Can you put the resistor at the end of the LED instead of the beginning ?

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Bracer

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I am using a typical 16F628A chip to power 2 LEDs among other things.

Let's just say space is something that's scares in this project and I have am using DIP Components.

One of the Pin will drive two LEDs.

I know that the "safest" approach is to put the resistor between the pin and the respective LED's anode.

The configuration is:
Pin->Resistor->Anode of LED.

This single pattern is repeated with the other LED [there are only two LEDs].

Could I get that one pin to drive the two LEDs like this ?

Pin->Anode of LED :: Cathode of LED to -> Resistor ->Ground.

The other LED will have its anode to the Pin and Cathode to the same resistor shared by the First LED. [resulting in one pin driving two LEDs with a SINGLE resistor]

Is this ok ?
 
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It doesn't make any difference. It's just a simple series circuit. Put them in any order you like.

So it really doesn't matter if you put the resistor between the Output Pin and the Anode or Between the ground and the Cathode ?
 
That could be a problem, as there is no guarantee that the current will be the same in both LEDs. It is best to use one resistor for each LED, or to put the LEDs in series if there is enough voltage.
 
When both LEDs are lit they will be dimmer than when one LED is lit. Apart from that it will work fine.

Edit, as Diver pointed out, when both are lit they may be different brightnesses. To avoid this probems you could time splice the two pins. I.E alternate each mS which pin is on.

Mike.
 
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Thank you Pommie, Diver300 and carbonzit [you sound angry, I apologize for this].
 
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