J-Log.eu - Forum

Re: Advice on how to power luxeon alarm

2011-11-02T13:59:16+01:00

It is not directly unsafe w/o opto. So you may go on the no-opto version with the 10k resistor providing that the LED is off with an open input.

Statistik: Verfasst von dl7uae — 2. Nov 2011, 13:59

Re: Advice on how to power luxeon alarm

2011-11-02T12:18:58+01:00

Thanks for quick reply !

dl7uae hat geschrieben:

You know that? http://62.153.249.80/jlog/jlog2-en/alarm-devices-for-jlog2-v3-2up

Yes, this is what I am trying to understand

dl7uae hat geschrieben:

Do you have a series resistor in the path collector-BD243--Luxeon?
Well, also with no resistor the pulses in mode "flash" are short enough (32ms) to be able to drive a power LED by overcurrent without burning it.

No, but I think I will add one to be safe

dl7uae hat geschrieben:

But: Attention! If you flash the logger or disconnect the alarm device (control) from the logger the LED will be ON continuously! If you do not use an optocoupler in front then the 10k resistor here http://62.153.249.80/jlog/wp-content/uploads/2011/06/Alarm_device_JLog2_v3.2up-2.jpg will avoid that. The only disadvantage is that there is no alarm in case of the alarm device is disconnected from the logger or the logger lost its supply voltage. But that kind of alarm w'd be anyway a continuously burning LED, "burning" in the meaning of the word, perhaps. (In case there is no current delimiting series resistor.)

So we cannot avoid 'open-input=steady on' unless we skip the opto ?
How safe is it to skip the opto ?

Statistik: Verfasst von zmooth — 2. Nov 2011, 12:18

Re: Advice on how to power luxeon alarm

2011-11-02T11:05:42+01:00

Modified the above post, --> refresh!

Statistik: Verfasst von dl7uae — 2. Nov 2011, 11:05

Re: Advice on how to power luxeon alarm

2011-11-02T10:52:16+01:00

Hi!

You know that? http://62.153.249.80/jlog/jlog2-en/alarm-devices-for-jlog2-v3-2up

Do you have a series resistor in the path collector-BD243--Luxeon?
Well, also with no resistor the pulses in mode "flash" are short enough (32ms) to be able to drive a power LED by overcurrent without burning it.

But: Attention! If you flash the logger or disconnect the alarm device (control) from the logger the LED will be ON continuously! If you do not use an optocoupler in front then the 10k resistor here http://62.153.249.80/jlog/wp-content/uploads/2011/06/Alarm_device_JLog2_v3.2up-2.jpg will avoid that. The only disadvantage is that there is no alarm in case of the alarm device is disconnected from the logger or the logger lost its supply voltage. But that kind of alarm w'd be anyway a continuously burning LED, "burning" in the meaning of the word, perhaps. (In case there is no current delimiting series resistor.)

Series resistor collector-LED: Ualarm (supply voltage of the alarm device) minus forward voltage (of the LED, check out the specs) devided by LED current (R=U/I).

Statistik: Verfasst von dl7uae — 2. Nov 2011, 10:52

Advice on how to power luxeon alarm

2011-11-02T10:31:55+01:00

Hello, nice forums ! I am very happy with my J-Log and with normal LED alarms.

But I've got some Luxeon LED that I wanted to use instead, so I made a new alarm device circuit with stronger transistor (BD243) and it works great with my current-limiting bench PSU and this Luxeon LED (this one: http://www.sparkfun.com/products/9634), current is about 300mA at 3.7V and visibility is much better. Now, how to power it on-board ? If I was to use a single-cell LiIon/LiPo do I need current limiting circuit so I won't blow out my LED ? Or am I safe because of the pulses ?

Thanks in advance

Statistik: Verfasst von zmooth — 2. Nov 2011, 10:31