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

On a sunny day (27 May 2007 14:05:44 -0700) it happened snipped-for-privacy@mail.dk wrote in :

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

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

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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

On a sunny day (27 May 2007 14:45:03 -0700) it happened snipped-for-privacy@mail.dk wrote in :

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

On a sunny day (Mon, 28 May 2007 07:09:25 GMT) it happened ehsjr wrote in :

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 :-)

On a sunny day (Sun, 27 May 2007 18:30:22 -0500) it happened Spehro Pefhany wrote in :

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

On a sunny day (28 May 2007 09:47:23 -0700) it happened snipped-for-privacy@mail.dk wrote in :

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.

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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 ...

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Bye.
   Jasen