Pinball machine - switch closure timing problem

"Huh huh huh! Very good, Will Robinson."

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

Cheers.

Ken

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

If it's Saturday? Neither!

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.

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?

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

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

Get Better! :-) Rich

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.

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

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

?!

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

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

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

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

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 > 0 so (Vin-Vcap) < Vin therefore I < Imax.

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

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 s***ty 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 =

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) =

Cheers Terry

Not always, but I try.

Thanks. :)