>>>READER-TO-READER QUESTIONS AND ANSWERS
12 volts. If the fuse blows, the LED will become part of the
load and glow. To monitor the 120 VAC fuse, place an appropriate 120 VAC bulb (preferably a neon bulb with 22K resistor) directly across the fuse. If the fuse blows, the bulb will
light. Run the wires for the LED and AC bulb from some convenient panel for monitoring, if appropriate to your cause.
John F. Mastromoro
Saint Johnsville, NY
#2 The standard way to detect a blown fuse is with an
indicator circuit wired across the fuse. For 120 VAC, use a
neon indicator lamp with a built-in resistor, as shown in
Figure 1. For a 12 VDC circuit, use an LED with a resistor,
as shown in Figure 2.
These indicators
require that a load Figure 1
be connected and
turned on in order to
indicate a blown
fuse. If you want to
detect a blown fuse Figure 2
with or without such
a load, use the
circuit in Figure 3.
This circuit holds the transistor off by applying + to the base
through the diode and the 1K resistor. If the fuse blows, the
transistor is turned on by the 27K resistor from base to
ground, and the LED lights. The diode from + 12 to the emitter compensates for the voltage drop in the diode between
104
May 2007
the 1K base resistor
and the fuse.
For a 120 VAC
blown fuse indication without a load,
use an NE- 2 neon
Figure 3 bulb without a built-in resistor (like part
#36NE002 from Mouser). Connect the bulb in-series with a
100K resistor and install that across the fuse as shown in the
first diagram. Then add a 100K 1/2 watt resistor from the
fuse to the other side of the AC line.
Ed Schick
Harrison, NY
#3 I designed
this circuit (Figure
1) back in 1990
for a competition
car stereo fuse
panel. It is quite
clever as it gives
an instant visual
indication of the
fuse condition.
GREEN means
GOOD and RED
means BAD. Later, I modified it slightly, (Figure 2) for use on
AC fuses. I have also used it with the new dual LEDs, super
bright LEDs, and even jumbo 10 mm LEDs.
The circuits
(Figure 1 and
Figure 2) are
based on
Kirchhoff's voltage law where, if
a circuit is
opened up
(blown fuse,
open switch,
etc.), the source
voltage will be seen across the open. Normal voltage
through the fuse fires LED2 (green) as long as the fuse is
good. D1, a standard rectifier diode, prevents LED1 from
firing except when a fuse blows. R1, in both circuits, serves
as a current limiter for about 3. 5 mA. I have also utilized this
same circuit as a toggle switch on/off indicator, as well. Just
substitute a switch into the circuit where the fuse is.
Note: The circuit was designed using whatever LEDs I
had laying around the shop at the time. These just happened
to illuminate brightest at 3. 5 mA so I used Ohm's law to
calculate 2.7K ohms as a current limiting resistor. Also, on the
AC circuit, since the LEDs are rectifying the signal, they will
require about 10 VAC @ 3. 5 mA to Illuminate, therefore a 30K
resistor is needed for 120 VAC. You will need to adjust the
resistance values slightly for modern LEDs, especially if you
use a jumbo LED or any of the super bright LEDs.
KI4NNH - Jerry Ginn
Franklin Springs, GA
