know what determines their size.
A. The simple answer is: capacitance is used to reduce the
amount of ripple voltage that is ever
present on a DC power supply output.
Unfortunately, the formula isn't as
simple, but I think I've at least made it
less math intensive. Here is the simplified equation:
C = [(ILOAD x t)/P-PRIPPLE] x 106
ILOAD = the DC output current; it is calculated using ILOAD = EOUT/RLOAD
keep the ship afloat after
battle damage by pumping
water out of the hull. I've
been playing with a circuit
built around a TIP120 transistor and a relay to start
the pump when water is
detected. The problem I
have with it right now is that it works
in most lakes, but not all. I'm guessing this is due to lower mineral content or salinity. I would be most interested in a simple circuit that would be
more reliable, but still compact
enough to fit in very confined spaces.
sensor more sensitive. (If the sensor
appears too sensitive, increase the
size of the resistor.) I also swapped the
TIP120 for a smaller MPSA14 unit and
recommend you use a compact reed
relay, like the RadioShack 275-232
(don't worry that it's rated five volts;
relays are current, not voltage operated, so no will be damage done).
P-PRIPPLE = the acceptable peak-to-peak DC output ripple voltage
So far, so good. Now comes t,
which is the hard part because it
depends on the rectification method
used and the line frequency. The
capacitor requirements are less strict
if you use full-wave rectification
t = 1/(2 x line frequency); t = 0.0083
for 60 Hz and 0.01 for 50 Hz
How about an example? Let's
design for five volts out at one amp,
with 100-mV peak-to-peak ripple, running from a 60-Hz wall-wart. Plugging
these values into the equation
nets a capacitance value of 8300
uF. An off-the-shelf 10,000uF cap
will work perfectly.
A. I looked over the circuit you sent
and can tell you that your connection of the level sensor to the collector of the TIP120 (Figure 6)
reduces the detector's sensitivity.
Think about it: you're using the transistor as an on/off switch. The sensors are used to provide base current
to the transistor to turn the switch on,
but when the transistor switch turns
on, the top sensor no longer has any
voltage to apply to the base, which
causes the transistor to turn off.
The answer is to connect the sensor to the + 6 volt source through a 10
k resistor. This way, the base current
isn't dependent on the transistor's collector voltage, only the conductivity
of the lake water, which makes the
Q. I am curious about your
thoughts on using modified sine
wave inverters with switching power
supplies, like those used in PC computers. I've looked at the waveform on
a scope, and noticed a big rush of
current at the step from 0 to 170 volts
that decays exponentially. With a true
sine wave, the current influx is less
sharp at the beginning, and the whole
thing looks like a rectified half-wave.
I'm using a Tripp-Lite modified sine-wave inverter on my old computers
with no apparent harm, but would
appreciate an expert opinion.
John M. Buzby
C = [(1A x .0083)/0.1V] x 106
C = [.0083/0.1] x 106
C = [.083] x 106 = 8300uF
Sink the Bismark!
Q. I spend many of my weekends sinking ships — really!
The hobby is called R/C Warship
Combat, and it is heavly into electronics for just about every aspect
of the sport. One of the tools of
the trade is a bilge pump that
pumps water out of the hull to