frenchy said:
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
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