Problem switching a 5V mini relay from a low power LED (1,2v 2 mA) input

I am trying to make a 5V mini relay switch when the HDMI indicator
LED
in my DVD player lights up.

I have unsuccesfully tried to trigger the mini relay (5V, 40mA) via
an
optocoupler (4N33 - infrared/darlington) "piggybagged" on the
indicator LED (1,2V 2mA), but the relay doesn't switch, only "hums"
slightly. Supplying an input of 1,5V from before the indicator LED
resistor (220ohm) generates the same result. I even tried using the
(normally unused) basis of the optocoupler as input, hoping it was
more "sensitive" - but still only "humming"


The setup works when "piggybacking" a standard 3V LED (>10mA), where
the relay swithes fine. What have I overlooked ?


Any suggestions for (simple) curcuitry I could use to trigger the 5V
relay from the 1,5V 2 mA input (I only have 5V supply power
available) ?

 

LEDs in parallel do not share current equally.
1.2V is enough to open a si transistor.
Do you need to be opto-isolated?

 

Numbers, numbers!
2mA<< 40mA
1.5V<< 5V
Do something about those inequalities, may start working.

Stanislaw.

 

Hi Jan,

The solution does not need to be opto isolated. I just thought it had
the most sensitive input, and I didn't have to worry about any
"feedback" currents.

I think I need another alternative; anything simple that can make a
5V, 40mA relay switch from 1,2V 2mA.

Stanislaw: I don't understand your point ?

Regards,
Ole

 

A comparator that has enough drive on the output?
May not be the simplest solution, however.
Plus, this suggestion comes well into a Memorial Day 6-pack...

I can't remember the part#, but it was from TI.
LMC7215? (Or whatever your favorite is.)

 

Maybe it's not getting DC, but rather some short duty cycle pulses.
You could try using a CMOS 555 as a monostable and drive the relay
with that. Pull pin 2 low with the optotransistor.


Best regards,
Spehro Pefhany

 

Jan's suggestion would be something like this:

+5 ------------------+------+
| |
[Relay] [D1]
| |a
+------+
| |
| /c
+---[220R]---| NPN
| \e
[220R] |
| |
[HDMI LED] |
| |
Gnd ---+-------------+


Most any NPN you have on hand will work.

Ed

 

Hi, I understand you _do_ have a separate 5V supply available?

There are several ways, this oen for example using your current setup:


+5V
-------|
| |----------
| k |
R diode [ \ ] relay coil
10 | a |
| |----------
--- c ----------- |
opto coupler --- b c
--- e --------------- b NPN
e
|
///

The resistor is for current limit.

An other cool way would be to siply use a LDR (CdS cell - light dependent
resistor) glued to the LED.
It perhaps will be low enough R to switch the relay when the LED is on.


If you have 5V DC on the set itself, try to fin what drives the LED, likely
a 5 or 3 V output from a micro controller or some chip, and use that voltage
to drive a transistor via resistor, or via a FET.

c ---- as above.

--- 1k ------- b

------------------e
GND

 

Right
One of them.
Problem why I did not publish that is that often the LED is on the +
and pulled down via a resistor...
Then we should use a PNP

 

You are so smart, yes I missed a clue 'buzzing'...

Maybe just capacitor parallel to the relay would work too?

 

Any easy way to determine if the LED is being "pulsed", before trying
something else ?

Ole

 

Sure, make room dark, move head while looking in direction of LED.
If you see many dots it is pulsed.
Capacitor will smooth it.

 

How many uF ? Don't want it to hang open forever ...

 

stick a cap across it and see if it gets brighter.
100 uf sounds good.

 

Indicator LED was indeed pulsed (which also explains the low power
measurements)
100 uF capacitor in parallel with relay solved problem

Thanks all,
Ole

 

hook it to an oscilloscope, or through a capacitor to headphones, or
through a capacitor to an AC voltmeter ...