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Old 18 June 2003, 05:25 PM
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Recaro
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Hi,

I need a circuit that will basically operate 5 - 7 LEDs when a PLC output is on. So the LEDs have to come on when the plc outputs 5v. Someone said I could use a transistor with the base connected to the PLC output. But I have no idea how to design the circuit for this operation. A 9 volt battery is used for powering the LEDs.
Old 18 June 2003, 05:46 PM
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Jay m A
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your friend is right

Use a BC337 transistor, the base driven by a 1K resistor connected to the PLC o/p. The emitter to ground, the LED's parralelled together (depending on LED type an inline resistor will probably be needed) between 9V and the collector, make sure they point the right way.

A picture is worth a thousand words, well about 30 in this case

Justin
Old 18 June 2003, 06:05 PM
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Recaro
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could i use this transistor:
http://www.maplin.co.uk/products/module.asp?CartID=030618170849890&moduleno=32952&m odulecode=

theres a load there but theres one called BC337-16
Old 18 June 2003, 07:00 PM
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AndiThompson
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What, you mean like robotic? Charge up at night? No then, I'm 100% human...
Old 18 June 2003, 07:04 PM
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Dougster
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Ask John Banks.

Usually found lurking around Drivetrain and Gen. Tech.!!
Old 19 June 2003, 07:23 PM
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Recaro
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I need 5 LEDs to operate on a 9 volt battery, can this be done? I'm sure I need a reistor, but I have no idea what size. any help?

I have no idea how many LEDs I can run on a 9v battery or what resistor i need. The LEDs will be just a normal cheap red LED.

[Edited by Recaro - 19/06/2003 19:24:16]

edited to see if I could attatch a picture, but it didnt like it.

[Edited by Recaro - 19/06/2003 19:25:14]
Old 19 June 2003, 09:13 PM
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AndyC_772
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5 LEDs off a 9V battery has to be done by splitting them into a group of 2 in series and another group of 3. The voltage drop across one LED is 2V, so five in series would require a 10V (minimum) supply.

Assuming a current of 10mA, the group of 3 requires a 300 Ohm resistor and the group of two requires 500 Ohms (the nearest you'll be able to buy are 510, which will make the LEDs slightly dimmer, or 470 which will make them a bit brighter). Join the two groups across the battery in parallel.

Andy.
Old 20 June 2003, 12:56 PM
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Recaro
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Its actually 4 leds, not 5. My mistake.
Old 20 June 2003, 01:07 PM
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LG John
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Wink

Any electronic people on here?
No but my dad was a borg, does that count.


<I'll get me coat>
Old 20 June 2003, 01:14 PM
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suprabeast
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well 4 leds across a 9 v battery is enough (2v per led) therefore 1v left for approx 15ma would be 1/0.015 = 66 ohms.

or for 10ma its gonna be 100 ohms

Think thats right
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