LED Indicator on 48VDC Alarm Lead?

[Wayne R.]

I've got several dry contact alarm leads (telecom environment) that I
need to piggy-back onto each other into a single reporting pair of
leads. I think that, with each as normally open, the one that closes
could be identified if that particular line has an LED included.

In other words, I want to tie in parallel several relay closures to a
single detection circuit. And I'd like to put an LED on each
individual relay line so we'd know which one is the guilty party when
something is tripped.

The voltage on each detection circuit is the available voltage from
our 48VDC plant - usually at 54VDC when on commercial AC, but at 48VDC
(or lower) when on batteries.

Can someone define for me and LED/resistor combination that will work?

Or, better yet, is there a way to do this simply with supervised/NC
circuits?

 

[TimPerry]

I've got several dry contact alarm leads (telecom environment) that I
need to piggy-back onto each other into a single reporting pair of
leads. I think that, with each as normally open, the one that closes
could be identified if that particular line has an LED included.

In other words, I want to tie in parallel several relay closures to a
single detection circuit. And I'd like to put an LED on each
individual relay line so we'd know which one is the guilty party when
something is tripped.

The voltage on each detection circuit is the available voltage from
our 48VDC plant - usually at 54VDC when on commercial AC, but at 48VDC
(or lower) when on batteries.

Can someone define for me and LED/resistor combination that will work?

Or, better yet, is there a way to do this simply with supervised/NC
circuits?

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)

 

[[email protected]]

I've got several dry contact alarm leads (telecom environment) that I
need to piggy-back onto each other into a single reporting pair of
leads. I think that, with each as normally open, the one that closes
could be identified if that particular line has an LED included. [...]
Can someone define for me and LED/resistor combination that will work?

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 http://members.misty.com/don/ledd.html 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.

Or, better yet, is there a way to do this simply with supervised/NC
circuits?

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

 

[Wayne R.]

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