ChadMan said:
Will this type of circuit do this work? Does it have too much overhead
in parts? I am a newbie and would like to know. This looks like it
should light up the LED on event failure (blown fuse). Could another
device like a FET or Transistor do the same thing (allow no current
to flow until the base looses power) as the relay?
o----------------------------o---------o
| |
| LED/LAMP |
| ___ .---. |
GND>--o---|___|--| X |o-----o |
R1 '---' | |
| | | Relay
o /o _|_
/ -|_\_|
/ |
o |
| |
o------------------o |
| |
| ____ |
+DC>------o--|_--_|-o---------------o---------o
Fuse
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ChadMan
Well it would work, only the relay should be activated at all time. Only
when the fuse blows it goes off and switches on the LED. In normal operation
(fuse ok) the relay draws some current and generates heat. I guess you do
not want that.
So at any time this circuit will draw a current.
The idea with a FET that John Fields drawed is a good one (last posting of
him). Only some diodes must be placed to prevent that the FET goes up in
smoke when the fuse blows, beteen G-S and between the Gate/resistor
connection and the load.
When the load has a huge capacitance (buffer) it is possible that the load
voltage drops slower when the fuse just went. This means a high voltage
across the G-S of the FET appears and can damage it. The G-S diode protects
it from high voltage spikes caused by interrupting a high current.
In this schematic the resistor between the gate/load and GND defines the
current that will be drawn through the fuse. This is much less then a coil
of a relay.
When a Mosfet is turned off the resistance between Drain and Source is very
high (>10meg), so no current goes through. (Be aware of the body diode of
the mosfet, between Drain-Source)
So in total only the current through the resistor OR through the LED is
drawn.
Hope this helps.
Al