LED Indicator on 48VDC Alarm Lead?

obtain LED of the size, shape, color, and brightness you wish to use. use the rated current for the device (or measure in using a variable supply and meter) then use ohms law to calculate resistor value.

(V - Vd) / I = R where Vd is the voltage of the diode (usually about 2 volts)

then I squared times R = P

example: if the diode needs 10 mA 54 Vdc - 2 / .01 = 5200 ohms

.01 * .01 * 5200 = .52 watts

i'd use two 1/2W resisters in series of the nearest standard value available to 2600 ohm (probably 2.7K)

Depending on how the detection circuit works, probably the simplest way is to diode-OR all of the dry contact closures together. View in a fixed-width font. + is a connection, --|-- is _no_ connection.

fuse

+------+------+---~~~------------> + 54 V DC | | | | | | \ \ \ Dry NO contacts \ \ \ on each device | | | | | | Diodes | | | (1N4002 or | | | better) | | +------>|---+-----> + To detection circuit | +------|------>|---+ +53 V = Fault +------|------|------>|---+ +---> 0 0 V = OK | | | | R R R Resistors | | | | | V V V LEDs |

- - - | | | | |

+------+------+-------------+---> 0 V DC

The LED associated with each contact will come on when that contact is closed. The 1N4002 diodes keep the contacts on each device isolated from one another. "Or better" means you can use a higher

1N400x part number, but not a 1N4001.

Read

for how to work out the resistor values. Figure out the fuse value by assuming that all of the LEDs will be on at once and multiply that current by 2. Make sure the fuse you use is rated to interrupt DC.

If your detection circuit wants a contact closure, instead of a voltage, follow the above circuit with a relay that has a 48 VDC coil, or a relay with a lower DC coil voltage and a series resistor. You can work out the resistor value in the same way that you did for the LEDs. The relay coil is the LED and its coil voltage is the LED voltage.

This circuit does have the problem that if the fuse blows, or any of the dry contacts, 1N4002s, or wiring fail open, you'll never know.

fuse

+---~~~------------------> + 54 V DC | | +---> 0 V DC | | I Dry NC contacts | I on device 1 | | | +---R--->|-----------+ Resistor and LED 1 | | I Dry NC contacts | I on device 2 | | | +---R--->|-----------+ Resistor and LED 2 | | I Dry NC contacts | I on device 3 | | | +---R--->|-----------+ Resistor and LED 3 | | | +---> 0 To detection circuit | +54 V = OK +------------------------> + 0 V = Fault

For this one, the FIRST LED to go out indicates the faulted device. For example, if the contacts on device 2 open, both LEDs 2 and 3 will be out.

Determine the fuse and resistor values in the same way as for the first circuit. Again, if the detection circuit wants a contact closure, follow this circuit with a relay (and series resistor, if required).

As you mentioned, this circuit is supervised: if the fuse blows, or any of the dry contacts or wiring fail open, you will know.

Matt Roberds

environment) that I

pair of

that closes

included.

closures to a

party when

voltage from

but at 48VDC

will work?

supervised/NC

I am just finishing up job with lighted Idec relays... red neon. very nice. Lately ive gotten non lighted by mistake though, a factory screw up maybe. The neon lights when the coil is energized...you should be able to get those in 48 vdc..but check.

You can mount the relays so they light behind indicator graphics. $3.50 each in STDP...plus a din rail mount base $7.50.

If you want LED calcs. see led vendor, they will have charts with resistors required for each voltage and light intensity desired. and can sell you both.

Without intensity specified, and a specific LED a resistor cannot be specified.

Search googles web tab for 'led' click on the adds in the left margin. those will have what you want in chart form.

Phil Scott

Exactly the detailed response I was hoping for. Perfect, and I appreciate the effort in the diagrams too. Thanks!