75 ohm Coax Connections
Seems like I always have reliability
issues with 75 ohm coax connections.
I’m certain the problem is the
center conductor. I have roughed it up
with serrated pliers, bent the end a
little, etc., but still have a problem.
I’m thinking about applying solder
paste to the end, to maybe seal out
oxygen, and the little solder balls make
All the quality parts/crimps are
just not reliable.
#1111 Frederick M. Raposa
I need to show when an electric
fence ( 6 KVa to 16 KVa) is active by
flashing a simple LED with every pulse.
Ideally, the circuit would simply attach
between high V and ground strands of
the fence. A perfect solution would be
able to cope with an inadvertent
It would be good to be able to use
multiple such devices at regular
intervals along the fence. I am looking
for lowest cost/complexity.
#1112 Kevin Dickinson
[#9108 - September 2010]
I have two circuits in the same
enclosure: one works on 12 VDC and
the other on 9 VDC. I have a 12 VDC
transformer for one, but need to lower
the output to 9 VDC for the other. I
tried with resistors to lower the voltage
to 9 VDC with no success.
76 January 2011
No mention was made of the
current requirements, but if they aren't
too demanding you can get 9 VDC
quite easily from the 12 VDC using a
three-terminal regulator such as the
7809, plus a small capacitor (0.01 µF)
on the output. Use the 7809 for up to
1A with a heatsink. For lower currents
(up to 100 mA), you might get by with
the 78L09. Alternatively, you can use
diodes in series to give you your
voltage drop. Depending on the type
of diode used, four to six in series
should be sufficient. If your circuit
does not already include one, add a
capacitor to ground after the last
diode for filtering. An electrolytic with
few to a few hundred µF (depending
on the current) should be sufficient.
[#10101 - October 2010]
How does one calculate the value
for the reactance capacitor in an AC to
DC reactive power supply so as not to
over-work the zener diode? I‘ve used
from . 68 µf to 4. 7 µf for various
voltages out, but it has always been
trial and error.
#1 Before starting, some advice:
These supplies have no isolation from
the line and should only be used to
power loads that are completely
isolated from the user. Have the
bridge rectifier end of the supply
connected to the neutral side of the
line. Use an isolation transformer
when testing the circuit; be careful
even with the isolation transformer,
there are still lethal voltages present. It
is strongly advised to include a fuse in
series with the limiting capacitor or
use "across the line" capacitors — also
called "X" capacitors.
A very good approach is to start
from the output and work back to the
source. Determine the maximum
All questions AND answers are submitted
by Nuts & Volts readers and are intended
to promote the exchange of ideas and
provide assistance for solving technical
problems. Questions are subject to
editing and will be published on a
space available basis if deemed suitable
by the publisher. Answers are submitted
output current desired. Add 3 to 5 mA
for the zener, to keep it at the zener
voltage even at full load. This is the
average current needed out of the full-wave bridge rectifier feeding the
zener. Subtract the zener voltage plus
two diode drops from the minimum
line voltage; I usually use 105 or 106V.
For example, a 12V zener plus 1.4V
(that is, two times 0.7V) from 105V
leaves 91.6V across the capacitor.
The average current is about 90%
of the RMS current, so multiply
average current by 1.111 to get
the RMS current in the capacitor.
The equivalent reactance Xc =
Vcapacitor/Irms. At 60 Hz, we know
that Xc = 1/377C, so C = 1/377Xc. Or,
C = Iav x 1.111/(377 x Vcapacitor).
Using our 12V example and a 20
mA maximum load current, we get Iav
= 20+ 3 = 23 mA and Irms = 25.553
mA. Using Iav and Vcapacitor in the
formula, C = ~0.74 µF. Note that this
makes no allowances for capacitor
tolerance. A "turnkey" solution would
add 10% for capacitors with a 10%
tolerance, so all values would work. In
this case, the nearest standard value
would be 0.82 µF. If a one-off circuit is
built, one can select a capacitor or
parallel combination very close to the
desired value and the worst case
current at high line voltage will be a
bit less. Also, the zener losses at
minimum load will be less.
Next, the zener wattage has to be
determined. This one assumes all
tolerances and line voltage go to the
worst case: capacitor is 0.82+10%;
line voltage is ~10% high; and load
current is at minimum. For the
example, assume .902 µF, 130V and
minimum load current is 5 mA.
by readers and NO GUARANTEES
WHATSOEVER are made by the publisher.
The implementation of any answer printed
in this column may require varying degrees
of technical experience and should only be
attempted by qualified individuals.
Always use common sense and good