Re: Circuit to detect blown fuse
- From: Jan Panteltje <pNaonStpealmtje@xxxxxxxxx>
- Date: Mon, 25 Dec 2006 14:38:20 GMT
On a sunny day (24 Dec 2006 10:45:49 -0800) it happened timmacrina@xxxxxxxxx
wrote in <1166985949.708407.151820@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:
I am looking for a circuit that will detect whether or not a 120v fuse
has blown. I would like a green LED to show the fuse is not blown and a
red LED to show when the fuse has blown. I am new to circuit designs
but I have assembled many circuits in the past.
LEDs need something like 10mA to glow.
So a LED circuit across the fuse to show 'open' will have 10mA flowing
still, and 10mA is deadly 9assuming you are playing with main fuses this
can be a nasty surprise.
So in that case you need it powered from before the fuse.
This design should do, but of course 2 Neons would work too :-)
I have worked in a place where some units had a neon that came on when
the fuse was blown, and some would be on to indicate power.. and there
were several hundered of these modules in racks....
Use something with high beta for the transistors, for example BC109C.
Ib peak will be 120 x sqrt(2) / 470 000 = 361 uA, Ic = x say beta 500 = 180mA,
but only 10mA can flow, this should saturate T2, light RED LED, and discharge
C3.
Between half periods C3 will charge again via the 330 Ohm resistor with about
10mA.
To keep the charge voltage below .7 V so T1 does not conduct, in 1/120 sec,
C3 = i . t / U = .01 x .0083 / .7 = 120uF.
120V AC ------------------------------------------------------ fuse ---------------
| | | | | |a |
| | e| NPN --- e| NPN diode D2 |
|a | \| === \| |k |
5V | C2 |----| C3 |----------===-------------
zener --- /| | /| T2 470k
|k === c| T1 |___c|
| D1 |+ | |
|- a diode k -| [ ] 330 [ ] 330
| | | |
| | k k
=== C1 | GREEN LED RED LED
| | a a
| |______|___________|
| C1:
| 12mA @ 120V -> Z = 10kOhm,
| @ 60Hz 1/jwC = 10000 = 1 / (6.28 x 60 x C) ->
| 4080000 C = 1 -> C = 1/ 4080000 = .000 000 242F -> 330nF
| (rest of the current into zener).
[ ] C1 should be specified at about 400 VAC.
|R1 D2 is Vbe protection.
|
------------------------------------------------------------------------------------
Now calculate C2, voltage rating about 10 V:
for 1 V discharge in 1/120 second (half period) at 10mA load (LED),
C x U = i . t -> C = i . t / U = .01 x .0083 / 1 = .000 083F = 83uF.
So use 100uF / 10V.
R1 is a fusible resistor to limit current in case C1 shorts, maybe 10 Ohm.
For DC somebody else :-)
Make sure you use gold plated fuses in case of audio :-)
Circuit is not tested, so use at your own risk.
.
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