period of 100ms (ie on + off) then

    fswitch = 1/0.1s = 10Hz (ie ten switch closures per second)

    and

    Pdischarge = E*fswitch = 338.4uJ*10Hz = 3.4mW ie bugger all.

    In a real machine, the average frequency would be tiny, and the power
    dissipation would likewise be negligible.


    the power dissipated in the charge circuitry is given by

    Pcharge = E*fswitch = 338.4uJ*10Hz = 3.4mW

    and this power is shared between Rdischarge and the charge circuitry
    resistance, which is usually a lot higher then Rdischarge and so
    dissipates the bulk of the power.

    I think the numbers stack up pretty well for 4.7uF and 150R. In a 5V
    system the numbers are:

    Idis_max = 5V/150R = 33.33mA

    Tau_dis = 150R*4.7uF = 705us

    Tau_chg = 4.85k*4.7uF = 22.795ms

    Q = 5V*4.7uF = 23.5uC

    E = 0.5*23.5uC*5V = 58.75uJ

    Pdis = 58.75uJ*10Hz = 0.5875mW

    Pchg = 58.75uJ*10Hz = 0.5875mW


    So the time constants stayed the same, but the power dissipation dropped
    - not that it mattered in the first place. assuming Rchg = 4k7 of course...

    Cheers
    Terry
     
  14. frenchy

    frenchy Guest

    Hmmm about 99.7% of that is over my head but.... am I to understand
    then that basically, the size of the cap NOT that important as to
    potential damage or over-current to say, a transistor that the charge
    would eventually be going to when the cap discharges, and that it's the
    RESISTOR that is the key in limiting this potential damage, regardless
    of the cap size? I am also deducing this from what another dude said
    about sometimes using caps up to 100 uf in this kind of application in
    pins and videogame switches and buttons. Thanks for going into all
    this detail (even if I will never understand it!).....Frenchy
     
  15. Terry Given

    Terry Given Guest

    exactly correct.

    when a cap is fully discharged, it "looks" like a short-circuit. So at
    the instant you apply (dc) voltage to an R-C series circuit (cap
    discharged) then it "looks" like the R alone.

    Ohms law applied to series RC circuit:

    At any instant in time, (Vin - Vcap) = I*R

    at time t=0, Vcap = 0 so I = (Vin-0)/R = Vin/R = Imax

    this I charges up the cap a little bit, so at some other time t>0, Vcap
    Ultimately the cap charges up to Vin, at which point the current into
    the RC circuit becomes (Vin-Vin)/R = 0 amps.


    Likewise a fully charged cap "looks" like a constant voltage source (if
    we look over a small enough slice of time). If we take our fully charged
    RC circuit, and switch the R to 0V with a npn transistor, then at the
    instant the transistor turns on, the cap looks like a voltage source of
    V volts, and the current thru the resistor (and hence switch) is
    (Vcap-Vswitch)/R and if the switch is any good, Vswitch is very small
    (hey, 300mV out of 12V is bugger all) and can be ignored.


    if you are really keen, you can turn this hand-wavey argument into a
    derivation of the cap charge & discharge equations. Or just remember
    them....


    All caps have some internal resistance (often called ESR, Equivalent
    Series Resistance). This is what makes caps get hot when you bang
    current through them. In the case of a smps cap, the ripple current is
    quite high and so ESR is designed to be very low - 20mOhms is not
    unusual. If we get say a 1000uF cap with ESR=0.02 Ohms, charge it to 12V
    then discharge it with a screwdriver, the peak current will be
    12V/0.02Ohms = 600A. A large spark will result, and quite possibly will
    blow a bit off the side of the screwdriver. If you short this cap with a
    switch, expect the switch contacts to wear out very quickly (hell, they
    might even weld straight away).

    OTOH if we get a crappy DSE 1000uF 16V cap, the ESR is more like 1-2
    Ohms, so the peak current will be 6-12A, a *huge* difference.

    If we bung say 10R in series with the good cap, it reduces the current
    from 600A to 1.2A. If we stuck the 10R in series with the shitty cap, it
    reduces the peak current from 6-12A down to 1-1.1A.


    The second part of the damage issue relates to how much energy is stored
    in the cap, which is directly proportional to the cap size.

    E = 0.5*C*V^2

    A cap with a large series R lets this energy out slowly, a cap with a
    small series R lets it out very quickly. A 100nF cap at 12V has 7.2uJ of
    energy, a 1000uF cap at 12V has 72mJ, ie 10,000 times more energy.

    While the 100nF cap has a very low ESR (so peak discharge current is
    very high) there is not much energy, so it cant heat up the switch
    contacts (or transistor) very much. This is why you often see 100nF caps
    placed directly across switches etc.

    I just recently designed a circuit with a motion sensor that is a little
    ball-bearing sitting in a concave seat, making contact between a couple
    of points. Move it and the switch opens & closes. *BUT* the switch has
    an absolute maximum current rating of 25mA - this is to prevent welding
    the ball bearing onto the contacts. In my case, I slapped a 510R
    resistor in series with it, then a 100nF cap across the whole lot, with
    a 1Meg pullup to +3.3V. max cap discharge current is thus 3.3V/510R =
    6.5mA. The contact resistance is 5R, so without the series resistor the
    switch current would have been 3.3V/5R = 660mA, about 26 times greater
    than its rated value.


    Another example of cap charging is the "thunk" a big audio amp makes at
    turn on - it has a transformer/rectifier/cap filter, with a *LOT* of
    capacitance - >= 10mF is not uncommon. When you first turn it on, the
    cap behaves like a short-circuit, and a *very* large current is drawn
    from the supply to charge it - the current is limited basically by the
    transformer resistance. Good amp's have a delayed turn-on circuit and a
    large series R (say 100R 10W) that gets shunted by a relay after the cap
    has charged up - often called a "soft-charge" circuit.

    Back when I was a videogame tech (great background for an engineer) I
    used to repair a *lot* of cheap korean smps. One of the most common
    failures was a blown-up input rectifier, and its the bus cap charge
    current that destroys the rectifier. A lot of these smps had 1 Ohm 5W
    series resistors in the AC line, supposedly to soft-charge the bus cap.

    Does it help? not really.....at 230Vac input, Vpeak = 230*sqrt(2) =
    325Vpeak. If you *happen* to turn the switch on at the peak of ac line
    voltage, then the peak current flowing into the power supply is (325V -
    1.4V)/1R = 324A peak (assuming a bridge rectifier). If the Ifsm rating
    (single shot current thump) of the diode is < 325A, it will go *bang* -
    if not sooner then later. Simply replacing the rectifier bridges with
    gruntier ones (having Ifsm > 400A) completely solved the problem.

    Cheers
    Terry
     

  16. Not always, but I try.
    Thanks. :)
     
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